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| author | Trupti Kini | 2016-09-09 23:30:25 +0600 |
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| committer | Trupti Kini | 2016-09-09 23:30:25 +0600 |
| commit | 881c3e39d046002e9910d5c518c20fe000e63b37 (patch) | |
| tree | c6f84e1956eb501ff64b872dafaa2184443e14c2 | |
| parent | 28bb57cacd0c8bd76a5c86d7e99e3583f02f0b6c (diff) | |
| download | Python-Textbook-Companions-881c3e39d046002e9910d5c518c20fe000e63b37.tar.gz Python-Textbook-Companions-881c3e39d046002e9910d5c518c20fe000e63b37.tar.bz2 Python-Textbook-Companions-881c3e39d046002e9910d5c518c20fe000e63b37.zip | |
Added(A)/Deleted(D) following books
A Heat_Transfer_Principles_And_Applications_by_Dutta/README.txt
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch10.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch11.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch5.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch6.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch7.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb
A Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/10.png
A Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/5.png
A Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/51.png
A Heat_Transfer_in_SI_units_by_Holman/Chapter1.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter11.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter2.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter3.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter4.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter5.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter7.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter8.ipynb
A Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb
A Heat_Transfer_in_SI_units_by_Holman/README.txt
A Heat_Transfer_in_SI_units_by_Holman/screenshots/9.1.png
A Heat_Transfer_in_SI_units_by_Holman/screenshots/9.2.png
A Heat_Transfer_in_SI_units_by_Holman/screenshots/9.4.png
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter1.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter2.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter3.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter4.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter5.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter6.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter7.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter8.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter9.ipynb
A Power_Electronics_Principles_and_Applications_by_Jacob/README.txt
A Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/4.png
A Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/5.png
A Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/6.png
A sample_notebooks/AviralYadav/Chapter5.ipynb
43 files changed, 16587 insertions, 0 deletions
diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/README.txt b/Heat_Transfer_Principles_And_Applications_by_Dutta/README.txt new file mode 100644 index 00000000..1af1e4ae --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/README.txt @@ -0,0 +1,10 @@ +Contributed By: Manish Punjabi +Course: mtech +College/Institute/Organization: IIT +Department/Designation: Contributor Operations Research +Book Title: Heat Transfer Principles And Applications +Author: Dutta +Publisher: PHI Learning Pvt. Ltd., New Delhi +Year of publication: 2006 +Isbn: 8120316258 +Edition: 1
\ No newline at end of file diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch10.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch10.ipynb new file mode 100644 index 00000000..6ff17007 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch10.ipynb @@ -0,0 +1,148 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 : Unsteady State And Multidimensional Heat Conduction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.8 Page No : 444" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The bottom surface temperature of given slab is 10.3 C\n", + "The top surface temperature of given slab is 19.4 C\n", + "The mid plane temperature of given slab is 12.6 C\n" + ] + } + ], + "source": [ + "# Variables\n", + "l = 0.05 \t\t\t#m,thickness of margarine slab\n", + "ro = 990. \t\t\t#Kg/m**3, density of margarine slab \n", + "cp = 0.55 \t\t\t#Kcal/kg C, ddpecific heat of slab\n", + "k = 0.143 \t\t\t#kcal/h mC, thermal conductivity of slab\n", + "Ti = 4. \t\t\t#C, initial temp\n", + "To = 25. \t\t\t#C, ambient temp.\n", + "t = 4. \t\t\t#hours, time\n", + "h = 8. \t\t\t#kcal/h m**2 C\n", + "\n", + "#calculation\n", + "Fo = k*t/(ro*cp*l**2) \t\t\t#, fourier no.\n", + "Bi = h*l/k \t\t\t#Biot no.\n", + "#from fig. 10.6 a\n", + "Tcbar = 0.7 \t\t\t#Tcbar = (Tc-To)/(Ti-To)\n", + "Tc = To+Tcbar*(Ti-To) \t\t\t#C, centre temp.\n", + "#from fig 10.6 b\n", + "#(T-To)/(Tc-To) = 0.382\n", + "T = 0.382*(Tc-To)+To \t\t\t#c,top surface temp.\n", + "#again from fig. 10.6 b\n", + "Tm = 0.842*(Tc-To)+To \t\t\t#, mid plane temp.\n", + "\n", + "# Results\n", + "print \"The bottom surface temperature of given slab is %.1f C\"%(Tc);\n", + "print \"The top surface temperature of given slab is %.1f C\"%(T);\n", + "print \"The mid plane temperature of given slab is %.1f C\"%(Tm);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.9 Page No : 449" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i) time required for the cantre-line temp.to drop down to 200 C is 229 s\n", + "ii)the temp. at half radius at that moment is 161 C \n", + "iii)the amount of heat that has been transfered to the liquid is 19647 Kj\n" + ] + } + ], + "source": [ + "import math \n", + "# Variables\n", + "Ti = 870. \t\t\t#C, initial temp.\n", + "To = 30. \t\t\t#C, ambient temp.\n", + "Tc = 200. \t\t\t#C, centre line temp.\n", + "h = 2000. \t\t\t#W/m**2 C, surface heat transfer coefficient\n", + "a = 0.05 \t\t\t#m, radius of cylinder \n", + "k = 20. \t\t\t#W/m C, thermal conductivity\n", + "ro = 7800. \t\t\t#kg/m**3, density\n", + "cp = 0.46*10**3 \t\t\t#j/kg C, specific heat\n", + "\n", + "#calculation\n", + "#i\n", + "Bi = h*a/k \t\t\t#Biot no.\n", + "alpha = k/(ro*cp) \t\t\t#m**2/C, thermal diffusivity\n", + "Tcbar = (Tc-To)/(Ti-To) \t\t\t# dimensionless centre line temp.\n", + "#from fig 10.7 a\n", + "fo = 0.51 \t\t\t#fourier no. fo = alpha*t/a**2\n", + "t = fo*a**2/alpha \t\t\t#s, time\n", + "\n", + "#ii\n", + "#at the half radius, r/a = 0.5 & Bi = 5\n", + "T = To+0.77*(Tc-To) \t\t\t#from fig. 10.7 b\n", + "\n", + "#iii\n", + "x = Bi**2*fo\n", + "#for x = 12.75 & Bi = 5.0. fig.10.9 b gives\n", + "#q/qi = 0.83\n", + "qi = math.pi*a**2*(1)*ro*cp*(Ti-To) \t\t\t#kj, initial amount of heat energy \n", + " #present in 1 m length of shaft\n", + "q = 0.83*qi \t\t\t#j, amount of heat transfered \n", + "\n", + "# Results\n", + "print \"i) time required for the cantre-line temp.to drop down to 200 C is %.0f s\"%(t);\n", + "print \"ii)the temp. at half radius at that moment is %.0f C \"%(T);\n", + "print \"iii)the amount of heat that has been transfered to the liquid is %d Kj\"%(q*10**-3)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch11.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch11.ipynb new file mode 100644 index 00000000..b98ede25 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch11.ipynb @@ -0,0 +1,322 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 : Boundary layer heat transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1 Page No : 478" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i) Boundary layer thickness is 0.0033 m\n", + "Local drag coefficient is 8.72e-04 \n", + "total drag force is 0.615 N \n", + "Shear stress is 0.285 N/m**2\n" + ] + } + ], + "source": [ + "import math \n", + "#Variable\n", + "v = 1. \t\t\t#m/s\n", + "#temprature\n", + "T = 25. \t\t\t# degree celcius\n", + "#length of plate,l = 1m\n", + "l = 1. \t\t\t#m\n", + "#width of plate,w = 0.5m\n", + "w = 0.5 \t\t\t#m\n", + "#angle of incidence,theta = 0 degree\n", + "theta = 0. \t\t\t#degree\n", + "\n", + "#Calculation\n", + "#for water at 25 degree celcius ,momentum diffusivity,\n", + "MD = 8.63*(10**-7) \t\t\t# m**2/s\n", + "#local Reynold no.\n", + "x = 0.5 \t\t\t#m\n", + "Re = x*v/MD \n", + "#from Eq. 11.39,the boundary layer thickness is\n", + "t = 5*x/(Re**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"i) Boundary layer thickness is %.4f m\"%(t)\n", + "\n", + "#local drag coefficient\n", + "#CD = local drag force per unit area (F)/kinetic energy per unit volume(KE)\n", + "#F = 0.332*rho*v**2*Re**0.5 and KE = 0.5*rho*v**2\n", + "CD = 0.332*v**2*(Re**-0.5)/(0.5)*v**2\n", + "\n", + "print \"Local drag coefficient is %.2e \"%(CD)\n", + "\n", + "#From eq 11.44, the drag force acting on one side of the plate is\n", + "#kinetic viscocity\n", + "mu = 8.6*(10**-4)\n", + "fd = 0.664*mu*v*(l*v/MD)**0.5*w\n", + "#the total force acting on both sides of the plate\n", + "\n", + "tfd = 2*fd\n", + "print \"total drag force is %.3f N \"%(tfd)\n", + "\n", + "#shear stress at any point in the boundary layer\n", + "#at a point in the boundary layer,\n", + "x = 0.5 \t\t\t#m\n", + "y = t/2\n", + "# n = blasius dimensionless variable\n", + "n = y/(MD*x/v)**0.5\n", + "#From table 11.1, at n = 2.5,f\"(n) = 0.218\n", + "#shear stress = tau\n", + "fn = 0.218 \t\t\t#f\"(n) = fn\n", + "tau = (mu*v*(v/(MD*x))**0.5)*fn\n", + "print \"Shear stress is %.3f N/m**2\"%(tau)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2 Page No : 488" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal boundary layer thickness is 8.7 mm \n", + "heat transfer coeff is 6.9 W/m**2 C\n" + ] + } + ], + "source": [ + "#Variable\n", + "Ts = 200. \t\t\t# C,temp. of air\n", + "Ta = 30. \t\t\t#C, temp .of surface\n", + "Va = 8. \t\t\t#m/s, velocity of air\n", + "d = 0.75 \t\t\t#m, dismath.tant from leading edge\n", + "\n", + "#Calculation and Results\n", + "Tm = (Ts+Ta)/2 \t\t\t#C, Mean temp. of boundary layer\n", + "mu = 2.5*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "P = 0.69 \t\t\t#prndatl no.\n", + "k = 0.036 \t\t\t#W/m c, thermal conductivity\n", + "Re = d*Va/mu \t\t\t#reynold no.\n", + "t = 5*d/(Re**0.5*P**(1./3)) \t\t\t#m, thermal boundary layer thickness\n", + "print \"Thermal boundary layer thickness is %.1f mm \"%(t*10**3)\n", + "\n", + "N = (0.332*Re**(0.5)*P**(1./3)) \t\t\t#Nusslet no.\n", + "h = k*N/d \t\t\t#heat transfer coefficent\n", + "print \"heat transfer coeff is %.1f W/m**2 C\"%(h)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.3 Page No : 489" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Local rate of heat exchange is 235 W/m2\n", + "Plate temperature is :108 Celsius \n" + ] + } + ], + "source": [ + "# Variables\n", + "#Free strean velocity (v1) and temp.(t1) on side 1\n", + "v1 = 6. \t\t\t#m/s\n", + "t1 = 150. \t\t\t#degree celcius\n", + "#same on side 2\n", + "v2 = 3. \t\t\t#m/s\n", + "t2 = 50. \t\t\t#degree celcius\n", + "#dismath.tant\n", + "x = 0.7 \t\t\t#m\n", + "#The plate temp. is assumed to be equal to the mean of the bulk air temp on the two sides of the plates\n", + "T = 100. \t\t\t#degree celcius\n", + "\n", + "# Calculations\n", + "#Side 1\n", + "#mean air temp.\n", + "tm1 = (T+t1)/2\n", + "#From thermophysical properties:kinetic vismath.cosity (kv),Prandtl no.(P), thermal conductivity (k)\n", + "kv1 = 2.6*10**-5 \t\t\t#m**2/s\n", + "P1 = 0.69\n", + "k1 = 0.0336 \t\t\t#W/m degree celcius\n", + "#Reynold no.\n", + "Re1 = x*v1/kv1\n", + "#Nusslet no(N1)\n", + "a = 1/3.\n", + "N1 = 0.332*(Re1)**0.5*P1**a\n", + "h1 = k1*N1/x\n", + "#Side 2 of the plate\n", + "tm2 = (T+t2)/2\n", + "#Similarly\n", + "kv2 = 2.076*(10)**-5 \t\t\t#m**2/s\n", + "P2 = 0.70\n", + "k2 = 0.03 \t\t\t#W/m degree celcius\n", + "Re2 = x*v2/kv2\n", + "N2 = 0.332*(Re2)**0.5*P2**a\n", + "h2 = k2*N2/x\n", + "#overall heat transfer coeff. \n", + "U = h1*h2/(h1+h2)\n", + "#The local rate of heat exchange\n", + "RH = U*(t1-t2)\n", + "\n", + "# Results\n", + "print \"Local rate of heat exchange is %.0f W/m2\"%(RH)\n", + "#the plate temp is given by\n", + "TP = t2+(t1-t2)*U/h2\n", + "print \"Plate temperature is :%.0f Celsius \"%(TP)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.4 Page No : 490" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The temprature of plate after 1 hour is 82 C\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "T1 = 120. \t\t\t#C, initial temp.\n", + "T2 = 25. \t\t\t#C, Final temp.\n", + "Tm = (T1+T2)/2 \t\t\t#C, mean temp.\n", + "rho = 8880. \t\t\t#kg/m**3, density of plate\n", + "#Properties of air at mean temp.\n", + "mu = 2.07*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "Pr = 0.7 \t\t\t#Prandtl no.\n", + "k = 0.03 \t\t\t#W/m C, thermal conductivity\n", + "l = 0.4 \t\t\t#m, length of plate\n", + "w = 0.3 \t\t\t#m, width of plate\n", + "d = 0.0254 \t\t\t#m, thickness of plate\n", + "Vinf = 1. \t\t\t#m/s, air velocity\n", + "Re = l*Vinf/mu \t\t\t#REynold no.\n", + "\n", + "#from eq. 11.90 (b)\n", + "Nu = 0.664*(Re)**(1./2)*(Pr)**(1./3) \t\t\t#average Nusslet no.\n", + "#Nu = l*h/k\n", + "h = Nu*k/l \t\t\t#W/m**2 C, heat transfer coefficient\n", + "#Rate of change of temp. is given by\n", + "A = 2*l*w \t\t\t#m**2. area of plate\n", + "t = 1*3600. \t\t\t#s, time\n", + "cp = 0.385*10.**3 \t\t\t#j/kg K, specific heat\n", + "m = l*w*d*rho \t\t\t#kg, mass of plate\n", + "\n", + "#-d/dt(m*cp8dt) = A*hv*(T1-T2)\n", + "#appling the boundary condition \n", + "T = (T1-T2)*math.exp(-A*h*t/(m*cp))+T2\n", + "print \"The temprature of plate after 1 hour is %.0f C\"%(T)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.5 Page No : 508" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nusslet no is: 388 \n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "#Reynold no (Re),friction factor(f),Prandlt no. (P)\n", + "Re = 7.44*(10**4)\n", + "f = 0.00485\n", + "P = 5.12\n", + "x = P-1 \t\t\t#assume\n", + "\n", + "# Calculations\n", + "#according to Von Karmen anamath.logy\n", + "N = ((f/2)*Re*P)/(1+(5*math.sqrt(f/2))*(x+math.log(1+(5./6)*x)))\n", + "\n", + "# Results\n", + "print \"Nusslet no is: %.0f \"%(N)\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb new file mode 100644 index 00000000..5fcf0a36 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb @@ -0,0 +1,580 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 :Steady State conduction In one dimension" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1 Page No : 14" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rate of heat gain is 16.27 W\n", + "interface temp. between brick and cork is 24.2 C\n", + "interface temp. between cement and cork is -13.6 C\n", + "thermal resistance offered by brick layer is 12.9 percent\n", + "thermal resistance offered by cork layer is 84.1 percent\n", + "thermal resistance offered by cement layer is 3.0 percent\n", + "Additional thickness of cork to be provided = 5.1 cm\n" + ] + } + ], + "source": [ + "# Variables\n", + "A = 1. \t\t\t#m**2, area\n", + "#for inner layer (cement)\n", + "ti = 0.06 \t\t\t#m, thickness\n", + "ki = 0.72 \t\t\t#W/m C, thermal conductivity\n", + "Ti = -15. \t\t\t#C, temprature\n", + "#for middle layer (cork)\n", + "tm = 0.1 \t\t\t#m, thickness\n", + "km = 0.043 \t\t\t#W/m C, thermal conductivity\n", + "#for outer layer(brick)\n", + "to = 0.25 \t\t\t#m, thickness\n", + "ko = 0.7 \t\t\t#W/m C, thermal conductivity\n", + "To = 30. \t\t\t#C, temprature\n", + "\n", + "# Calculation and Results\n", + "#Thermal resistance of outer layer \t\t\t#C/W\n", + "Ro = to/(ko*A) \n", + "#Thermal resistance of middle layer \t\t\t#C/W\n", + "Rm = tm/(km*A) \n", + "#Thermal resistance of inner layer \t\t\t#C/W\n", + "Ri = ti/(ki*A)\n", + "Rt = Ro+Rm+Ri\n", + "tdf = To-Ti \t\t\t#temp driving force\n", + "#(a)\n", + "Q = tdf/Rt \t\t\t#rate of heat gain\n", + "print \"the rate of heat gain is %.2f W\"%(Q)\n", + "\n", + "#(b)\n", + "#from fig. 2.4\n", + "td1 = Q*to/(ko*A) \t\t\t#C temp. drop across the brick layer\n", + "T1 = To-td1 \t\t\t#interface temp. between brick and cork\n", + "#similarly\n", + "td2 = Q*tm/(km*A) \t\t\t#C temp. drop across the cork layer\n", + "T2 = T1-td2 \t\t\t#C, interface temp. between cement and cork\n", + "print \"interface temp. between brick and cork is %.1f C\"%(T1)\n", + "print \"interface temp. between cement and cork is %.1f C\"%(T2)\n", + "\n", + "\n", + "#(c)\n", + "Rpo = Ro/Rt \t\t\t#thermal resistance offered by brick layer\n", + "Rpm = Rm/Rt \t\t\t#thermal resistance offered by cork layer\n", + "Rpi = Ri/Rt \t\t\t#thermal resistance offered by cement layer\n", + "print \"thermal resistance offered by brick layer is %.1f percent\"%(Rpo*100)\n", + "print \"thermal resistance offered by cork layer is %.1f percent\"%(Rpm*100)\n", + "print \"thermal resistance offered by cement layer is %.1f percent\"%(Rpi*100)\n", + "\n", + "#second part\n", + "x = 30. \t\t\t#percentage dec in heat transfer \n", + "Q1 = Q*(1-x/100) \t\t\t#W, desired rate of heat flow\n", + "Rth = tdf/Q1 \t\t\t#C/W, required thermal resistance\n", + "Rad = Rth-Rt \t\t\t#additional thermal resistance\n", + "Tad = Rad*km*A\n", + "print \"Additional thickness of cork to be provided = %.1f cm\"%(Tad*100)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2 Page No : 15" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss is 50.7 W\n", + "thermal conductivities of insulating layer is 0.1633 W/m C\n" + ] + } + ], + "source": [ + "# Variables\n", + "#outer thickness of brickwork (to) & inner thickness (ti)\n", + "to = 0.15 \t\t\t#m thickness\n", + "ti = 0.012 \t\t\t#m thickness\n", + "#thickness of intermediate layer(til)\n", + "til = 0.07 \t\t\t#m thick\n", + "#thermal conductivities of brick and wood\n", + "kb = 0.70 \t\t\t#W/m celcius\n", + "kw = 0.18 \t\t\t#W/m celcius\n", + "#temp. of outside and inside wall\n", + "To = -15 \t\t\t#celcius\n", + "Ti = 21 \t\t\t#celcius\n", + "#area\n", + "A = 1 \t\t\t#m**2\n", + "\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "#Thermal resistance of brick , wood and insulating layer\n", + "TRb = to/(kb*A) \t\t\t#C/W\n", + "TRw = ti/(kw*A) \t\t\t#C/W\n", + "TRi = 2*TRb \t\t\t#C/W\n", + "#Total thermal resistance\n", + "TR = TRb+TRw+TRi \t\t\t#C/W\n", + "#Temp. driving force\n", + "T = Ti-To \t\t\t#C\n", + "#Rate of heat loss\n", + "Q = T/TR\n", + "print \"Rate of heat loss is %.1f W\"%(Q)\n", + "#(b)thermal conductivities of insulating layer\n", + "k = til/(A*TRi)\n", + "print \"thermal conductivities of insulating layer is %.4f W/m C\"%(k)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3 Page No : 19" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss is 4095 W\n", + "interface temp.is 183 C\n", + "Fractional resistance offered by the special brick layer is 0.353 \n" + ] + } + ], + "source": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "#Length & Inside rdius of gas duct\n", + "L = 1. \t\t\t#m\n", + "ri = 0.5 \t\t\t#m radius\n", + "#Properties of inner and outer layer\n", + "ki = 1.3 \t\t\t#W/m C, thermal conductivity of inner bricks\n", + "ti = 0.27 \t\t\t#m, inner layer thickness \n", + "ko = 0.92 \t\t\t#W/m C, thermal conductivity of special bricks \n", + "to = 0.14 \t\t\t#m, outer layer thickness\n", + "Ti = 400. \t\t\t#C, inner layer temp.\n", + "To = 65. \t\t\t#C, outer layer temp.\n", + "\n", + "#calculation and Results\n", + "r_ = ri+ti \t\t\t#m, outer radius of fireclay brick layer\n", + "ro = r_+to \t\t\t#m, outer radius of special brick layer\n", + "#Heat transfer resistance\n", + "#Heat transfer resistance of fireclay brick\n", + "R1 = (math.log(r_/ri))/(2*math.pi*L*ki)\n", + "#Heat transfer resistance of special brick\n", + "R2 = (math.log(ro/r_))/(2*math.pi*L*ko)\n", + "#Total resistance\n", + "R = round(R1+R2,4)\n", + "#Driving force\n", + "T = Ti-To\n", + "#Rate of heat loss\n", + "Q = T/(R)\n", + "print \"Rate of heat loss is %d W\"%(Q)\n", + "#interface temp.\n", + "Tif = Ti-(Q*R1)\n", + "print \"interface temp.is %d C\"%(Tif)\n", + "#Fractional resistance offered by the special brick layer\n", + "FR = R2/(R1+R2)\n", + "print \"Fractional resistance offered by the special brick layer is %.3f \"%(FR)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4 Page No : 20" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The hot end temp. is 148 C\n", + "The temprature gradient at hot end is -294.7 C/m\n", + "The temprature gradient at cold end is -1179 C/m\n", + "the temprature at 0.15m away from the cold end is 131 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "d1 = 0.06 \t\t\t#m, one end diameter of steel rod\n", + "d2 = 0.12 \t\t\t#m,other end diameter of steel rod\n", + "l = 0.2 \t\t\t#m length of rod\n", + "T2 = 30. \t\t\t#C, temp. at end 2\n", + "Q = 50. \t\t\t#W, heat loss\n", + "k = 15. \t\t\t#W/m c, thermal conductivity of rod\n", + "\n", + "# Calculation and Results\n", + "#T = 265.8-(7.07/(0.06-0.15*x))........(a)\n", + "#(a)\n", + "x1 = 0\n", + "#from eq. (a)\n", + "T1 = 265.8-(7.07/(0.06-0.15*x1))\n", + "print \"The hot end temp. is %.0f C\"%(T1)\n", + "#(b) from eq. (i)\n", + "C = 50 \t\t\t#integration consmath.tant\n", + "#from eq. (i)\n", + "D1 = -C/(math.pi*d1**2*k) \t\t\t#D = dT/dx, temprature gradient\n", + "print \"The temprature gradient at hot end is %.1f C/m\"%(D1)\n", + "#similarly\n", + "D2 = -1179 \t\t\t#at x = 0.2m\n", + "print \"The temprature gradient at cold end is %.0f C/m\"%(D2)\n", + "\n", + "#(c)\n", + "x2 = 0.15 \t\t\t#m, given,\n", + "x3 = l-x2 \t\t\t#m, section away from the cold end\n", + "#from eq. (a)\n", + "T2 = 265.8-(7.07/(0.06-0.15*x3))\n", + "print \"the temprature at 0.15m away from the cold end is %.0f C\"%(T2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5 Page No : 24" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rate of heat transfer is -3746 W\n", + "Refrigeration capacity is 1.07 tons\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "#inside and outside diameter and Temp. of sphorical vessel\n", + "do = 16. \t\t\t#m, diameter \n", + "t = 0.1 \t\t\t#m, thick \n", + "Ri = do/2 \t\t\t#m, inside radius \n", + "Ro = Ri+t \t\t\t#m. outside radius\n", + "To = 27. \t\t\t#C, temperature\n", + "Ti = 4. \t\t\t#C ammonia\n", + "k = 0.02 \t\t\t#W/m C, thermal conductivity of foam layer \n", + "\n", + "# Calculations and Results\n", + "#from eq. 2.23 the rate of heat transfer\n", + "Q = (Ti-To)*(4*math.pi*k*Ro*Ri)/(Ro-Ri)\n", + "print \"the rate of heat transfer is %.0f W\"%(Q)\n", + "#Refrigeration capacity(RC)\n", + "#3516 Watt = 1 ton\n", + "RC = -Q/3516\n", + "print \"Refrigeration capacity is %.2f tons\"%(RC)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6 Page No : 28" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the temprature midway in the rod at steady state is 167.3 C\n", + "Temprature gradient at one end of the rod is 559 C/W\n", + "Temprature gradient at other end of the rod is 521.8 C/W\n" + ] + } + ], + "source": [ + "import math \n", + "# Variables\n", + "d = 0.05 \t\t\t#m, diameter of rod\n", + "l = 0.5 \t\t\t#m, length of rod\n", + "T1 = 30. \t\t\t#CTemp. at one end (1)\n", + "T2 = 300. \t\t\t#C, temp at other end (2)\n", + "\n", + "# Calculations and Results\n", + "x1 = l/2 \t\t\t#m, at mid plane\n", + "#temprature distribution ,\n", + "#comparing with quadratic eq. ax**2+bx+c \n", + "#and its solution as x = (-b+math.sqrt(b**2-4*a*c))/2*a\n", + "a = 1.35*10**-4\n", + "b = 1\n", + "c = -(564*x1+30.1)\n", + "T = (-b+math.sqrt(b**2-4*a*c))/(2*a)\n", + "print \"the temprature midway in the rod at steady state is %.1f C\"%(T)\n", + "\n", + "#Temprature gradient at the ends of the rod\n", + "x2 = 0 \t\t\t#m, at one end\n", + "a1 = 1.35*10**-4\n", + "b1 = 1\n", + "c1 = -(564*x2+30.1)\n", + "T1 = (-b1+math.sqrt(b1**2-4*a1*c1))/(2*a1)\n", + "k1 = 202+0.0545*T1 \n", + "C1 = 113930 \t\t\t#integration consmath.tant from eq. (1)\n", + "TG1 = C1/k1 \t\t\t#C/W, temprature gradient, dT/dx\n", + "#similarly\n", + "x3 = 0.5\n", + "a2 = 1.35*10**-4\n", + "b2 = 1\n", + "c2 = -(564*x3+30.1)\n", + "T2 = (-b2+math.sqrt(b2**2-4*a2*c2))/(2*a2)\n", + "k2 = 202+0.0545*T2\n", + "TG2 = C1/k2\n", + "print \"Temprature gradient at one end of the rod is %.0f C/W\"%(TG1)\n", + "print \"Temprature gradient at other end of the rod is %.1f C/W\"%(TG2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7 Page No : 29" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At the surface x = 0, the temp. is 600 C\n", + "At the surface x = 0.3m, the temp. is 270 C\n", + "Rhe average temprature of the wall is 615 C\n", + "The maximum temprature occurs at 0.104 m\n", + "The maximum temp. is 730 C\n", + "heat flux at left surface is -58750 W/m**2\n", + "heat flux at right surface is 110450 W/m**2\n", + "The average volumetric rate if heat genaration is 564000 W/m**3 \n" + ] + } + ], + "source": [ + "import math \n", + "from scipy.integrate import quad \n", + "# Variables\n", + "#temprature distribution in wall\n", + "\n", + "t = 0.3 \t\t\t#m, thickness of wall\n", + "k = 23.5 \t\t\t#W/m c thermal conductivity of wall\n", + "\n", + "# Calculations and Results\n", + "x1 = 0\n", + "T1 = 600+2500*x1-12000*x1**2 \t\t\t#C, at surface\n", + "x2 = 0.3\n", + "T2 = 600+2500*x2-12000*x2**2 \t\t\t#C, at x = 0.3\n", + "\n", + "def f3(x): \n", + " return 600+2500*x-12000*x**2\n", + "\n", + "Tav = 1/t* quad(f3,0,0.3)[0]\n", + "\n", + "print \"At the surface x = 0, the temp. is %.0f C\"%(T1)\n", + "print \"At the surface x = 0.3m, the temp. is %.0f C\"%(T2)\n", + "print \"Rhe average temprature of the wall is %.0f C\"%(Tav)\n", + "\n", + "#(b)\n", + "\n", + "#for maximum temprature D = 0\n", + "x3 = 2500/24000.\n", + "print \"The maximum temprature occurs at %.3f m\"%(x3)\n", + "Tmax = 600+2500*x3-12000*x3**2\n", + "print \"The maximum temp. is %.0f C\"%(Tmax)\n", + "\n", + "#(c)\n", + "D1 = 2500-24000*x1 \t\t\t#at x = 0, temprature gradient\n", + "Hf1 = -k*D1 \t\t\t#W/m**2, heat flux at left surface(x = 0)\n", + "D2 = 2500-24000*x2 \t\t\t#at x = 0.3, temprature gradient\n", + "Hf2 = -k*D2 \t\t\t#W/m**2, heat flux at right surface(x = 0.3)\n", + "print \"heat flux at left surface is %.0f W/m**2\"%(Hf1)\n", + "print \"heat flux at right surface is %.0f W/m**2\"%(Hf2)\n", + "\n", + "#(d)\n", + "Qt = Hf2-Hf1 \t\t\t#W/m**2, total rate of heat loss\n", + "Vw = 0.3 \t\t\t#m**3/m**2, volume of wall per unit surface area\n", + "Hav = Qt/Vw \t\t\t#W/m**3, average volumetric rate\n", + "print \"The average volumetric rate if heat genaration is %.0f W/m**3 \"%(Hav) \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.8 Page No : 30" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum temp. will occur at a position 0.209 m\n", + "The maximum temprature is 152.6 C\n" + ] + } + ], + "source": [ + "import math \n", + "# Variables\n", + "ka = 24 \t\t\t#W/mC thermal conductivitiy of material A\n", + "tA = 0.1 \t\t\t#m, thickness of A material\n", + "kB = 230 \t\t\t#W/mC thermal conductivity of metl B\n", + "kC = 200 \t\t\t#W/mC thermal conductivity of metal C\n", + "tB = 0.1 \t\t\t#m, thickness of B metal\n", + "tC = 0.1 \t\t\t#m, thickness of C metal\n", + "TBo = 100 \t\t\t#C, outer surface temp. of B wall\n", + "TCo = 100 \t\t\t#C, outer surface temp. of C wall\n", + "Q = 2.5*10**5 \t\t\t#W/m**3, heat generated\n", + "\n", + "# Calculations and Results\n", + "#At D = 0\n", + "x = 2175./10416\n", + "print \"The maximum temp. will occur at a position %.3f m\"%(x)\n", + "x1 = x\n", + "TA = -5208*x1**2+2175*x1-74.5\n", + "print \"The maximum temprature is %.1f C\"%(TA)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.9 Page No : 36" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer.\n", + " Similarly no temprature maximum occurs in layer 2.\n", + "The maximum temprature at the outer boundary is 200 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "di = 0.15 \t\t\t#m, inner diameter\n", + "do = 0.3 \t\t\t#m, outer diameter\n", + "Q1 = 100.*10**3 \t\t\t#W/,m**3,inner rate of heat generation\n", + "Q2 = 40.*10**3 \t\t\t#W/m**3, outer rate of heat generation\n", + "Ti = 100. \t\t\t#C, temp.at inside surface\n", + "To = 200. \t\t\t#C, temp. at outside surface\n", + "k1 = 30. \t\t\t#W/m C, thermal conductivity of material for inner layer\n", + "k2 = 10. \t\t\t#W/m C, thermal conductivity of material for outer layer\n", + "\n", + "# Calculations and Results\n", + "#T1 = 364+100*math.log(r)-833.3*r**2 (1)\n", + "#T2 = 718+216*math.log(r)-1000*r**2 (2)\n", + "#(b)from eq. 1\n", + "r = math.sqrt(100./2*833.3)\n", + "print \"This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer.\"\n", + "#similarly\n", + "print \" Similarly no temprature maximum occurs in layer 2.\"\n", + "ro = di \t\t # m, outer boundary\n", + "Tmax = To\n", + "print \"The maximum temprature at the outer boundary is %.0f C\"%(Tmax)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb new file mode 100644 index 00000000..ba532e83 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb @@ -0,0 +1,873 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Heat transfer coefficient" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 Page No : 53" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time required for melting the ice is 4274 s\n" + ] + } + ], + "source": [ + "# Variables\n", + "di = 0.06 \t\t\t#m,initial diameter of iceball\n", + "T1 = 30. \t\t\t#C, room temp.\n", + "T2 = 0. \t\t\t#ice ball temp.\n", + "h = 11.4 \t\t\t#W/m**2 C, heat transfer coefficient\n", + "x = 40. \t\t\t#% for reduction\n", + "rho = 929. \t\t\t#kg/m**3, density of ice\n", + "Lv = 3.35*10**5 \t\t\t#j/kg, latent heat of fusion\n", + "\n", + "# Calculations\n", + "# m = 4/3*math.pi*r**3 \t\t\t#kg,mass of ice ball\n", + "#rate of melting = -dm/dt\n", + "#rate of heat adsorption = -4*math.pi*r**2*rho*dr/dt*lamda\n", + "#at initial time t = 0\n", + "C1 = di/2 \t\t\t#consmath.tant of integration\n", + "#if the volume of the ball is reduced by 40% of the original volume \n", + "r = ((1-x/100)*(di/2)**3)**(1./3)\n", + "#time required for melting umath.sing eq. 1\n", + "t = (di/2-r)/(h*(T1-T2)/(rho*Lv))\n", + "\n", + "# Results\n", + "print \"The time required for melting the ice is %.0f s\"%(t)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 Page No : 54" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time required for the heating coil is 4.9 s\n" + ] + } + ], + "source": [ + "import math\n", + "from scipy.integrate import quad \n", + "#calculate the time required for the heating coil.\n", + "\n", + "# Variables\n", + "P = 1.*10**3 \t\t\t#W, electrical heating capacity\n", + "V = 220. \t\t\t#V, applied voltage\n", + "d = 0.574*10**-3 \t\t\t#m, diameter of wire\n", + "R = 4.167 \t\t\t#ohm, electrical resistance\n", + "Tr = 21. \t\t\t#C, room temp.\n", + "h = 100. \t\t\t#W/m**2 C, heat transfer coefficient\n", + "rho = 8920. \t\t\t#kg/m**3, density of wire\n", + "cp = 384. \t\t\t#j/kg C, specific heat of wire\n", + "percent = 63. \t\t\t#%, percent of the steady state\n", + "\n", + "#Calculation\n", + "R_ = V**2/P \t\t\t#ohm, total electrical resistance\n", + "l = R_/R \t\t\t#m, length of wire\n", + "A = math.pi*d*l \t\t\t#m**2, area of wire\n", + "Tf = P/(h*A)+Tr \t\t\t#final temp.\n", + "dtf = Tf-Tr \t\t\t#C. steady state temp. rise\n", + "#temp. of wire after 63% rise\n", + "T = Tr+(percent/100)*dtf \n", + "#rate of heat accumulation on the wire\n", + "#d/dt(m*cp*T) (1)\n", + "#rate of heat loss\n", + "#h*A*(T-Tr).........................(2)\n", + "#heat balance eq. (1) = (2)\n", + "m = math.pi*d**2*l*rho/4 \t\t\t#kg. mass of wire\n", + "#integrating heat balance eq.\n", + "\n", + "def f6(T): \n", + " return 1/((P/(m*cp))-((h*A)/(m*cp))*(T-Tr))\n", + "\n", + "t = quad(f6,21,322)[0]\n", + "\n", + "# Results\n", + "print \"The time required for the heating coil is %.1f s\"%(t)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 Page No : 56" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " the heat transfer coefficient is 11.63 W/m**2 C \n", + "So there is no heat flow at other surface of the wall \n", + "average volumetric rate of heat generation is 6396 W/m**3\n" + ] + } + ], + "source": [ + "# Variables\n", + "t = 0.2 \t\t\t#m, thickness of wall\n", + "k = 1.163 \t\t\t#W/m C, thermal conductivity of material\n", + "Ta = 30 \t\t\t#C, ambient temp\n", + "\n", + "# Calculations and Results\n", + "#(a) at x = 0.2 let T = T1 at x = x1\n", + "x1 = 0.2\n", + "T1 = 250-2750*x1**2\n", + "#let D = dT/dx\n", + "D = -5500*0.2 \t\t\t#C/m, at x = 0.2\n", + "h = -k*D/(T1-Ta)\n", + "print \" the heat transfer coefficient is %.2f W/m**2 C \"%(h)\n", + "\n", + "#(b)at other surface of wall, x = 0 = x2 (say)\n", + "x2 = 0\n", + "a = -5500*0\n", + "print \"So there is no heat flow at other surface of the wall \"\n", + "\n", + "#(c)\n", + "A = 1 \t\t\t#m**2, area\n", + "Vw = A*x1 \t\t\t#m**3, volume of wall\n", + "HL = h*(T1-Ta) \t\t\t#W, heat loss from unit area\n", + "Vav = HL/x1\n", + "print \"average volumetric rate of heat generation is %.0f W/m**3\"%(Vav)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 Page No : 61" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The rate of heat loss is 150.9 W\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve \n", + "import math\n", + "# Variables\n", + "id_ = 97.*10**-3 \t\t\t#m,internal diameter of steam pipe\n", + "od = 114.*10**-3 \t\t\t#m,outer diameter of steam pipe\n", + "pr = 30. \t\t\t#bar, absolute pressure os saturated steam\n", + "Ti = 234. \t\t\t#C, temp. at 30 bar absolute pressure\n", + "Ts = 55. \t\t\t#C, skin temp.\n", + "To = 30. \t\t\t#C, ambient temp.\n", + "kc = 0.1 \t\t\t#W/m C, thermal conductivity of wool\n", + "kw = 43. \t\t\t#W/m C, thermal conductivity of pipe\n", + "h = 8. \t\t\t#W/m**2 C, external air film coefficient \n", + "L = 1. \t\t\t#m, assume length\n", + "\n", + "#Calculation\n", + "ri = id_/2 \t\t\t#m, \n", + "r1 = (114.*10**-3)/2 \t\t\t#m,outer radius of steam pipe\n", + "\n", + "#thermal resistance of insulation\n", + "#Ri = math.log(ro/r1)/(2*math.pi*L*kc)\n", + "#Thermal resistance of pipe wall\n", + "Rp = math.log(r1/ri)/(2*math.pi*L*kw)\n", + "#RT = Ri+Rp\n", + "DF = Ti-Ts \t\t\t#C, driving force\n", + "#At steady state the rate of heat flow through the insulation\n", + "# and the outer air film are equ\n", + "\n", + "#by trial and error method :\n", + "def f(ro): \n", + " return (Ti-Ts)/(math.log(ro/r1)/kc+math.log(r1/ri)/kw)-(h*ro*(Ts-To))\n", + "ro = fsolve(f,0.1)\n", + "th = ro-r1 \t\t\t#m, required thickness of insulation\n", + "Q = 2*math.pi*ro*h*L*(Ts-To)\n", + "\n", + "# Results\n", + "print \"The rate of heat loss is %.1f W\"%(Q)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 Page No : 62" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "effective thickness of air is 0.75 mm\n", + "effective thickness of liquid films is 2.6 mm.\n", + "the overall heat transfer coefficient based on i.d of pipe is 2.707 W/m**2 C\n", + "the overall heat transfer coefficient based on od of pipe is 1.025 W/m**2 C\n", + "the percentage of total resistance offered by air film. is 10.25 percent\n", + "Rate of heat loss is 21.2 W\n", + "insulation skin temp.is 32.8 C\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "w1 = 8. \t\t\t#%, solubility of alcohol\n", + "w2 = 92. \t\t\t#%, solubility of water\n", + "k1 = 0.155 \t\t\t#W/m C, thermal conductivity of alcohol\n", + "k2 = 0.67 \t\t\t#W/m C thermal conductivity of water\n", + "ka = 0.0263 \t\t\t#W/m C thermal conductivity of air\n", + "kw = 45. \t\t\t#W/m Cthermal conductivity of pipe wall\n", + "ki = 0.068 \t\t\t#W/m C , thermal cond. of glass\n", + "id_ = 53.*10**-3 \t\t\t#m, internal diameter of pipe\n", + "od = 60.*10**-3 \t\t\t#m, outer diameter of pipe\n", + "t = 0.04 \t\t\t#m, thickness of insulation\n", + "hi = 800. \t\t\t#W/m**2 C, liquid film coefficient\n", + "ho = 10. \t\t\t#W/m**2 C, air film coefficient\n", + "L = 1. \t\t\t#m, length of pipe\n", + "T1 = 75. \t\t\t#C, initial temp.\n", + "T2 = 28. \t\t\t#C, ambient air temp.\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "km = (w1/100)*k1+(w2/100)*k2-0.72*(w1/100)*(w2/100)*(-(k1-k2))\n", + "deli = km/hi \t\t\t#m, effective thickness of liquid film\n", + "delo = ka/ho \t\t\t#m, effective thickness of air film\n", + "print \"effective thickness of air is %.2f mm\"%(deli*10**3)\n", + "print \"effective thickness of liquid films is %.1f mm.\"%(delo*10**3)\n", + "\n", + "#(b)\n", + "Ai = 2*math.pi*id_/2*L \t\t\t#m**2, inside area\n", + "ri = id_/2 \t\t\t#m,inside radius of pipe\n", + "r_ = od/2 \t\t\t#m, outside radius of pipe\n", + "ro = r_+t \t \t\t#m, outer radius of insulation\n", + "Ao = 2*math.pi*ro*L \t\t \t#m**2, outer area\n", + "#from eq. 3.11, overall heat transfer coefficient\n", + "Ui = 1/(1/hi+(Ai*math.log(r_/ri))/(2*math.pi*L*kw)+(Ai*math.log(ro/r_))/(2*math.pi*L*ki)+Ai/(Ao*ho))\n", + "print \"the overall heat transfer coefficient based on i.d of pipe is %.3f W/m**2 C\"%(Ui)\n", + "\n", + "#(c)\n", + "#frim eq. 3.14\n", + "Uo = Ui*Ai/Ao \n", + "print \"the overall heat transfer coefficient based on od of pipe is %.3f W/m**2 C\"%(Uo)\n", + "\n", + "#(d)\n", + "R = 1/(Ui*Ai) \t\t\t#C/W, total heat transfer resistance\n", + "Rair = 1/(Ao*ho) \t\t\t#C/W, heat transfer resistance of air film\n", + "p = Rair/R\n", + "print \"the percentage of total resistance offered by air film. is %.2f percent\"%(p*100)\n", + "\n", + "#(e)\n", + "Q = Ui*Ai*(T1-T2)\n", + "print \"Rate of heat loss is %.1f W\"%(Q)\n", + "\n", + "#(f)\n", + "Ts = Uo*Ao*(T1-T2)/(ho*Ao)+T2\n", + "print \"insulation skin temp.is %.1f C\"%(Ts)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 Page No : 64" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inlet liquid temp. should be 82 C \n", + " the insulation skin temp. at the flat top surface is 35 C \n", + "similarly the insulation skin temp at cylindrical surface is 38 C\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "id_ = 1.5 \t\t\t#m, internal diameter of math.tank\n", + "h = 2.5 \t\t\t#m, height of math.tank\n", + "t1 = 0.006 \t\t\t#m, thickness of wall\n", + "t2 = 0.04 \t\t\t#m, thickness of insulation\n", + "Ta = 25. \t\t\t#C, ambient temp.\n", + "T1 = 80. \t\t\t#C, outlet temp. of liquid\n", + "cp = 2000. \t\t\t#j/kg C, specific heat of liquid\n", + "FR = 700./3600 \t\t\t#KG/s, Liquid flow rate\n", + "\n", + "# Calculations and Results\n", + "ri = id_/2+t1 \t\t\t#m, inner radius of insulation\n", + "ro = ri+t2 \t\t\t#m, outer radius of insulation\n", + "ki = 0.05 \t\t\t#W/m C, thermal conductivity of insulation\n", + "hc = 4 \t\t\t#W/m**2 C, heat transfer coefficient at cylindrical surface\n", + "ht = 5.5 \t\t\t#W/m**2 C, heat transfer coefficient at flat surface\n", + "l = h+t1+t2 \t\t\t#m, height of the top of insulation\n", + "#fromm eq. 3.10\n", + "#heat transfer resistance of cylindrical wall\n", + "Rc = math.log(ro/ri)/(2*math.pi*l*ki)+1/(2*math.pi*ro*l*hc)\n", + "#heat transfer resistance of flat insulated top surface\n", + "Ri = (1/(math.pi*ro**2))*((ro-ri)/ki+1/ht)\n", + "tdf = T1-Ta \t\t\t#C, temp. driving force\n", + "Q = tdf/Rc + tdf/Ri \t\t\t#W, total rate of heat loss\n", + "Tt = Q/(FR*cp)+T1 \t\t\t#C, inlet temp. of liquid\n", + "print \"Inlet liquid temp. should be %.0f C \"%(Tt)\n", + "Q1 = tdf/Ri \t\t\t#W, rate of heat loss from flat surface\n", + "T1 = Q1/(math.pi*ro**2*ht)+Ta \n", + "print \" the insulation skin temp. at the flat top surface is %.0f C \"%(T1)\n", + "#similarly\n", + "T2 = 38\n", + "print \"similarly the insulation skin temp at cylindrical surface is %.0f C\"%(T2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7 Page No : 66" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the heat imput to the boiling.is 191.2 W\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "id_ = 2.5*10**-2 \t\t\t#m, internal diameter of glass tube\n", + "t = 0.3*10**-2 \t\t\t#m, thickness of wall\n", + "l = 2.5 \t\t\t#m, length of nichrome wire\n", + "L = 0.12 \t\t\t#m, length of steel covered with heating coil\n", + "Re = 16.7 \t\t\t#ohm, electrical resistance\n", + "ti = 2.5*10**-2 \t\t\t#m, thickness of layer of insulation\n", + "kg = 1.4 \t\t\t#W/m C, thermal conductivity of glass\n", + "ki = 0.041 \t\t\t#W/m C, thermal conductivity of insulation\n", + "T1 = 91. \t\t\t#C, boiling temp. of liquid\n", + "T2 = 27. \t\t\t#C, ambient temp.\n", + "ho = 5.8 \t\t\t#W/m **2 C outside air film coefficient\n", + "V = 90. \t\t\t#V, voltage\n", + "\n", + "#Calculation\n", + "Rc = Re*l \t\t\t#ohm, resistance of heating coil\n", + "Q = V**2/Rc \t\t\t#W, rate of heat generation\n", + "ri = id_/2 \t\t\t#m, inner radius of glass tube\n", + "r_ = ri+t \t\t\t#m, outer radius of glass tube\n", + "ro = r_+ti \t\t\t#m,outer radius of insulation\n", + "#heat transfer resistance of glass wall\n", + "Rg = math.log(r_/ri)/(2*math.pi*L*kg)\n", + "#combined resistance of insulation and outer air film\n", + "Rt = math.log(ro/r_)/(2*math.pi*L*ki)+1/(2*math.pi*ro*L*ho)\n", + "#Rate of heat input to the boiling liquid in steel = Q1 = (Ts-T1)/Rg\n", + "#Rate of heat loss through insulation ,Q2 = (Ts-To)/(Rt)\n", + "#Q1+Q2 = Q\n", + "Ts = (Q+ T1/Rg +T2/Rt)/(1/Rg +1/Rt)\n", + "Q1 = (Ts-T1)/Rg\n", + "Q2 = Q-Q1\n", + "\n", + "# Results\n", + "print \"the heat imput to the boiling.is %.1f W\"%(Q1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.8 Page No : 68" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum allowable current is 54.04 A\n", + "remp. at the centre of wire is 90.005 C\n", + "The temprature at the outer surface of insulation is 80.3 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "ri = 1.3*10**-3 \t\t\t#m, radius of 10 gauge wire\n", + "t = 1.3*10**-3 \t\t\t#m, thickness of rubber insulation\n", + "Ti = 90. \t\t\t#C, temp. 0f insulation\n", + "To = 30. \t\t\t#C, ambient temp.\n", + "h = 15. \t\t\t#W/m**2 C, air film coefficient\n", + "km = 380. \t\t\t#W/m C, thermal cond. of copper\n", + "kc = 0.14 \t\t\t#W/m C, thermal cond. of rubber(insulation)\n", + "Rc = 0.422/100 \t\t\t#ohm/m, eletrical resistance of copper wire\n", + "\n", + "# Calculations and Results\n", + "Tcmax = 90. \t\t\t#X, the maximum temp. in insulation\n", + "ro = ri+t \t\t\t#m, outside radius of 10 gauge wire\n", + "Sv = ((Tcmax-To)*(2*kc/ri**2))/(math.log(ro/ri)+kc/(h*ro))\n", + "I = (math.pi*ri**2*Sv/Rc)**0.5 \t\t\t#A, Current strength\n", + "print \"maximum allowable current is %.2f A\"%(I)\n", + "\n", + "#(b) at r = 0\n", + "Tm = To+(ri**2*Sv/2)*(1/km+(math.log(ro/ri))/kc+1/(h*ro))\n", + "print \"remp. at the centre of wire is %.3f C\"%(Tm)\n", + "\n", + "#at r = ro\n", + "Tc = 30+(ri**2*Sv/(2*kc))*(kc/(h*ro))\n", + "print \"The temprature at the outer surface of insulation is %.1f C\"%(Tc)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9 Page No : 72" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At x = 0.1 the temp. at the surface of slab A is 430 C\n", + "At x = 0.35 the temp. at the surface of slab A is 318 C\n", + " the maximum Temp. in A occurs at 0.2045 m\n", + " the maximum Temp. in A is 550.2 TAmax \n", + "temp. gradient at interface 2 of the slabs A is 2300 C/W\n", + "temp. gradient at interface 3 of the slabs A is -3200 C/W\n", + "temp. gradient at interface 2 of the slabs B is 3450 C/W\n", + "temp. gradient at interface 1 of the slabs B is 3450 C/W\n", + "temp. gradient at interface 3 of the slabs C is -1600 C/W\n", + "temp. gradient at interface 4 of the slabs C is -1600 C/W\n", + "The heat transfer coefficient at one surface of solid fluid interface is 766.7 W/m**2 C\n", + "The heat transfer coefficient at other surface of solid fluid interface is 1129 W/m**2 C\n" + ] + } + ], + "source": [ + "# Variables\n", + "tA = 0.25 \t\t\t#m, thickness of slab A\n", + "tB = 0.1 \t\t\t#m, thickness of slab B\n", + "tC = 0.15 \t\t\t#m, thickness of slab C\n", + "kA = 15. \t\t\t#W/m C, thermal comductivity of slab A\n", + "kB = 10. \t\t\t#W/m C, thermal comductivity of slab B\n", + "kC = 30. \t\t\t#W/m C, thermal comductivity of slab C\n", + "#Temprature distribution in slab A\n", + "T1 = 40. \t\t\t#C, fluid temp.\n", + "T2 = 35. \t\t\t#C, medium temp.\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "x1 = tB \n", + "TA1 = 90.+4500*x1-11000*x1**2\n", + "#similarly at the right surface\n", + "x2 = tA+tB\n", + "TA2 = 90.+4500*x2-11000*x2**2\n", + "#let dTA/dx = D\n", + "D = 0 \t\t\t#for maximum temp.\n", + "x3 = 4500./22000\n", + "TAmax = 90.+4500*x3-11000*x3**2\n", + "print \"At x = 0.1 the temp. at the surface of slab A is %.0f C\"%(TA1)\n", + "print \"At x = 0.35 the temp. at the surface of slab A is %.0f C\"%(TA2)\n", + "print \" the maximum Temp. in A occurs at %.4f m\"%(x3)\n", + "print \" the maximum Temp. in A is %.1f TAmax \"%(TAmax)\n", + "\n", + "#(b)\n", + "#At the interface 2\n", + "D1 = 4500-2.*11000*x1 \t\t\t#C/W, D1 = dTA/dx, at x = 0.1\n", + "#At the interface 3\n", + "D2 = 4500-2.*11000*x2 \t\t\t#D12 = dTA/dx, at x = 0.35\n", + "#Temprature gradient in slab B and C\n", + "#by umath.sing the continuity of heat flux at interface (2)\n", + "D3 = -kA*D1/(-kB) \t\t\t#D3 = dTB/dx, at x = 0.1\n", + "#at interface (1)\n", + "D4 = D3 \t\t\t#D4 = dTB/dx at x = 0\n", + "#similarly \n", + "D5 = -1600. \t\t\t#C/W, dTB/dx, x = 0.35\n", + "D6 = D5 \t\t\t#at interface 4\n", + "print \"temp. gradient at interface 2 of the slabs A is %.0f C/W\"%(D1)\n", + "print \"temp. gradient at interface 3 of the slabs A is %.0f C/W\"%(D2)\n", + "print \"temp. gradient at interface 2 of the slabs B is %.0f C/W\"%(D3)\n", + "print \"temp. gradient at interface 1 of the slabs B is %.0f C/W\"%(D4)\n", + "print \"temp. gradient at interface 3 of the slabs C is %.0f C/W\"%(D5)\n", + "print \"temp. gradient at interface 4 of the slabs C is %.0f C/W\"%(D6)\n", + "\n", + "#(c)\n", + "#from D3 = 3450 and TB = beeta1*x+beeta2\n", + "beeta1 = 3450.\n", + "beeta2 = 85.\n", + "x = 0.\n", + "TB = beeta1*x+beeta2\n", + "#similary\n", + "TC = 877.5-1600*x\n", + "h1 = -kB*D4/(T1-TB)\n", + "#similarly\n", + "h2 = 1129.\n", + "print \"The heat transfer coefficient at one surface of solid fluid interface is %.1f W/m**2 C\"%(h1)\n", + "print \"The heat transfer coefficient at other surface of solid fluid interface is %.0f W/m**2 C\"%(h2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.10 Page No : 79" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the percentage increase in the rate of heat transfer is 103.6 percent \n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "id_ = 78.*10**-3 \t\t\t#m, actual internal dia of pipe\n", + "tw = 5.5*10**-3 \t\t\t#m, wall thickness\n", + "nl = 8. \t\t\t#no. of longitudinal fins\n", + "tf = 1.5*10**-3 \t\t\t#m, thickness of fin\n", + "w = 30.*10**-3 \t\t\t#m,breadth of fin\n", + "kf = 45. \t\t\t#W/m C, thermal conductivity of fin \n", + "Tw = 150. \t\t\t#C, wall temp.\n", + "To = 28. \t\t\t#C, ambient temp.\n", + "h = 75. \t\t\t#W/m**2C, surface heat transfer coefficient\n", + "\n", + "#Calculation\n", + "#from eq. 3.27\n", + "e = math.sqrt(2*h/(kf*tf)) \n", + "n = (1./(e*w))*math.tanh(e*w) \t\t\t#efficiency of fin\n", + "L = 1. \t\t\t#m, length of fin\n", + "Af = 2.*L*w \t\t\t#m**2, area of math.single fin\n", + "Atf = nl*Af \t\t\t#m**2 total area of fin\n", + "Qmax = h*Atf*(Tw-To) \t\t\t#W, maximum rate of heat transfer\n", + "Qa = n*Qmax \t\t\t#W, actual rate of heat transfer\n", + "Afw = L*tf \t\t\t#m**2, area of contact of fin with pipe wall\n", + "Atfw = Afw*nl \t\t\t#m**2 , area of contact of all fin with pipe wall\n", + "ro = id_/2+tw \t\t\t#m, outer pipe radius\n", + "A = 2*math.pi*L*ro \t\t\t#m**2 area per meter\n", + "Afree = A-Atfw \t\t\t#m**2, free outside area of finned pipe\n", + "#Rate of heat transfer from free area of pipe wall\n", + "Q1 = h*Afree*(Tw-To) \t\t\t#W, \n", + "#total rate of hewat gtransfer from finned pipe\n", + "Qtotal = Qa+Q1 \t\t\t#W\n", + "#Rate of heat transfer fromm unfinned pipe\n", + "Q2 = h*A*(Tw-To)\n", + "per = (Qtotal-Q2)/Q2\n", + "\n", + "# Results\n", + "print \"the percentage increase in the rate of heat transfer is %.1f percent \"%(per*100)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.11 Page No : 80" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer in the absence of contact resistance is 11.585 KW\n", + "The actual rate of heat loss is 5.18kW is much less than this value. So there is a thermal contact resistance at the interface between the layers \n", + "The contact resistance is 0.001067 C/W \n", + "contact heat transfer coefficient is 298.2 W/m**2 C \n", + "The temprature jump is 5.5 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "id_ = 90.*10**-2 \t\t\t#m, internal diameter of steel\n", + "od = 110.*10**-2 \t\t\t#m, outer diameter of steel\n", + "Ti = 180. \t\t\t#C, inside temp. of steel\n", + "To = 170. \t\t\t#C, outside temp. of steel\n", + "k = 37. \t\t\t#W/m C, thermal conductivity of alloy\n", + "Q = 5.18*10**3 \t\t\t#W, Rate of heat loss\n", + "\n", + "# Calculations and Results\n", + "ri = id_/2 \t\t\t#m, inside radius of shell\n", + "ro = od/2 \t\t\t#m, outside radius of shell\n", + "r_ = 0.5 \t\t\t#m, boundary between the layers\n", + "L = 1 \t\t\t#m, length of shell\n", + "#Rate of heat transfer in the absence of contact resistance\n", + "Q1 = 2*math.pi*L*k*(Ti-To)/(math.log(ro/ri)) \n", + "print \"Rate of heat transfer in the absence of contact resistance is %.3f KW\"%(Q1/1000)\n", + "print \"The actual rate of heat loss is 5.18kW is much less than this value\\\n", + ". So there is a thermal contact resistance at the interface between the layers \"\n", + "\n", + "#(b)\n", + "Ri = (math.log(r_/ri)/(2*math.pi*L*k)) \t\t\t#C/W, resistance of inner layer\n", + "Ro = (math.log(ro/r_)/(2*math.pi*L*k)) \t\t\t#C/W, resistance of outer layer\n", + "Rc = ((Ti-To)/(Q))-(Ri+Ro) \t\t\t#C/W, contact resistance\n", + "print \"The contact resistance is %f C/W \"%(Rc)\n", + "Ac = 2*math.pi*L*r_ \t\t\t#m**2, area of contact surface of shell\n", + "hc = 1/(Ac*Rc) \t\t\t #W/m**2 c, contact heat transfer coefficient\n", + "print \"contact heat transfer coefficient is %.1f W/m**2 C \"%(hc)\n", + "\n", + "#(c)\n", + "dt = Q/(hc*Ac)\n", + "print \"The temprature jump is %.1f C\"%(dt)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.12 Page No : 84" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the critical thickness is 35.29 mm\n" + ] + } + ], + "source": [ + "# Variables\n", + "d = 5.2*10**-3 \t\t\t#m, diameter of copper wire\n", + "ri = d/2 \t\t\t#inner radius of insulation\n", + "kc = 0.43 \t\t\t#W/m C, thermal conductivity of PVC\n", + "Tw = 60. \t\t\t#C, temp. 0f wire\n", + "h = 11.35 \t\t\t#W/m**2 C, film coefficient\n", + "To = 21. \t\t\t#C, ambient temp.\n", + "\n", + "#calculation\n", + "Ro = kc/h \t\t\t#m,critical outer radius of insulation\n", + "t = Ro-ri\n", + "\n", + "# Results\n", + "print \"the critical thickness is %.2f mm\"%(t*10**3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.13 Page No : 85" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ro = 3.5 cm \n", + "Radius of bare pipe is larger than outer radius of insulation So critical insulation thickness does not exist \n" + ] + } + ], + "source": [ + "# calculate the critical insulation thickness.\n", + "\n", + "# Variables\n", + "d = 15.*10**-2 \t\t\t#m, length of steam main\n", + "t = 10.*10**-2 \t\t\t#m, thickness of insulation\n", + "ki = 0.035 \t\t\t#W/m C, thermal conductivity of insulation\n", + "h = 10. \t\t\t#W/m**2 C, heat transfer coefficient\n", + "\n", + "#calculation\n", + "#from eq. 3.29\n", + "ro = ki/h\n", + "\n", + "# Results\n", + "print \"ro = %.1f cm \"%(ro*10**3)\n", + "print \"Radius of bare pipe is larger than outer radius of insulation So critical \\\n", + " insulation thickness does not exist \"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.14 Page No : 87" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The optimum insulation thickness is 71 mm\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve\n", + "import math\n", + "\n", + "# Variables\n", + "Ti = 172. \t\t\t#C, saturation temp.\n", + "To = 20. \t\t\t#C, ambient temp.\n", + "Cs = 700. \t\t\t#per ton, math.cost of steam\n", + "Lv = 487. \t\t\t#kcal/kg, latent heat of steam\n", + "ho = 10.32 \t\t\t#kcal/h m**2 C, outer heat transfer coefficient\n", + "kc = 0.031 \t\t\t#W/m C, thermal conductivity of insulation\n", + "n = 5. \t\t\t#yr, service life of insulation\n", + "i = 0.18 \t\t\t#Re/(yr)(Re), interest rate\n", + "\n", + "#Calculation\n", + "di = 0.168 \t\t\t#m, inner diameter of insulation\n", + "#Cost of insulation\n", + "Ci = 17360.-(1.91*10**4)*di \t\t\t#Rs/m**3\n", + "Ch = Cs/(1000*Lv) \t\t\t#Rs/cal, math.cost of heat energy in steam\n", + "sm = 1./(1+i)+1/(1+i)**2+1/(1+i)**3+1/(1+i)**4+1/(1+i)**n\n", + "#from eq. 3.33\n", + "ri = di/2 \t\t\t#m inner radius of insulation\n", + "L = 1 \t\t\t#m, length of pipe\n", + "#Pt = Ch*sm*2*math.pi*ri*L*( 1/(((ri/kc)*('math.log(ro/ri)'))+ri/(ho*ro)))*7.2*10**3*(Ti-To)+math.pi*(ro**2-ri**2)*L*Ci\n", + "#On differentiating , dpt/dro = -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2\n", + "def f(ro): \n", + " return -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2+98960*ro\n", + "ro = fsolve(f,0.1)\n", + "t = ro-ri\n", + "\n", + "# Results\n", + "print \"The optimum insulation thickness is %.0f mm\"%(t*1000)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb new file mode 100644 index 00000000..38ba4576 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb @@ -0,0 +1,771 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 : Forced Convection" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2 Page No : 112" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The rate of heat loss is 5328 W\n" + ] + } + ], + "source": [ + "# Variables\n", + "l = 2. \t\t\t#m, length of flat surface\n", + "T1 = 150. \t\t\t#C, surface temp.\n", + "p = 1. \t\t\t#atm, pressure\n", + "T2 = 30. \t\t\t#C, bulk air temp.\n", + "V = 12. \t\t\t#m/s, air velocity\n", + "\n", + "#Calculation\n", + "Tf = (T1+T2)/2 \t\t\t#C, mean air film temp.\n", + "mu = 2.131*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "k = 0.031 \t\t\t#W/m C, thermal conductivity\n", + "rho = 0.962 \t\t\t#kg/m**3, density of air\n", + "cp = 1.01 \t\t\t#kj/kg C, specific heat of air\n", + "Pr = cp*10**3*mu/k \t\t\t#Prandtl no.\n", + "Remax = l*V*rho/mu \t\t\t#maximum Reynold no.\n", + "Re = 5.*10**5 \t\t\t#Reynold no. during transition to turbulent flow \n", + "L_ = (Re*mu)/(V*rho) \t\t\t#m,dismath.tance from the leading edge\n", + "#for laminar flow heat transfer coefficient h, \n", + "#h16.707*x**-(1/2)\n", + "#(a)\n", + "#h2 = 31.4*x**(-1/5)\n", + "#b\n", + "hav = 22.2\n", + "#c\n", + "Q = hav*l*p*(T1-T2)\n", + "\n", + "# Results\n", + "print \"The rate of heat loss is %.0f W\"%(Q)\n", + "\n", + "# rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3 Page No : 114" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The steady state temprature is 230 C\n", + "The recalculated value is almost equal to previous one.\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "d = 7.24*10**-4 \t\t\t#m, diameter of wire\n", + "l = 1. \t\t\t#m, length of wire\n", + "I = 8.3 \t\t\t#A, current in a wire\n", + "R = 2.625 \t\t\t#ohm/m, electrical resistance\n", + "V = 10. \t\t\t#m/s, air velocity\n", + "Tb = 27. \t\t\t#C, bulk air temp.\n", + "#the properties at bulk temp.\n", + "mu = 1.983*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "k = 0.02624 \t\t\t#W/m C, thermal conductivity\n", + "rho = 1.1774 \t\t\t#kg/m**3, density of air\n", + "cp = 1.0057 \t\t\t#kj/kg C, specific heat of air\n", + "\n", + "# Calculations and Results\n", + "Pr = cp*10**3*mu/k \t\t\t#Prandtl no.\n", + "Re = d*V*rho/mu \t\t\t# Reynold no.\n", + "#from eq. 4.19, nusslet no.\n", + "Nu = 0.3+(0.62*Re**(1./2)*Pr**(1./3)/(1+(0.4/Pr)**(2./3))**(1./4))*(1+(Re/(2.82*10**5))**(5./8))**(4./5)\n", + "hav = Nu*k/d \t\t\t#W/m**2 C, average heat transfer coefficient\n", + "Q = I**2*R \t\t\t#W, rate of electrical heat generation\n", + "A = math.pi*d*l\n", + "dt = Q/(hav*A) \t\t\t#C,temp. difference\n", + "T = dt+Tb \t\t\t#C, steady state temp.\n", + "print \"The steady state temprature is %.0f C\"%(T)\n", + "\n", + "Tm = (T+Tb)/2 \t\t\t#C, mean air film temp.\n", + "#the properties at Tm temp.\n", + "mu1 = 2.30*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "k1 = 0.0338 \t\t\t#W/m C, thermal conductivity\n", + "rho1 = 0.878 \t\t\t#kg/m**3, density of air\n", + "cp1 = 1.014 \t\t\t#kj/kg C, specific heat of air\n", + "Re1 = d*V*rho1/mu1 \t\t\t# Reynold no.\n", + "Pr1 = (1.014*10**3*2.30*10**-5)/k1 \t\t\t#Prandtl no.\n", + "#from eq. 4.19, nusslet no.\n", + "Nu1 = 0.3+(0.62*Re1**(1./2)*Pr1**(1./3)/(1+(0.4/Pr1)**(2./3))**(1./4))*(1+(Re1/(2.82*10**5))**(5./8))**(4./5)\n", + "hav1 = Nu1*k1/d \t\t\t#W/m**2 C, average heat transfer coefficient\n", + "dt1 = Q/(hav1*A) \t\t\t#C,temp. difference\n", + "T1 = dt1+Tb \t\t\t#C, steady state temp.\n", + "print \"The recalculated value is almost equal to previous one.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4 Page No : 116" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "initial rate of melting of ice is 0.0109 g/s\n", + "The required time is is 1665 s\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "di = 0.04 \t\t\t#m, diameter of ice ball\n", + "V = 2. \t\t\t#m/s, air velocity\n", + "T1 = 25. \t\t\t#C, steam temp.\n", + "T2 = 0.\n", + "#the properties of air\n", + "mu = 1.69*10**-5 \t\t\t#kg/ms, vismath.cosity\n", + "k = 0.026 \t\t\t#W/m C, thermal conductivity\n", + "rho = 1.248 \t\t\t#kg/m**3, density \n", + "cp = 1.005 \t\t\t#kj/kg C, specific heat \n", + "#propertice of ice\n", + "lamda = 334. \t\t\t#kj/kg, heat of fusion\n", + "rhoice = 920. \t\t\t#kg/m**3 density of ice\n", + "\n", + "# Calculations and Results\n", + "Pr = cp*10**3*mu/k \t\t\t#Prandtl no.\n", + "Re = di*V*rho/mu \t\t\t# Reynold no.\n", + "#from eq. 4.19, nusslet no.\n", + "Nu = 2+(0.4*Re**0.5+0.06*Re**(2./3))*Pr**0.4\n", + "hav = Nu*k/di \t\t\t#W/m**2 C, average heat transfer coefficient\n", + "Ai = math.pi*di**2 \t\t\t#initial area of sphere\n", + "Qi = Ai*hav*(T1-T2) \t\t\t#W = J/s, initial rate of heat transfer\n", + "Ri = Qi/lamda \t\t\t#initial rate of melting of ice\n", + "print \"initial rate of melting of ice is %.4f g/s\"%(Ri)\n", + "\n", + "#(b)\n", + "#mass of ice ball 4/3*math.pi*r**3\n", + "#Rate of melting = Rm = -d/dt(m)\n", + "#Rate of heat input required = -lamda*Rate of melting\n", + "#heat balance equation\n", + "# -lamda*(Rm) = h*4*math.pi*r**2*dt\n", + "#integrating and solving\n", + "rf = ((di/2)**3/2.)**(1./3)\n", + "#solving eq. 3\n", + "t1 = 1.355*10**-4/(8.136*10**-8)\n", + "print \"The required time is is %.0f s\"%(t1)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5 Page No : 121" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the required contact time is 1.43 s\n" + ] + } + ], + "source": [ + "from scipy.integrate import quad \n", + "# Variables\n", + "Vo = 0.5 \t\t\t#m/s air velocity\n", + "T1 = 800. \t\t\t#C, initial temp.\n", + "T2 = 550. \t\t\t#C, final temp.\n", + "Tam = 500. \t\t\t#C, air mean temp.\n", + "P = 1.2 \t\t\t#atm, pressure\n", + "#the properties of solid particles.\n", + "dp = 0.65*10**-3 \t\t\t#m, average particle diameter\n", + "cps = 0.196 \t\t\t#kcal/kg C, specific heat\n", + "rhos = 2550. \t\t\t#kg/m**3, density \n", + "#Properties of air\n", + "mu = 3.6*10**-5 \t\t\t#kg/ms, vismath.cosity\n", + "k = 0.05 \t\t\t#kcal/hm C, thermal conductivity\n", + "rho = 0.545 \t\t\t#kg/m**3, density of air\n", + "cp = 0.263 \t\t\t#kcal/kg C, specific heat of air\n", + "\n", + "#calculation\n", + "Pr = cp*mu*3600/k \t\t\t#Prandtl no.\n", + "Redp = dp*Vo*rho/mu \t\t\t# Reynold no.\n", + "#from eq. 4.29(b) heat transfer coefficient\n", + "h = (k/dp)*(2+0.6*(Redp)**(1./2)*(Pr)**(1./3))\n", + "Tg = 500 \t\t\t#C, gas temp.\n", + "#from heat balance equation\n", + "# -(dTs/dt) = 6h/(dp*rhos*cps)*(Ts-Tg)\n", + "\n", + "def f2(Ts): \n", + " return (1/(Ts-Tg))\n", + "\n", + "t = (dp*rhos*cps/(6*h))* quad(f2,550,800)[0]\n", + "\n", + "# Results\n", + "print \"the required contact time is %.2f s\"%(t*3600)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6 Page No : 126" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the required rate of flow of water is 1053 kg/h \n", + "the overall heat transfer coefficient is 300 W/m**2 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "mo_ = 1000. \t\t\t#kg/h, cooling rate of oil\n", + "cpo = 2.05 \t\t\t#kj/kg C, specific heat of oil\n", + "T1 = 70. \t\t\t#C, initial temp. of oil\n", + "T2 = 40. \t\t\t#C, temp. of oil after cooling\n", + "cpw = 4.17 \t\t\t#kj/kg C, specific heat of water\n", + "T3 = 42. \t\t\t#C, initial temp. of water\n", + "T4 = 28. \t\t\t#C, temp. of oil after cooling\n", + "A = 3. \t\t\t#m**2, heat exchange area\n", + "\n", + "# Calculation and Results\n", + "mw_ = mo_*cpo*(T1-T2)/(cpw*(T3-T4))\n", + "print \"the required rate of flow of water is %.0f kg/h \"%(mw_)\n", + "Q = mo_*cpo*(T1-T2)/3600 \t\t\t#kw, heat duty\n", + "dt1 = T1-T3 \t\t\t#C, hot end temp. difference\n", + "dt2 = T2-T4 \t\t\t#C, cold end temp. difference\n", + "LMTD = (dt1-dt2)/(math.log(dt1/dt2)) \t\t\t#math.log mean temp. difference\n", + "dtm = LMTD\n", + "U = Q*10**3/(A*dtm)\n", + "print \"the overall heat transfer coefficient is %.0f W/m**2 C\"%(round(U,-1))\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7 Page No : 126" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The inlet temprature is Ti = 26 C\n", + "The outlet temprature is To = 107 C\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve \n", + "import math\n", + "\n", + "# Variables\n", + "Q = 38700. \t\t\t#kcal/h, heat duty\n", + "W = 2000. \t\t\t#kg/h gas flow rate\n", + "cp = 0.239 \t\t\t#kcal/kg C, specific heat of nitrogen\n", + "A = 10. \t\t\t#m**2 ,heat exchanger area\n", + "U = 70. \t\t\t#kcal/hm**2 C, overall heat transfer coefficient\n", + "n = 0.63 \t\t\t#fin efficiency\n", + "\n", + "#Calculation\n", + "dt = Q/(W*cp) \t\t\t#C, temp. difference\n", + "#To-Ti = dt.........................(i)\n", + "dtm = Q/(U*A*n)\n", + "#(To-Ti)/(math.log((160-Ti)/(160-To))) = 87.8........(2)\n", + "#solving 1 and 2\n", + "def f(To): \n", + " return (To-(To-dt))/(math.log((160-(To-dt))/(160-To)))-87.8\n", + "\n", + "To = fsolve(f,100)\n", + "Ti = To-dt\n", + "\n", + "# Results\n", + "print \"The inlet temprature is Ti = %.0f C\"%(Ti)\n", + "print \"The outlet temprature is To = %.0f C\"%(To)\n", + "\n", + "# note : answers are slightly different because of fsolve function of python." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8 Page No : 127" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The outlet eater temp. is 109.8 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "V = 1.8 \t\t\t#m/s, velocity of hot water\n", + "T1 = 110. \t\t\t#C, initial temp.\n", + "l = 15. \t\t\t#m, length of pipe\n", + "t = 0.02 \t\t\t#m, thickness of insulation\n", + "kc = 0.12 \t\t\t#W/mC,thermal conductivity of insulating layer\n", + "ho = 10. \t\t\t#Wm**2 C, outside film coefficient\n", + "T2 = 20. \t\t\t#C, ambient temp.\n", + "#the properties of water at 110 C\n", + "mu = 2.55*10**-4 \t\t\t#m**2/s, vismath.cosity\n", + "k = 0.685 \t\t\t#W/m C, thermal conductivity\n", + "rho = 950. \t\t\t#kg/m**3, density of air\n", + "cp = 4.23 \t\t\t#kj/kg C, specific heat of air\n", + "di = 0.035 \t\t\t#m, actual internal dia. of pipe\n", + "ri = di/2. \t\t\t#m,internal radius\n", + "t1 = 0.0036 \t\t\t#m, actual thickness of 1-1/4 schedule 40 pipe\n", + "ro = ri+t1 \t\t\t#m, outer radius of pipe\n", + "r_ = ro+t \t\t\t#m, outer radius of insulation\n", + "kw = 43. \t\t\t#W/mC, thermal conductivity of steel\n", + "\n", + "#calculation\n", + "Pr = cp*10**3*mu/k \t\t\t#Prandtl no.\n", + "Re = di*V*rho/mu \t\t\t# Reynold no.\n", + "#from eq. 4.9, Nusslet no.\n", + "Nu = 0.023*(Re)**0.88*Pr**0.3\n", + "hi = Nu*k/di \t\t\t#W/m**2 C, average heat transfer coefficient\n", + "#the overall coefficient inside area basis Ui\n", + "Ui = 1./(1/hi+(ri*math.log(ro/ri))/kw+(ri*math.log(r_/ro))/kc+ri/(r_*ho)) \n", + "Ai = math.pi*di*l \t\t\t#m**2, inside area basis\n", + "W = math.pi*ri**2*V*rho \t\t\t#kg/s, water flow rate\n", + "#from the relation b/w LMTD and rate of heat loss\n", + "\n", + "def f(To): \n", + " return (W*cp*10**3)/(Ui*Ai)*(T1-To)-((T1-To)/math.log((T1-T2)/(To-T2)))\n", + "To = fsolve(f,100)\n", + "\n", + "# Results\n", + "print \"The outlet eater temp. is %.1f C\"%(To)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9 Page No : 129" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The exit water temp is 36 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "T1 = 28. \t\t\t#C, inlet temp. \n", + "T2 = 250. \t\t\t#C,bulk temp.\n", + "V = 10. \t\t\t#m/s, gas velocity\n", + "l = 20. \t\t\t#m, length of pipe\n", + "mw = 1.*3600 \t\t\t#kg/h, water flow rate\n", + "di = 4.1*10**-2 \t\t\t#m, inlet diameter\n", + "Tm = (T1+T2)/2 \t\t\t#C, mean temp.\n", + "ro = 0.0484 \t\t\t#m, outside radius\n", + "#properties of water\n", + "mu = 8.6*10**-4 \t\t\t#kg/ms, vismath.cosity\n", + "kw = 0.528 \t\t\t#kcal/h m C, thermal conductivity\n", + "kw_ = 0.528*1.162 \t\t\t#W/ m C, thermal conductivity\n", + "rho = 996. \t\t\t#kg/m**3, density of air\n", + "cp = 1*4.18 \t\t\t#kj/kg C, specific heat of air\n", + "cp_ = 1. \t\t\t#kcal/kg C\n", + "#properties of flue gas\n", + "mu1 = 2.33*10**-5 \t\t\t#kg/ms, vismath.cosity\n", + "ka = 0.0292 \t\t\t#kcal/h m C, thermal conductivity\n", + "rho1 = 0.891 \t\t\t#kg/m**3, density of air\n", + "cp1 = 0.243 \t\t\t#kcal/kg C, specific heat of air\n", + "Pr = 0.69\n", + "\n", + "#calculation\n", + "A = math.pi/4*di**2 \t\t\t#m**2, cross section of pipe\n", + "Vw = 1/(rho*A) \t\t\t#m/s, velocity of warer\n", + "Re = di*Vw*rho/mu \t\t\t# Reynold no.\n", + "Pr1 = cp*10**3*mu/kw_ \t\t\t#Prandtl no. for water\n", + "Nu = 0.023*Re**0.8*Pr1**0.4 \t\t\t#Nusslet no.\n", + "#water side heat transfer coefficient hi\n", + "hi = 206*kw/di\n", + "#gas side heat transfer coefficient ho\n", + "a = 41 \t\t\t#mm, i.d. schedule\n", + "Tw = 3.7 \t\t\t#mm, wall thickness\n", + "do = a+2*Tw \t\t\t#mm, outer diameter of pipe\n", + "Re1 = do*10**-3*V*rho1/mu1 \t\t\t# Reynold no\n", + "#from eq. 4.19, nusslet no.\n", + "Nu1 = 0.3+(0.62*Re1**(1./2)*Pr**(1./3)/(1+(0.4/Pr)**(2./3))**(1/4.))*(1+(Re1/(2.82*10**5))**(5./8))**(4/5.)\n", + "ho = (Nu1*ka/do)*10**3 \t\t\t#kcal/h m**2 C\n", + "Uo = 1/(ro/(di/2*hi)+1/ho) \t\t\t#kcal/h m**2 C, overall heat transfer coefficient\n", + "\n", + "#Heat balance\n", + "A1 = math.pi*ro*l \t\t\t#m62, outside area of pipe\n", + "#from the formula of LMTD\n", + "def f(T2_): \n", + " return mw*cp_*(T2_-T1)-Uo*A1*((T2_-T1)/math.log((T2-T1)/(T2-T2_)))\n", + "T2_ = fsolve(f,1)\n", + "\n", + "# Results\n", + "print \"The exit water temp is %.0f C\"%(T2_)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.10 Page No : 131" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The tube length is 123 m\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "dti = 0.0212 \t\t\t#m inner tube\n", + "dto = 0.0254 \t\t\t#cm, outer tube\n", + "dpi = 0.035 \t\t\t#cm, outer pipe\n", + "mo_ = 500. \t\t\t#kh/h, cooling rate of oil\n", + "To2 = 110. \t\t\t#C, initial temo. of oil\n", + "To1 = 70. \t\t\t#C, temp. after cooling of oil\n", + "Tw2 = 40. \t\t\t#C, inlet temp. of water\n", + "Tw1 = 29. \t\t\t#C, outlet temp. of water\n", + "#properties of oil\n", + "cpo = 0.478 \t\t\t#kcal/kg C\n", + "ko = 0.12 \t\t\t#kcal/h m C, thermal conductivity\n", + "rho = 850. \t\t\t#kg/m**3, density of oil\n", + "#properties of water\n", + "kw = 0.542 \t\t\t#kcal/h m C, thermal conductivity\n", + "kw_ = (kw*1.162) \t\t\t#kj/kg C\n", + "muw = 7.1*10**-4 \t\t\t#kg/ms, vismath.cosity of water\n", + "cpw = 1. \t\t\t#kcal/kg C\n", + "cpw_ = cpw*4.17 \t\t\t#kcal/kg C\n", + "rhow = 1000. \t\t\t#kg/m**3, density\n", + "\n", + "#calculation\n", + "HL = mo_*cpo*(To2-To1) \t\t\t#kcal/h, heat load of exchanger\n", + "mw_ = HL/(cpw*(Tw2-Tw1)) \t\t\t#kg/h water flow rate\n", + "mw_1 = mw_/(3600*10**3) \t\t\t#m**3/s water flow rate\n", + "A1 = (math.pi/4)*(dti)**2 \t\t\t#m**2, flow area of tube\n", + "Vw = mw_1/A1 \t\t\t#m/s water velocity\n", + "Rew = dti*Vw*rhow/muw \t\t\t#Reynold no.\n", + "Prw = cpw_*10**3*muw/kw_ \t\t\t#Prandtl no.\n", + "Nuw = 0.023*Rew**0.8*Prw**0.4 \t\t\t#nusslet no.\n", + "#water side heat transfer coefficient hi\n", + "hi = Nuw*kw/dti\n", + "\n", + "#oil side heat transfer coefficient\n", + "A2 = math.pi/4*(dpi**2-dto**2) \t\t\t#m**2, flow area of annulus\n", + "Vo = mo_/(3600*rho*A2) \t\t\t#m/s velocity of oil\n", + "de = (dpi**2-dto**2)/dto \t\t\t#m, equivalent dia of annulus\n", + "Tmo = (To2+To1)/2 \t\t\t#C,mean oil temp.\n", + "muoil = math.exp((5550./(Tmo+273))-19) \t\t\t#kg/ms, vismath.cosity of oil\n", + "Reo = de*Vo*rho/muoil\n", + "Pro = cpo*muoil*3600/ko \t\t\t#prandtl no. for oil \n", + "\n", + "#assume (1st approximation)\n", + "Nuo = 3.66\n", + "ho = Nuo*ko/de \t\t\t#kcal/h m**2 c\n", + "L = 1 \t\t\t#assume length of tube\n", + "Ai = math.pi*dti*L\n", + "Ao = math.pi*dto*L\n", + "#overall heat transfer coefficient 1st approximation\n", + "Uo = 1/(1/ho+Ao/(Ai*hi))\n", + "LMTD = ((To2-Tw2)-(To1-Tw1))/(math.log((To2-Tw2)/(To1-Tw1)))\n", + "Ao1 = HL/(Uo*LMTD) \t\t\t #m**2, heat transfer area\n", + "Lt = Ao1/(math.pi*dto) \t\t\t #m, tube length\n", + "#from eq. 4.8\n", + "Nuo1 = 1.86*(Reo*Pro/(Lt/de))**(1./3) \t\t\t#Nusslet no. \n", + "ho1 = Nuo1*ko/de\n", + "Tmw = (Tw1+Tw2)/2 \t\t\t#C, mean water temp.\n", + "#balancing heat transfer rate of oil and water\n", + "\n", + "#average wall temp. Twall\n", + "Twall = ((hi*dti*(-Tmw))-(ho1*dto*Tmo))/(-65.71216)\n", + "#vismath.cosity of oil at this temp.\n", + "muwall = math.exp((5550/(Twall+273))-19) \t\t\t#kg/ms, vismath.cosity of oil\n", + "#Nusslet no. \n", + "Nuo2 = 1.86*(Reo*Pro/(Lt/de))**(1./3)*(muoil/muwall)**0.14\n", + "ho2 = Nuo2*ko/de\n", + "Uo2 = 1/((1/ho2)+(Ao/(Ai*hi)))\n", + "Ao2 = HL/(Uo2*LMTD)\n", + "Lt_ = Ao2/(math.pi*dto)\n", + "\n", + "# Results\n", + "print \"The tube length is %d m\"%(Lt_)\n", + "\n", + "# rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.11 Page No : 135" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " the rate of heat transfer to water.is 6.93e+05 kcal/h\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "Ti = 260. \t\t\t#C, initial temp.\n", + "Ts = 70. \t\t\t#C, skin temp.\n", + "St = 0.15 \t\t\t#m,space between tubes in equilateral triangular arrangement\n", + "Sd = St \t\t\t#space between tubes\n", + "mu = 4.43*10**-5 \t\t\t#m**2/s, momentum diffusity\n", + "k = 0.0375 \t\t\t#W/m C, thermal conductivity\n", + "rho = 0.73 \t\t\t#kg/m**3, density of air\n", + "cp = 0.248 \t\t\t#kj/kg C, specific heat of air\n", + "V = 16. \t\t\t#m/s, velociity \n", + "d = 0.06 \t\t\t#m, outside diameter of tube\n", + "Nt = 15. \t\t\t#no. of tubes in transverse row\n", + "Nl = 14. \t\t\t#no. of tubes in longitudinal row\n", + "N = Nl*Nt \t\t\t#total no. of tubes\n", + "L = 1. \t\t\t#m, length\n", + "#Calculation\n", + "Sl = (math.sqrt(3)/2)*St\n", + "Pr = cp*mu*3600*rho/k \t\t\t#Prandtl no. of bulk air\n", + "Pr = 0.62\n", + "Prw = 0.70 \t\t\t#Prandtl no. of air at wall temp. 70 C\n", + "#from eq. 4.25\n", + "Vmax = (St/(St-d))*V\n", + "#from eq. 4.26\n", + "Vmax1 = (St/(2*(St-d)))*V\n", + "Redmax = d*Vmax/mu\n", + "p = St/Sl \t\t\t#pitch ratio\n", + "#from table 4.3\n", + "m = 0.6\n", + "C = 0.35*(St/Sl)**0.2\n", + "h = round((k/d)*C*(36163)**m*(Pr)**(0.36)*(Pr/Prw)**(0.25))\n", + "#from eq. 4.28\n", + "dt = round(190*math.exp(-math.pi*d*N*h/(rho*V*3600*Nt*St*cp)))\n", + "LMTD = ((Ti-Ts)-(dt))/math.log((Ti-Ts)/dt)\n", + "A = round(math.pi*d*L*N,1) \t\t\t#m**2, heat transfer area\n", + "Q = h*A*LMTD\n", + "\n", + "# Results\n", + "print \" the rate of heat transfer to water.is %.2e kcal/h\"%(Q)\n", + "\n", + "# Note : Value of LMTD is wrong in book please check." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.12 Page No : 140" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Outlet temp. of water for one pass through the tubes is 51 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "W = 0.057 \t\t\t#m**3/min/tube, flow rate of water\n", + "W_ = W*16.66 \t\t\t#kg/s. water flow rate\n", + "di = 0.0212 \t\t\t#m,inside diameter\n", + "Ti = 32. \t\t\t#C, inlet water temp.\n", + "Tw = 80. \t\t\t#C, wall temp.\n", + "L = 3. \t\t\t#m, length of pip\n", + "\n", + "#Calculation\n", + "V = (W/60)*(1/((math.pi/4)*di**2)) \t\t\t#m/s, water velocity\n", + "#the properties of water at mean liquid temp..\n", + "mu = 7.65*10**-4 \t\t\t#m**2/s, vismath.cosity\n", + "k = 0.623 \t\t\t#W/m C, thermal conductivity\n", + "rho = 995. \t\t\t#kg/m**3, density of air\n", + "cp = 4.17 \t\t\t#kj/kg C, specific heat of air\n", + "\n", + "Pr = cp*10**3*mu/k \t\t\t#Prandtl no.\n", + "Re = di*V*rho/mu \t\t\t# Reynold no.\n", + "#from eq. 4.19, nusslet no.\n", + "#from dittus boelter eq.\n", + "Nu = 0.023*Re**0.8*Pr**0.4 \t\t\t#Prandtl no.\n", + "f = 0.0014+0.125*Re**-0.32 \t\t\t#friction factor\n", + "#Reynold anamath.logy\n", + "St = f/2 \t\t\t#Smath.tanton no.\n", + "Nu1 = Re*Pr*St\n", + "#Prandtl anamath.logy\n", + "St1 = (f/2)/(1+5*(Pr-1)*math.sqrt(f/2))\n", + "Nu2 = St1*Re*Pr \n", + "#colburn analogy\n", + "Nu3 = Re*Pr**(1./3)*(f/2)\n", + "h = Nu3*k/(di) \t\t\t#W/m**2 C av heat transfer coefficient\n", + "#Q = W_*cp*10**3*(To-Ti) = h*A*LMTD\n", + "A = math.pi*di*L \t\t\t#m**2\n", + "def f(To): \n", + " return W_*cp*10**3*(To-Ti)-h*A*((To-Ti)/math.log((Tw-Ti)/(Tw-To)))\n", + "To = fsolve(f,1)\n", + "#Revised calculation\n", + "Tm = (Ti+To)/2 \t\t\t#C, mean liquid temp.\n", + "#the properties of water at new mean liquid temp..\n", + "mu1 = 6.2*10**-4 \t\t\t#m**2/s, vismath.cosity\n", + "k1 = 0.623 \t\t\t#W/m C, thermal conductivity\n", + "rho1 = 991. \t\t\t#kg/m**3, density of air\n", + "cp1 = 4.17 \t\t\t#kj/kg C, specific heat of air\n", + "\n", + "Pr1 = cp1*10**3*mu1/k1 \t\t\t#Prandtl no.\n", + "Re1 = di*V*rho1/mu1 \t\t\t# Reynold no.\n", + "#from dittus boelter eq.\n", + "f1 = 0.0014+0.125*Re1**(-0.32) \t\t\t#friction factor\n", + "#colburn anamath.logy\n", + "Nu4 = Re1*Pr1**(1./3)*(f1/2)\n", + "h1 = Nu4*k1/(di) \t\t\t#W/m**2 C av heat transfer coefficient\n", + "def f(To_): \n", + " return W_*cp*10**3*(To_-Ti)-h1*A*((To_-Ti)/math.log((Tw-Ti)/(Tw-To_)))\n", + "To_ = fsolve(f,1)\n", + "\n", + "print \"Outlet temp. of water for one pass through the tubes is %.0f C\"%(To_)\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch5.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch5.ipynb new file mode 100644 index 00000000..77c2b9da --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch5.ipynb @@ -0,0 +1,431 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 : free convection" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1 Page No : 153" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The rate of heat loss is 267 W\n" + ] + } + ], + "source": [ + "# Variables\n", + "T1 = 65. \t\t\t#C, furnace temp.\n", + "T2 = 25. \t\t\t#C, ambient temp.\n", + "h = 1.5 \t\t\t#m, height of door\n", + "w = 1. \t\t\t#m, width of door\n", + "Tf = (T1+T2)/2 \t\t\t#c, average air film temp.\n", + "\n", + "# Calculations\n", + "Pr = 0.695 \t\t\t#Prandtl no.\n", + "mu = 1.85*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "beeta = 1/(Tf+273) \t\t\t#K**-1. coefficient of volumetric expension\n", + "k = 0.028 \t\t\t#W/m C, thermal conductivity\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "Grl = g*beeta*(T1-T2)*h**3/(mu**2) \t\t\t#Grashof no.\n", + "Ral = Grl*Pr \t\t\t#Rayleigh no.\n", + "#Nusslet no.\n", + "Nul = (0.825+(0.387*(Ral)**(1./6))/(1+(0.492/Pr)**(9./16))**(8./27))**2 \n", + "hav = Nul*k/h \t\t\t#average heat transfer coefficient\n", + "Ad = h*w \t\t\t#m**2, door area\n", + "dt = T1-T2 \t\t\t#temp. driving force\n", + "q = hav*Ad*dt \t\t\t#W,rate of heat loss\n", + "\n", + "# Results\n", + "print \"The rate of heat loss is %.0f W\"%(q)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2 Page No : 154" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the steady state temp. of the plate is 61.6 C\n" + ] + } + ], + "source": [ + "# Variables\n", + "T1 = 60. \t\t\t#C, plate temp.\n", + "T2 = 25. \t\t\t#C, ambient temp.\n", + "h = 1.\n", + "w = 1. \t\t\t#m, width of door\n", + "q = 170. \t\t\t#W, rate of heat transfer\n", + "Tf = (T1+T2)/2 \t\t\t#c, average air film temp.\n", + "#Properties of air at Tf\n", + "Pr = 0.7 \t\t\t#Prandtl no.\n", + "mu = 1.85*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "beeta = 1./(Tf+273) \t\t\t#K**-1. coefficient of volumetric expension\n", + "k = 0.028 \t\t\t#W/m C, thermal conductivity\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "\n", + "#Calculation\n", + "A = h*w \t\t\t#m**2, plate area\n", + "P = 2*(h+w) \t\t\t#m,perimeter of plate \n", + "L = A/P \t\t\t#m characteristic length\n", + "Grl = g*beeta*(T1-T2)*L**3/(mu**2) \t\t\t#Grashof no.\n", + "Ral = Grl*Pr \t\t\t#Rayleigh no.\n", + "#Nusslet no.\n", + "Nul = 0.54*(Ral)**(1./4) \t\t\t#Nusslet no.\n", + "hav = Nul*k/L \t\t\t#average heat transfer coefficient\n", + "Ts = q/(hav*A)+T2\n", + "\n", + "# Results\n", + "print \"the steady state temp. of the plate is %.1f C\"%(Ts)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3 Page No : 156" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required time for cooling is 2.30 hr\n" + ] + } + ], + "source": [ + "import math \n", + "from scipy.integrate import quad \n", + "# Variables\n", + "d = 0.0254 \t\t\t#m, diameter of steel rod\n", + "l = 0.4 \t\t\t#m, length of rod\n", + "T1 = 80. \t\t\t#C, initial temp.\n", + "T2 = 30. \t\t\t#C, ambient temp.\n", + "T3 = 35. \t\t\t#c, temp. after cooling\n", + "rho = 7800. \t\t\t#kg/m**3 ,density of steel rod\n", + "cp = 0.473 \t\t\t#kj/kg C. specific heat\n", + "\n", + "#Calculation\n", + "m = math.pi/4*d**2*l*rho \t\t\t#kg. mass of cylinder\n", + "A = math.pi*d*l \t\t\t#m**2, area of cylinder\n", + "dt = T1-T2 \t\t\t#c, insmath.tanmath.taneous temp. difference\n", + "h = 1.32*(dt/d)**0.25 \t\t\t#W/m**2 C, heat transfer coefficient\n", + "\n", + "def f0(T): \n", + " return 1./(T**(5./4))\n", + "\n", + "i = quad(f0,5,50)[0]\n", + "\n", + "t = i/(3.306*A/(m*cp*10**3))\n", + "\n", + "# Results\n", + "print \"The required time for cooling is %.2f hr\"%(t/3600.)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4 Page No : 157" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rate of heat loss by free convection per meter length of pipe. is 107 W\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "id_ = 78.*10**-3 \t\t\t#m, internal diameter\n", + "od = 89.*10**-3 \t\t\t#m, outer diameter\n", + "Pg = 15. \t\t\t#kg/cm**2, gauge pressure\n", + "t = 2.*10**-2 \t\t\t#m, thickness of preformed mineral fibre\n", + "k = 0.05 \t\t\t#W/m C. thermal conductivity\n", + "Ta = 25. \t\t\t#C, ambient air temp.\n", + "Pr = 0.705 \t\t\t#Prandtl no.\n", + "#assume\n", + "Ts = 50. \t\t\t#C, skin temp.\n", + "l = 1. \t\t\t#m, length\n", + "Ti = 200.5 \t\t\t#C, initial temp.\n", + "rs = od/2+t \t\t\t#m, outer radius of insulation\n", + "ri = od/2 \t\t\t#m, inner radius of insulation\n", + "\n", + "# Calculations\n", + "Q = 2*math.pi*l*k*(Ti-Ts)/(math.log(rs/ri)) \t\t\t#W\n", + "#properties of air at taken at the mean film temp.\n", + "Tf = (Ta+Ts)/2 \t\t\t#C\n", + "mu = 1.76*10**-5 \t\t\t#m**2/s. vismath.cosity\n", + "beeta = (1/(Tf+273)) \t\t\t#K**-1, coefficient of volumetric expansion\n", + "k1 = 0.027 \t\t\t#W/m C, thermal conductivity\n", + "ds = 2*rs \t\t\t#m, outer dia. of insulated pipe\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "Grd = g*beeta*(Ts-Ta)*ds**3/(mu**2) \t\t\t#Grashof no.\n", + "Rad = Grd*Pr \t\t\t#Rayleigh no.\n", + "#from eq. 5.9\n", + "#Nusslet no. \n", + "Nu = (0.60+(0.387*(Rad)**(1./6))/(1+(0.559/Pr)**(9./16))**(8./27))**2 \n", + "hav = Nu*k1/ds \t\t\t#W/ m**2 C, average heat transfer coefficient\n", + "Ts = (Q/(math.pi*ds*l*hav))+Ta \t\t\t#C, skin temp.\n", + "#revised calculation by assuming\n", + "Ts1 = 70. \t\t\t#C, skin temp.\n", + "#Rate of heat transfer through insulation\n", + "Q1 = 2*math.pi*l*k*(Ti-Ts1)/(math.log(rs/ri))\n", + "Tf1 = (Ta+Ts1)/2 \t\t\t#C, average aie mean film temp.\n", + "mu1 = 1.8*10**-5 \t\t\t#m**2/s. vismath.cosity\n", + "beeta1 = (1/(Tf1+273)) \t\t\t#K**-1, coefficient of volumetric expansion\n", + "k1 = 0.0275 \t\t\t#W/m C, thermal conductivity\n", + "Pr1 = 0.703 \t\t\t#Prandtl no.\n", + "Grd1 = g*beeta1*(Ts1-Ta)*ds**3/(mu1**2) \t\t\t#Grashof no.\n", + "Rad = Grd1*Pr1 \t\t\t#Rayleigh no.\n", + "#from eq. 5.9\n", + "# average heat transfer coefficient, in \t\t\t#W/ m**2 C,\n", + "hav1 = (0.60+(0.387*(Rad)**(1./6))/(1+(0.559/Pr)**(9./16))**(8./27))**2*(k1/ds)\n", + "Ts2 = (Q1/(math.pi*ds*l*hav1))+Ta\n", + "#again assume skin temp. = 74\n", + "Ts2 = 74 \t\t\t#C, assumed skin temp.\n", + "Q3 = 2*math.pi*l*k*(Ti-Ts2)/(math.log(rs/ri))\n", + "\n", + "# Results\n", + "print \"the rate of heat loss by free convection per meter length of pipe. is %.0f W\"%(Q3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5 Page No : 159" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required insulation thickness is 0.188 m\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables\n", + "Ts = 65. \t\t\t#C, skin temp.\n", + "To = 30. \t\t\t#C, ambient temp.\n", + "Tw = 460. \t\t\t#C, wall temp.\n", + "Tf = (Ts+To)/2 \t\t\t#C,mean air film temp.\n", + "beeta = (1./(Tf+273)) \t\t\t#K**-1, coefficient of volumetric expansion\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "mu = 1.84*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "L = 10.5 \t\t\t#m, height of converter\n", + "di = 4. \t\t\t#m,diameter of converter\n", + "Pr = 0.705 \t\t\t#Prandtl no.\n", + "k = 0.0241 \t\t\t#kcal/h m C, thermal conductivity\n", + "\n", + "#Calculation\n", + "Grl = g*beeta*(Ts-To)*L**3/(mu**2) \t\t\t#Grashof no.\n", + "x = di/L \t\t\t#assume di/l = x\n", + "y = 35/(Grl)**(1./4) \t\t\t#assume 35/(Grl)**(3/4) = y\n", + "#for a verticla flat plate, from eq. 5.3\n", + "Ral = Grl*Pr \t\t\t#Rayleigh no.\n", + "#nusslet no.\n", + "Nu = (0.825+(0.387*(Ral)**(1./6))/(1+(0.496/Pr)**(9./16))**(8./27))**2\n", + "hav = Nu*k/L \t\t\t#kcal/h m**2 C, average heat transfer coefficient\n", + "#w = poly(0,\"w\")\n", + "#Dav = (4+(4+2*w))/2 \t\t\t#average diameter\n", + "#Aav = math.pi*Dav*L \t\t\t#average heat transfer area\n", + "#Qi = math.pi*Dav*L*0.0602*(Tw-Ts)/w \t\t\t#Rate of heat transfer through insulation\n", + "#rate of heat transfer from the outer surface of the insulation by free convection\n", + "#Qc = hav*math.pi*Dav*L*(Ts-To) \n", + "#Qi = Qc\n", + "def f(w): \n", + " return math.pi*(4+w)*L*0.0602*(Tw-Ts)/w-hav*math.pi*(4+2*w)*L*(Ts-To)\n", + "w = fsolve(f,0.1)\n", + "\n", + "# Results\n", + "print \"The required insulation thickness is %.3f m\"%(w)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6 Page No : 162" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rate of heat transfer is 13.4 W\n" + ] + } + ], + "source": [ + "# Variables\n", + "L = 1.6 \t\t\t#m,height of enclosure\n", + "w = 0.04 \t\t\t#m, width of enclosure\n", + "b = 0.8 \t\t\t#m, breath\n", + "T1 = 22. \t\t\t#C,surface temp.\n", + "T2 = 30. \t\t\t#C, wall temp.\n", + "Tm = (T1+T2)/2 \t\t\t#C, Mean air temp.\n", + "Pr = 0.7 \t\t\t#Prandtl no.\n", + "\n", + "# Calculations\n", + "#fpr air at 26 C\n", + "beeta = 1./(Tm+273) \t\t\t#K**-1. coefficient of volumetric expension\n", + "mu = 1.684*10**-5 \t\t\t#m**2/s, vismath.cosity\n", + "k = 0.026 \t\t\t#W/m C, thermal conductivity\n", + "alpha = 2.21*10**-5 \t\t\t#m**2/s, thermal diffusity\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "Raw = g*beeta*(T2-T1)*w**3/(mu*alpha) \t\t\t#Rayleigh no.\n", + "Nuw = 0.42*(Raw)**0.25*Pr**0.012*(L/w)**-0.3 \t\t\t#Nusslet no.\n", + "h = Nuw*k/w \t\t\t#kcal/h m**2 C, heat transfer coefficient\n", + "q = h*(T2-T1)*(L*b) \t\t\t#W,the rate of heat transfer\n", + "\n", + "# Results\n", + "print \"the rate of heat transfer is %.1f W\"%(q)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7 Page No : 163" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rate of heat loss per meter length is 39.7 kcal/h\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "Ts = 60. \t\t\t#C, surface temp\n", + "To = 30. \t\t\t#C, bulk temp.\n", + "d = 0.06 \t\t\t#m, diameter of pipe\n", + "l = 1. \t\t\t#m, length\n", + "Tm = (Ts+To)/2\n", + "#for air at Tm\n", + "rho = 1.105 \t\t\t#kg/m**3, density\n", + "cp = 0.24 \t\t\t#kcal/kg C. specific heat\n", + "mu = 1.95*10**-5 \t\t\t#kg/m s. vismath.cosity\n", + "P = 0.7 \t\t\t#Prandtl no. \n", + "kv = 1.85*10**-5 \t\t\t#m**2/s, kinetic vismath.cosity\n", + "k = 0.0241 \t\t\t#kcal/f m C, thermal conductivity\n", + "beeta = (1./(Tm+273)) \t\t\t#K**-1. coefficient of volumetric expension\n", + "V = 0.3 \t\t\t#m/s, velocity\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "\n", + "#Calculation\n", + "Rad = g*beeta*(Ts-To)*d**3*P/(kv**2) \t\t\t#Rayleigh no.\n", + "#from eq. 5.9\n", + "Nufree = (0.60+(0.387*Rad**(1./6))/(1+(0.559/P)**(9./16))**(8./27))**2\n", + "#calculation of forced convection nusslet no.\n", + "#from eq. 4.19\n", + "Re = d*V/(kv)\n", + "Nuforced = 0.3+(0.62*Re**(1./2)*P**(1./3)/(1+(0.4/P)**(2./3))**(1./4))*(1.+(Re/(2.82*10**5))**(5./8))**(4./5)\n", + "Nu = (Nuforced**3+Nufree**3)**(1./3) \t\t\t#nusslet no. for mixed convection\n", + "#Nu = h*d/k\n", + "h = Nu*k/d \t\t\t#kcal/h m**2 C, heat transfer corfficient\n", + "q = h*math.pi*d*l*(Ts-To)\n", + "\n", + "# Results\n", + "print \"the rate of heat loss per meter length is %.1f kcal/h\"%(q)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch6.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch6.ipynb new file mode 100644 index 00000000..362c8be7 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch6.ipynb @@ -0,0 +1,560 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 : Boiling and condensation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1 Page No : 177" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "So a bubble nucleus that has been detached from a cavity will not collapse in the liquid if it is larger than 1.89 micrometer \n", + "The superheat of the liquid is 9 C\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "import warnings\n", + "warnings.filterwarnings('ignore', 'The iteration is not making good progress')\n", + "# Variables\n", + "#(a)\n", + "Tsat = 350 \t\t\t#K, saturated temp.\n", + "Tl = Tsat+5 \t\t\t#K, liquid temp.\n", + "#By antoine eqn.\n", + "T = Tl-273 \t\t\t#C, \n", + "\n", + "# Calculations and Results\n", + "pl = math.exp(4.22658-(1244.95/(T+217.88)))\n", + "ST = 26.29-0.1161*T \t\t\t#dyne/cm, Surface tension of liquid\n", + "ST_ = ST*10**-3 \t\t\t#N/m Surface tension of liquid\n", + "Lv = 33605 \t\t\t#kj/kgmol, molar heat of vaporization\n", + "R = 0.08314 \t\t\t#m**3 bar/kgmol K, gas math.cosmath.tant\n", + "r = (2*ST_*R*Tsat**2)/((Tl-Tsat)*pl*(Lv*10**3))\n", + "print \"So a bubble nucleus that has been detached from a cavity will not collapse in \\\n", + "the liquid if it is larger than %.2f micrometer \"%(r*10**6)\n", + "\n", + "#(b)\n", + "r1 = 10**-6 \t\t\t#m\n", + "#pl1 = exp(4.22658-(1244.95/(Tl_-273+217.88))) \t\t\t#vapour pressure\n", + "#ST1 = 0.02629-1.161*10**-4(Tl_-273) \t\t\t#surface tension\n", + "\n", + "def f(Tl): \n", + " return (Tl-Tsat)-2*(0.02629-1.161*10**-4*(Tl-273))*R*Tsat**2/(r1*Lv*10**3)\n", + "Tl = fsolve(f,0.1)\n", + "T_ = (Tl-273.5)-(Tsat-273)\n", + "print \"The superheat of the liquid is %d C\"%(T_)\n", + "\n", + "# note : answers are slightly different because of rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2 Page No : 180" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "total rate of boiling of water is 69 kg/h \n", + "Qs2 compares reasonably well with the Qs1\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "d = 0.35 \t\t\t#m, diameter of pan\n", + "p = 1.013 \t\t\t#bar, pressure\n", + "T1 = 115. \t\t\t#C, bottom temp.\n", + "T2 = 100. \t\t\t#C, boiling temp.\n", + "Te = T1-T2 \t\t\t#C, excess temp.\n", + "#For Water\n", + "mu1 = 2.70*10**-4 \t\t\t#Ns/m**2, vismath.cosity\n", + "cp1 = 4.22 \t\t\t#kj/kg C, specific heat\n", + "rho1 = 958. \t\t\t#kg/m63. density\n", + "Lv1 = 2257. \t\t\t#kj/kg, enthalpy of vaporization \n", + "s1 = 0.059 \t\t\t#N/m , surface tension\n", + "Pr1 = 1.76 \t\t\t#Prandtl no.\n", + "#For saturated steam\n", + "rho2 = 0.5955\n", + "#For the pan\n", + "Csf = 0.013 \t\t\t#consmath.tant\n", + "n = 1. \t\t\t#exponent\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "\n", + "# Calculations and Results\n", + "#from eq. 6.6 \t\t\t#heat flux\n", + "Qs1 = mu1*Lv1*(g*(rho1-rho2)/s1)**(1./2)*(cp1*Te/(Csf*Lv1*(Pr1)**n))**3\n", + "Rate = Qs1/Lv1 \t\t\t#kg/m**2 s. rate of boiling\n", + "Ap = math.pi/4*d**2 \t\t\t#m**2, pan area\n", + "Trate = Rate*Ap \t\t\t#kg/s, Total rate of boiling\n", + "Trate_ = Trate*3600.5 \t\t\t#kg/h. Total rate of boiling\n", + "print \"total rate of boiling of water is %.0f kg/h \"%(Trate_)\n", + "\n", + "#umath.sing Lienhard's eq., \t\t\t#critical heat flux\n", + "Qmax = 0.149*Lv1*rho2*(s1*g*(rho1-rho2)/(rho2)**2)**(1/4)\n", + "#by Mostinski eq.\n", + "Pc = 221.2 \t\t\t#critical pressure\n", + "Pr = p/Pc \t\t\t#reduced pressure\n", + "hb = 0.00341*(Pc)**(2.3)*Te**(2.33)*Pr**(0.566) \t\t\t#boiling heat transfer coefficient\n", + "hb_ = hb/1000 \t\t\t#kW/m**2 C boiling heat transfer coefficient\n", + "Qs2 = hb_*(Te)\n", + "print \"Qs2 compares reasonably well with the Qs1\"\n", + "\n", + "# note: rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 Page No : 181" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The boilins rate is 63 kg/m**2 h\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "A = 12.5673\n", + "B = 4234.6\n", + "pv = 1.813\n", + "T1 = 200. \t\t\t#C, tube wall temp.\n", + "#For methanol\n", + "Tc = 512.6 \t\t\t#K, critical temp.\n", + "w = 0.556 \t\t\t#acentric factor\n", + "Zra = 0.29056-0.08775*w\n", + "R = 0.08314 \t\t\t#m**3bar/gmol K, universal gas consmath.tant\n", + "Pc = 80.9 \t\t\t#bar, critical temp.\n", + "Mw = 32. \t\t\t#g, molecular wt\n", + "\n", + "#Calculation\n", + "#Estimation of liquid and vapour properties \n", + "#from antoine eq.\n", + "T = B/(A-math.log(pv)) \t\t\t#K, boiling point\n", + "Te = (T1+273)-T \t\t\t#K, excess temp.\n", + "Tm = ((T1+273)+T)/2 \t\t\t#K, mean temp.\n", + "#Liquid properties\n", + "#(a)\n", + "Tr = T/Tc \t\t\t#K, reduced temp.\n", + "#from Rackett technique\n", + "Vm = R*Tc*(Zra)**(1+(1-Tr)**(2/7))/Pc \t\t\t#m**3/kg mol, molar volume\n", + "rhol = Mw/Vm \t\t\t#kg/m**3, density of satorated liquid density\n", + "#(b)\n", + "#from Missenard technique\n", + "T2 = 348. \t\t\t#K,given data temp.\n", + "T3 = 373. \t\t\t#K,given data temp.\n", + "Cp2 = 107.5 \t\t\t#j/g mol K specific heat at T2\n", + "Cp3 = 119.4 \t\t\t#j/g mol K specific heat at T3\n", + "#By linear interpolation at T = 353.7 K\n", + "Cp = Cp2+(Cp3-Cp2)*((T-T2)/(T3-T2)) \t\t\t#kj/kg mol C, specific heat at T = 353.7 K\n", + "Cp_ = Cp*0.03125 \t\t\t#kj/kg C\n", + "#(c)Surface tension at given temp.(K)\n", + "T4 = 313.\n", + "St4 = 20.96\n", + "T5 = 333.\n", + "St5 = 19.4\n", + "#By linear interpolation at T = 353.7 K\n", + "S = 17.8 \t\t\t#dyne/cm, surface temp.\n", + "#(d) liquid vismath.cosity\n", + "T6 = 298. \n", + "MUt6 = 0.55 \t\t\t#cP, liquid vismath.cosity at temp = 298\n", + "MU = ((MUt6)**-0.2661+((T-T6)/233))**(-1/0.2661) \t\t\t#cP\n", + "#(e)Prandtl no. a,b,c are consmath.tant\n", + "a = 0.3225\n", + "b = -4.785*10**-4\n", + "c = 1.168*10**-7\n", + "kl = a+b*T+c*T**2 \t\t\t#W/m C, thermal conductivity\n", + "Prl = Cp_*1000*MU*10**-3/kl \t\t\t#Prandtl no.\n", + "#(f)heat of vaporization at 337.5 K\n", + "Lv = 1100. \t\t\t#kj/kg, enthalpy of vaporization\n", + "\n", + "#Properties of methanol vapour at Tm\n", + "#(a)\n", + "Vm1 = R*Tm/pv \t\t\t#m**3/kg mol, molar volume\n", + "rhov = Mw/Vm1 \t\t\t#kg/m**3, density of vapour\n", + "#(b) a1,b1,c1,d1 are math.cosmath.tants\n", + "a1 = -7.797*10**-3\n", + "b1 = 4.167*10**-5\n", + "c1 = 1.214*10**-7\n", + "d1 = -5.184*10**-11\n", + "#thermal conductivity of vapour\n", + "kv = a1+b1*Tm+c1*Tm**2+d1*Tm**3 \t\t\t#W/m C\n", + "#(c)heat capacity of vapour, a2,b2,c2,d2 are math.cosmath.tants\n", + "a2 = 21.15\n", + "b2 = 7.092*10**-2\n", + "c2 = 2.589*10**-5\n", + "d2 = -2.852*10**-8\n", + "#heat capacity of vapour, in kj/kh mol K\n", + "Cpv = a2+b2*Tm+c2*Tm**2+d2*Tm**3\n", + "\n", + "#(d)vismath.cosity of vapour\n", + "T7 = 67.\n", + "MUt7 = 112.\n", + "T8 = 127.\n", + "MUt8 = 132.\n", + "#from linear inter polation at Tm\n", + "MUv = 1.364*10**-5 \t\t\t#kg/m s\n", + "\n", + "#from Rohsenow's eq.\n", + "Csf = 0.027 \t\t\t#consmath.tant\n", + "n = 1.7 \t\t\t#exponent value\n", + "#from eq. 6.6\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "#heat flux \t\t\t#kW/m**2\n", + "Q = MU*10**-3*Lv*(g*(rhol-rhov)/S*10**-3)**(1./2)*(Cp_*Te/(Csf*Lv*(Prl)**n))**3\n", + "#from eq. 6.11\n", + "#from eq 6.11, critical heat flux\n", + "Qmax = 0.131*Lv*(rhov)**(1./2)*(S*10**-3*g*(rhol-rhov))**(1./4)\n", + "#dimensionless radius r_\n", + "r = 0.016\n", + "r_ = r*(g*(rhol-rhov)/(S*10**-3))**(1./2)\n", + "#peak heat flux\n", + "Qmax1 = Qmax*(0.89+2.27*math.exp(-3.44*math.sqrt(r_)))\n", + "#from eq. 6.12\n", + "#heat transfer coefficient hb\n", + "d = 0.032 \t\t\t#m, tube diameter\n", + "hb = 0.62*((kv**3)*rhov*(694-rhov)*g*(Lv*10**3+0.4*Cpv*Te)/(d*MUv*Te))**(1./4)\n", + "Qb = hb*Te \t\t\t#kw/m**2, heat flux\n", + "BR = Qb*10**-3/Lv \t\t\t#kg/m**2s, boilng rate \n", + "\n", + "# Results\n", + "print \"The boilins rate is %.0f kg/m**2 h\"%(BR*3600)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4 Page No : 188" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total tube length is 0.393 m\n" + ] + } + ], + "source": [ + "import math \n", + "\n", + "# Variables\n", + "W1 = 200. \t\t\t#kg/h, rate of entering toluene\n", + "muv = 10.**-5 \t\t\t#kg/m s, vismath.cosity of toluene vapour\n", + "mul = 2.31*10**-4 \t\t\t#kg/m s, vismath.cosity of benzene\n", + "rhol = 753. \t\t\t#kg/m**3, density of benzene\n", + "rhov = 3.7 \t\t\t#kg/m**3, density of toluene vapour\n", + "Cpl = 1968. \t\t\t#j/kg C, specific heat of benzene\n", + "kl = 0.112 \t\t\t#W/m C, thermal conductivity of benzene\n", + "T1 = 160. \t\t\t#C tube wall temp.\n", + "T2 = 120. \t\t\t#C , saturated temp.\n", + "Te = T1-T2 \t\t\t#C, excess temp.\n", + "Lv = 3.63*10**5 \t\t\t#j/kg, enthalpy of vaporization\n", + "s = 1.66*10**-2 \t\t\t#N/m, surface tension\n", + "\n", + "#Calculation of hc & hb\n", + "w = 0.125 \t\t\t#m, mean step size\n", + "d = 0.0211 \t\t\t#, internal diameter of tube\n", + "G = W1/(3600*math.pi/4*(d**2)) \t\t\t#kg/m**2 s, mass flow rate\n", + "Re1 = G*(1-w)*d/mul \t\t\t#Reynold no. \n", + "Prl = Cpl*mul/kl \t\t\t#Prandtl no.\n", + "#from eq. 6.23\n", + "x = (w/(1-w))**(0.9)*(rhol/rhov)**(0.5)*(muv/mul)**0.1 \t\t\t#let x = 1/succepsibility\n", + "#from eq. 6.22 \n", + "F = 2.35*(x+0.231)**0.736 \t\t\t#factor signifies 'liquid only reynold no.' to a two phase reynold no.\n", + "#from eq. 7.21\n", + "Re2 = 10**-4*Re1*F**1.25 \t\t\t#Reynold no.\n", + "#from eq. 6.18\n", + "S = (1+0.12*Re2**1.14)**-1 \t\t\t#boiling supression factor\n", + "#from eq. 6.15\n", + "hc = 0.023*Re1**(0.8)*Prl**(0.4)*(kl/d)*F \t\t\t#W/m**2 C, forced convection boiling part\n", + "#from eq. 6.16\n", + "mulv = (1/rhov)-(1/rhol) \t\t\t#m**3/kg, kinetic vismath.cosity of liquid vpaour\n", + "dpsat = Te*Lv/((T2+273)*mulv) \t\t\t#N/m**2, change in saturated presssure \n", + "#nucleate boiling part hb\n", + "hb = 1.218*10**-3*(kl**0.79*Cpl**0.45*rhol**0.49*Te**0.24*dpsat**0.75*S/(s**0.5*mul**0.29*Lv**0.24*rhov**0.24))\n", + "h = hc+hb \t\t\t#W/m**2 C, total heat transfer coefficient\n", + "\n", + "#calculation of required heat transfer area\n", + "a = 5. \t\t\t#%, persentage change in rate of vaporization\n", + "W2 = W1*a/100 \t\t\t#kg/h, rate of vaporization\n", + "W2_ = W2/3600 \t\t\t#kg/s\n", + "Q = W2_*Lv \t\t\t#W,heat load\n", + "A = Q/(h*Te) \t\t\t#m**2, area of heat transfer\n", + "l = A/(math.pi*d) \t\t\t#m, required length of tube\n", + "#from table 6.2\n", + "Tl = 0.393\n", + "\n", + "# Results\n", + "print \"The total tube length is %.3f m\"%(Tl)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5 Page No : 195" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total rate of condensation is 33.08 kg/h\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables\n", + "rhol = 483. \t\t\t#kg/m**3, density of liquid propane\n", + "mul = 9.1*10**-5 \t\t\t#P ,vismath.cosity of liquid propane\n", + "kl = 0.09 \t\t\t#W/m K, thermal conductivity of liquid propane\n", + "Lv = 326. \t\t\t#kj/kg. enthalpy of vaporization\n", + "Cpl = 2.61 \t\t\t#kj/kg K, specific heat of liquid propane\n", + "T1 = 32.\n", + "T2 = 25. \t\t\t#C, surface temp.\n", + "p1 = 11.2\n", + "rhov = 24.7 \t\t\t#kg/m**3, density of vapour\n", + "g = 9.8\n", + "h = 0.3\n", + "\n", + "#Calculation\n", + "Lv1 = Lv+0.68*Cpl*(T1-T2)\n", + "#h = 0.943*(g*Lv1*10**3*rhol*(rhol-rhov)*kl**3/(mul*L*(T1-T2)))**(1/4)\n", + "#Q = h*(L*1)*(T1-T2)\n", + "#m = Q/(Lv1*10**3) = 1.867*10**-2*L**(3/4)\n", + "Ref = 30.\n", + "#from the relation 4*m/mu = Re\n", + "L = (Ref*mul/(4*1.867*10**-2))**(4./3)\n", + "m = 1.867*10**-2*L**(3./4) \t\t\t#rate of condensation for laminar flow\n", + "#from eq. 6.32\n", + "#Nu1 = h_/kl*(mul**2/(rhol*(rhol-rhov)*g))**(1/3) = Ref/(1.08*(Ref)**(1.22)-5.2)\n", + "Lp = h-L \t\t\t#length of plate over which flow is wavy\n", + "A = Lp*1 \t\t\t#m**2 area of condensation\n", + "\n", + "\n", + "def f(h1): \n", + " return h1/kl*(mul**2/(rhol*(rhol-rhov)*g))**(1./3)-(29.76+0.262*h1)/(1.08*(29.76+0.262*h1)**(1.22)-5.2)\n", + "h1 = fsolve(f,1000)\n", + "m2 = m+h1*A*(T1-T2)/(Lv1*10**3)\n", + "Ref1 = 4*m2/mul\n", + "m2 = m+h1*A*(T1-T2)/(Lv1*10**3)\n", + "\n", + "# Results\n", + "print \"Total rate of condensation is %.2f kg/h\"%(m2*3600)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6 Page No : 199" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of condensation is 45.7 kg/h \n", + "Rate of condensation is 1052 kg/h \n", + "thus there will be increase in the calculated rate of heat transfer and in rate of condensation as 1.188 percent\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "#data fot TCE\n", + "T1 = 87.4 \t\t\t#C, normal boiling point\n", + "T2 = 25. \t\t\t#C, surface temp.\n", + "Lv = 320.8 \t\t\t#kj/kg, heat of vaporization\n", + "cp = 1.105 \t\t\t#kj/kg C, specific heat\n", + "mu = 0.45*10**-3 \t\t\t#P. liquid vismath.cosity\n", + "k = 0.1064 \t\t\t#W/m C, thermal conductivity\n", + "rhol = 1375. \t\t\t#kg/m**3, liquid density\n", + "rhov = 4.44 \t\t\t#kg/m**3, density of vapour\n", + "Tm = (T1+T2)/2. \t\t\t#C, mean film temp.\n", + "d = 0.0254 \t\t\t#m, outside diameter of tube\n", + "l = 0.7 \t\t\t#m, length\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "\n", + "# Calculations and Results\n", + "#(a) from eq. 6.34\n", + "Lv1 = Lv+0.68*cp*(T1-T2)\n", + "h = 0.728*(g*Lv1*10**3*rhol*(rhol-rhov)*k**3/(mu*d*(T1-T2)))**(1./4)\n", + "A = math.pi*d*l \t\t\t#m**2, area of tube\n", + "Q = h*A*(T1-T2) \t\t\t#W, rate of heat transfer\n", + "m = (Q/Lv1)/1000 \t\t\t#kg/s rate of condensation\n", + "print \"Rate of condensation is %.1f kg/h \"%(m*3600)\n", + "\n", + "#(b) from eq. 6.35\n", + "N = 6. \t\t\t#No. of tubes in vertical tire\n", + "h1 = 0.728*(g*Lv1*10**3*rhol*(rhol-rhov)*k**3/(N*mu*d*(T1-T2)))**(1./4)\n", + "TN = 36. \t\t\t#total no. of tubes\n", + "TA = TN*math.pi*d*l \t\t\t#m**2, total area\n", + "Q1 = h1*TA*(T1-T2) \t\t\t#W, rate of heat transfer\n", + "m1 = (Q1/Lv1)/1000. \t\t\t#kg/s rate of condensation\n", + "print \"Rate of condensation is %.0f kg/h \"%(m1*3600)\n", + "#from chail's corelation\n", + "h2 = (1+0.2*cp*(T1-T2)*(N-1)/(Lv1))\n", + "print \"thus there will be increase in the calculated rate of\\\n", + " heat transfer and in rate of condensation as %.3f percent\"%(h2)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7 Page No : 201" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fraction of input vapour condensed is 52.7\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "Gv = 20. \t\t\t#kg/m**2 s, mass flow rate of benzene\n", + "di = 0.016 \t\t\t#m, tube diameter\n", + "muv = 8.9*(10**-6) \t\t\t#P, vismath.cosity\n", + "Lv = 391. \t\t\t#kj/kg., enthalpy of vaporization\n", + "cpl = 1.94 \t\t\t#kj/kg C, specific heat\n", + "Tv = 80. \t\t\t#C, normal boiling point of benzene\n", + "Tw = 55. \t\t\t#C, wall temp.\n", + "g = 9.8 \t\t\t#m/s**2, gravitational consmath.tant\n", + "rhol = 815. \t\t\t#kg/m**3, density of benzene\n", + "rhov = 2.7 \t\t\t#kg/m**3, density of benzene vapour\n", + "kl = 0.13 \t\t\t#W/m C, thermal conductivity\n", + "mu = 3.81*10**-4 \t\t\t#P, vismath.cosity of benzene\n", + "l = 0.5 \t\t\t#m, length of tube\n", + "\n", + "#calculation\n", + "Rev = di*Gv/muv \t\t\t#Reynold no. of vapour\n", + "#from eq. 6.38\n", + "Lv1 = Lv+(3./8)*cpl*(Tv-Tw)\n", + "#heat transfer corfficient , h\n", + "h = 0.555*(g*rhol*(rhol-rhov)*kl**3*Lv1*10**3/(di*mu*(Tv-Tw)))**(1./4)\n", + "Aavl = math.pi*di*l \t\t\t#m**2, available area\n", + "Q = Aavl*h*(Tv-Tw) \t\t\t#W, rate of heat transfer\n", + "m = Q/(Lv1*10**3) \t\t\t#kg/s, rate of condensation of benzene\n", + "Ratei = Gv*(math.pi/4)*di**2 \t\t\t#kg/s rate of input of benzene vapour\n", + "n = m/Ratei \n", + "\n", + "# Results\n", + "print \"fraction of input vapour condensed is %.1f\"%(n*100)\n", + "\n", + "# note : rouding off error." + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch7.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch7.ipynb new file mode 100644 index 00000000..19ff7097 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch7.ipynb @@ -0,0 +1,996 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 : radiation heat transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3 Page No : 215" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the fraction of radiation falls in visible range is 0.366 \n", + "the fraction of radiation on the left of visible range is 0.1229 \n", + "the fraction in right of visible range is 0.5111 \n", + "The maximum wavelength is 0.5014 micrometer is\n", + "The frequency is 5.98e+08 s**-1\n", + "the maximum spectral emissive power is 8.298e+13 W/m**2\n", + "the hemispherical total emissive power is 6.326e+07 W/m**2\n" + ] + } + ], + "source": [ + "import math \n", + "# Variables\n", + "Ts = 5780. \t\t\t#K, surface temp.\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "lamda1 = 0.4 \t\t\t#micrometer, starting visible spectrum range \n", + "lamda2 = 0.7 \t\t\t#micrometer,ending visible spectrum range\n", + "E1 = lamda1*Ts \t\t\t#micrometer K, \n", + "E2 = lamda2*Ts \t\t\t#micrometer K, \n", + "#from table 7.2\n", + "#fraction of radiation lying between 0 and lamda1\n", + "F1 = 0.1229\n", + "#fraction of radiation lying between 0 and lamda2\n", + "F2 = 0.4889\n", + "#the fraction of radiation falls betweem lamda1 & lamda 2\n", + "F3 = F2-F1\n", + "print \"the fraction of radiation falls in visible range is %.3f \"%(F3)\n", + "#(b)\n", + "F4 = F1\n", + "print \"the fraction of radiation on the left of visible range is %.4f \"%(F4)\n", + "#(c)\n", + "F5 = 1-F2\n", + "print \"the fraction in right of visible range is %.4f \"%(F5)\n", + "#(d)\n", + "#from wein's print lacement law\n", + "lmax = 2898/Ts\n", + "print \"The maximum wavelength is %.4f micrometer is\"%(lmax)\n", + "c = 2.998*10**8 \t\t\t#m/s, speed of light\n", + "mu = c/lmax\n", + "print \"The frequency is %1.2e s**-1\"%(mu)\n", + "#(e)\n", + "#from eq. 7.4\n", + "h = 6.6256*10**-34 \t\t\t#Js planck's consmath.tant\n", + "k = 1.3805*10**-23 \t\t\t#J/K, boltzman consmath.tant\n", + "Eblmax = (2*math.pi*h*c**2*(lmax*10**-6)**-5)/((math.exp(h*c/(lmax*10**-6*k*Ts)))-1)\n", + "print \"the maximum spectral emissive power is %1.3e W/m**2\"%(Eblmax)\n", + "#(f)\n", + "s = 5.668*10**-8 \t\t\t#stephen math.cosmath.tant\n", + "Eb = s*Ts**4\n", + "print \"the hemispherical total emissive power is %1.3e W/m**2\"%(Eb)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4 Page No : 216" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Surface temp. is 515 C\n", + "wavength is 5.62 micrometer \n", + " from fig 7.1 it falls in the infrared region of spectrum.\n" + ] + } + ], + "source": [ + "#Variables\n", + "Eb = 4000. \t\t\t#W/m sq, Total emmisive power\n", + "s = 5.669*10**-8 \t\t\t#Stephen boltzman consmath.tant\n", + "\n", + "#Calculation\n", + "T = (Eb/s)**0.25 \t\t\t#k, surface temp. of black body\n", + "ym = 2898./T \t\t\t#micro meter,\n", + "#By weins law : Max. wavelength of emmision is inversaly proportional \n", + "#to temprature. and consmath.tant is 2898 micrometer.\n", + "\n", + "#Result\n", + "print \"Surface temp. is %.0f C\"%(T)\n", + "print \"wavength is %.2f micrometer \"%(ym)\n", + "print \" from fig 7.1 it falls in the infrared region of spectrum.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5 Page No : 219" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "total hemispherical) emissive power is 1.241e+05 W/m**2\n", + "total hemispherical) emissivity of the surface is 0.4326\n" + ] + } + ], + "source": [ + "# Variables\n", + "T = 1500. \t\t\t#K, surface temprature\n", + "#from fig 7.7\n", + "e1 = 0.2 \t\t\t#emissivity ,when wavelength(l1) is 0<l1<2 micrometer\n", + "e2 = 0.6 \t\t\t#emissivity ,when wavelength(l2) is 2<l2<6 micrometer\n", + "e3 = 0.1 \t\t\t#emissivity ,when wavelength(l3) is 6<l3<10 micrometer\n", + "e4 = 0 \t\t\t#emissivity ,when wavelength(l4) is l4>10 micrometer\n", + "#from table 7.2\n", + "F1 = 0.2733 \t\t\t#fraction of energy in wavelength (l1)\n", + "F2 = 0.89-F1 \t\t\t#fraction of energy in wavelength (l2)\n", + "F3 = 0.9689-0.89 \t\t\t#fraction of energy in wavelength (l3)\n", + "\n", + "#Calculation and Result\n", + "s = 5.669*10**-8 \t\t\t#stephen's consmath.tant\n", + "Eb = s*T**4 \t\t\t#emissive power \n", + "E = (e1*F1+e2*F2+e3*F3)*Eb\n", + "print \"total hemispherical) emissive power is %1.3e W/m**2\"%(E)\n", + "#(b)\n", + "e = E/(s*T**4)\n", + "print \"total hemispherical) emissivity of the surface is %.4f\"%(e)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6 Page No : 226" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fraction of radiation passes through hole 0.0588 \n", + "fraction of radiation intercepted by the ring 0.0791 \n" + ] + } + ], + "source": [ + "from scipy.integrate import quad \n", + "\n", + "# Variables\n", + "ri = 5. \t\t\t#cm ,inside radius of ring\n", + "w = 3. \t\t\t#cm, width\n", + "ro = ri+w \t\t\t#cm, outside radius \n", + "L = 20. \t\t\t#cm, surface dismath.tance\n", + "\n", + "# Calculations\n", + "def f4(r): \n", + " return 20.**2*r/(20.**2+r**2)**2\n", + "\n", + "F1 = 2* quad(f4,0,ri)[0]\n", + "\n", + "#view factor along surface dA1-A2\"\n", + "\n", + "def f5(r): \n", + " return 20**2*r/(20**2+r**2)**2\n", + "\n", + "F2 = 2* quad(f5,ri,ro)[0]\n", + "\n", + "# Results\n", + "print \"fraction of radiation passes through hole %.4f \"%(F1)\n", + "print \"fraction of radiation intercepted by the ring %.4f \"%(F2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8 Page No : 232" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "view factor F11 = 0\n", + "view factor F22 = 1\n", + "view factor F21 = 0.5\n", + "view factor = 0.5\n" + ] + } + ], + "source": [ + "import math\n", + "#Variables\n", + "F11 = 0 \t\t\t#view factor\n", + "d = 1. \t\t\t#let it be\n", + "print \"view factor F11 = %.0f\" %(F11)\n", + "\n", + "#Calculation and Result\n", + "F12 = 1-F11 \t\t\t#view factor\n", + "print \"view factor F22 = %.0f\"%(F12)\n", + "\n", + "A1 = ((math.pi)*d**2)/4 \t\t\t#sq m, area\n", + "A2 = ((math.pi)*d**2)/2 \t\t\t#sq m, area\n", + "F21 = A1/A2 \t\t\t#from eq . 7.26\n", + "print \"view factor F21 = %.1f\"%( F21)\n", + "F22 = 1-F21 \n", + "#Results\n", + "print \"view factor = %.1f\"%(F22)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9 Page No : 233" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "view factor F11 = 0\n", + "view factor F33 = 0.5\n", + "view factor F14 = 0.056\n", + "view factor F13 = 0.056\n", + "view factor F12 = 0.944\n", + "view factor F31 = 0.028\n", + "view factor F32 = 0.472\n", + "view factor F21 = 0.118\n", + "view factor F23 = 0.118\n", + "view factor F22 = 0.764\n" + ] + } + ], + "source": [ + "import math\n", + "#Variable\n", + "s = 3. \t\t\t#no. of surface\n", + "tvf = s**2 \t\t\t#total view factor\n", + "\n", + "#using the result of example 7.8\n", + "F11 = 0 \n", + "F33 = 0.5\n", + "print \"view factor F11 = %.0f\"%(F11)\n", + "print \"view factor F33 = %.1f\"%(F33)\n", + "\n", + "#Calculation & Results\n", + "R1 = 0.25 \t\t\t#R = d/2*h &h = 2d\n", + "R2 = 0.25\n", + "X = 1+((1+R2**2)/(R1**2))\n", + "F14 = (0.5)*(X-math.sqrt((X**2)-4*(R2/R1)**2))\n", + "print \"view factor F14 = %.3f\"%(F14)\n", + "F13 = F14\n", + "print \"view factor F13 = %.3f\"%(F13)\n", + "F12 = 1-F11-F13 \t\t\t# from eq. 7.31 for surface 1\n", + "print \"view factor F12 = %.3f\"%(F12)\n", + "\n", + "d = 1. \t\t\t#say\n", + "A1 = (math.pi*(d**2))/4.\n", + "A3 = (math.pi*(d**2))/2.\n", + "F31 = A1*F13/(A3)\n", + "print \"view factor F31 = %.3f\"%(F31)\n", + "\n", + "# from eq. 7.31 for surface 3\n", + "F33 = 0.5\n", + "F32 = 1-F31-F33\n", + "print \"view factor F32 = %.3f\"%(F32)\n", + "\n", + "#for surface 2\n", + "A2 = 2*math.pi*d**2\n", + "F21 = A1*F12/A2\n", + "print \"view factor F21 = %.3f\"%(F21)\n", + "F23 = A3*F32/A2\n", + "print \"view factor F23 = %.3f\"%(F23)\n", + "F22 = 1-F21-F23\n", + "print \"view factor F22 = %.3f\"%(F22)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.10 Page No : 235" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The view factor from surface 1 to 2 is 1\n", + "The view factor from surface 2 to 1 is 0.167\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "ds = 0.3 \t\t\t#m, diameter of shell\n", + "r1 = 0.1 \t\t\t#m, dismath.tance from the centre\n", + "\n", + "#Calculation and Results\n", + "#by the defination of view factor\n", + "F12 = 1.\n", + "print \"The view factor from surface 1 to 2 is %.0f\"%(F12)\n", + "#F21\n", + "R = ds/2. \t\t\t#m, radius of sphere\n", + "r2 = math.sqrt(R**2-r1**2)\n", + "A1 = math.pi*r2**2 \t\t\t#m**2 area\n", + "A2 = 2*math.pi*R**2+2*math.pi*R*math.sqrt(R**2-r2**2)\n", + "#from reciprocity relation\n", + "F21 = (A1/A2)*F12\n", + "print \"The view factor from surface 2 to 1 is %.3f\"%(F21)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.12 Page No : 237" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time required for the ball to cool is 10.3 h\n" + ] + } + ], + "source": [ + "import math \n", + "from scipy.integrate import quad \n", + "# Variables\n", + "d = 0.3 \t\t\t#m, diameter of steel sphere\n", + "Ti = 800. \t\t\t#K, initial temp. of sphere\n", + "T2 = 303. \t\t\t#C,ambient temp.\n", + "T1 = 343. \t\t\t#C, final tempreture\n", + "rho = 7801. \t\t\t#kg/m**3, density of steel\n", + "cp = 0.473 \t\t\t#kj/kg C, specific heat of steel\n", + "#calculation\n", + "R = d/2 \t\t\t#m, radius of sphere\n", + "A1 = 4*math.pi*R**2 \t\t\t#m**2, area of sphere\n", + "m = 4./3*math.pi*R**3*rho \t\t\t#m**3, mass of sphere\n", + "F12 = 1. \t\t\t#view factor\n", + "s = 5.669*10**-8 \t\t\t#stephen Boltzman's consmath.tant\n", + "#-dT1/dt = A1*F12*s*(T**4-T2**4)/(m*cp)\n", + "\n", + "def f1(T1): \n", + " return (1/(T1**4-T2**4))\n", + "\n", + "I = quad(f1,343,800)[0]\n", + "\n", + "t = I/(A1*F12*s/(m*cp*10**3))\n", + "\n", + "# Results\n", + "print \"The time required for the ball to cool is %.1f h\"%(t/3600)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.13 Page No : 247" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Net rate of radiative heat loss Q12 = 596.6 W \n", + "b) Net rate of radiative heat loss Q12 = 441.5 W\n" + ] + } + ], + "source": [ + "import math\n", + "#Variables\n", + "d = 0.114 \t\t\t#m, dia.o f pipe\n", + "l = 1. \t\t\t#m, length of pipe\n", + "A = (math.pi)*d*l \t\t\t#m sq, area\n", + "e1 = 1. \t\t\t#emmisivity of black body\n", + "F12 = 1. \t\t\t#view factor, 1:pipe surface, 2:room walls\n", + "s = 5.67*10**-8 \t\t\t#stephen boltzman consmath.tant\n", + "T1 = 440. \t\t\t#K, steam temp.\n", + "T2 = 300. \t\t\t#K, wall temp.\n", + "#Caluclation\n", + "Q12 = A*e1*F12*s*(T1**4-T2**4) \t\t\t#net rate of radiative heat loss\n", + "\n", + "#Results\n", + "print \"a) Net rate of radiative heat loss Q12 = %.1f W \"%(Q12)\n", + "#Part-b\n", + "e2 = 0.74\n", + "Q12 = A*e2*F12*s*(T1**4-T2**4) \t\t\t#net rate of radiative heat loss\n", + "print \"b) Net rate of radiative heat loss Q12 = %.1f W\"%(Q12)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.14 Page No : 247" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a-i) View factor F12 = 1\n", + "view factor F21 = 0.937\n", + "ii) The net rate of heat gain Q1net = 4.0 J/s\n", + "b) Rate of nitrogen loss = 72 g/h\n" + ] + } + ], + "source": [ + "import math\n", + "#Variable declaration\n", + "F12 = 1. \t\t\t#view factor\n", + "r1 = 0.15 \t\t\t#m inner radius of phere\n", + "r2 = 0.155 \t\t\t#m , outer radius\n", + "\n", + "#Calculation\n", + "A1 = 4.*(math.pi)*r1**2 \t\t\t#sq m inner area\n", + "A2 = 4.*(math.pi)*r2**2 \t\t\t#sq m,outer area \n", + "F21 = A1/A2\n", + "h = 200. \t\t\t#J/g, heat of vaporization of nitrogen\n", + "s = 5.669*10**-8 \t\t\t# boltzman consmath.tant\n", + "T2 = 298. \t\t\t#K, temp. of outer wall\n", + "T1 = 77. \t\t\t#K, Temp. of inner wall\n", + "e1 = 0.06 \t\t\t#emmisivity\n", + "e2 = 0.06 \t\t\t#emmisivity\n", + "x = ((1-e1)/(e1*A1))+(1/(A1*F12))+((1-e2)/(e2*A2))\n", + "Q1net = (s*(T2**4-T1**4))/(x)\n", + "\n", + "#Result-a-i\n", + "print \"a-i) View factor F12 = %.0f\"%(F12)\n", + "print \"view factor F21 = %.3f\"%(F21)\n", + "#Result- b\n", + "print \"ii) The net rate of heat gain Q1net = %.1f J/s\"%(Q1net)\n", + "nl = Q1net/h\n", + "nl = nl*3600 \t\t\t#g/h\n", + "print \"b) Rate of nitrogen loss = %.0f g/h\"%(nl)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.15 Page No : 248" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the net rate of radiant heat transfer to the wall is 2900 W\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "x = 0.15 \t\t\t#m, length of opening on a furnace\n", + "y = 0.12 \t\t\t#m, width of opening on a furnace\n", + "x1 = 6. \t\t\t#m, width of wall\n", + "y1 = 5. \t\t\t#m, height of wall\n", + "e2 = 0.8 \t\t\t#emissivity of wall\n", + "T1 = 1400. \t\t\t#C, furnace temp.\n", + "T2 = 35. \t\t\t#C, wall temp.\n", + "T3 = 273. \t\t\t#C, smath.radians(numpy.arcmath.tan(ard temp.\n", + "s = 5.669*10**-8 \t\t\t#stephen boltzman's consmath.tant\n", + "#in fig. 7.29\n", + "l1 = 2. \t\t\t#m, l1 = AF\n", + "l2 = 1.5 \t\t\t#m, l2 = AH\n", + "h = 3. \t\t\t#m, E = dA1\n", + "\n", + "# Calculations\n", + "F1 = (1./(2*math.pi))*((l2/(math.sqrt(l2**2+h**2)))*math.tanh(l1/(math.sqrt(l2**2+h**2)))+(l1/(math.sqrt(l1**2+h**2)))*math.tan(l2/(math.sqrt(l1**2+h**2))))\n", + "#Similarly\n", + "#for the dA1-A3 pair the equation is\n", + "F2 = 0.1175\n", + "#for the dA1-A4 pair the equation is\n", + "F3 = 0.1641\n", + "#for the dA1-A5 pair the equation is\n", + "F4 = 0.0992\n", + "#view factor b/w the opening (dA1)and the wall (W) is \n", + "F5 = F1+F2+F3+F4\n", + "#Calculation of radient heat exchange\n", + "dA1 = x*y\n", + "Aw = x1*y1\n", + "Eb1 = s*(T1+T3)**4\n", + "Ebw = s*(T2+T3)**4\n", + "F6 = dA1*F5/Aw\n", + "Q = dA1*F5*e2*(Eb1*(1-(1-e2)*F6)-Ebw)\n", + "\n", + "# Results\n", + "print \"the net rate of radiant heat transfer to the wall is %.0f W\"%(round(Q,-2))\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.16 Page No : 250" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the net rate of radiant heat loss = 17.1 kW \n", + "convective heat transfer coeff. = 90 W/sq m C\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "l = 3. \t\t\t#m, length of wall\n", + "w = 2. \t\t\t#m, width of, wall\n", + "d = 3. \t\t\t#m\n", + "R1 = l/d\n", + "A1 = l*w \t\t\t#sq m,area 1: front part\n", + "A2 = A1 \t\t\t#sq m , area, 2\"back part\n", + "e1 = 0.7 \t\t\t#emmisivity\n", + "e2 = 0.7 \t\t\t#emmisivity\n", + "T1 = 673. \t\t\t#k\n", + "T2 = 523. \t\t\t#k\n", + "s = 5.669*10**-8 \t\t\t#stephen boltzman consmath.tant\n", + "\n", + "#Calculation\n", + "F12 = 0.148 \t\t\t#view factor ,from fig. 7.12\n", + "x = (A1+A2-2*A1*F12)/(A2-(A1*(F12**2)))+((1/e1)-1)+(A1/A2)*((1/e2)-1)\n", + "\n", + "#Results\n", + "Q1net = -1*A1*(s*(T2**4-T1**4))/(x)\n", + "print \"the net rate of radiant heat loss = %.1f kW \"%(Q1net/1000)\n", + "# (b)\n", + "F24 = 1. \t\t\t#from fig 7.12\n", + "T20 = 333. \t\t\t#K, outer surface temp. of surface 2\n", + "T4 = 303. \t\t\t#K, ambient temp\n", + "Q2rad = A2*e2*F24*s*(T20**4-T4**4)\n", + "q = Q1net-Q2rad\n", + "q1 = q/1000 \t\t\t# Kw\n", + "h = q/(A2*(T20-T4))\n", + "print \"convective heat transfer coeff. = %.0f W/sq m C\"%(h)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.17 Page No : 251" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "net rate of radiation exchange b/w disk 1 and 2 is 2286 W/m**2\n" + ] + } + ], + "source": [ + "from numpy import array, linalg\n", + "import math\n", + "\n", + "# Variables\n", + "r1i = 0.1 \t\t\t#m, inner radius of disk 1\n", + "r1o = 0.2 \t\t\t#m, outer radius of disk 1\n", + "r2i = 0.12 \t\t\t#m, inner radius of disk 2\n", + "r2o = 0.25 \t\t\t#m, outer radius of disk 2\n", + "h = 0.08 \t\t\t#m, dismath.tance between the disks\n", + "R2 = r2o/h\n", + "R1 = r1o/h\n", + "X = 1+(1+R1**2)/R2**2\n", + "F23_14 = 1./2*(X-math.sqrt(X**2-4*(R1/R2)**2))\n", + "\n", + "#calculation\n", + "R2_ = r2o/h\n", + "R1_ = r1i/h\n", + "X_ = 1+(1+R1_**2)/R2_**2\n", + "F23_4 = 1/2*(X_-math.sqrt(X_**2-4*(R1_/R2_)**2)) \t\t\t#view factor\n", + "#similarly\n", + "F3_14 = 0.815 \t\t\t#view factor\n", + "F34 = 0.4 \t\t\t#view factor\n", + "A23 = math.pi*r2o**2 \t\t\t#area\n", + "A3 = math.pi*r2i**2\n", + "A1 = math.pi*(r1o**2-r1i**2)\n", + "#from eq. 1\n", + "F12 = A23*(F23_14-F23_4)/A1-(A3*(F3_14-F34))/A1\n", + "\n", + "#calculation of the rate of radiative heat exchange\n", + "# Variables\n", + "T1 = 1000. \t\t\t#K, temprature of disk 1\n", + "T2 = 300. \t\t\t#K, temprature of disk 2\n", + "s = 5.669*10**-8 \t\t\t#stephen's Boltzman consmath.tant\n", + "e1 = 0.8 \t\t\t#emissivity\n", + "e2 = 0.7\n", + "A2 = math.pi*(r2o**2-r2i**2)\n", + "F1s = 1-F12\n", + "F2s = 1-(A1*F12/A2)\n", + "#calculation\n", + "#let some quantities equal to \n", + "a = (1-e1)/(e1*A1)\n", + "b = 1/(A1*F12)\n", + "c = (1-e2)/(e2*A2)\n", + "d = 1/(A1*F1s)\n", + "e = 1/(A2*F2s)\n", + "f = s*T1**4\n", + "g = s*T2**4\n", + "#from eq. 7.42(a)\n", + "#(f-J1)/a = (J1-J2)/b+J1/d\n", + "#(g-J2)/c = (J2-J1)/b+J1/e\n", + "#solving two eqns by matrix\n", + "A = array([[-0.0564,0.5036],[0.4712,-0.0564]])\n", + "B = array([[161.847],[21376.31]])\n", + "X = linalg.solve(A,B)\n", + "J1 = X[0]\n", + "J2 = X[1]\n", + "\n", + "#net rate of radiation exchange \n", + "Q12net = (J1-J2)/17.73\n", + "\n", + "# Results\n", + "print \"net rate of radiation exchange b/w disk 1 and 2 is %d W/m**2\"%(Q12net)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.18 Page No : 255" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The net rate of heat gain of tube is 0.30 W\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables\n", + "di = 0.0254 \t\t\t#m, inner diameter of tube\n", + "Ti = 77. \t\t\t#K, liquid temprature\n", + "do = 52.5*10**-3 \t\t\t#m, pipe internal diameter\n", + "To = 270. \t\t\t#K, wall temprature\n", + "l = 1. \t\t\t#m, length of tube\n", + "e1 = 0.05 \t\t\t#emissivity of tube wall\n", + "e2 = 0.1 \t\t\t#emissivity of pipe wall\n", + "e3 = 0.02 \t\t\t#emissivity for inner surface of radiation field\n", + "e4 = 0.03 \t\t\t#emissivity for outer surface of radiation field\n", + "s = 5.669*10**-8 \t\t\t#stephen boltzman math.cosmath.tantl\n", + "\n", + "#Calculation\n", + "ds = (do+di)/2 \t\t\t#m, diameter of radiation shield\n", + "Ao = math.pi*do*l \t\t\t#m**2, outer pipe area\n", + "As = math.pi*ds*l \t\t\t#m**2, shield area\n", + "Ai = math.pi*di*l \t\t\t#m**2, inner pipe area\n", + "#View factors\n", + "#for the long cylindrical enclosure made up of the outer pipe and the shield\n", + "Fso = 1. \t\t\t#because outer surface of shield cant see itself\n", + "Fos = As/Ao \n", + "Fsi = Ai/As\n", + "#now assume \n", + "#(1-e2)/e2+ 1/Fos +Ao*(1-e4)/(As*e4) = x\n", + "#(1-e3)/e3 +1/Fsi +(1/Fsi)*(1-e1)/e1 = y\n", + "x = (1-e2)/e2+ 1/Fos +Ao*(1-e4)/(As*e4)\n", + "y = (1-e3)/e3 +1/Fsi +(1/Fsi)*(1-e1)/e1\n", + "#solving the equations for heat transfer from the outer pipe and inner pipe\n", + "def f(Ts): \n", + " return (Ao*(To**4-Ts**4)/x)-(Ai*(Ts**4-Ti**4)/x)\n", + "Ts = fsolve(f,1)\n", + "Qos = (Ao*s*(To**4-Ts**4))/x\n", + "\n", + "# Results\n", + "print \"The net rate of heat gain of tube is %.2f W\"%(Qos)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.19 Page No : 258" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss from the top surface : 403 W\n", + "Rate of heat loss from the side walls : 1271 W\n", + "Total rate of heat loss : 1674 W\n" + ] + } + ], + "source": [ + "# Variables\n", + "T1 = 300 + 273. #K\n", + "Ta = 30. + 273. # K\n", + "w = 0.075 # m\n", + "k = 0.08 # W/m c\n", + "l = 1.075 # m\n", + "delta = 5.669 * 10**-8 # W/m**2 K**4\n", + "A1 = 1 * 1.5 # M**2\n", + "A2 = A1\n", + "A2m = (1.5 + 1.9)/2 # m**2\n", + "A3m = (5. + 6.02)/2 # m**2 \n", + "\n", + "# Calculations\n", + "T2 = 545.1 # k\n", + "T2_ = 322.6\n", + "T3 = 544.7\n", + "T3_ = 328.4\n", + "rate_of_heatloss1 = int(A2m*k/w*(T2-T2_))\n", + "rate_of_heatloss2 = int(A3m*k/w*(T3-T3_))\n", + "total = rate_of_heatloss1 + rate_of_heatloss2\n", + "\n", + "# results\n", + "print \"Rate of heat loss from the top surface : %d W\"%rate_of_heatloss1\n", + "print \"Rate of heat loss from the side walls : %d W\"%rate_of_heatloss2\n", + "print \"Total rate of heat loss : %d W\"%total" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.20 Page No : 264" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the spectral extinction coefficient is 24.08 m**-1\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "T = 300. \t\t\t#K, temprature\n", + "per = 91. \t\t\t#percent, adsorbed radiation\n", + "lam = 4.2 \t\t\t#micrometer, wavelength radiation\n", + "L = 0.1 \t\t\t#m, path length\n", + "\n", + "#calculation\n", + "# I2/I1 = f\n", + "f = 1-per/100. \t\t\t#fraction of incident radiation transmitted\n", + "#from eq. 7.69\n", + "a = -math.log(f)/L\n", + "\n", + "# Results\n", + "print \"the spectral extinction coefficient is %.2f m**-1\"%(a)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.21 Page No : 265" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total rate of heat transfer from the gas to the wall is 22.5 kW\n" + ] + } + ], + "source": [ + "# Variables\n", + "Ts = 800. \t\t\t#C, wall temp.\n", + "Tg = 1100. \t\t\t#C. burner temprature\n", + "CO2 = 8. \t\t\t#percent, composition of CO2 in flue gas\n", + "M = 15.2 \t\t\t#percent, composition of moisture in flue gas\n", + "a = 0.4 \t\t\t#m, length of duct\n", + "b = 0.4 \t\t\t#width of duct\n", + "h = 15. \t\t\t#W/m**2 C, heat transfer coefficient\n", + "P = 1. \t\t\t#atm pressure\n", + "#CAlCULATION of Eg(Tg)\n", + "pc = CO2/100.*P \t\t\t#atm, partial pressure of CO2\n", + "pw = M/100.*P \t\t\t#atm, partial pressure of moisture\n", + "l = 1. \t\t\t#m, length of duct\n", + "V = a*b*l \t\t\t#m**3, volume of duct\n", + "A = 1.6*l \t\t\t#m**2 area of duct\n", + "Le = 3.6*(V/A) \t\t\t#m, mean beam length\n", + "\n", + "pc*Le\n", + "pw*Le\n", + "Tg_ = Tg+273.\n", + "Ts_ = Ts+273.\n", + "#from fig 7.38\n", + "Ec = 0.06\n", + "Eg = 0.048 \t\t\t#from fig 7.39\n", + "#a correction dE need to be calculated\n", + "#pw/(pc+pw)\n", + "#pc*Le+pw*Le\n", + "#from fig. 7.39\n", + "dE = 0.003\n", + "Eg_Tg = Ec+Eg-dE \t\t\t#emissivity at temp. Tg\n", + "\n", + "#Calculation of alpha\n", + "#pc*Le*Ts/Tg\n", + "#from fig. 7.37\n", + "Ec1 = 0.068\n", + "#from fig. 7.38\n", + "Ew1 = 0.069\n", + "Cc = 1 \t\t\t#correction factor\n", + "Cw = 1 \t\t\t#correction factor\n", + "d_alpha = dE \t\t\t#AT 1 ATM TOTAL PRESSURE\n", + "alpha = Cc*Ec1*(Tg_/Ts_)**0.65+Cw*Ew1*(Tg_/Ts_)**0.45-dE\n", + "#radiant heat ransfer rate\n", + "s = 5.669*10**-8 \t\t\t#stephen's boltzman consmath.tant\n", + "Qrad = A*s*(Eg_Tg*Tg_**4-alpha*Ts_**4) \t\t\t#kW\n", + "Qconv = h*A*(Tg-Ts) \t\t\t#kW, convective heat transfer rate\n", + "Q = Qrad+Qconv\n", + "\n", + "# Results\n", + "print \"The total rate of heat transfer from the gas to the wall is %.1f kW\"%(Q/1000)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb new file mode 100644 index 00000000..ff0dc69f --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb @@ -0,0 +1,494 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 : Heat Exchanger" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1 Page No : 303" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the heat duty of the exchanger is 47000 kj/h\n", + "the water flow rate is 1122 kg/h\n", + "heat transfer coefficient based on inside area is 3560 W/m**2 C \n", + "heat transfer coefficient based on outside area is 880.3 W/m**2 C \n", + "overall heat transfer coefficient outside area basis is 663.1 W/m**2 C \n", + "overall heat transfer coefficient inside area basis is 802.0 W/m**2 C \n", + "The fouling factor is 0.000949 m**2 C/W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "#for Benzene\n", + "Mb = 1000. \t\t\t#Kg, mass of benzene\n", + "T1 = 75. \t\t\t#C initial temp. of benzene\n", + "T2 = 50. \t\t\t#C final temp. of benzene\n", + "Cp1 = 1.88 \t\t\t#Kj/Kg C. specific heat of benzene\n", + "mu1 = 0.37 \t\t\t#cP. vismath.cosity of benzene\n", + "rho1 = 860. \t\t\t#kg/m**3, density\n", + "k1 = 0.154 \t\t\t#W/m K. thermal conductivity\n", + "\n", + "#for water\n", + "Tav = 35. \t\t\t#C av, temp.\n", + "Cp2 = 4.187 \t\t\t#specific heat\n", + "mu2 = 0.8 \t\t\t#cP. vismath.cosity\n", + "k2 = 0.623 \t\t\t#W/m K. thermal conductivity\n", + "T3 = 30. \t\t\t#C. initial temp.\n", + "T4 = 40. \t\t\t#C final temp.\n", + "\n", + "#Calculation and Results\n", + "#(a)\n", + "HD = Mb*Cp1*(T1-T2) \t\t\t#Kj/h, heat duty\n", + "WR = HD/(Cp2*(T4-T3)) \t\t\t#kg/h Water rate\n", + "print \"the heat duty of the exchanger is %.0f kj/h\"%(HD)\n", + "print \"the water flow rate is %d kg/h\"%(WR)\n", + "\n", + "#(b)\n", + "#tube side (water) calculations\n", + "# Variables\n", + "di1 = 21. \t\t\t#mm, inner diameter of inner tube \n", + "do1 = 25.4 \t\t\t#mm, outer dia. of inner tube\n", + "t = 2.2 \t\t\t#mm/ wall thickness\n", + "kw = 74.5 \t\t\t#W/m K. thermal conductivity of the wall\n", + "di2 = 41. \t\t\t#mm, inner diameter of outer pipe\n", + "do2 = 48. \t\t\t#mm, outer diameter of outer pipe\n", + "\n", + "FA1 = (math.pi/4)*(di1*10**-3)**2 \t\t\t#m**2, flow area\n", + "FR1 = WR/1000.\n", + "v1 = FR1/(FA1*3600) \t\t\t#m/s, velocity\n", + "Re1 = (di1*10**-3)*v1*1000/(mu2*10**-3) \t\t\t#Reynold no.\n", + "Pr1 = Cp2*1000*(mu2*10**-3)/k2 \t\t\t#Prandtl no.\n", + "#umath.sing dittus boelter eq.\n", + "Nu1 = 0.023*(Re1)**(0.8)*(Pr1)**(0.3) \t\t\t#nusslet no.\n", + "h1 = round(Nu1*k2/(di1*10**-3),-1) \t\t\t#W/m**2 C, heat transfer coefficient\n", + "\n", + "#Outer side (benzene) calculation\n", + "FA2 = (math.pi/4)*(di2*10**-3)**2-(math.pi/4)*(do1*10**-3)**2 \t\t\t#flow area\n", + "wp = math.pi*(di2*10**-3+do1*10**-3) \t\t\t#wettwd perimeter\n", + "dh = 4*FA2/wp \t\t\t#hydrolic diameter\n", + "bfr = Mb/rho1 \t\t\t#m**3/h benzene flow rate\n", + "v2 = bfr/(FA2*3600) \t\t\t#m/s, velocity\n", + "Re2 = dh*v2*rho1/(mu1*10**-3) \t\t\t#Reynold no\n", + "Pr2 = Cp1*10**3*(mu1*10**-3)/k1 \t\t\t#Prandtl no.\n", + "Nu2 = 0.023*(Re2)**(0.8)*(Pr2)**(0.4) \t\t\t#nusslet no.\n", + "h2 = Nu2*k1/(dh) \t\t\t#W/m**2 C, heat transfer coefficient\n", + "\n", + "print \"heat transfer coefficient based on inside area is %.0f W/m**2 C \"%(h1)\n", + "print \"heat transfer coefficient based on outside area is %.1f W/m**2 C \"%(h2)\n", + "\n", + "#Calculation of clean overall heat transfer coefficient, outside area basis\n", + "#from eq. 8.28\n", + "# Variables\n", + "l = 1. \t\t\t#assume , length\n", + "Ao = math.pi*do1*10**-3*l\n", + "Ai = math.pi*di1*10**-3*l\n", + "Am = (do1*10**-3-di1*10**-3)*math.pi*l/(math.log(do1*10**-3/(di1*10**-3)))\n", + "\n", + "#overall heat transfer coefficient\n", + "Uo = 1/((1/h2)+(Ao/Am)*((do1*10**-3-di1*10**-3)/(2*kw))+(Ao/Ai)*(1/h1))\n", + "Ui = Uo*Ao/Ai\n", + "\n", + "#Calculation of LMTD\n", + "dt1 = T1-T4\n", + "dt2 = T2-T3\n", + "LMTD = (dt1-dt2)/math.log(dt1/dt2) \t\t\t#math.log mean temp. difference correction factor\n", + "Q = HD*1000/3600 \t\t\t#W, heat required\n", + "Ao_ = Q/(Uo*LMTD) \t\t\t#m**@, required area\n", + "len = Ao_/(math.pi*do1*10**(-3)) \t\t\t#m, tube length necessary\n", + "\n", + "#(c)\n", + "la = 15. \t\t\t#m ,actual length\n", + "Aht = (math.pi*do1*10**(-3)*la)\n", + "Udo = Q/(Aht*LMTD) \t\t\t#W/m**2 C, overall heat transfer coefficient with dirt factor\n", + "#from eq. 8.2\n", + "Rdo = (1/Udo)-(1/Uo) \t\t\t#m**2 C/W\n", + "print \"overall heat transfer coefficient outside area basis is %.1f W/m**2 C \"%(Uo)\n", + "print \"overall heat transfer coefficient inside area basis is %.1f W/m**2 C \"%(Ui)\n", + "print \"The fouling factor is %f m**2 C/W\"%(Rdo)\n", + "\n", + "# note : rounding off error. please check." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2 Page No : 309" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tube side Pressure drop is 1.118e+04 N/m**2 \n", + "Shell side Pressure drop is 120 N/m**2 \n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "Cp = 50. \t\t\t#tpd, plant capacity\n", + "T1 = 135. \t\t\t#C, Temp.\n", + "T2 = 40. \t\t\t#C temp.\n", + "T3 = 30. \t\t\t#C temp.\n", + "dt1 = (T1-T2) \t\t\t#C hot end temp. \n", + "dt2 = (T2-T3) \t\t\t#C cold end temp.\n", + "#Properties of ethylbenzene\n", + "rho1 = 840. \t\t\t#kg/m**3, density\n", + "cp1 = 2.093 \t\t\t#kj/kg K , specific heat\n", + "T = 87.5 \t\t\t#C\n", + "mu1 = math.exp(-6.106+1353/(T+273)+5.112*10**-3*(T+273)-4.552*10**-6*((T+273)**2))\n", + "k1 = 0.2142-(3.44*10**-4)*(T+273)+(1.947*10**-7)*(T+273)**2\n", + "k1_ = k1*0.86 \t\t\t#kcal/h m K\n", + "#properties of water\n", + "rho2 = 993. \t\t\t#kg/m**3, density\n", + "mu2 = 8*10.**-4 \t\t\t#kg/m s , vismath.cosity \n", + "cp2 = 4.175 \t\t\t#kj/kg K , specific heat\n", + "k2 = 0.623 \t\t\t#W/m K, thermal conductivity\n", + "k2_ = k2*0.8603 \t\t\t#kcal/h m**2 K\n", + "#Calculation\n", + "#(i) Energy balance\n", + "Cp = Cp*1000./24 \t\t\t#kg/h, plant capacity\n", + "Cp = 2083. \t\t\t#approx.\n", + "HD = Cp*cp1*dt1 \t\t\t#kj/h, Heat duty \n", + "HD_ = HD*0.238837 \t\t\t#kcal/h\n", + "wfr = HD/(cp2*dt2)\n", + "\n", + "#(ii)\n", + "mu1 = mu1 \t\t\t#cP, vismath.cosity of ethylbenzene\n", + "k1 = k1 \t\t\t#W/m K, thermal conductivity of ethylbenzene\n", + "\n", + "#(iii)\n", + "#LMTD calculation\n", + "LMTD = (dt1-dt2)/math.log(dt1/dt2)\n", + "#assume\n", + "Udo = 350. \t\t\t#kcal/h m**2 C, overall coefficient\n", + "A = HD_/(Udo*LMTD) \t\t\t#m**2, area required\n", + "\n", + "#(iv)\n", + "id_ = 15.7 \t\t\t#mm, internal diameter of tube\n", + "od = 19. \t\t\t#mm, outer diameter of tube\n", + "l = 3000. \t\t\t#mm, length\n", + "OSA = math.pi*(od*10**-3)*(l*10**-3) \t\t\t#m**2. outer surface area\n", + "n = A/OSA \t\t\t#no. of tubes required\n", + "fa = n*(math.pi/4)*(id_*10**-3)**2 \t\t\t#m**2, flow arae\n", + "lv = (wfr/1000)/(3600*fa) \t\t\t#m/s, linear velocity\n", + "\n", + "#(v)\n", + "n1 = 44. \t\t\t#total no. of tubes that can be accomodated in a 10 inch shell\n", + "np = 11. \t\t\t#no. of tubes in each pass\n", + "#(vi)\n", + "bf = 0.15 \t\t\t#m, baffel spacing\n", + "#(vii)\n", + "#estimation of heat transfer coefficient\n", + "#Tube side (water)\n", + "fa1 = (math.pi/4)*(id_*10**-3)**2*np \t\t\t#m**2, flow area\n", + "v1 = (wfr/1000.)/(3600*fa1) \t\t\t#m/s, velocity\n", + "Re = (id_*10**-3)*v1*rho2/mu2 \t\t\t#Reynold no.\n", + "#from fig . 8.11(a)\n", + "jh = 85. \t\t\t#colburn factor\n", + "#jh = (hi*di)/k*(cp*mu/k)**-1/3 \n", + "#assume, (cp*mu/k) = x\n", + "hi = jh*(k2_/(id_*10**-3))*(cp2*1000*mu2/k2)**(1/3) \t\t\t#kcal/h m**2 C\n", + "\n", + "#shell side(organic)\n", + "B = bf \t\t\t#m, baffel spacing\n", + "p = 0.0254 \t\t\t#m,radius of 1 tube\n", + "Ds = 0.254 \t\t\t#m, inside diameter of shell\n", + "c = 0.0064 \n", + "#from eq. 8.32\n", + "As = c*B*Ds/p \t\t\t#m**2, flow area\n", + "Gs = Cp/As \t\t\t#kg/m**2 h, mass flow rate of shell fluid\n", + "do = od/10 \t\t\t#cm, outside diameter of shell\n", + "#from eq. 8.31\n", + "Dh = 4*((0.5*p*100)*(0.86*p*100)-((math.pi*(do)**2)/8))/((math.pi*do)/2)\n", + "Dh_ = Dh*10**-2 \t\t\t#m, hydrolic diameter\n", + "Re1 = (Dh_*Gs)/(3600*(mu1*10**-3)) \t\t\t#Reynold no.\n", + "#from fig 8.11(b)\n", + "jh1 = 32 \t\t\t#colburn factor\n", + "ho = jh1*(k1_/Dh_)*((6)**(1./3))\n", + "#from eq. 8.28\n", + "ratio = od/id_ \t\t\t#ratio = Ao/Ai\n", + "Rdo = 0.21*10**-3 \t\t\t#outside dirt factor\n", + "Rdi = 0.35*10**-3 \t\t\t#inside dirt factor\n", + "Udo = 1/((1/ho)+Rdo+(ratio)*Rdi+(ratio)*(1/hi))\n", + "\n", + "#SECOND TRIAL\n", + "#estimation of heat transfer coefficient\n", + "#Tube side (water)\n", + "np1 = 12 \t\t\t#\n", + "fa2 = (math.pi/4)*(id_*10**-3)**2*np1 \t\t\t#m**2, flow area\n", + "v2 = (wfr/1000)/(3600*fa2) \t\t\t#m/s, velocity\n", + "Re2 = (id_*10**-3)*v2*rho2/mu2 \t\t\t#Reynold no.\n", + "#from fig . 8.11(a)\n", + "jht = 83. \t\t\t#colburn factor\n", + "#jh = (hi*di)/k*(cp*mu/k)**-1/3 \n", + "#assume, (cp*mu/k) = x\n", + "hit = jht*(k2_/(id_*10**-3))*(cp2*1000*mu2/k2)**(1./3) \t\t\t#kcal/h m**2 C\n", + "\n", + "#shell side\n", + "B2 = 0.1 \t\t\t#m, baffel spacing\n", + "p2 = 0.0254 \t\t\t#m,radius of 1 tube\n", + "Ds2 = 0.254 \t\t\t#m, inside diameter of shell\n", + "c2 = .0064\n", + "#from eq. 8.32\n", + "As2 = c2*B2*Ds2/p2 \t\t\t#m**2, flow area\n", + "Gs2 = Cp/As2 \t\t\t#kg/m**2 h, mass flow rate of shell fluid\n", + "do2 = od/10 \t\t\t#cm, outside diameter of shell\n", + "#from eq. 8.30\n", + "Dh2 = 4*((p2*100)**2-((math.pi*(do2)**2)/4))/((math.pi*do2))\n", + "Dh2_ = Dh2*10**-2 \t\t\t#m, hydrolic diameter\n", + "Re2 = (Dh2_*Gs2)/(3600*(mu1*10**-3))\n", + "#from fig 8.11(b)\n", + "jh2 = 48 \t\t\t#colburn factor\n", + "ho2 = jh2*(k1_/Dh2_)*((6)**(1./3))\n", + "#from eq. 8.28\n", + "ratio = od/id_ \t\t\t#ratio = Ao/Ai\n", + "Rdo2 = 0.21*10**-3 \t\t\t#outside dirt factor\n", + "Rdi2 = 0.35*10**-3 \t\t\t#inside dirt factor\n", + "Udo2 = 1/((1/ho2)+Rdo+(ratio)*Rdi+(ratio)*(1/hit))\n", + "\n", + "#from eq. 8.10(a)\n", + "tauc = (T2-T3)/(T1-T3) \t\t\t#Temprature ratio\n", + "R = (T1-T2)/(T2-T3) \t\t\t#Temprature ratio\n", + "Ft = 0.8 \t\t\t#LMTD correction ftor\n", + "Areq = HD_/(Udo2*Ft*LMTD) \t\t\t#area required\n", + "tubes = 48. \t\t\t#no. of tubes\n", + "lnt = 4.5 \t\t\t#length of 1 tube\n", + "Aavl = (math.pi*od*10**-3)*tubes*lnt \t\t\t#available area\n", + "excA = ((Aavl-Areq)/Areq)*100 \t\t\t#% excess area\n", + "\n", + "#Pressure drop calculation\n", + "#Tube side\n", + "#from eq. 8.33\n", + "Gt = wfr/(3600*fa2) \t\t\t#kg/m**2 s, mass flow rate of tube fluid\n", + "n2 = 4. \t\t\t#tube passes\n", + "fit = 1. \t\t\t#dimensionless vismath.cosity ratio\n", + "g = 9.8 \t\t\t#gravitational consmath.tant\n", + "f = 0.0037 \t\t\t#friction factor\n", + "dpt = f*Gt**2*lnt*n2/(2*g*rho2*id_*10**-3*fit) \t\t\t#kg/m**2, tube side pressure drop\n", + "\n", + "#eq.8.35\n", + "dpr = 4*n2*v2**2*rho2/(2*g) \t\t\t#kg/m**2, return tube pressure loss\n", + "dpr_ = dpr*9.801 \t\t\t#N/m**2\n", + "tpr = dpt+dpr \t\t\t#kg/m**2, total pressure drop\n", + "#shell side\n", + "fs = 0.052 \t\t\t#friction factor for shell\n", + "bf1 = 0.1 \t\t\t#m, baffel spacing\n", + "Nb = lnt/bf1-1 \t\t\t#no. of baffles\n", + "dps = fs*(Gs2/3600)**2*Ds*(Nb+1)/(2*g*rho1*Dh2_*fit) \t\t\t#kg/m**2, shell side pressure drop\n", + "dps_ = dps*9.81 \t\t\t#N/m**2, shell side pressure drop\n", + "print \"Tube side Pressure drop is %1.3e N/m**2 \"%(dpr_)\n", + "print \"Shell side Pressure drop is %.0f N/m**2 \"%(round(dps_,-1))\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3 Page No : 320" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Th2 = 49.5 C\n", + "The new rate of heat transfer : 161003 kcal/h\n", + "the heat teansfer rate will be affected by 1.3 percent \n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "#for hot stream\n", + "Wh = 10000. \t\t\t#kg/h, Rate of leaving a hydrolic system by the oil\n", + "Cph = 0.454 \t\t\t#Kcal/Kg C, specific heat of oil\n", + "Th1 = 85. \t\t\t#C initial temp. of oil\n", + "Th2 = 50. \t\t\t#C final temp. of oil \n", + "\n", + "#For cold stream\n", + "Cpc = 1. \t\t\t#Kcal/Kg C, specific heat of water\n", + "Tc2 = 30. \t\t\t#C final temp. of water\n", + "Tc1 = 38. \t\t\t#C initial temp. of water\n", + "\n", + "# Calculations\n", + "#from heat balance eq.\n", + "#kg/h, Rate of leaving a hydrolic system by the water\n", + "Wc = Wh*Cph*(Th1-Th2)/(Cpc*(Tc1-Tc2))\n", + "#For the hot stream\n", + "Cmin = Wh*Cph \t\t\t#Kcal/h C.Taking hot stream as min. stream\n", + "#For cold stream\n", + "Cmax = Wc*Cpc \t\t\t#Kcal/h C.Taking cold stream as max. stream\n", + "Cr = Cmin/Cmax \t\t\t#Capacity ratio\n", + "n = (Th1-Th2)/(Th1-Tc2) \t\t\t#effectiveness factor\n", + "#From eq. 8.57\n", + "#No. of transfer units\n", + "NTU = -(1+(Cr)**2)**-(1./2)*math.log(((2/n)-(1+Cr)-(1+(Cr)**2)**(1./2))/((2./n)-(1+Cr)+(1+(Cr)**2)**(1./2)))\n", + "Ud = 400. \t\t\t#kcal/h m**2C , overall dirty heat transfer coefficient\n", + "#from eq. 8.53\n", + "A = (NTU*Cmin)/Ud \t\t\t#Area required\n", + "#if the water rate is increased by 20 %,\n", + "a = 20.\n", + "Wc_ = Wc+(Wc*(a/100))\n", + "Cmax_ = Wc_*Cpc\n", + "Cr_ = Cmin/Cmax_\n", + "#From eq. 8.56\n", + "n_ = 2*((1+Cr_)+(1+(Cr_)**2)**(1./2)*(1+math.exp(-(1+(Cr_)**2)**(1./2)*NTU))/(1-math.exp(-(1+(Cr_)**2)**(1./2)*NTU)))**(-1)\n", + "Th2_ = Th1-(n_*(Th1-Tc2))\n", + "q1 = Wh*Cph*(Th1-Th2) \t\t\t#kcal/h previous rate of heat transfer\n", + "q2 = Wh*Cph*(Th1-Th2_) \t\t\t#kcal/h new rate of heat transfer\n", + "#increase in rate of heat transfer\n", + "dq = (q2-q1)/q1 \n", + "\n", + "# Results\n", + "print \"Th2 = %.1f C\"%Th2_\n", + "print \"The new rate of heat transfer : %d kcal/h\"%q2\n", + "print \"the heat teansfer rate will be affected by %.1f percent \"%(dq*100 )\n", + "\n", + "# note : rounding off error would be there." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4 Page No : 337" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the time required to heat the charge 22 min\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "p = 0.0795 \t\t\t#m. pitch of the coil\n", + "d1 = 0.0525 \t\t\t#m,coil diameter\n", + "h = 1.464 \t\t\t#m,height of the limpetted section\n", + "d2 = 1.5 \t\t\t#m,diameter of batch polymerization reactor\n", + "d3 = 0.5 \t\t\t#m,diameter of agitator\n", + "rpm = 150. \t\t\t#speed of agitator\n", + "rho = 850. \t\t\t#kg/m3,density of monomer\n", + "rho1 = 900. \t\t\t#kg/m3,density of fluid\n", + "mu = 0.7*10**-3 \t\t\t#poise, vismath.cosity of monomer\n", + "mu1 = 4*10.**-3 \t\t\t#poise, vismath.cosity of fluid\n", + "cp = 0.45 \t\t\t#kcal/kg C, specific heat of monomer\n", + "cp1 = 0.5 \t\t\t#kcal/kg C, specific heat of fluid\n", + "k = 0.15 \t\t\t#kcal/h mC, thermal conductivity of monomer\n", + "k1 = 0.28 \t\t\t#kcal/h mC, thermal conductivity of fluid\n", + "Rdi = 0.0002 \t\t\t#h m2 C/kcal, fouling factor for vessel\n", + "Rdc = 0.0002 \t\t\t#h m2 C/kcal, fouling factor for coil\n", + "Tci = 120. \t\t\t#C, initial temp. of coil liquid\n", + "Tvi = 25. \t\t\t#C, initial temp. of vessel liquid\n", + "Tvf = 80. \t\t\t#C, final temp. of vessel liquid\n", + "\n", + "#calculation\n", + "a = math.pi*d2*h \t\t\t#outside area of the vessel\n", + "x = 60. \t\t\t#%. added of the unwetted area to the wetted area\n", + "ao = ((d1+(x/100)*(p-d1))/p)*a \t\t\t#m**2,effective outside heat transfer area of vessel\n", + "ai = 6.9 \t\t\t#m**2,inside heat transfer area of vessel\n", + "#same as outside area , if thickness is very small\n", + "#vessel side heat transfer coefficient\n", + "Re = (d3**2*(rpm/60)*rho)/mu \t\t\t#reynold no.\n", + "Pr = ((cp*3600)*(mu))/k\n", + "#from eq. 8.66\n", + "y = 1 \t\t\t#x = mu/muw = 1\n", + "Nu = 0.74*(Re**(0.67))*(Pr**(0.33))*(y**(0.14)) \t\t\t#Nusslet no\n", + "hi = Nu*(k/d2) \t\t\t#heat transfer coefficient\n", + "\n", + "#coil side heat transfer coefficient\n", + "v = 1.5 \t\t\t#m/s, linear velocity of fluid\n", + "fa = ((math.pi/4)*d1**2) \t\t\t#m2, flow area of coil\n", + "fr = v*fa*3600 \t\t\t#m3/h , flow rate of the fluid\n", + "Wc = fr*rho \t\t\t#kg/h , flow rate\n", + "dh = (4*(math.pi/8)*d1**2)/(d1+(math.pi/2)*d1) \t\t\t#m,hydrolic diameter of limpet coil\n", + "Re1 = v*rho1*dh/mu1 \t\t\t#coil reynold no.\n", + "Pr1 = cp1*mu1*3600/k1 \t\t\t#prandtl no. of the coil fluid\n", + "#from eq. 8.68\n", + "d4 = 0.0321 \t\t\t#m, inside diameter of the tube\n", + "Nu1 = 0.021*(Re1**(0.85)*Pr1**(0.4)*(d4/d2)**(0.1)*y**0.14) \n", + "hc = Nu1*(k1/dh) \t\t\t#coil side coefficient\n", + "\n", + "U = 1/((1/hi)+(ai/(hc*ao))+Rdi+Rdc) \t\t\t#overall heat transfer corfficient\n", + "#from eq. 8.63\n", + "beeta = math.exp(U*ai/(Wc*cp1))\n", + "Wv = 2200. \t\t\t#kg, mass of fluid vessel\n", + "t = (beeta/(beeta-1))*((Wv*cp)/(Wc*cp1))*math.log((Tci-Tvi)/(Tci-Tvf)) \n", + "\n", + "# Results\n", + "print \"the time required to heat the charge %.0f min\"%(t*60)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb new file mode 100644 index 00000000..6a4e4ff0 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb @@ -0,0 +1,692 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 : Evaporation and Evaporators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1 Page No : 391" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The rate at which heat must be supplied at 1 atm pressure is 2.441e+08 kj/ day\n", + "The rate at which heat must be supplied at a pressure of 600 mm Hg is 2.373e+08 kj/day \n" + ] + } + ], + "source": [ + "# Variables\n", + "ro = 1020. \t\t\t# kg/m**3, density of feed\n", + "sf = 4.1 \t\t\t#kj/kg C,specific heat of the feed\n", + "sp = 3.9 \t\t\t#kj/kg C,specific heat of the product\n", + "ci = 5. \t\t\t#initial concentration\n", + "cw = 100.-ci \t\t\t#conc. of water\n", + "cf = 40. \t\t\t#final conc.\n", + "rate = 100. \t\t\t#m**3/day, rate of conc. of aq. solution\n", + "ft = 25. \t\t\t# C, feed temp.\n", + "\n", + "#calculation and results\n", + "#materiel balance\n", + "Wf = rate*ro \t\t\t#Kg. feed entering\n", + "Ms = ro*ci \t\t\t#Kg mass of solute\n", + "Mw = ro*cw \t\t\t#kg,mass of water\n", + "fc = cw/ci \t\t\t#kg,feed concentration\n", + "pc = (100-cf)/cf \t\t\t# kg,product concentration\n", + "wlwp = Ms*pc \t\t\t#Kg, water leaving with the product\n", + "Ws = Mw-wlwp \t\t\t#kg,water evaporated\n", + "Wp = wlwp+Ms \t\t\t# kg, product\n", + "#energy balance\n", + "rt = 0. \t\t\t#C reference temp.\n", + "ef = sf*(ft-rt) \t\t\t#kj/kg,enthlpy of the feed\n", + "#case i\n", + "Tp = 100. \t\t\t#temp. of the product (because the solute has a 'high molecular wt' the boiling pt elevation is neglected)\n", + "ip = sp*(Tp-rt) \t\t\t#kj/kg, enthalpy of the product\n", + "iv = 2680. \t\t\t#kj/kg, enthalpy of the vapour generated at 100 C and 1 atm pr. from the steam table\n", + "#refer to fig. 9.23\n", + "#from energy balance eq. (Wf*if+qs = Wv*iv+Wp*ip)\n", + "qs = Ws*iv+Wp*ip-Wf-ef \t\t\t#Wv = Ws\n", + "print \"The rate at which heat must be supplied at 1 atm pressure is %1.3e kj/ day\"%(qs)\n", + "\n", + "#case ii\n", + "#650 mm Hg vaccum = 110 mmHg pressure\n", + "bp = 53.5 \t\t\t#C, boiling point of water\n", + "ip2 = sp*(bp-rt) \t\t\t#kj/kg, enthalpy of the product\n", + "es = 2604. \t\t\t#kj/kg, enthalpy of the saturated steam (from steam table)\n", + "#from energy balnce eq.\n", + "qs2 = Wp*ip+Ws*es-Wf-ef\n", + "print \"The rate at which heat must be supplied at a pressure of 600 mm Hg is %1.3e kj/day \"%(qs2)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2 Page No : 393" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The steam required is 1737 kg/h\n", + "No. of tube are 102\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "ci = 10. \t\t\t#%,initial concentration\n", + "cf = 40. \t\t\t#%, final conc\n", + "Wf = 2000. \t\t\t#kg/h, feed rate\n", + "ft = 30. \t\t\t#C feed temp.\n", + "rp = 0.33 \t\t\t#kg/cm**2, reduced pressure\n", + "bt1 = 75. \t\t\t#C,boiling point temp.\n", + "sst = 115. \t\t\t#C, saturated steam temp.\n", + "l = 1.5 \t\t\t# m,height of calandria\n", + "sh = 0.946 \t\t\t#kcal/kg C, specific heat of liquir\n", + "lh = 556.5 \t\t\t#kcal/kg latent heat of steam\n", + "bt2 = 345. \t\t\t#K, boiling point of water \n", + "h = 2150. \t\t\t#kcal/h m**2 C, overall heat transfer coefficient\n", + "si = 2000.*(ci/100) \t\t\t#kg/h, solids in\n", + "wi = 1800. \t\t\t#kg/h,wate in\n", + "\n", + "# Calculations\n", + "Wp = si/(cf/100) \t\t\t#kg/h, product out\n", + "Wv = Wf-Wp \t\t\t#evaporation rate\n", + "ef = sh*(ft-bt1)\n", + "ip = 0\n", + "lamda_s = 529.5 \t\t\t#kcal/kg, lamda_s = is-il\n", + "bpe = (273+bt1)-345 \t\t\t#boiling point elevation.\n", + "#from eergy balance eq.\n", + "Ws = (Wp*ip+Wv*lh-Wf*ef)/lamda_s\n", + "q = Ws*lamda_s \t\t\t#kcal/h,rate of heat transfer\n", + "A = q/(h*(sst-bt1)) \t\t\t# m**2\n", + "di = 0.0221 \t\t\t#m,inside diameter\n", + "At = math.pi*l*di \t\t\t#m**2, area of a math.single tube\n", + "N = A/At \t\t\t#no. of tubes\n", + "\n", + "# Results\n", + "print \"The steam required is %.0f kg/h\"%(Ws)\n", + "print \"No. of tube are %d\"%(N)\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3 Page No : 393" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The steam pressure to be used in the calandria is 2.15 barabs)\n", + "The heat transfer rate required is 4.01e+06 Kj/h\n", + "Rate of steam supply is 1833 kg/h\n" + ] + } + ], + "source": [ + "# Variables\n", + "Wf = 2000. \t\t\t#kg/h, feed rate\n", + "ci = 8. \t\t\t#% initial conc.\n", + "cf = 40. \t\t\t#% final conc.\n", + "ft = 30. \t\t\t#C, feed temp.\n", + "vp = 660. \t\t\t#mm Hg, vaccum pressure\n", + "ssp = 8. \t\t\t# bar absolute, saturated steam pr.\n", + "\n", + "#calculation\n", + "sr = Wf*(ci/100) \t\t\t#kg/h, solid rate\n", + "Wp = sr/(cf/100) \t\t\t#kg/h,concentrated product rate\n", + "ap = 760-vp \t\t\t#mm Hg, absolute pressure in the evaporator\n", + "bt = 325. \t\t\t#K,boiling temp. of water\n", + "l_s = 2380. \t\t\t#kj/kg, latent heat\n", + "R = 8.303 \t\t\t#gas consmath.tant\n", + "w = 40. \t\t\t#g,mass of solute\n", + "M = 18. \t\t\t#g,molecular wt of solvent\n", + "W = 60. \t\t\t#g,mass of the solvent\n", + "m = 2000. \t\t\t#g,molecular wt of solute\n", + "dtb = (R*bt**2*w*M)/(l_s*W*m) \t\t\t#C, boiling point elevation\n", + "bp = bt+dtb \t\t\t#k,boiling point of 40% solution\n", + "dt = 70. \t\t\t#C, from given data flux becomes maximum at a temp. drop = 70 C\n", + "st = bp+dt \t\t\t#K,saturation temp. of steam in the steam chest\n", + "Sp = 2.15 \t\t\t# bar, from steam table, saturation lr. of steam at this temp.\n", + "\n", + "sh = 4.2 \t\t\t#kj/kg C, specific heat of product\n", + "rt = 0. \t\t\t#C reference teml.\n", + "ef = sh*(ft-rt) \t\t\t# kj/kg, enthalpy of the feed\n", + "ip = sh*(54-rt) \t\t\t#kj/kg, enthalpy of the product\n", + "iv = 2607. \t\t\t#kj/kg, enthalpy of vapour produced\n", + "#from eq 9.6\n", + "Wv = 1600. \t\t\t#enthalpy of evaporation\n", + "q = Wp*ip+Wv*iv-Wf*ef \t\t\t#kj/h, heat transfe rate required\n", + "hvp = 2188. \t\t\t#kj/kg, heat of vaporization of saturated steam at 397 K\n", + "rs = q/hvp \t\t\t#kg/h, rate of steam supply\n", + "\n", + "# Results\n", + "print \"The steam pressure to be used in the calandria is %.2f barabs)\"%(Sp);\n", + "print \"The heat transfer rate required is %.2e Kj/h\"%(q);\n", + "print \"Rate of steam supply is %.0f kg/h\"%(rs);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4 Page No : 402" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The evaporator area is 72 square metre \n", + "Steam economy is 1.79\n" + ] + } + ], + "source": [ + "import math \n", + "\n", + "from numpy import array, linalg\n", + "# Variables\n", + "Wf = 6000. \t\t\t#kg/h, feed rate\n", + "ci = 2. \t\t\t#%, initial concentration\n", + "cf = 35. \t\t\t#%, final conc.\n", + "ft = 50. \t\t\t#C,feed temp.\n", + "ssp = 2. \t\t\t#bar abs, saturated steaam pr.\n", + "sep = 0.0139 \t\t\t#bar abs, maintained temp. in second effect\n", + "h1 = 2000. \t\t\t#W/m**2 K,overall heat transfer coeffcient in 1st effect\n", + "h2 = 1500. \t\t\t#W/m**2 K, overall heat transfer coefficient in 2nd effect\n", + "cp = 4.1 \t\t\t#kj/kg k,specific heat\n", + "\n", + "#calculation\n", + "si = Wf*(ci/100) \t\t\t#kg/h, solid in\n", + "wi = 5880. \t\t\t#kg/h, water in\n", + "Wp = si/(cf/100) \t\t\t#kg/h product out\n", + "wo = Wp*(1-cf/100) \t\t\t#kg/h, water out with the product\n", + "ter = wi-wo \t\t\t#kg/h, total evaporation rate\n", + "\n", + "#boiling temp. in the first effect\n", + "T1 = 120. \t\t\t#C,Temprature\n", + "l_s1 = 2200. \t\t\t#kj/kg, latent heat\n", + "T2 = 12. \t\t\t#C,boiling point in second effect\n", + "l_s2 = 2470. \t\t\t# kj/kg in second effect\n", + "tatd = T1-T2 \t\t\t# C,tatd = dt1+dt2 = T1-T2 , total available temp. drop\n", + "#from eq. 9.20\n", + "#h1*dt1 = h2*dt2\n", + "#solving above two equations by matrix\n", + "A = array([[1,1],[2000,-1500]])\n", + "C = array([108,0])\n", + "X = linalg.solve(A,C)\n", + "#X = inv(A)*C\n", + "\n", + "dt1 = X[0]\n", + "dt2 = X[1]\n", + "t1 = T1-dt1 \t\t\t#temp. of steam leaving the first effect\n", + "t2 = T2-dt2 \t\t\t#temp. of steam leaving second effect\n", + "#energy balance over the 1st effect, from eq.9.14\n", + "rt1 = t1\n", + "ef = cp*(ft-t1) \t\t\t#kj/kg,enthalpy of feed\n", + "i1 = 0\n", + "lam_s1 = 2330. \t\t\t#kj/kg\n", + "is1 = lam_s1\n", + "#Wf*ef+Ws*l_s = (Wf-Ws1)*i1+Ws1*is1\n", + "#substituting we get,\n", + "#Ws1 = 0.9442*Ws-253.4..........(1)\n", + "#energy balance over second effect\n", + "#from eq 9.15\n", + "#(Wf-Ws1)*i1+Ws1*lam_s1 = (Wf-Ws1-Ws2)*i2+Ws2*is2\n", + "rt2 = t2\n", + "lam_s2 = 2470.\n", + "is2 = lam_s2\n", + "i2 = 0\n", + "# substituting we get\n", + "#Ws2 = 0.8404*Ws1+617.5............(2)\n", + "#ter,Ws1+Ws2 = 5657...............(3)\n", + "#solving by matrix method\n", + "A = array([[0.9442,-1,0],[0,0.8404,-1],[0,1,1]])\n", + "B = array([253.4,-617.5,5657])\n", + "X = linalg.solve(A,B)\n", + "#X = inv(A)*B\n", + "Ws = X[0]\n", + "Ws1 = X[1]\n", + "Ws2 = X[2]\n", + "\n", + "#evaporator area\n", + "A1 = Ws*l_s1/(h1*dt1) \t\t\t#for 1st effect\n", + "A2 = Ws1*lam_s1/(h2*dt2) \t\t\t#for second effect\n", + "\n", + "#revised calculation\n", + "#taking\n", + "dt1_ = 48.\n", + "dt2_ = 60.\n", + "T1_ = T1-dt1_\n", + "T2_ = T2-dt2_\n", + "ls1_ = 2335.\n", + "ls2_ = 2470.\n", + "# energy balance over first effect gives\n", + "#Ws1 = 0.9422Ws-231.8.........(4)\n", + "#energy balance over second effect gives\n", + "#Ws2 = 0.8457Ws1+579.5......(5)\n", + "#solving eq 3,4,5\n", + "P = array([[0.9422,-1,0],[0,0.8457,-1],[0,1,1]])\n", + "Q = array([231.8,-579.5,5657])\n", + "Y = linalg.solve(P,Q)\n", + "#Y = inv(P)*Q\n", + "Ws_ = Y[0]\n", + "Ws1_ = Y[1]\n", + "Ws2_ = Y[2]\n", + "\n", + "#eveporator area for 1st & 2nd effect in m**2\n", + "A1_ = Ws_*l_s1/(h1*dt1_)\n", + "A2_ = Ws1_*ls1_/(h2*dt2_)\n", + "EA = (A1_+A2_)/2\n", + "SE = (Ws1_+Ws2_)/Ws_\n", + "\n", + "# Results\n", + "print \"The evaporator area is %.0f square metre \"%(EA);\n", + "print \"Steam economy is %.2f\"%(SE);\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5 Page No : 404" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum no. of effects are 4\n" + ] + } + ], + "source": [ + "# Variables\n", + "ssp = 3.32 \t\t\t#bar abs, saturated steam pr.\n", + "rp = 0.195 \t\t\t# bar abs, residual pr. in the condenser\n", + "tl = 41. \t\t\t#K, sun of temp. losses because of BPE\n", + "mt = 8. \t\t\t#k,minimum available temp. driving force\n", + "#calculation\n", + "sst = 410. \t\t\t#K,saturated steam temp.\n", + "st = 333. \t\t\t#K,corresponding saturation temp. when pressure in the last effect is 0.195 bar\n", + "ttd = sst-st \t\t\t#K,total temp. difference\n", + "atd = ttd-tl \t\t\t# K,available temp. drop across the unit\n", + "n = atd/mt \t\t\t#maximum no. of effect\n", + "\n", + "# Results\n", + "print \"Maximum no. of effects are %.0f\"%(n);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6 Page No : 405" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The areas are now reasonably close \n", + "Steam Rate is 8854 Kg/h \n", + "Steam economy is 1.93\n" + ] + } + ], + "source": [ + "# Variables\n", + "fc = 9.5 \t\t\t#%,feed concentration\n", + "pc = 50. \t\t\t#%, product conc.\n", + "ft = 40. \t\t\t# C,feed temp.\n", + "er = 2000. \t\t\t#kg NaOH/h, evaporation rate\n", + "vp = 714. \t\t\t#mm Hg, vaccum pr. in last effect\n", + "#heat transfer coefficients, W/m**2 C\n", + "h1 = 6000. \t\t\t#for first effect\n", + "h2 = 3500. \t\t\t#for second effect\n", + "h3 = 2500. \t\t\t#for third effect\n", + "\n", + "#calculatiin\n", + "Wf = er/(fc/100) \t\t\t#kg/h, 2 tons NaOH per hour, feed rate\n", + "Wp = er/(pc/100) \t\t\t#kg/h, product rate\n", + "ter = Wf-Wp \t\t\t#kg/h, total evaporation rate\n", + "#steam\n", + "p = 3.3 \t\t\t#bar,assumed saturated\n", + "#from steam table\n", + "Ts = 137. \t\t\t#C,temp.\n", + "l_s = 2153. \t\t\t#kj/kg, latent heat\n", + "pl = 760.-vp \t\t\t#mm Hg,pressure in the last effect\n", + "bp = 37. \t\t\t#C,boiling point of water\n", + "#refer to fig. 9.24\n", + "attd = Ts-bp \t\t\t#C,apparent total temp. drop\n", + "#let assume the following evaporation rate for three effects in kg/h\n", + "ev1 = 5600.\n", + "ev2 = 5680.\n", + "ev3 = 5773.\n", + "#conc. in three effects\n", + "c1 = er/(Wf-ev1)\n", + "c2 = er/(Wf-ev1-ev2)\n", + "c3 = 0.5 \t\t\t# Variables\n", + "#boiling point elevations in three effects in C\n", + "bpe1 = 3.5\n", + "bpe2 = 8.\n", + "bpe3 = 39.\n", + "attda = attd-(bpe1+bpe2+bpe3) \t\t\t#actual total temp. drop available\n", + "#temp. drop in three effects\n", + "#from eq. 9.23\n", + "dt1 = attda*((1/h1)/((1/h1)+(1/h2)+(1/h3)))\n", + "dt2 = attda*((1/h2)/((1/h1)+(1/h2)+(1/h3)))\n", + "dt3 = attda*((1/h3)/((1/h1)+(1/h2)+(1/h3)))\n", + "\n", + "#from table 9.4\n", + "#enthalpy of solution in three effects in kj/kg\n", + "i1 = 486.\n", + "i2 = 385.\n", + "i3 = 460.\n", + "#enthalpy of vapour generated for three effects in kj/kg\n", + "is1 = 2729.\n", + "is2 = 2691.\n", + "is3 = 2646.\n", + "#Enthalpy of condensate over effect 1,2,3 in kj/kg\n", + "il1 = 0.\n", + "il2 = 519.\n", + "il3 = 418.\n", + "#Enthalpy balance over effect 1\n", + "ef = 145. \t\t\t#kj/kg,enthalpy of feed\n", + "#from energy balance eq.\n", + "#Ws1 = 0.96Ws-3200......(1)\n", + "#enthalpy balanc over effect 2\n", + "#Ws2 = 0.9146Ws1+922...........(2)\n", + "#enthalpy balanc over effet 3\n", + "#Ws3 = 1.073Ws2+0.0343Ws1-722........(3)\n", + "#ter = Ws1+Ws2+Ws3 = 17053..........(4)\n", + "\n", + "#Solving above four eqns by matrix\n", + "A = array([[0.96,-1,0,0],[0,0.9146,-1,0],[0,0.0343,1.073,-1],[0,1,1,1]])\n", + "B = array([3200,-922,722,17053])\n", + "X = linalg.solve(A,B)\n", + "#X = inv(A)*B\n", + "Ws = X[0]\n", + "Ws1 = X[1]\n", + "Ws2 = X[2]\n", + "Ws3 = X[3]\n", + "\n", + "#calculation of heat transfer areas iver effect 1, 2 ,3\n", + "A1 = Ws*l_s*10**3/(h1*dt1*3600)\n", + "A2 = Ws1*(is1-il2)*10**3/(h2*dt2*3600)\n", + "A3 = Ws2*(is2-il3)*10**3/(h3*dt3*3600)\n", + "\n", + "#Revised dt\n", + "avar = (A1+A2+A3)/3\n", + "dt1_ = (A1/avar)*dt1\n", + "dt2_ = (A2/avar)*dt2\n", + "dt3_ = attda-dt1_-dt2_\n", + "\n", + "#from table 9.5\n", + "#enthalpy of vapour generated over effect 1,2,3 in kj/kg\n", + "is1_ = 2720.\n", + "is2_ = 2685.\n", + "is3_ = 2646.\n", + "#enthalpy of soln on 1,2,3 in kj/kg\n", + "i1_ = 470.\n", + "i2_ = 380.\n", + "i3_ = 460.\n", + "#enthalpy of condensate over effect 1 ,2,3 in kj/kg\n", + "il1_ = 0.\n", + "il2_ = 513.\n", + "il3_ = 412.\n", + "#enthalpy balance ove effect 1,2,3 gives\n", + "Ws_ = 8854.\n", + "Ws1_ = 5432.\n", + "Ws2_ = 5812.\n", + "Ws3_ = 5809.\n", + "#revised heat transfer areas for effect 1 ,2,3 in m**2\n", + "A1_ = Ws_*l_s*1000/(h1*dt1_*3600)\n", + "A2_ = Ws1_*(is1_-il2_)*10**3/(h2*dt2_*3600)\n", + "A3_ = Ws2_*(is2_-il3_)*10**3/(h3*22.5*3600)\n", + "avar_ = (A1_+A2_+A3_)/3\n", + "SE = ter/Ws_\n", + "\n", + "# Results\n", + "print \"The areas are now reasonably close \"\n", + "print \"Steam Rate is %.0f Kg/h \"%(Ws_)\n", + "print \"Steam economy is %.2f\"%(SE)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7 Page No : 409" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The increase in evaporation capacity ic 113 percentage \n", + " The percentage change in the math.cost of concentrating a ton of feed is 15 percentage\n" + ] + } + ], + "source": [ + "from numpy import array, linalg\n", + "\n", + "# Variables\n", + "Wf = 3000. \t\t\t#kg/h,feed\n", + "fc = 8. \t\t\t#%, feed concentration\n", + "pc = 40. \t\t\t#% product concentration\n", + "si = Wf*(fc/100) \t\t\t#kg,solid in\n", + "pr = si/(40./100) \t\t\t#g/h, product rate\n", + "ft = 60. \t\t\t#C,feed temp.\n", + "er = Wf-pr \t\t\t#kg/h, evaporation rate\n", + "math.cost = 120000. \t\t\t#total math.cost per year\n", + "p1 = 4.5 \t\t\t#bar, low pressure steam\n", + "scpt = 700. \t\t\t#per ton. math.cost of steam\n", + "cp = 0.764 \t\t\t# kcal/kg, specific heat\n", + "\n", + "#from table 9.6\n", + "eep = 1. \t\t\t#atm existing evaporator pressure \n", + "oop = 400000. \t\t\t# peryear ,other operatingmath.cost\n", + "oop_ = 600000. \t\t\t#per yr, for proposed condition\n", + "wd = 300. \t\t\t#days per year.working days\n", + "wh = wd*24. \t\t\t#working hr\n", + "\n", + "# Calculations\n", + "#EXISTING OPERATING CONDITION \n", + "rt = 0 \t\t\t#C,reference temp.\n", + "ef = eep*(ft-rt) \t\t\t#kcal/kg, enthalpy of feed\n", + "pt = 100. \t\t\t#C,product temp.\n", + "i1 = cp*(pt-rt) \t\t\t#kcal/kg, enthalpy of soln\n", + "is1 = 639. \t\t\t#kcal/kg,enthalpy of vapour generated at 1 atm (from steam table)\n", + "l_s = 496. \t\t\t#kcal/kg,latent heat of steam at 4.5 bar\n", + "T = 425. \t\t\t#K\n", + "#heat balance\n", + "Ws = (er*is1+pr*i1-Wf*ef)/l_s \t\t\t#kg/h, steam required\n", + "q = Ws*l_s \t\t\t#ton/ hr,heat supplied\n", + "x = q/(T-(pt+273)) \t\t\t#x = Ud*A\n", + "#hourly math.cost\n", + "sc = Ws/1000*(scpt) \t\t\t# /perh, steam math.cost\n", + "lc = 100. \t\t\t#per h,labour math.cost\n", + "oc = oop/(wh) \t\t\t# per h,othe math.cost\n", + "tc = sc+lc+oc \t\t\t#total math.cost\n", + "C = tc/(Wf/1000) \t\t\t# per ton,math.cost per ton of feed\n", + "\n", + "#PROPOSED OPERATING CONDITION\n", + "bpl = 320. \t\t\t#K,boiling point of liquid\n", + "dt = T-bpl\n", + "q_ = x*dt \t\t\t#kcal/h,rate of heat supply\n", + "sr = q_/l_s \t\t\t#steam rate ton per hr\n", + "pt_ = 47. \t\t\t#C,product temp .\n", + "ep = cp*(pt_-rt) \t\t\t#kcal/kg. enthalpy of product\n", + "ev = 618. \t\t\t#kcal/kg, enthalpy of vapour generated\n", + "#heat balance\n", + "#24Wf_-582Ws1_ = 2825000 ..........(1)\n", + "#material balance\n", + "# 4Wf_-5Ws1_ = 0 .............(2)\n", + "#solving by matrix method\n", + "a = array([[24,-582],[4,-5]])\n", + "b = array([-2825000,0])\n", + "x_ = linalg.solve(a,b)\n", + "#x_ = inv(a)*b\n", + "Wf_ = x_[0]\n", + "Ws1_ = x_[1]\n", + "ic = (Wf_-Wf)/Wf\n", + "print \"The increase in evaporation capacity ic %d percentage \"%(ic*100)\n", + "sr_ = Ws1_/1000 \t\t\t#ton per hr ,steam rate \n", + "#hourly math.cost\n", + "sc_ = Ws1_*scpt \t\t\t#steam math.cost\n", + "lc_ = 200. \t\t\t#labour math.cost rs.200/ h\n", + "oc_ = oop_/wh \t\t\t# other math.cost\n", + "tc_ = sc_/1000+lc_+oc_\n", + "C_ = tc_/(Wf_/1000) \t\t\t#math.cost per ton of feed\n", + "ps = (C-C_)/C\n", + "print \" The percentage change in the math.cost of concentrating a ton of feed is %.0f percentage\"%(ps*100)\n", + "\n", + "# rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.8 Page No : 415" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Make up steam required is 1.302e+04 kg/day\n" + ] + } + ], + "source": [ + "# Variables\n", + "q = 2200. \t\t\t#kj/kg heat of condensation of steam \n", + "#from example 9.1\n", + "Qr = 2.337*10**8 \t\t\t#kj/day rate of heat supply\n", + "\n", + "#calculation\n", + "Rate = Qr/q \t\t\t#kg/day steam supply rate\n", + "Rate_ = 1.062*10**5 \t\t\t#approximate value\n", + "E = 2800. \t\t\t#kj/kg enthalpy of compressed vapour\n", + "T = 175.7 \t\t\t#C, temprature\n", + "Ts = 121. \t\t\t#C Saturation temprature\n", + "E1 = 2700. \t\t\t#enthalpy at saturation temprature\n", + "q1 = T-Ts \t\t\t#Superheat of vapour\n", + "T1 = 100. \t\t\t#C hot water temprature\n", + "E2 = 419. \t\t\t#Enthalpy at hot water temp.\n", + "x = (E-E1)/(E1-E2) \t\t\t#water supplied per kg of superheated steam\n", + "S = 1.044 \t\t\t#steam obtained after desuperheating\n", + "R1 = 8.925*10**4 \t\t\t#kg/day rate of vapour generation \n", + "R2 = S*R1 \t\t\t#Rate of recompressed sat. steam\n", + "R2_ = 9.318*10**4 \t\t\t#approximate value\n", + "SR = Rate_-R2_ \n", + "\n", + "# Results\n", + "print \"Make up steam required is %.3e kg/day\"%(SR)\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/10.png b/Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/10.png Binary files differnew file mode 100644 index 00000000..291f94e6 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/10.png diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/5.png 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"cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer per unit area is 3.7 MW/sq meter\n" + ] + } + ], + "source": [ + "#Example Number 1.1\n", + "# HOW MUCH HEAT IS TRANSFERRED THROUGH THE PLATE\n", + "\n", + "#VARIABLE DECLARATION\n", + "\n", + "k = 370 \t\t\t # [W/m] at 250 degree celsius\n", + "dt = 100-400 \t\t\t#[degree celsius] temperature difference\n", + "dx = 3*10**(-2) \t\t#[m] thickness of plate\n", + "\n", + "#CALCULATION\n", + "\n", + "q = -k*dt/dx \t\t\t#[MW/square meter]\n", + "\n", + "#RESULTS\n", + "\n", + "print \"Rate of heat transfer per unit area is\", q/1000000 ,\"MW/sq meter\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer is 2.156 kW\n" + ] + } + ], + "source": [ + "#Example Number 1.2\n", + "# CALCULATE THE HEAT TRANSFER\n", + "\n", + "#Variable declaration\n", + "\n", + "Twall = 250 \t\t\t#[degree celsius] wall temperature\n", + "Tair = 20 \t\t\t#[degree celsius] air temperature\n", + "h = 25 \t\t\t\t#[W/square meter] heat transfer coefficient\n", + "l = 75*10**(-2) \t\t#[m] length of plate\n", + "b = 50*10**(-2) \t\t#[m] width of plate\n", + "area = l*b \t\t\t#[square meter] area of plate\n", + "dt = 250-20 \t\t\t#[degree celsius]\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "q = h*area*dt \t\t\t# [W] from newton's law of cooling\n", + "\n", + "\n", + "#Result\n", + "\n", + "print\"Rate of heat transfer is\",round(q/1000,3),\"kW\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inside plate temperature is 253.05 degree C\n" + ] + } + ], + "source": [ + "#Example Number 1.3\n", + "# Calculate the inside plate temperature.\n", + "\n", + "#variable declaration\n", + "\n", + "Qconv = 2156 \t # [W] from previous problem\n", + "Qrad = 300\t\t # [W] given\n", + "dx = 0.02\t\t # [m] plate thicknesss\n", + "l = 0.75 \t\t # [m] length of plate \n", + "w = 0.5 \t\t # [m] width of plate\n", + "k = 43\t\t\t #[W/m] from table 1.1\n", + "area = l*w\t\t #[square meter] area of plate\n", + "\n", + "#Calculation\n", + "\n", + "Qcond = Qconv+Qrad \t # [W]\n", + "dt = Qcond*dx/(k*area) \t # [degree celsius] temperature difference\n", + "Ti = 250+dt \t\t # inside temperature\n", + "\n", + "#Results\n", + "\n", + "print\"Inside plate temperature is\",round(Ti,2),\"degree C\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer is: 22.0 W\n", + " This is equal to the electric power which must be applied\n" + ] + } + ], + "source": [ + "#Example Number 1.4\n", + "# Calculate electric power to be supplied to the wire \n", + "\n", + "#Variable declaration\n", + "\n", + "d = 1*10**(-3) \t\t#[m] diameter of wire\n", + "l = 10*10**(-2) \t#[m] length of wire\n", + "Sarea = 22*d*l/7 \t#[square meter] surface area of wire\n", + "h = 5000 \t\t#[W/square meter] heat transfer coefficient\n", + "Twall = 114 \t\t# [degree celsius]\n", + "Twater = 100\t # [degree celsius]\n", + "\n", + "#total convection loss is given by equation(1-8)\n", + "\n", + "#Calculation\n", + "\n", + "Q = h*Sarea*(Twall-Twater) # [W]\n", + "\n", + "#Results\n", + "\n", + "print\"Heat transfer is:\",Q,\"W\" \n", + "print\" This is equal to the electric power which must be applied\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer per unit area is: 69.03 kW/sq meter\n" + ] + } + ], + "source": [ + "#Example Number 1.5\n", + "# radiation heat transfer\n", + "# Calculate the heat transfer per unit area\n", + "\n", + "#Variable declaration\n", + "\n", + "sigma = 5.669*10**(-8) \t\t#[W/square meter*k^(4)] universal constant\n", + "T1 = 273+800 \t\t\t# [k] first plate temperature\n", + "T2 = 273+300 \t\t\t# [k] second plate temperature\n", + "\n", + "#equation(1-10) may be employed for this problem\n", + "\n", + "#Calculation\n", + "Q = sigma*(T1**4-T2**4) \t# [W/square meter]\n", + "\n", + "#Results\n", + "\n", + "print\"Heat transfer per unit area is:\",round(Q/1000,2),\" kW/sq meter\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total heat loss is: 55.67 W/m\n" + ] + } + ], + "source": [ + "#Example Number 1.6\n", + "# total heat loss by convection and radiation\n", + "\n", + "#Variable declaration\n", + "d = 0.05 #[m] diameter of pipe\n", + "Twall = 50 #[degree celsius] \n", + "Tair = 20 #[degree celsius]\n", + "emi = 0.8 #emissivity\n", + "h = 6.5 #[W/square meter] heat transfer coefficient for free convection\n", + "import math\n", + "Q1 = h*math.pi*d*(Twall-Tair) #[W/m] convection loss per unit length\n", + "sigma = 5.669*10**(-8) # [W/square meter*k^(4)] universal constant\n", + "T1 = 273+Twall # [k]\n", + "T2 = 273+Tair # [k]\n", + "Q2 = emi*math.pi*d*sigma*((T1**(4))-(T2**(4))) # [W/m] heat loss due to radiation per unit length\n", + "Qtotal = Q1+Q2 # [W/m] total heat loss per unit length\n", + "\n", + "print\"Total heat loss is:\",round(Qtotal,2),\"W/m\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb new file mode 100644 index 00000000..b6a504f0 --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb @@ -0,0 +1,775 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 Heat Exchangers " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.3" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "hi is: 1409.0\n", + "% reduction because of fouling factor is 28.0\n" + ] + } + ], + "source": [ + "#Example Number 10.3\n", + "# influence of fouling factor\n", + "\n", + "#Variable declaration\n", + "\t\n", + "\t\n", + "Rf = 0.0002 \n", + "\t# using Rf=(1/hi-1/h_clean)\n", + "h_clean = 1961.0 \t\t\t# [W/square meter degree celsius]\n", + "\t# we obtain \n", + "\n", + "#Calculation\n", + "\n", + "hi = 1/(Rf+(1/h_clean)) \t\t# [W/square meter degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"hi is:\",round(hi)\n", + "print \"% reduction because of fouling factor is \",round((h_clean-hi)*100/h_clean) \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.4" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area of heat-exchanger is 15.81 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.4\n", + "# calculation of heat exchanger size from known temperatures\n", + "\n", + "#Variable declaration\n", + "import math\n", + "m_dot = 68.0 \t\t\t# [kg/min] water flow rate \n", + "U = 320.0 \t\t\t# [W/sq m degree C] overall heat transfer coefficient\n", + "T1 = 35.0 \t\t\t# [degree celsius] initial temperature \n", + "T2 = 75.0 \t\t\t# [degree celsius] final temperature\n", + "Toe = 110.0 \t\t\t# [degree celsius] oil entering temperature \n", + "Tol = 75.0 \t\t\t# [degree celsius] oil leaving temperature\n", + "Cw = 4180.0 \t\t\t# [J/kg degree celsius] water specific heat capacity\n", + "\t# the total heat transfer is determined from the energy absorbed by the water:\n", + "\n", + "#Claculation\n", + "\n", + "q = m_dot*Cw*(T2-T1) \t\t# [J/min]\n", + "q = q/60 \t\t\t# [W]\n", + "\t# since all the fluid temperatures are known, the LMTD can be calculated by \t\tusing the temperature scheme in figure 10-7b\n", + "dT_m = ((Toe-Tol)-(T2-T1))/(math.log((Toe-Tol)/(T2-T1))) \t# [degree celsius]\n", + "\t\t\t\t# then, since q = U*A*dT_m\n", + "A = q/(U*dT_m) \t\t\t# [square meter] area of heat-exchanger\n", + "\n", + "#Result\n", + "\n", + "print \"Area of heat-exchanger is\",round(A,2),\"square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.5" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area required for this exchanger is 19.53 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.5\n", + "# shell-and-tube heat exchanger\n", + "\n", + "# Variable declaration\n", + "\t\n", + "\t# determine a correction factor from figure 10-8 to be used \n", + "\t# the parameters according to figure 10-8(page no.-532) are \n", + "T1 = 35 \t\t\t# [degree celsius]\n", + "T2 = 75 \t\t\t# [degree celsius]\n", + "t1 = 110 \t\t\t# [degree celsius]\n", + "t2 = 75 \t\t\t# [degree celsius]\n", + "P = (t2-t1)/(T1-t1) \n", + "R = (T1-T2)/(t2-t1) \n", + "\t# so the correction factor is \n", + "F = 0.81 \t\t\t# from figure 10-10(page no.-534)\n", + "\t# and the heat transfer is q = U*A*F*dT_m\n", + "\t# so that. from example 10-4 we have \n", + "U = 320 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n", + "q = 189493.33 \t\t\t# [W]\n", + "\n", + "#Calculation\n", + "\n", + "dT_m = 37.44 \t\t\t# [degree celsius]\n", + "A = q/(U*F*dT_m) \t\t# [square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Area required for this exchanger is\",round(A,2),\"square meter\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.6" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of tubes per pass 37.0\n", + "Number of passes = 2\n", + "Length of tube per pass = 1.708 m\n" + ] + } + ], + "source": [ + "#Example Number 10.6\n", + "# design of shell-and-tube heat exchanger\n", + "\n", + "# Variable declaration\n", + "\n", + "import math\n", + "m_dot_c = 3.8 \t\t\t# [kg/s] water flow rate\n", + "Ti = 38 \t\t\t# [degree celsius] initial temperature of water\n", + "Tf = 55 \t\t\t# [degree celsius] final temperature of water\n", + "m_dot_h = 1.9 \t\t\t# [kg/s] water flow rate entering the exchanger\n", + "Te = 93 \t\t\t# [degree celsius] entering water temperature\n", + "U = 1419 \t\t\t# [W/sq m degree C] overall heat transfer coefficient\n", + "d = 0.019 \t\t\t# [m] diameter of tube\n", + "v_avg = 0.366 \t\t\t# [m/s] average water velocity in exchanger\n", + "Cc = 4180 \t\t\t# [] specific heat of water\n", + "Ch = Cc \t\t\t# [] specific heat \n", + "rho = 1000 \t\t\t# [kg/cubic meter] density of water\n", + "\t# we first assume one tube pass and check to see if it satisfies the \t\t\tconditions of this problem. the exit temperature of the hot water is \t\t\tcalculated from\n", + "\n", + "#Calculation\n", + "\n", + "dTh = m_dot_c*Cc*(Tf-Ti)/(m_dot_h*Ch) \t# [degree celsius]\n", + "Th_exit = Te-dTh \t\t\t# [degree celsius]\n", + "\t# the total required heat transfer is obtained for the cold fluid is \n", + "q = m_dot_c*Cc*(Tf-Ti) \t\t\t# [W]\n", + "\t# for a counterflow exchanger, with the required temperature \n", + "LMTD = ((Te-Tf)-(Th_exit-Ti))/math.log((Te-Tf)/(Th_exit-Ti)) \t# [degree celsius]\n", + "dTm = LMTD \t\t\t\t# [degree celsius]\n", + "A = q/(U*dTm) \t\t\t\t# [square meter]\n", + "\n", + "\n", + "\t#calculate the total area with\n", + "A1 = m_dot_c/(rho*v_avg) \t\t# [square meter]\n", + "\t# this area is the product of number of tubes and the flow area per tube:\n", + "n = A1*4/(math.pi*d**(2)) \t\t# no. of tubes\n", + "n = round(n) \t\n", + "\t# rounding of value of n because no. of pipe is an integer value\n", + "\t# the surface area per tube per meter of length is \n", + "S = math.pi*d \t\t\t\t# [square meter/tube meter]\n", + "\t# total surface area required for a one tube pass exchanger\t\t was \t\tcalculated above .\n", + "\t# we may thus compute the length of tube for this type of exchanger from \n", + "L = A/(S*n) \t\t\t\t# [m]\n", + "\t# this length is greater than the allowable 2.438 m, so we must use more than \t\tone tube pass.\n", + "\t\n", + "\t# we next try two tube passes. from figure 10-8(page no.-532) \n", + "F = 0.88 \n", + "A_total = q/(U*F*dTm) \t\t\t# [square meter]\n", + "\t# the number of tubes per pass is still 37 because of the velocity \t\t\trequirement. for the two pass exchanger the total surface area is now related \t\tto the length by\n", + "L1 = A_total/(2*S*n) \t\t\t# [m]\n", + "\t# this length is within the 2.438 m requirement, so the final design choice is \n", + "\n", + "#Result\n", + "\n", + "print \"Number of tubes per pass\",n \n", + "print \"Number of passes = 2\" \n", + "print \"Length of tube per pass =\",round(L1,3),\"m\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Surface area of heat exchanger is 10.84 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.7\n", + "# cross flow exchanger with one fluid mixed \n", + "\n", + "# Variable declaration\n", + "import math\n", + "\n", + "m_dot = 5.2 \t\t\t# [kg/s] mass flow rate\n", + "T1 = 130.0 \t\t\t# [degree celsius] temperature of entering steam\n", + "T2 = 110.0 \t\t\t# [degree celsius] temperature of leaving steam\n", + "t1 = 15.0 \t\t\t# [degree celsius] temperature of entering oil\n", + "t2 = 85.0 \t\t\t# [degree celsius] temperature of leaving oil\n", + "c_oil = 1900.0 \t\t\t# [J/kg degree celsius] heat capacity of oil\n", + "c_steam = 1860.0\t\t# [J/kg degree celsius] heat capacity of steam\n", + "U = 275 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n", + "\t#the total heat transfer may be obtained from an energy balance on the steam \n", + "\n", + "#Calculation\n", + "\n", + "q = m_dot*c_steam*(T1-T2) \t\t\t\t# [W]\n", + "\t# we can solve for the area from equation (10-13). the value of dT_m is \t\tcalculated as if the exchanger were counterflow double pipe,thus\n", + "dT_m = ((T1-t2)-(T2-t1))/math.log((T1-t2)/(T2-t1)) \t# [degree celsius]\n", + "\n", + "\t# t1,t2 is representing the unmixed fluid(oil) and T1,T2 is representing the \t\tmixed fluid(steam) so that:\n", + "\t# we calculate \n", + "\n", + "R = (T1-T2)/(t2-t1) \n", + "P = (t2-t1)/(T1-t1) \n", + "\t# consulting figure 10-11(page no.-534) we find \n", + "F = 0.97 \n", + "\t# so the area is calculated from \n", + "A = q/(U*F*dT_m) \t\t\t\t\t# [square meter]\n", + "\n", + "#Result\n", + "print \"Surface area of heat exchanger is \",round(A,2),\"square meter\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.8" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the objective of this example is to show that an iterative procedure is required when the inlet and outlet temperatures are not known or easily calculated\n", + "there is no need to go through this iteration because it can be avoided by using the techniques described in section 10-6\n" + ] + } + ], + "source": [ + "#Example Number 10.8\n", + "# effects of off-design flow rates for exchanger in example 10-7 \n", + "# Variable declaration\n", + "\n", + "# we did not calculate the oil flow in example 10-7 but can do so now from \n", + "q = 193 # [kW]\n", + "c_oil = 1.9 # [J/kg degree celsius] heat capacity of oil\n", + "t1 = 15 # [degree celsius] temperature of entering oil\n", + "t2 = 85 # [degree celsius] temperature of leaving oil\n", + "m_dot_o = q/(c_oil*(t2-t1)) # [kg/s]\n", + "# the new flow rate will be half this value \n", + "m_dot_o = m_dot_o/2 # [kg/s]\n", + "# we are assuming the inlet temperatures remain the same at 130 degree celsius for the steam and 15 degree celsius for the oil.\n", + "# the new relation for the heat transfer is q = m_dot_o*c_oil*(Teo-15) = m_dot_s*cp*(130-Tes) (a)\n", + "# but the exit temperatures, Teo and Tes are unknown. furthermore, dT_m is unknown without these temperatures, as are the values of R and P from figure 10-11(page no.-535). this means we must use an iterative procedure to solve for the exit temperatures using equation (a) and q = U*A*F*dT_m (b)\n", + "# the general procedure is to assume values of the exit temperatures until the q's agree between equations(a) and (b).\n", + "print \"the objective of this example is to show that an iterative procedure is required when the inlet and outlet temperatures are not known or easily calculated\" \n", + "print \"there is no need to go through this iteration because it can be avoided by using the techniques described in section 10-6\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.9" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A Reduction in the oil flow rate of 50 % causes a reduction in heat transfer of only 34.0 percent\n" + ] + } + ], + "source": [ + "#Example Number 10.9\n", + "# off-design calculation using E-NTU method \n", + " \n", + "#Variable declaration\t\n", + "import math\n", + "m_dot_o = 0.725 \t\t\t# [kg/s] oil flow rate\n", + "m_dot_s = 5.2 \t\t\t\t# [kg/s] steam flow rate\n", + "t1 = 15 \t\t\t\t# [degree celsius] temperature of entering oil\n", + "T1 = 130 \t\t\t\t# [deg C] temperature of entering steam\n", + "c_oil = 1900 \t\t\t\t# [J/kg degree celsius] heat capacity of oil\n", + "c_steam = 1860 \t\t\t\t# [J/kg degree celsius] heat capacity of steam\n", + "\t# for the steam \n", + "Cs = m_dot_s*c_steam \t\t\t# [W/degree celsius]\n", + "\t# for the oil\n", + "Co = m_dot_o*c_oil \t\t\t# [W/degree celsius]\n", + "\t# so the oil is minium fluid. we thus have\n", + "C_min_by_C_max = Co/Cs \n", + "U = 275 \t\t\t\t# [W/sq m deg C] overall heat transfer \t\t\t\t\t\t\t coefficient\n", + "A = 10.83 \t\t\t\t# [sq meter] surface area of heat exchanger\n", + "NTU = U*A/Co \n", + "\t# we choose to use the table and note that Co(minimum) is unmixed and \t\t\tCs(maximum) is mixed so that the first relation in the table 10-3 applies.\n", + "\t# we therfore calculate E(effectiveness) as \n", + "\n", + "E = (1/C_min_by_C_max)*(1-math.exp(-C_min_by_C_max*(1-math.exp(-NTU)))) \n", + "\t# if we were using figure 10-14(page no.-544) we would have to evaluate \n", + "C_mixed_by_C_unmixed = Cs/Co \n", + "\t# and would still determine \n", + "E = 0.8 # approximately\n", + "\t# now, using the effectiveness we can determine the temperature difference of \t\tthe minimum fluid(oil as)\n", + "dT_o = E*(T1-t1) \t\t\t# [degree celsius]\n", + "\t# so that heat transfer is \n", + "q = m_dot_o*c_oil*(dT_o) \t\t# [W]\n", + "q_initial = 193440 \t\t\t# [W] heat transfer when oilrate is 100 %\n", + "print \"A Reduction in the oil flow rate of 50 % causes a reduction in heat transfer of only \",round((q_initial-q)*100/q_initial),\"percent\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Exit water temperature is 90.8 degree celcius\n", + "The total heat transfer under the new flow conditions is 155.5 kW\n" + ] + } + ], + "source": [ + "#Example Number 10.10\n", + "# off-design calculation of exchanger in example 10-4 \n", + "\n", + "# Variable declaration\n", + "\n", + "m_dot_c = 68 \t\t\t# [kg/min] water flow rate\n", + "\n", + "T1 = 35 \t\t\t# [degree celsius] initial temperature \n", + "T2 = 75 \t\t\t# [degree celsius] final temperature\n", + "Toe = 110 \t\t\t# [degree celsius] oil entering temperature \n", + "Tol = 75 \t\t\t# [degree celsius] oil leaving temperature\n", + "Cc = 4180 \t\t\t# [J/kg degree celsius] water specific heat capacity\n", + "Ch = 1900 \t\t\t# [J/kg degree celsius] heat capacity of oil\n", + "U = 320 \t\t\t# [W/squ m deg C] overall heat transfer coefficient\n", + "A = 15.814568 \t\t\t# [sq m] area of heat exchanger (from example 10-4)\n", + "\t# the flow rate of oil is calculated from the energy balance for the original \t\tproblem:\n", + "\n", + "#Calculation\n", + "m_dot_h = m_dot_c*Cc*(T2-T1)/(Ch*(Toe-Tol)) \t# [kg/min]\n", + "\t# the capacity rates for the new conditions are calculated as \n", + "C_h = m_dot_h*Ch/60 \t\t\t\t# [W/degree celsius]\n", + "C_c = m_dot_c*Cc/60 \t\t\t\t# [W/degree celsius]\n", + "\t# so that the water (cold fluid) is the minimum fluid, and \n", + "C_min_by_C_max = C_c/C_h \n", + "NTU_max = U*A/C_c \n", + "\t# from figure 10-13(page no.-542) or table 10-3(page no.-543) the \t\teffectiveness is \n", + "E = 0.744 \n", + "\t# and because the cold fluid is the minimum, we can write \n", + "dT_cold = E*(Toe-T1) \t\t\t\t# [degree celsius]\n", + "\t\t\t\t\t\t# and the exit water temperature is \n", + "Tw_exit = T1+dT_cold \t\t\t\t# [degree celsius]\n", + "\t# the total heat transfer under the new flow conditions is calculated as \n", + "m_dot_c = 40 \t\t\t\t\t# [kg/min]\n", + "q = m_dot_c*Cc*dT_cold/60 \t\t\t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Exit water temperature is\",Tw_exit,\"degree celcius\" \n", + "print \"The total heat transfer under the new flow conditions is\",round(q/1000,1),\"kW\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The exit water temperature is 21.1 degree celsius\n", + "the heat transfer is 40.33 kW\n" + ] + } + ], + "source": [ + "#Example Number 10.11\n", + "# cross-flow exchanger with both fluid unmixed \n", + " \n", + "# Variable declaration\n", + "\n", + "pa = 101325 \t\t\t# [Pa] pressure of air\n", + "Ti = 15.55 \t\t\t# [degree celsius] initial temperature of air\n", + "Tf = 29.44 \t\t\t# [degree celsius] final temperature of air\n", + "Thw = 82.22 \t\t\t# [degree celsius] hot water temperature\n", + "U = 227 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n", + "S = 9.29 \t\t\t# [square meter] total surface area of heat exchanger\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "Cc = 1006 \t\t\t# [J/kg degree celsius] specific heat of air \n", + "Ch = 4180 \t\t\t# [J/kg degree celsius] specific heat of water\n", + "\t# the heat transfer is calculated from the energy balance on the air. first, \t\tthe inlet air density is \n", + "rho = pa/(R*(Ti+273.15)) \t# [kg/cubic meter]\n", + "\t# so the mass flow of air (the cold fluid) is \n", + "mdot_c = 2.36*rho \t\t# [kg/s]\n", + "\t# the heat transfer is then \n", + "q = mdot_c*Cc*(Tf-Ti) \t\t# [W]\n", + "\t# from the statement of the problem we do not know whether the air or water is \tthe minimum fluid. a trial and error procedur must be used \n", + "\t# we assume that the air is the minimum fluid and then check out our \t\tassumption.then\n", + "Cmin = mdot_c*Cc \t\t# [W/degree celsius]\n", + "NTU_max = U*S/Cmin \n", + "\t# and the effectiveness based on the air as the minimum fluid is \n", + "E = (Tf-Ti)/(Thw-Ti) \n", + "\t# we must assume values for the water flow rate until we are able to match the \tperformance as given by figure 10-15 or table 10-3. we first note that\n", + "Cmax = mdot_c*Cc \t\t# [W/degree celsius] (a)\n", + "\t\t\t\t# NTU_max = U*S/Cmin (b)\n", + "\t\t\t\t# E = dT_h/(Thw-Ti) (c)\n", + "\t\t\t\t# dT_h = q/Cmin (d)\n", + "\n", + "\t# now we assume different values for Cmin abd calculate different-different \t\tvalues for NTU_max, dT_h, and E\n", + "\n", + "\t# for \n", + "Cmin_by_Cmax1 = 0.5 \n", + "Cmin1 = Cmin_by_Cmax1*Cmax \t\t\t# [W/degree celsius]\n", + "NTU_max1 = U*S/Cmin1 \n", + "dT_h1 = q/Cmin1 \t\t\t\t# [degree celsius]\n", + "E1_c1 = dT_h1/(Thw-Ti) \t\t\t\t# calculated\n", + "E1_t1 = 0.65 \t\t\t\t\t# from table \n", + "\n", + "\t# for \n", + "Cmin_by_Cmax2 = 0.25 \n", + "Cmin2 = Cmin_by_Cmax2*Cmax \t\t\t# [W/degree celsius]\n", + "NTU_max2 = U*S/Cmin2 \n", + "dT_h2 = q/Cmin2 \t\t\t\t# [degree celsius]\n", + "E1_c2 = dT_h2/(Thw-Ti) \t\t\t\t# calculated\n", + "E1_t2 = 0.89 \t\t\t\t\t# from table \n", + "\n", + "\t# for \n", + "Cmin_by_Cmax3 = 0.22 \n", + "Cmin3 = Cmin_by_Cmax3*Cmax \t\t\t# [W/degree celsius]\n", + "NTU_max3 = U*S/Cmin3 \n", + "dT_h3 = q/Cmin3 \t\t\t\t# [degree celsius]\n", + "E1_c3 = dT_h3/(Thw-Ti) \t\t\t\t# calculated\n", + "E1_t3 = 0.92 \t\t\t\t\t# from table \n", + "\n", + "\t# we estimate the water-flow rate as about\n", + "Cmin = 660 \t\t\t\t\t# [W/degree celsius]\n", + "mdot_h = Cmin/Ch \t\t\t\t# [kg/s]\n", + "\t# the exit water temperature is accordingly\n", + "Tw_exit = Thw-q/Cmin \t\t\t\t# [degree celsius]\n", + "\n", + "print \"The exit water temperature is\",round(Tw_exit,1),\"degree celsius\" \n", + "print \"the heat transfer is\",round(q/1000,2),\"kW\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area required for the heat exchanger is 20.09 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.13\n", + "# shell and tube exchangeras air heater\n", + " \n", + "# Variable declaration\n", + "\n", + "import math\n", + "To = 100.0 \t\t\t\t# [degree celsius] temperature of hot oil\n", + "m_dot_a = 2.0 \t\t\t\t# [kg/s] flow rate of air\n", + "T1 = 20.0 \t\t\t\t# [degree celsius] initial temperature of air \n", + "T2 = 80.0 \t\t\t\t# [degree celsius] final temperature of air\n", + "Cp_o = 2100.0 \t\t\t\t# [J/kg deg C] specific heat of the oil\n", + "Cp_a = 1009.0 \t\t\t\t# [J/kg deg C] specific heat of the air\n", + "m_dot_o = 3 \t\t\t\t# [kg/s] flow rate of oil\n", + "U = 200.0 \t\t\t\t# [W/sq m] overall heat transfer coefficient\n", + "\t# the basic energy balance is m_dot_o*Cp_o*(To-Toe) = m_dot_a*Cp_a*(T2-T1)\n", + "#Calculation\n", + "Toe = To-m_dot_a*Cp_a*(T2-T1)/(m_dot_o*Cp_o) # [degree celsius]\n", + "\n", + "\t# we have\n", + "m_dot_h_into_Ch = m_dot_o*Cp_o \t\t# [W/degree celsius]\n", + "m_dot_c_into_Cc = m_dot_a*Cp_a \t\t# [W/degree celsius]\n", + "\t# so the air is minimum fluid\n", + "C = m_dot_c_into_Cc/m_dot_h_into_Ch \n", + "\t# the effectiveness is \n", + "E = (T2-T1)/(To-T1) \n", + "\t# now we may use figure 10-16 to obtain NTU. \n", + "NTU = -(1+C**(2))**(-1.0/2.0)*math.log((2/E-1-C-(1+C**2)**(1.0/2.0))/(2/E-1-C+(1+C**2)**(1.0/2.0))) \n", + "\t# now, we calcuate the area as \n", + "A = NTU*m_dot_c_into_Cc/U \t\t# [square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Area required for the heat exchanger is\",round(A,2),\"square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.14" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area to achieve a heat exchanger effectiveness of 60% with an exit water temperature of 40 degree celsius is 9.16 square meter\n", + "by reducing the flow rate we have lowered the heat transfer by 37.0 percent\n" + ] + } + ], + "source": [ + "#Example Number 10.14\n", + "# ammonia condenser \n", + "\n", + "#Variable declaration\n", + "\n", + "Ta = 50 \t\t\t# [degree C] temperature of entering ammonia vapour\n", + "Tw1 = 20 \t\t\t# [degree celsius] temperature of entering water\n", + "q = 200 \t\t\t# [kW] total heat transfer required\n", + "U = 1 \t\t\t\t# [kW/sq m deg C] overall heat transfer coefficient\n", + "Tw2 = 40 \t\t\t# [deg C] temperature of exiting water\n", + "Cw = 4.18 \t\t\t# [kJ/kg degree celsius] specific heat of water\n", + "\t# the mass flow can be calculated from the heat transfer with\n", + "m_dot_w = q/(Cw*(Tw2-Tw1)) \t# [kg/s]\n", + "\t# because this is the condenser the water is the minimum fluid and \n", + "C_min = m_dot_w*Cw # [kW/degree celsius]\n", + "\t# the value of NTU is obtained from the last entry of table 10-4\n", + "E = 0.6 \t\t\t# effectiveness\n", + "\n", + "#Calculation\n", + "\n", + "import math\n", + "NTU = -math.log(1-E) \n", + "\n", + "\t# so that area is calculated as \n", + "A = C_min*NTU/U \t\t# [square meter]\n", + "\n", + "\t# when the flow rate is reduced in half the new value of NTU is \n", + "NTU1 = U*A/(C_min/2) \n", + "\n", + "\t# and the effectiveness is computed from the last entry of table 10-3\n", + "E1 = 1-math.exp(-NTU1) \n", + "\n", + "\t# the new water temperature difference is computed as \n", + "dT_w = E1*(Ta-Tw1) \t\t# [degree celsius]\n", + "\n", + "\t# so that the heat transfer is \n", + "q1 = C_min*dT_w/2 \t\t# [kW]\n", + "\n", + "#Result\n", + "\n", + "print \"Area to achieve a heat exchanger effectiveness of 60% with an exit water temperature of 40 degree celsius is\",round(A,2),\"square meter\" \n", + "print \"by reducing the flow rate we have lowered the heat transfer by\",(q-q1)*100/q,\" percent\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.16" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat-transfer coefficient is 174.0 W/square meter degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 10.16 \n", + "# heat-transfer coefficient in compact exchanger \n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "T = 300.0\t\t\t# [K] temperature of entering air\n", + "u = 15.0 \t\t\t\t# [m/s] velocity of air\n", + "\t# we obtain the air properties from table A-5\n", + "rho = 1.774 \t\t\t# [kg/cubic meter] density of air\n", + "Cp = 1005.7 \t\t\t# [J/kg degree celsius] specific heat of air\n", + "mu = 1.983*10**(-5) \t\t# [kg/m s] viscosity of air\n", + "Pr = 0.708 \t\t\t# prandtl number\n", + "\t# from figure 10-19 we have\n", + "Ac_by_A = 0.697 \n", + "sigma = 0.697 \n", + "\n", + "#Calculation\n", + "\n", + "Dh = 3.597*10**(-3) \t\t# [m] \n", + "\n", + "\t\t\t\t# the mass velocity is thus \n", + "G = ((rho*u)/sigma) \t\t# [kg/square meter s]\n", + "\n", + "\n", + "\t# and the reynolds number is \n", + "Re = Dh*G/mu \n", + "\n", + "\t# from figure 10-19(page no.-557) we can read\n", + "\n", + "St_into_Pr_exp_2_by_3 = 0.0036\n", + " \n", + "\t# and the heat transfer coefficient is \n", + "\n", + "h = St_into_Pr_exp_2_by_3*G*Cp*(Pr)**(-2.0/3.0) \t# [W/sq m deg C]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat-transfer coefficient is\",round(h),\"W/square meter degree celsius\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter11.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter11.ipynb new file mode 100644 index 00000000..764a2588 --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter11.ipynb @@ -0,0 +1,254 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 Mass Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "value of diffusion coefficient for co2 in air is 0.132 square centimeter/s\n", + "From Table A-8,D=0.164 sq cm/s\n", + " So,they are in fair agreement\n" + ] + } + ], + "source": [ + "#Example Number 11.1\n", + "# diffusion coefficient for co2\n", + "\n", + "# Variable declaration\n", + "\n", + "T =298.0\t\t\t# [K] temperature of air\n", + "Vco2 = 34.0 \t\t\t# molecular volume of co2\n", + "Vair = 29.9 \t\t\t# molecular volume of air\n", + "Mco2 = 44.0 \t\t\t# molecular weight of co2\n", + "Mair = 28.9 \t\t\t# molecular weight of air\n", + "P = 1.0132*10**(5) \t\t# [Pa] atmospheric pressure\n", + "\t# using equation (11-2)\n", + "#Calculation\n", + "D = 435.7*T**(3.0/2.0)*(((1/Mco2)+(1/Mair))**(1.0/2.0))/(P*(Vco2**(1.0/3.0)+Vair**(1.0/3.0))**(2)) \n", + "\n", + "#Result\n", + "print \"value of diffusion coefficient for co2 in air is\",round(D,3),\"square centimeter/s\" \n", + "print \"From Table A-8,D=0.164 sq cm/s\\n So,they are in fair agreement\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.3" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of dry air is 53.66 degree celsius\n", + " recalculate the density at the arithmetic-average temperature between wall and free-stream conditions\n", + "With this adjustments these results are RHO = 1.143 kg/m**(3) and Tinf = 55.8 degree celcius\n" + ] + } + ], + "source": [ + "#Example Number 11.3\n", + "# Wet-bulb temperature\n", + "\n", + "#Variable declaration\n", + "\n", + "Pg = 2107.0\t\t\t# [Pa] from steam table at 18.3 degree celcius\n", + "Pw = Pg*18.0 \t\t\t# [Pa]\n", + "Rw = 8315.0 \t\t\t# [J/mol K] gas constant\n", + "Tw = 273+18.3\t \t\t# [K]\n", + "\n", + "RHOw = Pw/(Rw*Tw) \t\t# [kg/cubic meter]\n", + "\n", + "\n", + "Cw = RHOw \t\t\t# [kg/cubic meter]\n", + "RHOinf = 0.0 \t\t\t# since the free stream is dry air\n", + "Cinf = 0.0 \n", + "P = 1.01325*10**(5) \t\t# [Pa]\n", + "R = 287 \t\t\t# [J /kg  K]\n", + "T = Tw \t\t\t\t# [K]\n", + "RHO = P/(R*T) \t\t\t# [kg/cubic meter]\n", + "\n", + "Cp = 1004.0 \t\t\t# [J/kg degree celsius]\n", + "Le = 0.845 \n", + "Hfg = 2.456*10**(6) \t\t# [J/kg]\n", + "#Calculations\n", + "# now using equation(11-31)\n", + "\n", + "Tinf = (((Cw-Cinf)*Hfg)/(RHO*Cp*(Le**(2.0/3.0))))+Tw \t# [K]\n", + "Tin = Tinf-273 \t\t\t\t\t# [degree celsius]\n", + "\n", + "print \"Temperature of dry air is\",round(Tin,2),\"degree celsius\" \n", + "print \" recalculate the density at the arithmetic-average temperature between wall and free-stream conditions\" \n", + "print \"With this adjustments these results are RHO = 1.143 kg/m**(3) and Tinf = 55.8 degree celcius\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.4" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Relative humidity is therefore 27.8 percentage\n" + ] + } + ], + "source": [ + "#Example Number 11.4\n", + "# relative humidity of air stream\n", + "\n", + "#Variable declaration\n", + "\n", + "\t\t# these data were taken from previous example\n", + "Rho = 1.212 \t\t\t\t\t# [kg/cubic meter]\n", + "Cp = 1004 \t\t\t\t\t# [J/kg]\n", + "Le = 0.845 \n", + "Tw = 18.3 \t\t\t\t\t# [degree celsius]\n", + "Tinf = 32.2 \t\t\t\t\t# [degree celsius]\n", + "Rhow = 0.015666 \t\t\t\t# [kg/cubic meter]\n", + "Cw = Rhow \t\t\t\t\t# [kg/cubic meter]\n", + "\n", + "#calculation\n", + "\n", + "Hfg = 2.456*10**(6) \t\t\t\t# [J/kg]\n", + "\t\t# we use eqn 11-31\n", + "Cinf = Cw-(Rho*Cp*Le**(2.0/3.0)*(Tinf-Tw)/Hfg) \t# [kg/cubic meter]\n", + "Rhoinf = Cinf \t\t\t\t\t# [kg/cubic meter]\n", + "Rhog = 0.0342 \t\t\t\t\t# [kg/cubic meter]\n", + "RH = (Rhoinf/Rhog)*100 \n", + "\n", + "#Result\n", + "\n", + "print \"Relative humidity is therefore\",round(RH,1),\"percentage\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.5" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Evaporation rate on the land under these conditions is 0.0028 kg/h square meter\n" + ] + } + ], + "source": [ + "#Example Number 11.5\n", + "# water evaporation rate\n", + "\n", + "# Variable declaration\n", + "\n", + "Ta = 38+273 \t\t\t# [K] temperature of atmospheric air\n", + "RH = 0.30 \t\t\t# relative humidity\n", + "u = 10.0\t\t\t# [mi/h] mean wind speed\n", + "R = 0.287 \t\t\t# universal gas constant\n", + "Dw = 0.256*10**(-4) \t\t# [square meter/s] from table A-8(page no.-610)\n", + "rho_w = 1000 \t\t\t# [kg/cubic meter]\n", + "\t# for this calculation we can make use of equation(11-36). from thermodynamic \tsteam tables\n", + "p_g = 6.545 \t\t\t# [kPa] at 38 degree celsius\n", + "p_s = p_g \t\t\t# [kPa]\n", + "p_w = RH*p_s \t\t\t# [kPa]\n", + "p_s = 1.933 \t\t\t# [in Hg]\n", + "p_w = 0.580 \t\t\t# [in Hg]\n", + "\t# also \n", + "u_bar = u*24 \t\t\t# [mi/day]\n", + "\t# equation(11-36) yields, with the application of the 0.7 factor\n", + "\n", + "E_lp = 0.7*(0.37+0.0041*u_bar)*(p_s-p_w)**(0.88) \t\t# [in/day]\n", + "E_lp = E_lp*2.54/100 \t\t\t\t\t\t# [m/day]\n", + "\n", + "\t# noting that standard pan has the diameter of 1.2m, we can use the figure to \t\tcalculate the mass evaporation rate per unit area as\n", + "m_dot_w_by_A = E_lp*rho_w/24 \t\t\t\t# [kg/h square meter]\n", + "\n", + "\n", + "\t# as a matter of interest, we might calculate the molecular-diffusion rate of \t\twater vapour from equation(11-35), taking z1 as the 1.5m dimension above the \t\tstandard pan.\n", + "z1 = 1.5 \t\t\t\t\t\t\t# [m]\n", + "\n", + "\t# since rho = p/(R*T)\n", + "\t# equation(11-35) can be written as \n", + "m_dot_w_by_A1 = 0.622*Dw*p_g*3600/(R*Ta*z1) \t\t\t# [kg/h square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Evaporation rate on the land under these conditions is\",round(m_dot_w_by_A1,4),\"kg/h square meter\"\n", + "\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter2.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter2.ipynb new file mode 100644 index 00000000..71fd76dc --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter2.ipynb @@ -0,0 +1,674 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 Steady State Conduction One Dimension" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of thickness is: 5.75 cm\n" + ] + } + ], + "source": [ + "#Example Number 2.1\n", + "#Calculate thickness of insulation to reduce heat loss/gain by 80 %\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "dx1 = 0.1\t # [m] thickness of layer of common brick\n", + "k1 = 0.7\t # [W/m degree celsius] heat transfer coefficient of common brick\n", + "dx2 = 0.0375 # [m] thickness of layer of gypsum plaster\n", + "k2 = 0.48 \t # [W/m degree celsius] heat transfer coefficient gypsum plaster\n", + "\n", + "#Calculation\n", + "\n", + "Rb = dx1/k1\t # [sq m degree C /W] thermal resistance of brick\n", + "Rp = dx2/k2 # [sq m degree C /W] thermal resistance of gypsum plaster\n", + "R = Rb+Rp \t # [sq m degree C /W] thermal resistance without insulation\n", + "R1 = R/0.2 \t # [sq m degree C /W] with insulation \n", + "\n", + "# heat loss with the rock-wool insulation is 20 percent \n", + "\n", + "Rrw = R1-R \t # [square meter degree celsius /W]\n", + "k3 = 0.065 \t # [W/m degree celsius] heat transfer coefficient\n", + "dx3 = Rrw*k3 # [m]\n", + "\n", + "#Result\n", + "\n", + "print \"Length of thickness is:\",round(dx3*100,2),\" cm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flow is given by: 680.0 W/m\n", + "Interface temperature is 596.0 degree celsius \n" + ] + } + ], + "source": [ + "#Example Number 2.2\n", + "#Calculate heat loss/metre length and tube-insulation interface temperature\n", + "\n", + "#Variable declaration\n", + "\n", + "ID = 0.02 \t\t# [m] inner diameter of steel\n", + "OD = 0.04 \t\t#[m] outer diameter of steel\n", + "t = 0.03 \t\t#[m] thickness of asbestos insulation\n", + "\t\t\t# system is like three concentric cylinders \n", + "T1 = 600 \t\t# [degree celsius] inside wall temperature\n", + "T2 = 100 \t\t# [degree celsius] outside insulation temperature\n", + "Ks = 19 \t\t#[W/m degree celsius] heat transfer coefficient of steel\n", + "Ka = 0.2 \t\t# [W/m degree celsius] heat transfer coefficient of asbestos\n", + "\n", + "\t\t\t# heat flow is given by per unit length\n", + "\n", + "#Calculation\n", + "\n", + "import math\n", + "Q_l = ((2*math.pi*(T1-T2))/((math.log(OD/ID)/Ks)+(math.log(0.1/OD)/Ka))) # [W/m]\n", + "\n", + "\t# above calculated heat flow is used to calculate the interface temperature\n", + "\t# between the outside wall and the insulation\n", + "\n", + "Ta = Q_l*(math.log(0.1/OD)/(2*math.pi*Ka))+T2 \n", + "\t# [degree C] Ta is interface temperature\n", + "\n", + "#Result\n", + "\n", + "print \"Heat flow is given by:\",round(Q_l),\" W/m\" \n", + "print \"Interface temperature is\",round(Ta),\"degree celsius \"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "TOTAL THERMAL RESISTANCE THROUGH,\n", + "wood stud is: 31.39 degree C\n", + "Insulation section is: 7.34 degree C/W\n", + "Overall heat transfer coefficient is: 0.414 W/sq m per deg C\n", + "R value is: 2.415 sq m/W\n" + ] + } + ], + "source": [ + "#Example Number 2.3 \n", + "# Calculate overall heat transfer coefficient and R Value of wall\n", + "\n", + "# 1. heat transfer through studs for unit depth\n", + "\n", + "#Variable declaration\n", + "\n", + "l = 0.0413 \t # [m] length of wood studs\n", + "b = 1.0 \t # [m] unit depth\n", + "A = l*b\t\t # [square meter] area of studs for unit depth\n", + "hi = 7.5\t # [W/sq meter/degr C] convectional heat transfer coefficient\n", + "ho = 15 \t # [W/sq m per deg C] convectional heat transfer coefficient\n", + "Kb = 0.69 \t # [W/m per deg celsius] heat transfer coefficient of brick\n", + "Kgi = 0.96 \t # [W/m per deg C] heat transfer coefficient of gypsum inner sheath\n", + "Ki = 0.04 \t # [W/m per deg C] heat transfer coefficient of insulation\n", + "Kws = 0.1\t # [W/m per deg C] heat transfer coefficient of wood stud\n", + "Kgo = 0.48 \t # [W/m per deg C] heat transfer coefficient of gypsum outer sheath\n", + "Rair = 1/(ho*A) # [degree C /W] convection resistance outside of brick\n", + "dx_b = 0.08 \t # [m] thickness of brick\n", + "dx_os = 0.019 \t #[m] thickness of outer sheet\n", + "dx_ws = 0.0921 # [m] thickness of wood stud\n", + "dx_is = 0.019 # [m] thickness of inner sheet\n", + "Rb = dx_b/(Kb*A) \t# [degr C /W] conduction resistance in brick\n", + "Ros = dx_os/(Kgi*A) \t# [deg C /W] conduction resistance through outer sheet\n", + "Rws = dx_ws/(Kws*A) \t# [deg C/W] conduction resistance through wood stud\n", + "Ris = dx_is/(Kgo*A) \t# [deg C/W] conduction resistance through inner sheet\n", + "\n", + "#Calculation\n", + "\n", + "Ri = 1/(hi*A) \t\t# [degree celsius /W] convection resistance on inside\n", + "\n", + "Rt = Rair+Rb+Ros+Rws+Ris+Ri \n", + "\t\t\t# [deg C/W] total thermal Res through the wood stud section\n", + "\n", + "\n", + "print \"TOTAL THERMAL RESISTANCE THROUGH,\"\n", + "print\"wood stud is:\",round(Rt,2),\"degree C\"\n", + "\n", + "\n", + "# 2. Heat transfer through insulation section \n", + "\n", + "#Calculation\n", + "\n", + "A1 = 0.406-A\t\t # [sq meter] area of insulation section for unit depth\n", + "dx_ins = 0.0921 # [m] thickness of insulation\n", + "Rins = dx_ins/(Ki*A1) # [deg C /W] conduction resistance through insulation section\n", + "\n", + "\t\t# five of the materials are same but resistance involve different area \n", + "\t\t# i.e. (40.6-4.13) cm instead of 4.13 cm \n", + "\t\t# so that each of the previous must be multiplied by a factor of \t\t\t#(4.13/(40.6-4.13)) = 0.113 \n", + "\n", + "#Calculation\n", + "\n", + "Rt_ins = (Rair+Rb+Ros+Ris+Ri)*0.113+Rins\n", + " \t\t# [deg C/W] total resistance through insulation section \n", + "\t\n", + "print\"Insulation section is:\",round(Rt_ins,3),\"degree C/W\"\n", + "\n", + "R_overall = 1/((1/Rt)+(1/Rt_ins)) \n", + "\t\t# [degree celsius /W] overall resistance for the section\n", + "\n", + "\t\t# the value is related to overall heat transfer coefficient by \n", + "\t\t# Q = U*A*dt = dt/R_overall \n", + "\t\t# where A is area of total section\n", + "\n", + "\n", + "A_ = 0.406 \t\t# [sq meter] area of total section\n", + "U = 1/(R_overall*A_) # [W/sq meter deg C] overall heat transfer coefficient\n", + "\t\t\t# R value is somewhat different from thermal resistance and is \t\t\t#given by\n", + "R_value = 1/U \t\t# [degree celsius square meter/W] R value of system\n", + "\n", + "\n", + "#Results\n", + "\n", + "print\"Overall heat transfer coefficient is:\",round(U,3),\"W/sq m per deg C\"\n", + "\n", + "print\"R value is:\",round(R_value,3),\"sq m/W\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall heat transfer coefficient is: 7.58 W/sq m degree C\n", + "Heat loss per unit length is: 19.0 W(for 1m length)\n" + ] + } + ], + "source": [ + "#Example Number 2.4\n", + "# Calculate overall heat transfer coeff and heat loss/unit length at 20 deg C\n", + "\n", + "#Variable declaration\n", + "\n", + "ID = 0.025 \t\t# [m] inner diameter of steel\n", + "OD = ID+2*0.0008 \t#[m] outer diameter of steel\n", + "hi = 3500 \t\t# [W/sq m per deg C] convectional heat transfer coefficient of \t\t\t#inside\n", + "ho = 7.6 \t\t# [W/sq m per deg C] convectional heat transfer coefficient of \t\t\t#outside \n", + "L = 1.0 \t\t# [m] tube length\n", + "import math\n", + "Ai = math.pi*ID*L \t# [sq meter] inside crossectional area \n", + "Ao = math.pi*OD*L \t# [sq meter] outside crossectional area \n", + "k = 16 \t\t\t# [W/sq meter per deg C] thermal conductivity of tube\n", + "\n", + "#Calculation\n", + "\n", + "Ri = 1/(hi*Ai)\t\t\t # [degree C /W] convection resistance inside tube\n", + "Rt = math.log(OD/ID)/(2*math.pi*k*L) # [degree C /W] thermal resistance \n", + "Ro = 1/(ho*Ao) \t\t\t # [deg C /W] convection resistance outside tube\n", + "R_total = Ri+Rt+Ro\t\t # [deg C/W] total thermal and convection \t\t\t\t \t\t #resistance \n", + "Uo = 1/(Ao*R_total) \t\t # [W/sq m deg C] overall heat transfer \t\t\t\t \t\t #coefficient\n", + "\n", + "Tw = 50\t\t\t\t # [degree C] water temperature\n", + "Ta = 20 \t\t\t # [degree C] surrounding air temperature\n", + "dt = Tw-Ta \t\t\t # [degree C] temperature difference\n", + "q = Uo*Ao*dt\t\t\t # [W] heat transfer \n", + "\n", + "#Results\n", + "\n", + "print\"Overall heat transfer coefficient is:\",round(Uo,2),\" W/sq m degree C\"\n", + "\n", + "\n", + "print\"Heat loss per unit length is:\",round(q),\" W(for 1m length)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Critical radius of insulation for asbestos is: 5.67 cm \n", + "Heat loss when covered with critical radius of insulation is: 105.7 W/m\n", + "Heat loss without insulation is: 84.8 W/m\n", + "Addition of 3.17 of insulation actually increases the heat transfer by: 25.0 %\n" + ] + } + ], + "source": [ + "#Example Number 2.5\n", + "#Calculate the heat loss \n", + "\n", + "#Variable declaration\n", + "\n", + "k = 0.17 \t\t# [W/m per deg C] heat transfer coefficient of asbestos\n", + "Tr = 20 \t\t# [degree celsius] temperature of room air\n", + "h = 3 \t\t\t# [W/sq m per deg C] convectional heat transfer coefficient\n", + "Tp = 200 \t\t# [degree celsius] temperature of pipe\n", + "d = 0.05 \t\t# [m] diameter of pipe\n", + "\n", + "\t\t\t# from equation (2-18) we calculate r_o as \n", + "\n", + "#Calculation\n", + "\n", + "r_o = k/h \t\t# [m] critical radius of insulation\n", + "print\"Critical radius of insulation for asbestos is:\",round(r_o*100,2),\"cm \"\n", + "\n", + "Ri = d/2\t # [m] inside radius of insulation\n", + "\t\t\t# heat transfer is calculated from equation (2-17)\n", + "import math\n", + "q_by_L = (2*math.pi*(Tp-Tr))/(((math.log(r_o/Ri))/0.17)+(1/(h*r_o)))\n", + "\t \t\t# [W/m] heat transfer per unit length\n", + "\n", + "\n", + "#Results\n", + "\n", + "print\"Heat loss when covered with critical radius of insulation is:\",round(q_by_L,1),\" W/m\"\n", + "\n", + "\t\t# without insulation the convection from the outer surface of pipe is \n", + "\n", + "q_by_L1 = h*2*math.pi*Ri*(Tp-Tr) \n", + "\t\t#[W/m] convection from outer surface without insulation\n", + "print\"Heat loss without insulation is:\",round(q_by_L1,1),\" W/m\"\n", + "per_inc = ((q_by_L-q_by_L1)/q_by_L1)*100 \t# percentage increase in heat transfer\n", + "\n", + "print\"Addition of 3.17 of insulation actually increases the heat transfer by:\",round(per_inc),\"%\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Center temperature of the wire is: 231.7 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 2.6\n", + "#Calculate the centre temperature of wire\n", + "\n", + "#Variable declaration\n", + "\n", + "\t\t# all the power generated in the wire must be dissipated by convection \t\tto the liquid \n", + "\t\t# P = i**(2)*R = q = h*A*dt\n", + "L = 100 \t# [cm] length of the wire \n", + "k = 19 \t\t# [W/m per deg C] heat transfer coefficient of steel wire\n", + "\n", + "#Calculations\n", + "\n", + "import math\n", + "A = math.pi*(0.15)**(2) \t# [sq m] crossectional area of wire\n", + "rho = 70*10**(-6) \t\t# [micro ohm cm] resistivity of steel \n", + "R = rho*L/A \t\t\t# [ohm] resistance of wire\n", + "i = 200 \t\t\t# [ampere] current in the wire\n", + "P = i**(2)*R \t\t\t# [W] power generated in the wire\n", + "Tl = 110 \t\t\t# [degree celsius] liquid temperature\n", + "d = 0.003 \t\t\t# [m] diameter of wire\n", + "l = 1 \t\t\t\t# [m] length of wire\n", + "Tw = (P/(4000*3.14*d*l))+110 \t# [degree celsius] wire temperature\n", + "\n", + "\t\t\t\t# heat generated per unit V q_dot is calculated as\n", + "\t\t\t\t# P = q_dot*V = q_dot*3.14*r**(2)*l\n", + "r = d/2 \t\t\t# [m] radius of wire\n", + "q_dot = P/(math.pi*r**(2)*l) \t# [W/m**(3)]\n", + "\t\t\t\t# finally the center temperature of the wire is \t\t\t\t\tcalculated from equation (2-26)\n", + "\t\n", + "To = ((q_dot*(r**(2)))/(4*k))+Tw\t # [degree celsius]\n", + "\n", + "\n", + "#Result\n", + "\n", + "print \"Center temperature of the wire is:\",round(To,1) ,\"degree celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.8" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss from the fin per unit length is 360.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 2.8\n", + "# Calculate heat loss per unit depth of the material\n", + "\n", + "#Variable declaration\n", + "\n", + "t = 0.003 \t\t# [m] thickness of fin\n", + "L = 0.075 \t\t# [m] length of fin\n", + "Tb = 300 \t\t# [degree celsius] base temperature\n", + "Tair = 50 \t\t# [degree celsius] ambient temperature\n", + "k = 200 \t\t# [W/m per deg C] heat transfer coefficient of aluminium fin\n", + "h = 10 \t\t\t# [W/sq m per deg C] convectional heat transfer coefficient\n", + "\t\t\t# We Will use the approximate method of solution by extending \t\t\tthe fin \n", + "\t\t\t# With a fictitious length t/2\n", + "\t\t\t# using equation(2-36)\n", + "\n", + "#Calculation\n", + "Lc = L+t/2 \t\t# [m] corrected length\n", + "z = 1 \t\t\t# [m] unit depth\n", + "p = (2*z+2*t) \t\t# [m] perimeter of fin\n", + "A = z*t \t\t# [square meter] crossectional area of fin\n", + "m = ((h*p)/(k*A))**(0.5)\n", + "\t \n", + "\t\t\t\t\t\t# from equation(2-36)\n", + "dt = Tb-Tair \t\t\t\t\t# [degree C] temperature difference\n", + "import math\n", + "q = math.tanh(m*Lc)*((h*p*k*A)**(0.5))*dt\t# [W/m] heat transfer per unit length \n", + "\n", + "#Results\n", + "print \"Heat loss from the fin per unit length is\",round(q),\"W/m\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.9" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The actual heat transferred is: 60.97 W\n" + ] + } + ], + "source": [ + "#Example Number 2.9\n", + "# Calculate the heat loss per fin\n", + "\n", + "#Variable declaration\n", + "\n", + "t = 0.001 \t\t\t# [m] thickness of fin\n", + "L = 0.015 \t\t\t# [m] length of fin\n", + "Ts = 170\t\t\t# [degree celsius] surface temperature\n", + "Tfluid = 25 \t\t\t# [degree celsius] fluid temperature\n", + "k = 200 \t\t\t# [W/m per deg C] heat transfer coefficient of \t\t\t\t\t\t aluminium fin\n", + "h = 130 \t\t\t# [W/sq m per deg C] \t\t\t\t convectional heat transfercoefficient\n", + "d = 0.025 \t\t\t# [m] tube diameter\n", + "Lc = L+t/2\t\t\t# [m] corrected length\n", + "r1 = d/2 \t\t\t# [m] radius of tube\n", + "r2_c = r1+Lc \t\t\t# [m] corrected radius\n", + "\n", + "#Calculation\n", + "\n", + "Am = t*(r2_c-r1) \t\t# [sq m] profile area \n", + "c = r2_c/r1 \t\t\t# constant to determine\tefficiency of fin from curve \n", + "\n", + "c1 = ((Lc)**(1.5))*((h/(k*Am))**(0.5)) \t# constant to determine efficiency of fin from \t\t\t\t\tcurve\n", + "\n", + "\t\t \t\t# using c and c1 to determine the efficiency \t\t\t\t\t\tof the fin from figure (2-12)\n", + "\t\t\t\t# we get nf = 82 percent\n", + "\t\t\t\t# heat would be transferred if the entire fin were at \t\t\t\t\tthe base temperature \n", + "\t\t\t\t# both sides of fin exchanging heat \n", + "import math\n", + "q_max = 2*math.pi*(r2_c**(2)-r1**(2))*h*(Ts-Tfluid) \t# [W] maximum heat transfer\n", + "q_act = 0.82*q_max \t\t\t\t\t#[W] actual heat transfer\n", + "\n", + "#Result\n", + "print\"The actual heat transferred is:\",round(q_act,2),\" W\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The expression for the temperature distribution in the rod is \n", + "theta`=(((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1))\n", + "for an infinitely long heat generating fin with the left end maintained at T1, the temperature distribution becomes \n", + "theta`/theta1 = exp^(-m*x)\n" + ] + } + ], + "source": [ + "#Example Number 2.10\n", + "#Obtain an expression for the temperature distribution in rod\n", + "\n", + "#Variable declaration & Calculations\n", + "\n", + "\t# q_dot is uniform heat source per unit volume \n", + "\t# h is convection coefficient\n", + "\t# k is heat transfer coefficient\n", + "\t# A is area of crossection\n", + "\t# P is perimeter \n", + "\t# Tinf is environment temperature \n", + "\t# we first make an energy balance on the element of the rod shown in \tfigure(2-10)\n", + "\t# energy in left place + heat generated in element = energy out right face + \tenergy lost by convection\n", + "\t# or \n", + "\t# -(k*A*dT_by_dx)+(q_dot*A*dx) = \t-(k*A(dT_by_dx+(d2T_by_dx2)*dx))+h*P*dx*(T-Tinf)\n", + "\t# simlifying we have \n", + "\t# d2T_by_dx2-((h*P)/(k*A))*(T-Tinf)+q_dot/k = 0\n", + "\t# replacing theta = (T-Tinf) and (square meter) = ((h*P)/(k*A))\n", + "\t# d2theta_by_dx2-(square meter)*theta+q_dot/k = 0\n", + "\t# we can make a further substitution as theta` = theta-(q_dot/(k*(square \tmeter)))\n", + "\t# so that our differential equation becomes \n", + "\t# d2theta`_by_dx2-(square meter)*theta`\n", + "\t# which has the general solution theta` = C1*exp^(-m*x)+C2*exp^(m*x)\n", + "\t# the two end temperatures are used to establish the boundary conditions:\n", + "\t# theta` = theta1` = T1-Tinf-q_dot/(k*(square meter)) = C1+C2\n", + "\t# theta` = theta2` = T2-Tinf-q_dot/(k*(square meter)) = \tC1*exp^(-m*L)+C2*exp^(m*L)\n", + "\t# solving for the constants C1 and C2 gives \n", + "\t\t#((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1))\n", + "\n", + "\n", + "#RESULTS\n", + "\n", + "print\"The expression for the temperature distribution in the rod is \" \n", + "\n", + "print\"theta`=(((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1))\" \n", + "\n", + "print\"for an infinitely long heat generating fin with the left end maintained at T1, the temperature distribution becomes \" \n", + "\n", + "print\"theta`/theta1 = exp^(-m*x)\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall heat flow is: 5.52 W\n", + "The temperature drop across the contact is: 4.3 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 2.11\n", + "# Calculate the axial heat flow and temperature drop across the contact surface\n", + "\n", + "#Variable declaration\n", + "\n", + "d = 0.03\t\t #[m] diameter of steel bar\n", + "l = 0.1 #[m] length of steel bar\n", + "import math\n", + "A = (math.pi*d**(2))/4\t # [square meter] crossectional area of bar \n", + "k = 16.3 \t\t # [W/sq m per degree celsius] thermal conductivity of tube \n", + "hc = 1893.93\t\t # [W/sq m per degree celsius] contact coefficient\n", + "\t\t # the overall heat flow is subjected to three thermal resistances \n", + "\t\t # one conduction resistance for each bar\n", + "\t\t # contact resistance \n", + "#Calculation\n", + "\n", + "Rth = l/(k*A) # [degree celsius /W]\n", + "\n", + "\t\t # from table(2-2) the contact resistance is \n", + "Rc = 1/(hc*A) \t # [degree celsius /W]\n", + "Rt = 2*Rth+Rc\t\t # [degree celsius /W] total resistance\n", + "dt = 100\t\t # [degree celsius] temperature difference\n", + "q = dt/Rt \t\t # [W] overall heat flow\n", + "\n", + "#Results\n", + "\n", + "print \"Overall heat flow is:\",round(q,2),\"W\" \n", + "\n", + "\t\t# temperature drop across the contact is found by taking the ratio \n", + "\t\t# of the contact resistance to the total thermal resistance \n", + "\n", + "dt_c = (Rc/(2*Rth))*dt\t\t\t # [degree celsius]\n", + "\n", + "\n", + "print \"The temperature drop across the contact is:\",round(dt_c,2),\"degree celsius\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter3.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter3.ipynb new file mode 100644 index 00000000..7dd6ca44 --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter3.ipynb @@ -0,0 +1,228 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 Steady State Conduction Multiple Dimension" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by the pipe is 859.9 W\n" + ] + } + ], + "source": [ + "#Example Number 3.1\n", + "# Calculate the heat loss by the pipe\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.15 \t\t\t# [m] diameter of pipe\n", + "r = d/2 \t\t\t# [m] radius of pipe\n", + "L = 4 \t\t\t\t# [m] length of pipe\n", + "Tp = 75\t\t\t\t# [degree celsius] pipe wall temperature\n", + "Tes = 5 \t\t\t# [degree celsius] earth surface temperature\n", + "k = 0.8\t\t\t\t# [W/m per deg C] thermal conductivity of earth \n", + "D = 0.20 \t\t\t# [m] depth of pipe inside earth\n", + "\n", + "\t# We may calculate the shape factor for this situation using equation given in \ttable 3-1 \n", + "\t\n", + "\t# since D<3*r\n", + "#Calculation\n", + "import math\n", + "S = (2*math.pi*L)/math.acosh(D/r) \t# [m] shape factor\n", + "\t# the heat flow is calculated from \n", + "q = k*S*(Tp-Tes) \t\t\t# [W]\n", + "\n", + "#Result\n", + "print\"Heat lost by the pipe is\",round(q,1),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost through the walls is: 8.592 kW\n" + ] + } + ], + "source": [ + "#Example Number 3.2 \n", + "# Calculate heat loss through the walls\n", + "\n", + "# VARIABLE DECLARATION\n", + "\n", + "a = 0.5 \t # [m] length of side of cubical furnace\n", + "Ti = 500 \t # [degree celsius] inside furnace temperature\n", + "To = 50 \t # [degree celsius] outside temperature\n", + "k = 1.04 \t # [W/m per degree celsius] thermal conductivity of fireclay brick \n", + "t = 0.10 \t # [m] wall thickness\n", + "A = a*a \t # [square meter] area of one face \n", + "\t\t # we compute the total shape factor by adding the shape factors \t\t \t for the walls, edges and corners\n", + "\n", + "#Calculation\n", + "Sw = A/t\t # [m] shape factor for wall\n", + "Se = 0.54*a \t # [m] shape factor for edges\n", + "Sc = 0.15*t\t # [m] shape factor for corners\n", + "\n", + "\t\t # there are six wall sections, twelve edges and eight corners, so \t\t\tthe total shape factor S is\n", + "\n", + "S = 6*Sw+12*Se+8*Sc \t# [m]\n", + "\t\t \n", + "\t\t# the heat flow is calculated as \n", + "\n", + "q = k*S*(Ti-To) \t# [W]\n", + "\n", + "#Result\n", + "print\"Heat lost through the walls is:\",round(q/1000,3),\"kW\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.3" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by disk is: 198.46 W\n" + ] + } + ], + "source": [ + "#Example Number 3.3\n", + "# Calculate the heat loss by the disk\n", + "\n", + "# Variable declaration\n", + "\n", + "import math\n", + "d = 0.30 \t# [m] diameter of disk\n", + "r = d/2 \t# [m] radius of disk\n", + "Td = 95 \t# [degree celsius] disk temperature\n", + "Ts = 20 \t# [degree celsius] isothermal surface temperature\n", + "k = 2.1 \t# [W/m per degree celsius] thermal conductivity of medium \n", + "D = 1.0 \t# [m] depth of disk in a semi-infinite medium\n", + "\t# We have to calculate shape factor using relation given in table (3-1) \n", + "\t# We select the relation for the shape factor is for the case D/(2*r)>1\n", + "\n", + "#Calculation\n", + "S = (4*math.pi*r)/((math.pi/2)-math.atan(r/(2*D)))\t # [m] shape factor\n", + "\t# heat lost by the disk is \n", + "q = k*S*(Td-Ts) \t\t\t\t\t # [W]\n", + "\n", + "#Result\n", + "print\"Heat lost by disk is:\",round(q,2),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.4" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer between the disks is: 308.4 W\n" + ] + } + ], + "source": [ + "#Example Number 3.4 \n", + "#Calculate the heat transfer betwwen the disks\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.50\t # [m] diameter of both disk\n", + "r = d/2 \t # [m] radius of disk\n", + "Td1 = 80\t # [degree celsius] first disk temperature\n", + "Td2 = 20 \t # [degree celsius] second disk temperature\n", + "k = 2.3 \t # [W/m per degree celsius] thermal conductivity of medium \n", + "D = 1.5\t\t # [m] seperation of disk in a infinite medium\n", + "\t# We have to calculate shape factor using relation given in table (3-1) \n", + "\t# We select the relation for the shape factor is for the case D>5*r\n", + "#Calculation\n", + "import math\n", + "\n", + "S = (4*math.pi*r)/((math.pi/2)-math.atan(r/D)) # [m] shape factor\n", + "q = k*S*(Td1-Td2) # [W]\n", + "\n", + "#Result\n", + "print\"Heat transfer between the disks is:\",round(q,1),\"W\" \n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter4.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter4.ipynb new file mode 100644 index 00000000..3363aee9 --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter4.ipynb @@ -0,0 +1,669 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 Unsteady State Conduction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.1" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "hV/A/k is: 0.0024 which is less than 0.1,So by eq 4.5\n", + "hA/pcV is 0.000334448160535 per second\n", + "Time is 5818.0 seconds\n" + ] + } + ], + "source": [ + "#Example 4.1\n", + "# Calculate the time required for ball to attain a temperature of 150 deg C.\n", + "\n", + "#variables:\n", + "\n", + "c=460.0 \t# kJ/kg\n", + "k=35.0 \t#W/m\n", + "d=0.05 \t# diameter in meter\n", + "r=d/2 \t\t# radius in meter\n", + "h=10.0 \t#convection heat transfer coeff in W/sq meter\n", + "T=150.0 \t#Temperature in deg C\n", + "p=7800 \t#Density in kg/cubic meter\n", + "Ti=100.0 \t# T,infinity ..in celsius\n", + "To=450.0 \t# in Celsius\n", + "\n", + "#CALCULATION\n", + "\n", + "import math \n", + "A=(4*math.pi) * (r**2) \t\t\t #Arear in sq meter\n", + "V=(A*r/3) \t\t\t\t # Volume in cubic meters\n", + "c1=(h*(V/A))/k \t \t\t #it is less than 0.1 So,\n", + "print \"hV/A/k is:\",round(c1,4),\"which is less than 0.1,So by eq 4.5\"\n", + "c2=((h*A)/(p*c*V)) \t\t # assumed variable for easiness\n", + "\n", + "#RESULTS\n", + "\n", + "print \"hA/pcV is\",c2,\"per second\"\n", + "t=(-1/c2)*(math.log((T-Ti)/(To-Ti))) \n", + "print \"Time is\",round(t),\"seconds\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.2" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at depth of 0 .025 m after 30 s for case 1 is 118.5 degree celsius \n", + "Temperature at depth of 0 . 0 2 5 m after 30 second for case 2 is 79.3 degree celsius\n", + "Surface temperature after 30 second is: 199.4 degree celsius \n" + ] + } + ], + "source": [ + "#Example 4.2\n", + "#Calculate the temperature at a depth of 2.5 cm after 0.5 min for both cases\n", + "import math\n", + "\n", + "#variable declaration\n", + "\n", + "k = 45 \t\t\t\t# [W/m per deg C ] thermal conductivity of steel block\n", + "alpha = 1.4*10**( -5) ; \t# [ square meter / s ] constant\n", + "Tb = 35 \t\t\t# [ degree celsius ] block temperature\n", + "x = 0.025 \t\t\t# [m] depth at which temperature is calculated\n", + "t = 30 \t\t\t\t# [s]time after which temperature is to be calculated\n", + "To = 250 \t\t\t# [ degree celsius]\n", + "\n", + "\n", + "#CALCULATION & rESULT\n", + "\n", + "er=(x/(2*math.sqrt(alpha*t)))\t \t # error function short form\n", + "T_x_t = (To +(Tb -To)*( math.erf(er))) \n", + "\t\t\t\t\t # for the constant heat flux case B we \t\t\t\t\t\t make use of eq4.13 a\n", + "\t\t\t # since qo/A is given\n", + "\n", + "print \"Temperature at depth of 0 .025 m after 30 s for case 1 is\",round(T_x_t,1),\" degree celsius \"\n", + "q_by_A = 3.2*10**(5) ; \t\t\t\t# [W/ s qua r e meter ]\n", + "\n", + "T_x_t1 =(Tb +(2* q_by_A*math.sqrt(alpha*t/math.pi)*math.exp(-(x**2)/(4*alpha *t))/k) -(q_by_A *x*(1 - math.erf(er))/k)) \t\t# [ degree celsius ]\n", + "\n", + "print \"Temperature at depth of 0 . 0 2 5 m after 30 second for case 2 is\",round(T_x_t1,1),\"degree celsius\" \n", + "\t\t\t\t# for the constant heat flux case the surface \t\t\t\t temperature after 30 s would be evaluated with x= 0 \t\t\t\t in equation(4.13 a )\n", + "x = 0 \t\t\t\t\t\t\t\t# [m] at the surface\n", + "\n", + "T_x_o = Tb +(2* q_by_A *math.sqrt(alpha*t/math.pi) * math.exp (-(x **2)/(4*alpha*t))/k) -(q_by_A*x*(1 -math.erf (er))/k) \t\t# [degree celsius]\n", + "\n", + "print \"Surface temperature after 30 second is:\",round(T_x_o,1) ,\"degree celsius \" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.3" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ASSUMPTION: Negligible radiation effects\n", + "At x=0\n", + "At x=2mm ie 0.002 m\n", + "0.002\n", + "T =: 513.0 Celsius\n" + ] + } + ], + "source": [ + "#Example 4.3\n", + "#Estimate the temperature at the surface and at a depth of 2 mm after 2 sec\n", + " \t\t\t#Qo/A=10^7 J/sq meter\n", + "import math \t \t# import math file\n", + "\n", + "#variable declaration\n", + "\n", + "c1=10 \t\t# =Qo/A laser pulse in MJ/sq meter\n", + "c1=(c1*10**6) \t\t# convert in joule from MJ\n", + "p=7800 \t\t\t# Density in kg/cubic meter\n", + "c=460\t\t\t#J/kg\n", + "a=(0.44*(10**(-5))) \t# sq m/sec\n", + "t=2 \t\t\t# time in seconds\n", + "Ti=40.0 \t\t# initial temp in deg C\n", + "\n", + "print \"ASSUMPTION: Negligible radiation effects\"\n", + "#From eqq 4.13,:\n", + "print \"At x=0\"\n", + "\n", + "#calculation\n", + "To=Ti+(c1/(p*c*math.sqrt((math.pi)*a*t)))\n", + "print \"At x=2mm ie 0.002 m\"\n", + "x=0.002\n", + "print x\n", + "T=(Ti+(c1/(p*c*math.sqrt((math.pi)*a*t)))*(math.exp(-(x**2)/(4*a*t))))\n", + "\n", + "#RESULTS\n", + "print \"T =:\", round(T),\"Celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.5" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The iteration are listed below\n", + " h*math.sqrt(alpha*t)/k x/(2*sqrt(alpha*t)) (T_x_t-Ts)/(Te-Ts)\n", + "1000 0.708 0.069 0.41\n", + "3000 1.226 0.04 0.61\n", + "4000 1.415 0.035 0.68\n", + "Consequently the time required is approximately 3000 second\n" + ] + } + ], + "source": [ + "#Example 4.5:Sudden exposure of semi-infinite solid slab to convection\n", + "# Calculate time req for the temp to reach 120 deg C at depth of 4 cm\n", + "\n", + "#VARIABLE DECLARATION\n", + "\n", + "alpha = 8.4*10**(-5) \t\t# [square meter/s] constant\n", + "Ts = 200\t\t # [degree celsius] initial temperature of of slab\n", + "Te = 70\t\t\t\t# [degree celsius] environment temperature \n", + "k = 215 \t\t\t# [W/m deg C] heat transfer coefficient of slab\n", + "h = 525 \t\t\t# [W/sq m degree celsius] heat transfer coefficient \n", + "x = 0.04\t\t\t# [m] depth at which temperature is calculated\n", + "T_x_t = 120 \t\t\t# [degree celsius] temperature at depth 0.04 m\n", + "\n", + "\t# using eq 4-15 or figure (4-5) for solution of this problem\n", + "\t# by using figure it is easier to calculate it involves iterative method to \tsolve because time appeares in both the variables \n", + "\n", + "\t# h*sqrt(alpha*t)/k and x/(2*sqrt(alpha*t))\n", + "K = (T_x_t-Ts)/(Te-Ts) \n", + "\t# Seek the values of t such that the above value of K is equal to the value of \tK which comes out from graph\n", + "\n", + "# values of t and obtain other readings\n", + "\n", + "#CALCULATION & RESULT\n", + "\n", + "print \"The iteration are listed below\"\n", + "\t# at t = 1000s\n", + "import math\n", + "t = 1000 # [s] time\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + "\n", + "print \" h*math.sqrt(alpha*t)/k x/(2*sqrt(alpha*t)) (T_x_t-Ts)/(Te-Ts)\"\n", + "\n", + "print t,\" \",round(A,3),\" \",round(B,3),\" \",\"0.41\"\n", + "\n", + "t = 3000 \t\t\t# [s] time\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + "\n", + "print t,\" \",round(A,3),\" \",round(B,3),\" \",\"0.61\"\n", + "\n", + "t = 4000 \t\t\t# [s] time\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + " \n", + "print t,\" \",round(A,3),\" \",round(B,3),\" \",\"0.68\"\n", + "print \"Consequently the time required is approximately 3000 second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.6" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at a depth of 1.25 cm from one of faces after 1 min of exposure of plate to the environment is 147.7 degree celsius\n", + "Energy removed per unit area from the plate in this time is 6475950.0 J/square meter\n" + ] + } + ], + "source": [ + "#Example 4.6-aluminium plate suddenly exposed to convection\n", + "#Calculate Energy removed /Area from the plate at this time\n", + "\n", + "#VARIABLE DECLARATION\n", + "alpha = 8.4*10**(-5) \t\t\t# [square meter/s] constant\n", + "Ts = 200 \t\t\t\t# [deg celsius] initial temperature of plate\n", + "Te = 70 \t\t\t\t# [degree celsius] environment temperature \n", + "k = 215 \t\t\t\t# [W/m deg C] heat transfer coeff of plate\n", + "h = 525 \t\t\t\t#[W/sq m deg C] heat transfer coefficient \n", + "x = 0.0125 \t\t\t\t#[m] depth at which temperature is calculated\n", + "t = 60\t\t\t\t\t#[s]time after which temperature is calculated\n", + "L = 0.025 \t\t\t\t# [m] thickness of plate\n", + "\n", + "#CALCULATION\n", + "\n", + "theta_i = Ts-Te\t\t\t\t # [degree celsius]\n", + "\n", + "\t\t\t# then \n", + "Z = alpha*t/L**2 \n", + "X = k/(h*L) \n", + "x_by_L = x/L \n", + "\t\n", + "\t\t\t# from figure 4-7(page no.-144-145)\n", + "\n", + "theta_o_by_theta_i = 0.61 \n", + "theta_o = theta_o_by_theta_i*theta_i # [degree celsius]\n", + "\t\t\t# from figure 4-10(page no.-149) at x/L = 0.5,\n", + "theta_by_theta_o = 0.98 \n", + "theta = theta_by_theta_o*theta_o \t# [degree celsius]\n", + "T = Te+theta \t\t\t\t# [degree celsius]\n", + "\n", + "\t\t\t# using Figure 4-14(page no.-152). For this calculation we \t\t\trequire the following properties of aluminium:\n", + "\t\n", + "rho = 2700 \t\t\t\t# [kg/cubic meter]\n", + "C = 900 \t\t\t\t# [J/kg degree celsius]\n", + "\n", + "\t\t\t# for figure 4-14(page no.-152) we need \n", + "\n", + "V = h**2*alpha*t/(k**2) \n", + "B = h*L/k \n", + "\n", + "\t\t\t# from figure 4-14(page no.-152)\n", + "\n", + "Q_by_Qo = 0.41 \n", + "\n", + "\t\t\t# for unit area \n", + "Qo_by_A = rho*C*2*L*theta_i \t\t# [J/square meter]\n", + "\n", + "# Now, heat removed per unit surface area is \n", + "Q_by_A = Qo_by_A*Q_by_Qo \t\t# [J/square meter]\n", + "\n", + "#RESULTS\n", + "\n", + "print\"Temperature at a depth of 1.25 cm from one of faces after 1 min of exposure of plate to the environment is\",round(T,1),\" degree celsius\" \n", + "\n", + "print\"Energy removed per unit area from the plate in this time is\",Q_by_A,\" J/square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.7" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at a radius of 1.25 cm is 118.4 degree celsius\n", + "Heat lost per unit length 1 minute after the cylinder is exposed to the environment is 403174.0 J/m\n" + ] + } + ], + "source": [ + "#Example4.7-Long cylinder suddenly exposed to convection \n", + "\n", + "#VARIABLE DECLARATION\n", + "\n", + "d = 0.05 \t\t\t# [m] diameter of cylinder\n", + "Ti = 200 \t\t\t# [degree C] initial temperature of aluminium cylinder\n", + "Tinf = 70 \t\t\t# [degree celsius] temperature of environment\n", + "h = 525 \t\t\t# [W/sq m degree celsius] heat transfer coefficient\n", + "\n", + "#CALCULATION\n", + "\n", + "# we have\n", + "theta_i = Ti-Tinf \t\t# [degree celsius]\n", + "alpha = 8.4*10**(-5) \t\t# [square meter/s]\n", + "ro = d/2 \t\t\t# [m]\n", + "t = 60 \t\t\t\t# [s]\n", + "k = 215\t\t # [W/m degree celsius]\n", + "r = 0.0125 \t\t\t# [m]\n", + "rho = 2700 \t\t\t# [kg/cubic meter]\n", + "C = 900 \t\t\t# [J/kg degree celsius]\n", + "\n", + "\t# we compute\n", + "\n", + "Z = alpha*t/ro**2 \n", + "X = k/(h*ro) \n", + "r_by_ro = r/ro \n", + "\n", + "\n", + "\t# from figure 4-8(page no.-146)\n", + "theta_o_by_theta_i = 0.38 \n", + "\n", + "\t# from figure 4-11(page no.-150) at r/ro = 0.5\n", + "theta_by_theta_o = 0.98 \n", + "\n", + "\t# so that \n", + "theta_by_theta_i = theta_o_by_theta_i*theta_by_theta_o \n", + "theta = theta_i*theta_by_theta_i \t# [degree celsius]\n", + "T = Tinf+theta \t\t\t\t# [degree celsius]\n", + "\n", + "\t# to compute the heat lost, we determine\n", + "V = h**2*alpha*t/k**2 \n", + "B = h*ro/k \n", + "\n", + "\t# from figure 4-15(page no.-153)\n", + "Q_by_Qo = 0.65 \n", + "\n", + "\t# for unit length\n", + "import math\n", + "Qo_by_L = rho*C*math.pi*ro**2*theta_i \t# [J/m]\n", + "\n", + "\t#actual heat lost per unit length is \n", + "Q_by_L = Qo_by_L*Q_by_Qo \t\t# [J/m]\n", + "\n", + "\n", + "\n", + "#RESULTS\n", + "\n", + "print\"Temperature at a radius of 1.25 cm is\",round(T,1),\" degree celsius\" \n", + "print\"Heat lost per unit length 1 minute after the cylinder is exposed to the environment is\",round(Q_by_L),\"J/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.8" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The temperature at the axis is: 117.6 degree celsius\n", + "The temperature at the surface is 116.2 degree celsius\n" + ] + } + ], + "source": [ + "#Example 4.8-\n", + "#Calculate temperature at axis and surface of th cylinder 10 cm from end 1 min after exposure\n", + "\n", + "#VARIABLES DECLARATION\n", + "import math\n", + "\n", + "d = 0.05 \t\t# [m] diameter of aluminium cylinder \n", + "Ti = 200 \t\t# [degree celsius] initial temperature of of cylinder\n", + "Te = 70 \t\t# [degree celsius] environment temperature\n", + "k = 215 \t\t# [W/m degree celsius] heat transfer coefficient of plate\n", + "h = 525 \t\t# [W/sq m degree celsius] convection heat transfer coefficient \n", + "alpha = 8.4*10**(-5) # [square meter/s] constant\n", + "x = 0.10 \t\t# [m] distance at which temperature is calculated from end\n", + "t = 60 \t\t\t# [s] time after which temperature is measured\n", + "\t# so that the parameters for use with figure(4-5)\n", + "\n", + "#CALCULATION\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + "# from figure (4-5)\n", + "z = 1-0.036 \n", + "S_of_X = z \n", + "\t# for the infinite cylinder we seek both the axis- and surface-temperature \tratios.\n", + "\t# the parameters for use with fig(4-8) are \n", + "r_o = d/2 \t\t\t# [m] radius of aluminium cylinder \n", + "r = d/2 \t\t\t# [m] for surface temperature ratio\n", + "C = k/(h*r_o) \n", + "D = (alpha*t/r_o**(2)) \n", + "y = 0.38 \n", + "\n", + "\t# this is the axis temperature ratio.\n", + "\t# to find the surface-temperature ratio,we enter figure (4-11),using \n", + "R = r/r_o \n", + "u = 0.97 \n", + "# thus \n", + "w = y \t\t\t# at r = 0\n", + "v = y*u \t\t# at r = r_o\n", + "C_of_O_axis = w \t# at r = 0\n", + "C_of_O_r_o = v \t\t# at r = r_o\n", + "\t# combining the solutions for the semi-infinite slab and infinite cylinder, we \thave \n", + "t = S_of_X*C_of_O_axis \n", + "s = S_of_X*C_of_O_r_o \n", + "\t# the corresponding temperatures are \n", + "T_axis = Te+t*(Ti-Te) \n", + "T_r_o = Te+s*(Ti-Te) \n", + "\n", + "\n", + "#RESULTS\n", + "\n", + "print\"The temperature at the axis is:\",round(T_axis,1),\" degree celsius\"\n", + "print\"The temperature at the surface is\",round(T_r_o,1),\" degree celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.9" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at a radial position of 0.0125 m and a distance of 0.00625m from one end of cylinder 60 second after exposure to environment is 104.5 degree celsius\n" + ] + } + ], + "source": [ + "#Example 4.9\n", + "# finite length cylinder suddenly exposed to convection \n", + "\n", + "#Variable declaration\n", + "\n", + "d = 0.05 \t\t\t# [m] diameter of aluminium cylinder \n", + "Ti = 200 \t\t\t# [degree celsius] initial temperature of of cylinder\n", + "Te = 70\t\t\t # [degree celsius] environment temperature\n", + "k = 215 \t\t\t# [W/m deg celsius] heat transfer coefficient of plate\n", + "h = 525 \t\t\t# [W/sq m deg C] convection heat transfer coefficient \n", + "alpha = 8.4*10**(-5) \t\t#[square meter/s] constant\n", + "x1 = 0.00625 \t\t\t#[m]dist at which temperature is calculated from end\n", + "t = 60 \t\t\t\t# [s] time after which temperature is measured\n", + "r = 0.0125 \t\t\t# [m] radius at which temperature is calculated\n", + "\t# to solve this problem we combine the solutions from heisler charts for an \t\tinfinite cylinder and an infinite plate in accordance with the combination \t\tshown in fig (4-18f)\n", + "\t# for the infinite plate problem \n", + "L = 0.05\t\t\t # [m]\n", + "\n", + "#CALCULATION\n", + "\n", + "\t# the x position is measured fromthe center of the plate so that\n", + "x = L-x1 \t\t\t# [m]\n", + "A = k/(h*L) \n", + "B = (alpha*t/L**(2)) \n", + "\t# from figures (4-17) and (4-10) respectively\n", + "thetha_o_by_i = 0.75 \n", + "thetha_by_i = 0.95 \n", + "\t# so that\n", + "thetha_by_i_plate = thetha_o_by_i*thetha_by_i \n", + "\t# for the cylinder \n", + "r_o = d/2 \t\t\t# [m] radius of the cylinder\n", + "R = r/r_o \n", + "C = k/(h*r_o) \n", + "D = (alpha*t/r_o**(2)) \n", + "\t# and from figures (4-8) and (4-11), respectively\n", + "thetha_o_by_i_cyl = 0.38 \n", + "thetha_by_o = 0.98 \n", + "\t# so that\n", + "thetha_by_i_cyl = thetha_o_by_i_cyl*thetha_by_o \n", + "\t# combining the solutions for the plate and cylinder gives\n", + "thetha_by_i_short_cyl = thetha_by_i_plate*thetha_by_i_cyl \n", + "\t#thus\n", + "T = Te+thetha_by_i_short_cyl*(Ti-Te) \n", + "\n", + "#RESULTS\n", + "print\"Temperature at a radial position of 0.0125 m and a distance of 0.00625m from one end of cylinder 60 second after exposure to environment is\",round(T,1),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.10" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The actual heat loss in the 1-minute: 40.2 kJ\n" + ] + } + ], + "source": [ + "#Example 4.10\n", + "# heat loss for finite-length cylinder \n", + "\n", + "#Variable declaration\n", + "d = 0.05 \t\t# [m] diameter of aluminium cylinder\n", + "l = 0.1 \t\t# [m] length of aluminium cylinder \n", + "Ti = 200\t\t# [degree celsius] initial temperature of of cylinder\n", + "Te = 70 \t\t# [degree celsius] environment temperature\n", + "k = 215 \t\t# [W/m degree celsius] heat transfer coefficient of plate\n", + "h = 525 \t\t# [W/sq m deg celsius] convection heat transfer coefficient \n", + "alpha = 8.4*10**(-5) \t#[square meter/s] constant\n", + "x1 = 0.00625 \t\t#[m] distance at which temperature is calculated from end\n", + "t = 60 \t\t\t#[s]time after which temperature is measured\n", + "r = 0.0125 #[m] radius at which temperature is calculated\n", + "\n", + "\n", + "#CALCULATION\n", + "\t\n", + "\t# calculate the dimensionless heat-loss ratio for the infinite plate and \tinfinite cylinder which make up the multidimensional body\n", + "\t# for the plate we have \n", + "L = 0.05 \t\t# [m]\n", + "A = h*L/k \n", + "B = h**(2)*alpha*t/k**(2) \n", + "\t# from figure (4-14), for the plate, we read \n", + "Q_by_Q_o_plate = 0.22 \n", + "\n", + "\t# for the cylinder \n", + "r_o = 0.025 \t\t# [m]\n", + "\t# so we calculate \n", + "C = h*r_o/k \n", + "\t# and from figure(4-15) we have \n", + "Q_by_Q_o_cyl = 0.55 \n", + "\n", + "\t# the two heat ratios may be inserted in equation(4-22) to give \n", + "Q_by_Q_o_tot = Q_by_Q_o_plate+Q_by_Q_o_cyl*(1-Q_by_Q_o_plate) \n", + "c = 896 \t\t# [J/kg degree celsius] specific heat of aluminium\n", + "rho = 2707 \t\t# [kg/cubic meter] density of aluminium\n", + "\n", + "import math\n", + "\n", + "V = math.pi*r_o**(2)*l \t# [cubic meter]\n", + "Qo = rho*c*V*(Ti-Te) \t# [J]\n", + "Q = Qo*Q_by_Q_o_tot \t# [J] the actual heat loss in the 1-minute \n", + "\n", + "#RESULTS\n", + "\n", + "print \"The actual heat loss in the 1-minute:\",round((Q/1000),1),\" kJ\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter5.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter5.ipynb new file mode 100644 index 00000000..ec39bab5 --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter5.ipynb @@ -0,0 +1,552 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 Principles of Convection" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the increase in static pressure between sections 1 and 2 is: 61.88 kPa\n" + ] + } + ], + "source": [ + "#Example 5.1\n", + "# water flow in a diffuser \n", + "\n", + "#VARIABLE DECLARATION\n", + "Tw = 20 \t\t\t# [degree celcius] water temperature \n", + "m_dot = 8 \t\t\t# [kg/s] water flow rate \n", + "d1 = 0.03 \t\t\t#[m] diameter at section 1\n", + "d2 = 0.07 \t\t\t# [m] diameter at section 2\n", + "\n", + "#CALCULATION\n", + "import math\n", + "A1 = math.pi*d1**(2)/4 \t\t# [square meter] cross-sectional area at section 1\n", + "A2 = math.pi*d2**(2)/4 \t\t# [square meter] cross-sectional area at section 2\n", + "gc = 1 \t\t\t\t# [m/s**(2)] acceleration due to gravity\n", + "rho = 1000 \t\t\t# [kg/cubic m] density of water at 20 degree celcius\n", + "\n", + "\t# calculate the velocities from the mass-continuity relation\n", + "u1 = m_dot/(rho*A1) \t\t# [m/s]\n", + "u2 = m_dot/(rho*A2) \t\t# [m/s]\n", + "\t# the pressure difference is obtained by Bernoulli equation(5-7a)\n", + "p2_minus_p1 = rho*(u1**(2)-u2**(2))/(2*gc)\t # [Pa] \n", + "\n", + "#RESULTS\n", + "print\"the increase in static pressure between sections 1 and 2 is:\",round(p2_minus_p1/1000,2),\" kPa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The static temperature is: 529.0 K\n", + "Static pressure is: 0.5 MPa\n", + "Mach number is: 0.651\n" + ] + } + ], + "source": [ + "#Example Number 5.2\n", + "# isentropic expansion of air \n", + "\n", + "# Variable declaration\n", + "\n", + "Ta = 300.0+273.0\t \t\t# [K] air temperature\n", + "Pa = 0.7 \t\t\t\t# [MPa] pressure of air\n", + "u2 = 300 \t\t\t\t# [m/s] final velocity\n", + "gc = 1 \t\t\t\t\t# [m/s^(2)] acceleration due to gravity\n", + "Y = 1.4 \t\t\t\t# gama value for air \n", + "Cp = 1005 \t\t\t\t# [J/kg degree celsius]\n", + "\t#the initial velocity is small and the process is adiabatic. in terms of \t\ttemperature \n", + "\n", + "\n", + "#Calculation\n", + "\n", + "T2 = Ta-u2**(2)/(2*gc*Cp) \n", + "\n", + "#Result\n", + "print \"The static temperature is:\",round(T2,1),\"K\" \n", + "\n", + "\t# we may calculate the pressure difference from the isentropic relation \n", + "\n", + "p2 = Pa*((T2/Ta)**(Y/(Y-1))) \n", + "\n", + "print \"Static pressure is:\",round(p2,1),\"MPa\"\n", + "\n", + "\t# the velocity of sound at condition 2 is \n", + "a2 = (20.045*(T2**(0.5))) \t\t# [m/s] \n", + "\t#so that the mach no. is \n", + "M2 = u2/a2 \n", + "print \"Mach number is:\",round(M2,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat transfered in first case of the plate is 81.2 W\n", + "and the heat transfered in second case of the plate is: 114.8 W\n" + ] + } + ], + "source": [ + "#Example Number 5.4\n", + "#Calculate the heat transfereed in first 20 cm of the plate and the first 40 cm of the plate\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# total heat transfer over a certain length of the plate is desired, so we \t\twish to calculate average heat transfer coefficients. \n", + "\t# for this purpose we use equations (5-44) and (5-45), evaluating the \t\tproperties at the film temperature :\n", + "Tp = 60+273.15 \t\t\t\t# [K] plate temperature \n", + "Ta = 27+273.15 \t\t\t\t# [K] air temperature\n", + "Tf = (Tp+Ta)/2 \t\t\t\t# [K]\n", + "u = 2 \t\t\t\t\t# [m/s] air velocity\n", + "\n", + "\t# from appendix A the properties are \n", + "\n", + "v = 17.36*(10**(-6)) \t\t\t# [square meter/s] kinematic viscosity\n", + "x1 = 0.2 \t\t\t\t# [m] distance from the leading edge of plate\n", + "x2 = 0.4 \t\t\t\t# [m] distance from the leading edge of plate\n", + "k = 0.02749 \t\t\t\t# [W/m K] heat transfer coefficient\n", + "Pr = 0.7 \t\t\t\t# prandtl number\n", + "Cp = 1006 \t\t\t\t# [J/kg K]\n", + "\n", + "\t# at x = 0.2m\n", + "\n", + "#Calculation\n", + "\n", + "Re_x1 =(u*x1/v) \t\t\t\t# reynolds number\n", + "Nu_x1 = 0.332*(Re_x1**(0.5))*(Pr**(0.333)) \t# nusselt number\n", + "hx1 = Nu_x1*k/x1 \t\t\t\t# [W/square meter K] \n", + "\n", + "\t# the average value of the heat transfer coefficient is twice this value, or\n", + "\n", + "h_bar1 = 2*hx1 \t\t\t\t\t# [W/square meter K] \n", + "\n", + "\t# the heat flow is \n", + "\n", + "A1 = x1*1 \t\t\t\t\t# [square meter] area for unit depth\n", + "q1 = h_bar1*A1*(Tp-Ta) \t\t\t\t# [W]\n", + "\n", + "\t# at x = 0.4m\n", + "\n", + "Re_x2 = u*x2/v \t\t\t\t\t# reynolds number\n", + "Nu_x2 = 0.332*Re_x2**(0.5)*Pr**(0.333) \t\t# nusselt number\n", + "hx2 = Nu_x2*k/x2 \t\t\t\t# [W/square meter K] \n", + "\n", + "\t# the average value of the heat transfer coefficient is twice this value, or\n", + "\n", + "h_bar2 = 2*hx2 \t\t\t\t\t# [W/square meter K] \n", + "\t# the heat flow is \n", + "A2 = x2*1 \t\t\t\t\t# [square meter] area for unit depth\n", + "q2 = h_bar2*A2*(Tp-Ta) \t\t\t\t# [W] \n", + "\n", + "#Result\n", + " \n", + "print\"The heat transfered in first case of the plate is\",round(q1,2),\"W\"\n", + "print\"and the heat transfered in second case of the plate is:\",round(q2,1),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average temperature difference along the plate is: 241.0 degree celsius\n", + "Temperature difference at the trailing edge is: 365.3 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 5.5\n", + "# Calculate the av temperature difference along the plate & Temperature diff at the trailing edge\n", + "\n", + "#Variable declaration\n", + "\n", + "u = 5 \t\t\t\t\t# [m/s] air velocity\n", + "l = 0.6 \t\t\t\t# [m] plate length\n", + "Ta = 27+273.15 \t\t\t\t# [K] temperature of airstream\n", + "\n", + "\t# properties should be evaluated at the film temperature, but we do not know \tthe plate temperature so for an initial calculation we take the properties at \tthe free-stream conditions of\n", + "\n", + "v = 15.69*10**(-6)\t \t\t#[square meter/s] kinematic viscosity\n", + "k = 0.02624 \t\t\t\t#[W/m deg celsius] heat transfer coefficient\n", + "Pr = 0.7 \t\t\t\t# prandtl number\n", + "Re_l = l*u/v \t\t\t\t# reynolds number\n", + "P = 1000 \t\t\t\t# [W] power of heater\n", + "qw = P/l**(2) \t\t\t\t# [W/square meter] heat flux per unit area \n", + "\n", + "\t# from equation (5-50) the average temperature difference is \n", + "\n", + "#Calculation\n", + "\n", + "Tw_minus_Tinf_bar = qw*l/(0.6795*k*(Re_l)**(.5)*(Pr)**(0.333)) \t # [degree celsius]\n", + "\n", + "\t# now, we go back and evaluate properties at \n", + "Tf = (Tw_minus_Tinf_bar+Ta+Ta)/2 \t# [degree celsius]\n", + "\n", + "\t# and obtain\n", + "\n", + "v1 = 28.22*10**(-6) \t\t\t# [square meter/s] kinematic viscosity\n", + "k1 = 0.035 \t\t\t\t# [W/m deg celsius] heat transfer coefficient\n", + "Pr1 = 0.687 \t\t\t\t# prandtl number\n", + "Re_l1 = l*u/v1 \t\t\t\t# reynolds number\n", + "Tw_minus_Tinf_bar1 = qw*l/(0.6795*k1*(Re_l1)**(0.5)*(Pr1)**(0.333)) #[degree celsius]\n", + "\n", + "\t# at the end of the plate(x = l = 0.6m) the temperature difference is obtained \tfrom equation (5-48) and (5-50) with the constant of 0.453\n", + "\n", + "Tw_minus_Tinf_x_equal_l = Tw_minus_Tinf_bar1*0.6795/0.453 \t# [degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"Average temperature difference along the plate is:\",round(Tw_minus_Tinf_bar),\" degree celsius\"\n", + "print \"Temperature difference at the trailing edge is:\",round(Tw_minus_Tinf_x_equal_l,1),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average value of the convection coefficient is 219.1 W/sq meter degree celsius\n", + " and the heat lost by the plate is 350.6 W\n" + ] + } + ], + "source": [ + "#Example Number 5.7\n", + "# Calculate the heat lost by the plate\n", + "\n", + "# Variable declaration\n", + "\n", + "u = 1.2 \t\t\t# [m/s] oil velocity\n", + "l = 0.2 \t\t\t# [m] plate length as well as width (square) \n", + "To = 20+273.15 \t\t\t# [K] temperature of engine oil\n", + "Tu = 60+273.15 \t\t\t# [K] uniform temperature of plate \n", + "\t# First we evaluate the film temperature \n", + "T = (To+Tu)/2 \t\t\t# [K]\n", + "\t# and obtain the properties of engine oil are \n", + "rho = 876 \t\t\t# [kg/cubic meter] density of oil\n", + "v = 0.00024 \t\t\t# [square meter/s] kinematic viscosity\n", + "k = 0.144 \t\t\t# [W/m degree celsius] heat transfer coefficient\n", + "Pr = 2870 \t\t\t# prandtl number\n", + "\t# at the trailing edge of the plate the reynolds number is \n", + "\n", + "#Calculation\n", + "\n", + "Re = l*u/v \t\t\t# reynolds number\n", + "\n", + "\n", + "\n", + "\t# because the prandtl no. is so large we will employ equation(5-51) for the \t\tsolution. \n", + "\n", + "\t# we see that hx varies with x in the same fashion as in equation(5-44) , i.e. \thx is inversely proportional to the square root of x ,\n", + "\t# so that we get the same solution as in equation(5-45) for the average heat \t\ttransfer coefficient. \n", + "\n", + "\t# evaluating equation(5-51) at x = 0.2m gives\n", + "\n", + "Nux = (0.3387*(Re**(1.0/2.0))*(Pr**(1.0/3.0)))/((1+(0.0468/Pr)**(2.0/3.0))**(1.0/4.0))\n", + "\n", + "\n", + "hx = Nux*k/l \t\t\t# [W/sq m degree celsius] heat transfer coefficient\n", + "\n", + "\t# the average value of the convection coefficient is \n", + "\n", + "h = 2*hx \t\t\t# [W/square meter degree celsius] \n", + "\n", + "\t# so that total heat transfer is \n", + "\n", + "A = l**(2) \t\t\t# [square meter] area of the plate \n", + "q = h*A*(Tu-To) \t\t#[W] \n", + "\n", + "print \"Average value of the convection coefficient is\",round(h,1),\"W/sq meter degree celsius\"\n", + "print \" and the heat lost by the plate is\",round(q,1),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.8" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drag force exerted on the first 0.4 m of the plate is 5.45 mN\n" + ] + } + ], + "source": [ + "#Example Number 5.8\n", + "# Compute the drag force on the first 40 cm of the plate \n", + "\n", + "# Variable declaration\n", + "\n", + "\n", + "\t# data is used from example 5.4 \n", + "\t# we use equation (5-56) to compute the friction coefficient and then \t\tcalculate the drag force .\n", + "\t# an average friction coefficient is desired, so st_bar*pr**(2/3) = Cf_bar/2\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "x = 0.4\t\t\t\t#[m] drag force is computed on first 0.4 m of the \t\t\t plate \n", + "R = 287 \t\t\t# []\n", + "Tf = 316.5 \t\t\t# [K]\n", + "u = 2 \t\t\t\t# [m/s] air velocity\n", + "Cp = 1006 \t\t\t# [J/kg K]\n", + "Pr = 0.7 \t\t\t# prandtl no.\n", + "rho = p/(R*Tf) \t\t\t# [kg/cubic meter] density at 316.5 K \n", + "h_bar = 8.698 \t\t\t# [W/square meter K] heat transfer coefficient\n", + "\n", + "\n", + "#Calculation\n", + "\t# for the 0.4m length\n", + "\n", + "st_bar = h_bar/(rho*Cp*u) \n", + "\n", + "\t# then from equation (5-56)\n", + "\n", + "Cf_bar = st_bar*Pr**(2.0/3.0)*2 \n", + "\n", + "\t# the average shear stress at the wall is computed from equation(5-52)\n", + "\n", + "tau_w_bar = Cf_bar*rho*u**(2)/2 \t# [N/square meter]\n", + "A = x*1 \t\t\t\t# [square meter] area per unit length \n", + "\n", + "\t# the drag force is the product of this shear stress and the area,\n", + "\n", + "D = tau_w_bar*A \t\t\t# [N] \n", + "\n", + "#Result\n", + "\n", + "print \"Drag force exerted on the first 0.4 m of the plate is\",round(D*1000,2),\"mN\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.9" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nul_bar is 2175.0\n", + "Heat transfer from plate is 2369.0 W\n" + ] + } + ], + "source": [ + "#Example Number 5.9\n", + "# Calculate the heat transfer from the plate\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101320.0 \t\t\t\t# [Pa] pressure of air\n", + "R = 287.0 \t\t\t\t# []\n", + "Ta = 20+273 \t\t\t\t# [K] temperature of air \n", + "u = 35 \t\t\t\t\t# [m/s] air velocity\n", + "L = 0.75 \t\t\t\t# [m] length of plate \n", + "Tp = 60+273 \t\t\t\t# [K] plate temperature \n", + "\n", + "\n", + "\t\t# we evaluate properties at the film temperature \n", + "\n", + "#Calculations\n", + "\n", + "Tf = (Ta+Tp)/2 \t\t\t\t# [K]\n", + "\n", + "\n", + "rho = (p/(R*Tf)) \t\t\t# [kg/cubic meter]\n", + "\n", + "\n", + "mu = 1.906*(10**(-5)) \t\t\t# [kg/m s] viscosity \n", + "k = 0.02723 \t\t\t\t# [W/m degree celsius]\n", + "Cp = 1007 \t\t\t\t# [J/kg K]\n", + "Pr = 0.7 \t\t\t\t# prandtl no.\n", + "\n", + "\t\t# the reynolds number is \n", + "\n", + "Rel = (rho*u*L)/mu \n", + "Rel=round(Rel)\n", + "\n", + "\t\t# and the boundary layer is turbulent because the reynolds number is \t\t\tgreater than 5*10**(5).\n", + "\t\t# therefore, we use equation(5-85) to calculate the average heat \t\t\ttransfer over the plate:\n", + "\n", + "Nul_bar = (Pr**(1.0/3.0))*(0.037*(Rel**(0.8))-871) \n", + "\n", + "print \"Nul_bar is\",round(Nul_bar)\n", + "A = L*1 \t\t\t\t# [square meter] area of plate per unit depth\n", + "h_bar = Nul_bar*k/L \t\t\t# [W/square meter degree celsius]\n", + "q = h_bar*A*(Tp-Ta) \t\t\t# [W] heat transfer from plate\n", + "\n", + "#Result\n", + "\n", + "print \"Heat transfer from plate is\",round(q),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.10" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Turbulent-boundary-layer thickness at the end of the plate from the leading edge of the plate is 16.5 mm\n", + "Turbulent-boundary-layer thickness at the end of the plate from the transition point at Re_crit = 5*10**(5) is 9.9 mm\n" + ] + } + ], + "source": [ + "#Example Number 5.10\n", + "# Calculate turbulent-boundary-layer thickness at the end of plate \n", + "\n", + "# Variable declaration\n", + "\n", + "\t\t# we have to use the data from example 5.8 and 5.9\n", + "Rel = 1.553*10**6 \t\t\t\t\t# from previous example\n", + "L = 0.75 \t\t\t\t\t\t# [m] length of plate\n", + "\t\t# it is a simple matter to insert this value in equations(5-91) and \t\t(5-95) \talong with\n", + "x = L \t\t\t\t\t\t\t# [m]\n", + "\t\t# turbulent-boundary-layer thickness are\n", + "\t\t# part a. from the leading edge of the plate \n", + "\n", + "#Calculation\n", + "\n", + "del_a = x*0.381*Rel**(-0.2) \t\t\t\t# [m] \n", + "\t\t# part b from the transition point at Recrit = 5*10**(5)\n", + "\n", + "del_b = x*0.381*Rel**(-0.2)-10256*Rel**(-1) \t\t# [m]\n", + "\n", + "#Result\n", + "\n", + "print \"Turbulent-boundary-layer thickness at the end of the plate from the leading edge of the plate is\",round(del_a*1000,1),\"mm\" \n", + "print \"Turbulent-boundary-layer thickness at the end of the plate from the transition point at Re_crit = 5*10**(5) is\",round(del_b*1000,1),\" mm\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb new file mode 100644 index 00000000..92b1940b --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb @@ -0,0 +1,734 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 Empirical and Practical Relations for Forced Convection Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.1" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 14749.0\n", + "so that the flow is turbulent\n", + "Heat transfer per unit length is 103.5 W/m\n", + "Bulk temperature increase over the length of 3 m on tube is 40.04 degree C\n" + ] + } + ], + "source": [ + "#Example Number 6.1\n", + "# turbulent heat transfer in a tube \n", + "# Variable declaration\n", + "\n", + "p = 2*101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 200+273.15\t\t\t# [K] temperature of air \n", + "d = 0.0254 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 10 \t\t\t\t# [m/s] velocity of air\n", + "dT = 20 \t\t\t# [deg C] temperature difference between wall and air \n", + "\n", + "\t# we first calculate the reynolds number to determine if the flow is laminar \t\tor turbulent, and then select the appropriate empirical correlation to \t\t\tcalculate the heat transfer \n", + "\n", + "\t# the properties of air at a bulk temperature of 473 K are\n", + "\n", + "#Calculaiton\n", + "\n", + "rho = p/(R*Ta) \t\t\t# [kg/cubic meter] density of gas\n", + "mu = 2.57*10**(-5)\t\t# [kg/m s] viscosity \n", + "k = 0.0386 \t\t\t# [W/m degree celsius]\n", + "Cp = 1025 \t\t\t# [J/kg K]\n", + "Pr = 0.681 \t\t\t# prandtl no.\n", + "Re_d = rho*u*d/mu \t\t# reynolds number\n", + "\n", + "print \"Reynolds number is\",round(Re_d) \n", + "print\"so that the flow is turbulent\"\n", + " \n", + "\t# we therefore use equation (6-4a) to calculate the heat transfer coefficient\n", + "\n", + "Nu_d = 0.023*Re_d**(0.8)*Pr**(0.4) \t# nusselt no.\n", + "h = Nu_d*k/d \t\t\t\t# [W/m**2 deg C] heat transfer coefficient\n", + "\n", + "\t# the heat transfer per unit length is then\n", + "\n", + "import math\n", + "q_by_L = h*math.pi*d*(dT) \t\t# [W/m]\n", + "L = 3 \t\t\t\t\t# [m] \n", + "\t# we can now make an energy balance to calculate the increase in bulk \t\ttemperature in a 3 m length of tube :\n", + "\t# q = m_dot*Cp*dT_b = L*(q_byL)\n", + "m_dot = rho*u*math.pi*d**(2)/4 \t\t# [kg/s] gas flow rate\n", + "\t# so that we insert the numerical values in the energy balance to obtain \n", + "dT_b = L*q_by_L/(m_dot*Cp) \t\t# [degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print\"Heat transfer per unit length is\",round(q_by_L,1),\" W/m\"\n", + "print\"Bulk temperature increase over the length of 3 m on tube is\",round(dT_b,2),\" degree C\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.3" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 823.0\n", + "so that the flow is laminar\n", + "\n", + "Total heat transfer is 3.49 W\n", + "\n", + "Exit wall temperature is 161.0 degree celsius\n", + "\n", + "Heat transfer coefficient is 26.45 W/sq meter degree C\n" + ] + } + ], + "source": [ + "#Example Number 6.3\n", + "# heating of air in laminar tube flow for constant heat flux\n", + "\n", + "#Variable declaration\n", + "\n", + "import math\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 27 \t\t\t# [degree celsius] temperature of air \n", + "d = 0.005 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 3 \t\t\t\t# [m/s] velocity of air\n", + "L = 0.1 \t\t\t# [m] length of tube\n", + "Tb = 77 \t\t\t# [degree celsius] exit bulk temperature \n", + "\t# we first must evaluate the flow regime and do so by taking properties at the \taverage bulk temperature \n", + "Tb_bar = (Ta+Tb)/2 \t\t# [degree celsius]\n", + "v = 18.22*10**(-6) \t\t# [square meter/s] kinematic viscosity\n", + "k = 0.02814 \t\t\t# [W/m degree celsius]\n", + "Cp = 1006 \t\t\t# [J/kg K]\n", + "Pr = 0.703 \t\t\t# prandtl no.\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "print \"Reynolds number is\",round(Re_d) \n", + "print\"so that the flow is laminar\\n\" \n", + "\t\n", + "\t#the tube length is short, so we expect a thermal entrance effect and shall \t\tconsult figure(6-5)\n", + "\t# the inverse Graetz number is computed as \n", + "Gz_inverse = L/(Re_d*Pr*d) \n", + "\t# therefore, for qw = constant, we obtain the nusselt number at exit from \t\tfigure (6-5) as\n", + "Nu = 4.7 \n", + "\t# the total heat transfer is obtained in terms of the overall energy balance \n", + "\t# at entrance \n", + "rho = 1.1774 \t\t\t # [kg/cubic meter] density\n", + "\t# mass flow is\n", + "m_dot = rho*math.pi*d**(2)*u/4 \t # [kg/s]\n", + "q = m_dot*Cp*(Tb-Ta) \t\t # [W]\n", + "\t# thus we may find the heat transfer without the actually determining wall \t\ttemperatures or values of h. However, to determine Tw we must compute qw for \t\tinsertion in equation(b). we have\n", + "qw = q/(math.pi*d*L) \t\t # [W]\n", + "\t# now\n", + "Tw = Tb+(qw*d/(Nu*k)) \t\t # [degree celsius]\n", + "\t# and the heat transfer coefficient is\n", + "h = qw/(Tw-Tb) \t\t\t # [W/square meter degree celsius]\n", + "print \"Total heat transfer is\",round(q,2),\"W\"\n", + "print \"\\nExit wall temperature is\",round(Tw),\" degree celsius\" \n", + "print \"\\nHeat transfer coefficient is\",round(h,2),\" W/sq meter degree C\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 823.0\n", + "so that the flow is laminar\n", + "Exit wall temperature is 128.66 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 6.4\n", + "# heating of air with isothermal tube wall\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 27\t\t\t\t# [degree celsius] temperature of air \n", + "d = 0.005 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 3 \t\t\t\t# [m/s] velocity of air\n", + "L = 0.1 \t\t\t# [m] length of tube\n", + "Tb = 77 \t\t\t# [degree celsius] exit bulk temperature \n", + "\n", + "\t# we first must evaluate the flow regime and do so by taking properties at the \taverage bulk temperature \n", + "\n", + "Tb_bar = (Ta+Tb)/2 \t\t# [degree celsius]\n", + "v = 18.22*10**(-6) \t\t# [square meter/s] kinematic viscosity\n", + "k = 0.02814 \t\t\t# [W/m degree celsius]\n", + "Cp = 1006 \t\t\t# [J/kg K]\n", + "Pr = 0.703 \t\t\t# prandtl no.\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "print \"Reynolds number is\",round(Re_d) \n", + "print \"so that the flow is laminar\" \n", + "\t# so that the flow is laminar\n", + "\t# now we determine Nu_d_bar for Tw = constant. for Gz_inverse = 0.0346 we read \n", + "Nu_d = 5.15 \n", + "\t# we thus calculate the average heat transfer coefficient as \n", + "\n", + "h_bar = Nu_d*k/d \t\t# [W/square meter degree celsius]\n", + "\t# we base the heat transfer on a mean bulk temperature of Tb_bar, so that\n", + "import math\n", + "Tw = 3.49/(h_bar*math.pi*d*L)+Tb_bar \t# [degree celsius]\n", + "\n", + "\n", + "print \"Exit wall temperature is\",round(Tw,2),\"degree celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The length of tube necessary to accomplish the heating is 1.4 m\n" + ] + } + ], + "source": [ + "#Example Number 6.5\n", + "# heat transfer in a rough tube \n", + "\n", + "# Variable declaration\n", + "\n", + "Tw = 90 \t\t\t# [degree celsius] temperature of tube wall \n", + "d = 0.02 \t\t\t# [m] diameter of tube \n", + "u = 3 \t\t\t\t# [m/s] velocity of air\n", + "Tw_i = 40 \t\t\t# [degree celsius] entering water temperature \n", + "Tw_f = 60 \t\t\t# [degree celsius] leaving water temperature\n", + "Cp = 4.174*10**3 \t\t# [J/kg K]\n", + "\n", + "#Calculation\n", + "\n", + "Tb_bar = (Tw_i+Tw_f)/2 \t\t# [degree celsius]\n", + "\n", + "\t#we first calculate the heat transfer from q = m_dot*Cp*dTb\n", + "\t\n", + "import math\n", + "q = 989*3*math.pi*0.01**(2)*4174*(Tw_f-Tw_i) \t# [W]\n", + "\n", + "\t# for the rough tube condition, we may employ the Petukhov relation, equation \t\t (6-7) The mean film temperaturee is \n", + "\n", + "Tf = (Tw+Tb_bar)/2 \t\t# [degree celsius]\n", + "\n", + "\t# and the fluid properties are \n", + "\n", + "rho = 978 \t\t\t# [kg/cubic meter] density of gas\n", + "mu = 4.0*10**(-4) \t\t# [kg/m s] viscosity \n", + "k = 0.664 \t\t\t# [W/m degree celsius]\n", + "Pr = 2.54 \t\t\t# prandtl no.\n", + "\n", + "\t# also\n", + "\n", + "mu_b = 5.55*10**(-4) \t\t# [kg/m s] viscosity \n", + "mu_w = 2.81*10**(-4) \t\t# [kg/m s] viscosity \n", + "\n", + "\t# the reynolds number is thus \n", + "\n", + "Re_d = rho*u*d/mu \n", + "\n", + "\t# consulting figure(6-14), we find the friction factor as \n", + "\n", + "f_f = 0.0218 \n", + "\n", + "\t# because Tw>Tb, we take \n", + "\n", + "n = 0.11 \n", + "\n", + "\t# and obtain\n", + "\n", + "Nu_d=((f_f*Re_d*2.54)/((1.07+12.7*(f_f/8)**(0.5)*(2.54**(2.0/3.0)-1))*8))*(mu_b/mu_w)**(n) \n", + "h = Nu_d*k/d \t\t\t# [W/square meter degree celsius]\n", + "\n", + "\t# the tube length is obtained from energy balance \n", + "\n", + "L = q/(h*math.pi*d*(Tw-Tb_bar)) # [m]\n", + "\n", + "print \"The length of tube necessary to accomplish the heating is\",round(L,2),\"m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 50988.0\n", + "so the flow is turbulent\n", + "The constant heat flux that must be applied at the tube surface to result in an air temperature rise of 5 degree celsius is 11870.8 W/square meter\n", + "average wall temperature is 370.0 K\n" + ] + } + ], + "source": [ + "#Example Number 6.6\n", + "# turbulent heat transfer in a short tube \n", + "\n", + "#Variable declaration\n", + "\n", + "p = 101325\t\t\t # [Pa] pressure of air\n", + "Ta = 300 \t\t\t # [K] temperature of air \n", + "d = 0.02 \t\t\t # [m] diameter of tube \n", + "u = 40 \t\t\t\t # [m/s] velocity of air\n", + "L = 0.1\t\t\t\t # [m] length of tube\n", + "dT = 5.0 \t\t\t\t # [degree celsius] rise in temperature \n", + "#\t we first must evaluate the air properties at 300 K \n", + "v = 15.69*10**(-6) \t\t # [square meter/s] kinematic viscosity\n", + "k = 0.02624 \t\t\t # [W/m degree celsius]\n", + "Cp = 1006.0\t\t\t # [J/kg K]\n", + "Pr = 0.70 \t\t\t# prandtl no.\n", + "rho = 1.18 \t\t\t# [kg/cubic meter]\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "\n", + "print \"Reynolds number is\",round(Re_d)\n", + "print\"so the flow is turbulent\"\n", + "\n", + "\t# consulting figure (6-6) for this value of Re_d and L/d = 5 we find\n", + "Nu_x_by_Nu_inf = 1.15 \n", + "\t# or the heat transfer coefficient is about 15 percent higher that it would be \tfor thermally developed flow.\n", + "\t# we calculate heat-transfer for developed flow using \n", + "Nu_d = 0.023*Re_d**(0.8)*Pr**(0.4) \n", + "\t# and \n", + "h = k*Nu_d/d \t\t\t # [W/square meter degree celsius]\n", + "\t# increasing this value by 15 percent\n", + "h = 1.15*h \t\t\t # [W/square meter degree celsius]\n", + "\t# the mass flow is\n", + "import math\n", + "Ac = math.pi*d**(2)/4 \t\t # [square meter] \n", + "m_dot = rho*u*Ac \t\t # [kg/s]\n", + "\t# so the total heat transfer is\n", + "A = math.pi*d*L\t\t\t # [square meter] \n", + "q_by_A = m_dot*Cp*dT/A\t\t # [W/square meter]\n", + "print \"The constant heat flux that must be applied at the tube surface to result in an air temperature rise of 5 degree celsius is\",q_by_A,\" W/square meter\"\n", + "Tb_bar = (Ta+(Ta+dT))/2 \t # [K]\n", + "Tw_bar = Tb_bar+q_by_A/h \t # [K] \n", + "print \"average wall temperature is\",round(Tw_bar),\"K\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per unit length of cylinder is 3100.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 6.7\n", + "# airflow across isothermal cylinder\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 35+273.15 \t\t\t# [K] temperature of air \n", + "d = 0.05 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 50 \t\t\t\t# [m/s] velocity of air\n", + "Tc = 150+273.15 \t\t# [degree celsius] cylinder temperature\n", + "\n", + "\t# we first find the reynolds number and then find the applicable constants \tfrom table(6-2) for use with equation (6-17) \n", + "\t# the properties of air are evaluated at the film temperature:\n", + "\n", + "#Calculation\n", + "\n", + "Tf = (Ta+Tc)/2 \t\t\t# [K]\n", + "rho_f = p/(R*Tf) \t\t# [kg/cubic meter]\n", + "mu_f = 2.14*10**(-5) \t\t# [kg/m s]\n", + "k_f = 0.0312 \t\t\t# [W/m degree celsius]\n", + "Pr_f = 0.695 \t\t\t# prandtl number\n", + "Re_f = rho_f*u*d/mu_f \t\t# reynolds number\n", + "\n", + "\t# from table (6-2) table\n", + "\n", + "C = 0.0266 \n", + "n = 0.805 \n", + "\n", + "\t# so from equation (6-17)\n", + "\n", + "h = C*(Re_f)**(n)*(Pr_f)**(1.0/3.0)*k_f/d # [W/sq m deg C] heat transfer coefficient\n", + "\n", + "\t# the heat transfer per unit length is \n", + "import math\n", + "\n", + "q_by_L = h*math.pi*d*(Tc-Ta) \t\t# [W/m]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat loss per unit length of cylinder is\",round(q_by_L),\"W/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.8" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost per unit length by the wire is 11.88 W/m\n" + ] + } + ], + "source": [ + "#Example Number 6.8\n", + "# heat transfer from electrically heated\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Tw = 25+273.15 \t\t\t# [K] temperature of air \n", + "d = 3.94*10**(-5) \t\t# [m] diameter of wire\n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 50 \t\t\t\t# [m/s] velocity of air perpendicular to the air\n", + "Tr = 50+273.15 \t\t\t# [degree celsius] rise in surface temperature\n", + "\t# we first obtain the properties at the film temperature :\n", + "\n", + "#Calculation\n", + "\n", + "Tf = (Tw+Tr)/2 \t\t\t# [K]\n", + "v_f = 16.7*10**(-6) \t\t# [square meter/s]\n", + "k = 0.02704 \t\t\t# [W/m degree celsius]\n", + "Pr_f = 0.706 \t\t\t# prandtl number\n", + "Re_d = u*d/v_f \t\t\t# reynolds number\n", + "\t# the Peclet number is \n", + "Pe = Re_d*Pr_f \n", + "\t# and we find that equations (6-17),(6-21), or (6-19) apply.\n", + "\t# let us make the calculation with both the simplest expression, (6-17),and \t\tthe most complex,(6-21), and compare results.\n", + "\t# using equation (6-17) with \n", + "C = 0.683 \n", + "n = 0.466 \n", + "\t# we have\n", + "Nu_d = 0.683*Re_d**(n)*Pr_f**(1/3) \n", + "\t# the value of heat transfer coefficient is\n", + "h = Nu_d*k/d \t\t\t # [W/square meter degree celsius]\n", + "\t# the heat transfer per unit length is then \n", + "import math\n", + "q_by_L = math.pi*d*h*(Tr-Tw)\t # [W/m]\n", + "\t# using equation (6-21), we calculate the nusselt no as \n", + "Nu_d1=0.3+((0.62*Re_d**(1.0/2.0)*Pr_f**(1.0/3.0))/((1+(0.4/Pr_f)**(2.0/3.0))**(1.0/4.0))*((1+(Re_d/282000)**(5.0/8.0))**(4.0/5.0))) \n", + "h1 = Nu_d1*k/d \t\t\t # [W/square meter degree celsius]\n", + "\t# and\n", + "q_by_L1 = h1*math.pi*d*(Tr-Tw) \t # [W/m]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat lost per unit length by the wire is\",round(q_by_L1,2),\"W/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.9" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by the sphere is 1.554 W\n" + ] + } + ], + "source": [ + "#Example Number 6.9\n", + "# heat transfer from sphere \n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 27+273.15 \t\t\t# [K] temperature of air \n", + "d = 0.012 \t\t\t# [m] diameter of sphere\n", + "u = 4 \t\t\t\t# [m/s] velocity of air \n", + "Ts = 77+273.15 \t\t\t# [degree celsius] surface temperature of sphere\n", + "\t# consulting equation (6-30) we find that the reynolds number is evaluated at \t\tthe free-stream temperature.\n", + "\t# we therefore need the following properties at Ta = 300.15K\n", + "v = 15.69*10**(-6) \t\t# [square meter/s]\n", + "k = 0.02624 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.708 \t\t\t# prandtl number\n", + "mu_inf = 1.8462*10**(-5) \t# [kg/m s]\n", + "\t# at Ts = 350K\n", + "mu_w = 2.075*10**(-5) \t\t# [kg/m s]\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "\n", + "#Calculation\n", + "\t# from equation (6-30),\n", + "Nu_bar=2+((0.4)*(Re_d)**(1.0/2.0)+0.06*(Re_d)**(2.0/3.0))*(Pr**(0.4))*((mu_inf/mu_w)**(1.0/4.0)) \n", + "\n", + "\n", + "\t# and\n", + "\n", + "h_bar = Nu_bar*k/d \t\t# [W/sq m degree celsius] heat transfer coefficient\n", + "\n", + "\t# the heat transfer is then \n", + "\n", + "import math\n", + "\n", + "A = (4*math.pi*d**(2))/4 \t# [square meter] area of sphere\n", + "\n", + "q = h_bar*A*(Ts-Ta) \t\t# [W]\n", + "\n", + "\n", + "\t# for comparison purposes let us also calculate the heat-transfer coefficient \t\t using equation(6-25). the film temperature is \n", + "Tf = (Ta+Ts)/2 \t\t\t# [K]\n", + "v_f = 18.23*10**(-6) \t\t# [square meter/s]\n", + "k_f = 0.02814 \t\t\t# [W/m degree celsius] \n", + "\t# reynolds number is \n", + "Re_d1 = u*d/v_f \n", + "\t# from equation (6-25)\n", + "Nu_f = 0.37*(u*d/v_f)**(0.6) \n", + "\t# and h_bar is calculated as\n", + "\n", + "h_bar = Nu_f*k_f/d \t\t# [W/sq m degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat lost by the sphere is\",round(q,3),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.11" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer coefficient is 185.6 W/square meter degree celsius\n", + "Heat transfer coefficient for previous problem is 163.5 W/sq meter degree C\n", + "Percentage increase in value of h is 14.0\n" + ] + } + ], + "source": [ + "#Example Number 6.11\n", + "# alternate calculation method \n", + "\n", + "# Variable declaration\n", + "\n", + "\t# data for this example is taken from previous example (6-10)\n", + "\t# properties for use in equation (6-34) are evaluated at free-atream \tconditions of 10 degree celsius\n", + "v = 14.2*10**(-6)\t\t\t # [square meter/s]\n", + "k = 0.0249 \t\t\t\t # [W/m degree celsius]\n", + "Pr = 0.712 \t\t\t\t # prandtl number\n", + "Pr_w = 0.70 \t\t\t\t# prandtl number\n", + "u = 7 \t\t\t\t\t# [m/s] velocity of air \n", + "Sp = 0.0381 \t\t\t\t# [m] spacing between normal and parallel \t\t\t\t\t\tdirection to the flow\n", + "Sn = 0.0381 \t\t\t\t# spacing between normal and parallel \t\t\t\t\t direction to the flow\n", + "d = 0.0254 \t\t\t\t# [m] diameter of tube\n", + "\t#maximum velocity is \n", + "u_max = u*(Sn/(Sn-d)) \t\t\t# [m/s]\n", + "\t# the reynolds number is \n", + "Re_d_max = u_max*d/v \n", + "\t# so that the constants for equation (6-34) are\n", + "C = 0.27 \n", + "n = 0.63 \n", + "\t# inserting values we obtain\n", + "h = C*Re_d_max**(n)*(Pr/Pr_w)**(1/4)*k/d \t# [W/sq m degree C] heat transfer \t\t\t\t\t\t\tcoefficient\n", + "\t# multiplying by 0.92 from table 6-7 (page no.-300) to correct for only five \ttube rows gives\n", + "h = 0.92*h \t\t\t\t\t# [W/square meter degree celsius]\n", + "print \"Heat transfer coefficient is\",round(h,1),\"W/square meter degree celsius\"\n", + "\n", + "h_in = 163.46432 \t\t\t# [W/sq m deg C] from previous example\n", + "\n", + "print \"Heat transfer coefficient for previous problem is\",round(h_in,1),\"W/sq meter degree C\" \n", + "P = (h-h_in)*100/h_in \n", + "print \"Percentage increase in value of h is\",round(P) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.12" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of tube required to effect the heat transfer is 1.56 m\n" + ] + } + ], + "source": [ + "#Example Number 6.12\n", + "# heating of liquid bismuth in tube \n", + "\n", + "# variable declaration\n", + "\n", + "m_dot = 4.5 \t\t\t# [Kg/s] flow rate of bismuth\n", + "d = 0.05 \t\t\t# [m] diameter of steel tube\n", + "\n", + "Ti = 415 \t\t\t# [degree celsius] initial temperature of bismuth\n", + "Tf = 440 \t\t\t# [degree celsius] final temperature of bismuth\n", + "\t# because a constant heat flux is maintained, we may use equation 6-47 to \t\tcalculate the heat transfer coefficient.\n", + "\t# the properties of bismuth are evaluated at the average bulk temperature of \n", + "#Calculation\n", + "\n", + "Ta = (Ti+Tf)/2 \t\t\t# [degree celsius]\n", + "mu = 1.34*10**(-3) \t\t# [Kg/m s] viscosity\n", + "Cp = 149 \t\t\t# [J/Kg degree celsius] heat \n", + "k = 15.6 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.013 \t\t\t# prandtl number\n", + "\t# the total transfer is calculated from\n", + "q = m_dot*Cp*(Tf-Ti) \t\t# [W]\n", + "\t# we calculate reynolds and peclet number as \n", + "\n", + "import math\n", + "G = m_dot/(math.pi*d**(2)/4) \n", + "Re_d = d*G/mu \n", + "Pe = Re_d*Pr \n", + "\t# the heat transfer coefficient is calculated from equation 6-47\n", + "Nu_d = 4.82+0.0185*Pe**(0.827) \n", + "h = Nu_d*k/d \t\t\t\t# [W/square meter degree celsius]\n", + "\t# the total required surface area of the tube may now be computed from q=h*A*DT\n", + "\t# where we use the temperature difference of\n", + "DT = 20 \t\t\t# [degree celsius] \n", + "A = q/(h*DT) \t\t\t# [square meter] \n", + "\t# the area in turn can be expressed in terms of tube length \n", + "L = A/(math.pi*d) \t\t# [m]\n", + "\n", + "#Result\n", + "print\"Length of tube required to effect the heat transfer is\",round(L,2),\"m\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter7.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter7.ipynb new file mode 100644 index 00000000..7793c6da --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter7.ipynb @@ -0,0 +1,794 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 Natural Convection Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average wall temperature is 185.0 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 7.1\n", + "# constant heat flux from vertical plate \n", + "\n", + "# Variable declaration\n", + "\n", + "q_w = 800.0 \t\t\t# [W/square meter] radiant energy flux\n", + "H = 3.5 \t\t\t# [m] height of metal plate surface\n", + "W = 2 \t\t\t\t# [m] width of metal plate\n", + "Ta = 30 \t\t\t# [degree celsius] surrounding air temperature \n", + "\t# we treat this problem as one with constant heat flux on the surface since we \tdo not know the surface temperature, we must make an estimate for determining \t\tTf and the air properties.\n", + "\t# an approximate value of h for free convection problems is \n", + "h = 10 \t\t\t\t# [W/square meter degree celsius]\n", + "dT = q_w/h \t\t\t# [degree celsius]\n", + "\t# then\n", + "Tf = (dT/2)+Ta \t\t\t# [degree celsius] approximately \n", + "\n", + "#Calculation\n", + "\n", + "\t# at Tf the properties of air are \n", + "v = 2.043*10**(-5) \t\t# [square meter/s]\n", + "k = 0.0295 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.7 \t\t\t# prandtl number\n", + "Beta = 1.0/(Tf+273) \t\t# [K**(-1)]\n", + "\t# from equation (7-30), with\n", + "x = 3.5 \t\t\t# [m]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity \n", + "Gr_x = (g*Beta*q_w*x**(4))/(k*v**(2)) \n", + "\n", + "\n", + "\t# we may therefore use equation (7-32) to evaluate h_x\n", + "\n", + "h_x = (k*0.17*(Gr_x*Pr)**(1.0/4.0))/x \t# [W/square meter degree celsius]\n", + "\n", + "\t# in the turbulent heat transfer governed by equation (7-32), we note that\n", + "\t# Nu_x = h*x/k ~ (Gr_x)**(1/4) ~ x\n", + "\t# or h_x doest noy vary with x, and we may take this as the average value. the \tvalue of h\n", + "\n", + "\n", + "h = 5.41 \t\t\t# [W/square meter degree celsius]\n", + "\t# is less than the approximate value we used to estimate Tf, recalculating dT, \t we obtain\n", + "dT1 = q_w/h_x \t\t\t# [degree celsius]\n", + "\n", + "\n", + "\t# our new film temperature would be\n", + "Tf1 = Ta+dT1/2 \t\t\t# [degree celsius]\n", + "\t# at Tf the properties of air are\n", + "v1 = 2.354*10**(-5) \t\t# [square meter/s]\n", + "k1 = 0.0320 \t\t\t# [W/m degree celsius]\n", + "Pr1 = 0.695 \t\t\t# prandtl number\n", + "Beta1 = 1/(Tf1+273) \t\t# [K**(-1)]\n", + "\n", + "\t# then \n", + "Gr_x1 = (g*Beta1*q_w*x**(4))/(k1*v1**(2)) \n", + "\t# and h_x is caalculated from\n", + "h_x1 = (k1*0.17*(Gr_x1*Pr1)**(1.0/4.0))/x \t# [W/square meter degree celsius]\n", + "\n", + "\n", + "\t# our new temperature difference is calculated as \n", + "dT2 = q_w/h_x1 \t\t\t# [degree celsius]\n", + "\n", + "\n", + "\t# the average wall temperature is therefore\n", + "T_w_avg = dT2+Ta \t\t# [degree celsius]\n", + "\n", + "#Result\n", + "print \"The average wall temperature is\",round(T_w_avg),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer if the plate is 10 m wide is 9603.0 W\n" + ] + } + ], + "source": [ + "#Example Number 7.2\n", + "# heat transfer from isothermal vertical plate\n", + "\n", + "# Variable declaration\n", + "\n", + "H = 4.0 \t\t\t\t# [m] height of vertical plate\n", + "Tp = 60.0 \t\t\t# [degree celsius] plate temperature\n", + "Ta = 10.0 \t\t\t# [degree celsius] atmospheric temperature\n", + "\t# we first determine the film temperature as\n", + "Tf = (Tp+Ta)/2 \t\t\t# [degree celsius]\n", + "\t# the properties of interest are thus\n", + "v = 16.5*10**(-6) \t\t# [square meter/s]\n", + "k = 0.02685 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.7 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "\n", + "#Calculation\n", + "\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity \n", + "\t# and\n", + "Gr_into_Pr = (g*Beta*(Tp-Ta)*H**(3)*Pr)/(v**(2)) \n", + "\n", + "\t# we then may use equation (7-29) to obtain\n", + "Nu_bar_root = (0.825+(0.387*(Gr_into_Pr)**(1.0/6.0))/(1+(0.492/Pr)**(9.0/16.0))**(8.0/27.0) )\n", + "\n", + "\n", + "Nu_bar = (Nu_bar_root)**(2) \n", + "\t# the heat transfer coefficient is \n", + "h_bar = Nu_bar*k/H \t\t# [W/square meter degree celsius]\n", + "\n", + "\n", + "\t# the heat transfer is \n", + "A = H*10 \t\t\t# [square meter] for 10 m wide plate\n", + "q = h_bar*A*(Tp-Ta) \t\t# [W]\n", + "\n", + "\t# as an alternative, we could employ the simpler relation \n", + "Nu = 0.1*(Gr_into_Pr)**(1/3) \n", + "\n", + "#Result\n", + "print \"Heat transfer if the plate is 10 m wide is\",round(q),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "free-convection heat loss per unit length of heater is 443.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 7.3\n", + "# heat transfer from horizontal tube in water\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.02 \t\t\t# [m] diameter of heater\n", + "Ts = 38 \t\t\t# [degree celsius] surface temperature of heater\n", + "Tw = 27 \t\t\t# [degree celsius] water temperature\n", + "\t# the film temperature is \n", + "Tf = (Ts+Tw)/2 \t\t\t# [degree celsius]\n", + "\t# from appendix A the properties of water are \n", + "k = 0.630 \t\t\t# [W/m degree celsius] thermal conductivity\n", + "\t# and the following term is particularly useful in obtaining the product GrPr \t\tproduct when it is multiplied by d**(3)*DT\n", + "\t# g*Beta*rho**(2)*Cp/(mu*k) = 2.48*10**(10) [1/m**(3) degree celsius]\n", + "\n", + "K = 2.48*10**(10) \t\t# [1/m**(3) degree celsius]\n", + "Gr_into_Pr = K*(Ts-Tw)*d**(3) \n", + "\t\n", + "\t# using table 7-1 (page number -328), we get \n", + "\n", + "C = 0.53 \n", + "m = 1/4 \n", + "\t# so that\n", + "\n", + "Nu = C*(Gr_into_Pr)**(1.0/4.0) \n", + "h = Nu*k/d \t\t\t# [W/sq m deg C] convection heat transfer coefficient\n", + "\t# the heat transfer is thus\n", + "import math\n", + "\n", + "q_by_L = h*math.pi*d*(Ts-Tw) \t# [W/m]\n", + "print\"free-convection heat loss per unit length of heater is\",round(q_by_L),\"W/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.4" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "#Example Number 7.4\n", + "# heat transfer from fine wire in air\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.00002 \t\t\t# [m] diameter of wire\n", + "L = 0.5 \t\t\t# [m] length of wire whose temperature is maintained\n", + "Ts = 54.0 \t\t\t# [degree celsius] surface temperature of wire \n", + "Pa = 101325.0 \t\t\t# [Pa] pressure of air\n", + "Ta = 0 \t\t\t\t# [degree celsius] temperature of air \n", + "\t# we first determine the film temperature as\n", + "Tf = (Ts+Ta)/2 \t\t\t# [degree celsius]\n", + "\t# the properties of interest are thus\n", + "v = 15.69*10**(-6) \t\t# [square meter/s]\n", + "k = 0.02624 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.708 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K^(-1)]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity \n", + "\t# and\n", + "Gr_into_Pr = (g*Beta*(Ts-Ta)*d**(3)*Pr)/(v**(2)) \n", + "\t# from table 7-1 we find\n", + "C = 0.675 \n", + "m = 0.058 \n", + "\t# so that\n", + "Nu_bar = C*(Gr_into_Pr)**(m) \n", + "h_bar = Nu_bar*k/d \t\t# [W/square meter degree celsius]\n", + "\t# the heat required is \n", + "import math\n", + "A = math.pi*d*L \t\t# [square meter] surface area of wire \n", + "q = h_bar*A*(Ts-Ta) \t\t# [W]\n", + "print \"Electric power to maintain the the wire temperature if the length is 0.5 m is\",round(q,3),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "free-convection heat loss per unit length is 1.5 kW/m\n" + ] + } + ], + "source": [ + "#Example Number 7.5\n", + "# heated horizontal pipe in air \n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.3048 \t\t\t# [m] diameter of pipe\n", + "Ts = 250.0 \t\t\t# [degree celsius] surface temperature of pipe \n", + "Ta = 15.0 \t\t\t# [degree celsius] temperature of air \n", + "\t# we first determine the Grashof-prandtl number product and then select the \tappropriate constants from table 7-1(page no.-328) for use with \tequation (7-25) \n", + "\t# the properties of air are evaluated at the film temperature:\n", + "Tf = (Ts+Ta)/2 \t\t\t# [degree celsius]\n", + "\t# the properties of interest are thus\n", + "v = 26.54*10**(-6) \t\t# [square meter/s]\n", + "k = 0.03406 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.687 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity\n", + "Gr_d_into_Pr = g*Beta*(Ts-Ta)*d**(3)*Pr/(v**(2)) \n", + "\t# from table 7-1 \n", + "C = 0.53 \n", + "m = 1.0/4.0 \n", + "Nu_d = C*(Gr_d_into_Pr)**(m) \n", + "h = Nu_d*k/d \t\t\t# [W/square meter degree celsius]\n", + "\t# the heat transfer per unit length is then calculated from \n", + "import math\n", + "\n", + "q_by_L = h*math.pi*d*(Ts-Ta) \t# [W/m]\n", + "print \"free-convection heat loss per unit length is\",round(q_by_L/1000,1),\"kW/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer is 51.8 W\n" + ] + } + ], + "source": [ + "#Example Number 7.6 \n", + "# cube cooling in air\n", + "\n", + "# Variable declaration\n", + "\n", + "L = 0.2 \t\t\t\t# [m] side length of cube\n", + "Ts = 60 \t\t\t\t# [degree celsius] surface temperature of cube\n", + "Ta = 10 \t\t\t\t# [degree celsius] air temperature\n", + "\t# this is an irregular solid so we use the information in the last entry of \ttable 7-1(page no.-328) in the absence of a specific correlation for this \tgeometry. \n", + "\t# the properties were evaluated as\n", + "v = 17.47*10**(-6)\t\t\t # [square meter/s]\n", + "k = 0.02685 \t\t\t\t# [W/m degree celsius]\n", + "Pr = 0.70 \t\t\t\t# prandtl number\n", + "Beta = 3.25*10**(-3) \t\t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t\t# [square meter/s] acceleration due to gravity \n", + "\t# the characteristic length is the distance a particle travels in the boundary \tlayer, which is L/2 along the bottom plus L along the side plus L/2 on the \ttop or\n", + "\n", + "\n", + "#Calculation\n", + "Gr_into_Pr = (g*Beta*(Ts-Ta)*(2*L)**(3)*Pr)/(v**(2)) \n", + "\t# from the last entry in table 7-1 we find\n", + "C = 0.52 \n", + "n = 1.0/4.0 \n", + "\t# so that\n", + "Nu = C*(Gr_into_Pr)**(n) \n", + "h_bar = Nu*k/(2*L) \t\t\t# [W/square meter degree celsius]\n", + "\t# the cube has six sides so the area is \n", + "A = 6*L**(2) \t\t\t\t# [square meter]\n", + "\t# the heat required is \n", + "q = h_bar*A*(Ts-Ta) \t\t\t# [W]\n", + "\n", + "#Result\n", + "print \"Heat transfer is\",round(q,1),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "heat transfer is 1.57 kW/m\n" + ] + } + ], + "source": [ + "#Example Number 7.7\n", + "# calculation with simplified relations \n", + "\n", + "#Variable declaration\n", + "\n", + "\t# this example is calculation of heat transfer with simplified relations for \texample (7.5) so we use the data of example 7.5\n", + "\n", + "d = 0.3048 \t\t\t# [m] diameter of pipe\n", + "Ts = 250 \t\t\t# [degree celsius] surface temperature of pipe \n", + "Ta = 15 \t\t\t# [degree celsius] temperature of air \n", + "\t# we first determine the Grashof-prandtl number product and then select the \tappropriate constants from table 7-1 for use with equation (7-25) \n", + "\t# the properties of air are evaluated at the film temperature:\n", + "\n", + "#Calculation\n", + "\n", + "Tf = (Ts+Ta)/2 \t\t\t\t# [degree celsius]\n", + "\t\t\t\t\t# the properties of interest are thus\n", + "v = 26.54*10**(-6)\t\t\t# [square meter/s]\n", + "k = 0.03406 \t\t\t\t# [W/m degree celsius]\n", + "Pr = 0.687 \t\t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t\t# [square meter/s] acceleration due to gravity\n", + "\t# in example (7.5) we found that a rather large pipe with a substantial \t\ttemperature difference between the surface and air still had a GrPr product of \t1.57*10**(8)<10**(9), so laminar equation is selected from table 7-2(page \t\tno.-339). the heat transfer coefficient is given by \n", + "h = 1.32*((Ts-Ta)/d)**(1.0/4.0) \t\t# [W/square meter degree celsius]\n", + "\t# the heat transfer is then\n", + "import math\n", + " \n", + "q_by_L = h*math.pi*d*(Ts-Ta) \t\t# [W/m]\n", + "\n", + "#Result\n", + "\n", + "print \"heat transfer is\",round(q_by_L/1000,2),\"kW/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.8" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Free-convection heat transfer across the air space is 39.64 W\n", + "Radiation heat transfer across the air space is 15.37 W\n" + ] + } + ], + "source": [ + "#Example Number 7.8\n", + "# heat transfer across vertical air gap\n", + "\n", + "#Variable declaration\n", + "\n", + "L = 0.5 \t\t\t# [m] side length vertical square plate\n", + "d = 0.015 \t\t\t# [m] distance between plates\n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "T1 = 100.0 \t\t\t# [degree celsius] temperature of first plate\n", + "T2 = 40.0 \t\t\t# [degree celsius] temperature of second plate\n", + "E = 0.2 \t\t\t# emissivity of both surfaces\n", + "\t\t# the properties of air is evaluated at the mean temperature\n", + "Tf = (T1+T2)/2 \t\t\t# [degree celsius]\n", + "rho = p/(R*(Tf+273)) \t\t# [Kg/m**(3)] density\n", + "k = 0.0295 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.70 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "mu = 2.043*10**(-5) \t\t# [Kg/m s] viscosity\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity\n", + "\t\t# the Grashof-prandtl number product is now calculated as \n", + "\n", + "Gr_into_Pr = (g*rho**(2)*Beta*(T1-T2)*(d)**(3)*Pr)/(mu**(2)) \n", + "\n", + "\t\t# we may now use eq(7-64) to calculate the effective thermal \t\tconductivity, with\n", + "L = 0.5 \t\t\t# [m]\n", + "deli=0.015\t\t\t# [m]\n", + "\t\t\t\t# and the constants taken from table 7-3:\n", + "\n", + "#Calculation\n", + "\n", + "Ke_by_K = 0.197*(Gr_into_Pr)**(1.0/4.0)*(L/deli)**(-1.0/9.0) \n", + "\t\t# the heat transfer may now be calculated with eq(7-54). the area is \n", + "A = L**(2) \t\t\t# [square meter]\n", + "q = Ke_by_K*k*A*(T1-T2)/deli \t# [W]\n", + " \t\t# the radiation flux is calculated with equation(7-67), taking \n", + "T1 = 373 \t\t\t# [K]\n", + "T2 = 313 \t\t\t# [K]\n", + "E1 = E \n", + "E2 = E \n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "q_A = sigma*(T1**(4)-T2**(4))/((1/E1)+(1/E2)-1) \t# [W/square meter]\n", + "q_rad = A*q_A \t\t\t\t\t\t# [W]\n", + "\n", + "#Result\n", + "print \"Free-convection heat transfer across the air space is\",round(q,2),\"W\" \n", + "print \"Radiation heat transfer across the air space is\",round(q_rad,2),\"W\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.9" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer across the air space is 10.34 W\n" + ] + } + ], + "source": [ + "#Example Number 7.9 \n", + "# heat transfer across horizontal air gap\n", + "\n", + "# Variable declaration\n", + "\n", + "a = 0.2 \t\t\t# [m] side length of plate\n", + "d = 0.01 \t\t\t# [m] seperation between two plates \n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "T1 = 100.0 \t\t\t# [degree celsius] temperature of first plate\n", + "T2 = 40.0 \t\t\t# [degree celsius] temperature of second plate\n", + "\t# the properties are the same as given in example(7.8)\n", + "Tf = (T1+T2)/2 \t\t\t# [degree celsius]\n", + "rho = p/(R*(Tf+273)) \t\t# [Kg/m**(3)] density\n", + "k = 0.0295 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.70 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "mu = 2.043*10**(-5) \t\t# [Kg/m s] viscosity\n", + "g = 9.8 \t\t\t# [sq m/s] acceleration due to gravity\n", + "\t# the GrPr product is evaluated on the basis of the separating distance, so we \t have \n", + "\n", + "#Calculation\n", + "\n", + "Gr_into_Pr = (g*rho**(2)*Beta*(T1-T2)*(d)**(3)*Pr)/(mu**(2)) \n", + "\t# consulting table 7-3(page no.-344) we find\n", + "C = 0.059 \n", + "n = 0.4 \n", + "m = 0 \n", + "Ke_by_K = C*(Gr_into_Pr)**(n)*(a/d)**(m) \n", + "A = a**(2) \t\t\t# [square meter] area of plate \n", + "q = Ke_by_K*k*A*(T1-T2)/d \t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat transfer across the air space is\",round(q,2),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.10" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "3.47131600256\n", + "heat lost by the lower plate is 964.0 W\n" + ] + } + ], + "source": [ + "#Example Number 7.10\n", + "# heat transfer across water layer\n", + "\n", + "# Variable declaration\n", + "\n", + "L = 0.5 \t\t\t# [m] length of square plate\n", + "d = 0.01 \t\t\t# [m] seperation between square plates\n", + "T1 = 100\t\t\t# [degree F] temperature of lower plate\n", + "T2 = 80 \t\t\t# [degree F] temperature of upper plate\n", + "\n", + "dT=5.0*(T1-T2)/9.0\t\t\t#in Degree C\n", + "\t\t# we evaluate properties at mean temperature of 90 deg F and \t\t\t\tobtain, for water\n", + "k = 0.623 \t\t\t# [W/m degree celsus]\n", + "\t# and the following term is particularly useful in obtaining the product GrPr \n", + "\t# g*Beta*rho**(2)*Cp/(mu*k) = 2.48*10**(10) [1/m**(3) degree celsius]\n", + "\t# the Grashof-prandtl number product is now evaluated using the plate spacing \t\tof 0.01 m as the characterstic dimension\n", + "\n", + "#Calculation\n", + "\n", + "K = 2.48*10**(10.0) \t\t# [1/m**(3) degree celsius]\n", + "\n", + "Gr_into_Pr = K*(T1-T2)*(5.0/9.0)*d**(3.0) \n", + "\n", + "\t# now, using equation 7-64 and consulting table 7-3(page no.-344) we obtain\n", + "C = 0.13 \n", + "n = 0.3 \n", + "m = 0.0 \n", + "\t# therefore, equation (7-64) becomes\n", + "Ke_by_K = C*Gr_into_Pr**(n) \n", + "\n", + "\t# the effectve thermal conductivity is thus\n", + "ke = k*Ke_by_K \t\t\t# [W/m degree celsius]\n", + "\n", + "print ke\n", + "\t\t\t\t# and the heat transfer is\n", + "A = L**(2.0) \t\t\t# [square meter] area of plate\n", + "q = ke*A*(dT)/d \t# [W]\n", + "\n", + "#Result\n", + "print \"heat lost by the lower plate is \",round(q),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.11" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vacuum necessary for glass spacings of 1 cm is 13160.0 Pa\n", + "Vacuum necessary for glass spacings of 2 cm is 70154.0 Pa\n" + ] + } + ], + "source": [ + "#Example Number 7.11 \n", + "# reduction of convection in ar gap\n", + "\n", + "# Variable declaration\n", + "\n", + "Tm = 300.0 \t\t\t# [K] mean temperature of air\n", + "dT = 20.0 \t\t\t# [degree celsius] temperature difference\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "g = 9.8 \t\t\t# [m/s**(2)] acceleration due to gravity\n", + "p_atm = 101325.0 \t\t\t# [Pa] atmospheric pressure\n", + "\t# consulting table 7-13,we find that for gases, a value Grdeli_into_Pr<2000 is \tnecessary to reduce the system to one of pure \tconduction.\n", + "\t# at 300 K the properties of air are\n", + "k = 0.02624 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.7 \t\t\t# prandtl no.\n", + "mu = 1.846*10**(-5) \t\t# [Kg/m s]\n", + "Beta = 1.0/300.0 \n", + "\n", + "\t# we have\n", + "Grdel_into_Pr = 2000.0 \n", + "\n", + "\t# Part A for spacing of 1cm\n", + "import math\n", + "deli = 0.01 \t\t\t# [m] spacing between plate\n", + "p = math.sqrt((Grdel_into_Pr*((R*Tm)**(2))*mu**(2))/(g*Beta*dT*deli**(3)*Pr)) \t# [Pa]\n", + "p=math.sqrt(7773/deli**3)\n", + "\n", + "\t# or vacuum\n", + "vacuum = p_atm-p \t\t# [Pa]\n", + "\n", + "\n", + "\t# Part B for spacing of 2cm\n", + "\n", + "deli1 = 0.02 \t\t\t# [m] spacing between plate\n", + "p1 = math.sqrt(Grdel_into_Pr*(R*Tm)**(2)*mu**(2)/(g*Beta*dT*deli1**(3)*Pr)) # [Pa]\n", + "p1=math.sqrt(7773/deli1**3)\n", + "\t# or vacuum\n", + "vacuum1 = p_atm-p1 \t\t# [Pa]\n", + "\n", + "\n", + "#Result\n", + "print \"Vacuum necessary for glass spacings of 1 cm is\",round(vacuum),\"Pa\" \n", + "print \"Vacuum necessary for glass spacings of 2 cm is\",round(vacuum1),\"Pa\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "heat transfer coefficient is 9.4 W/square meter degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 7.13\n", + "# combined free and forced convection with air\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "Ta = 27.0 \t\t\t# [degree celsius] temperature of air\n", + "d = 0.025 \t\t\t# [m] diameter of tube\n", + "u = 0.3 \t\t\t# [m/s] velocity of air\n", + "Tw = 140.0 \t\t\t# [degree celcius] temperature of tube wall\n", + "L = 0.4 \t\t\t# [m] length of tube\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "\t# the properties of air are evaluated at the film temperature:\n", + "Tf = (Tw+Ta)/2 \t\t\t# [degree celcius]\n", + "\t# the properties of interest are thus\n", + "kf = 0.0305 \t\t\t# [W/m degree celcius]\n", + "Pr = 0.695 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity\n", + "mu_f = 2.102*10**(-5) \t\t# [Kg/m s]\n", + "mu_w = 2.337*10**(-5) \t\t# [Kg/m s]\n", + "\n", + "#Calculations\n", + "\n", + "rho_f = p/(R*(Tf+273)) \t\t# [Kg/cubic meter]\n", + "\t# let us take the bulk temperature as 27 degree celsius for evaluating mu_b \tthen\n", + "mu_b = 1.8462*10**(-5) \t\t# [Kg/m s]\n", + "\t# the significant parameters are calculated as \n", + "Re_f = rho_f*u*d/mu_f \n", + "Gr = rho_f**(2)*g*Beta*(Tw-Ta)*d**(3)/mu_f**(2) \n", + "Z = Gr*Pr*d/L \t\t\t# constant\n", + "\t# according to figure(7-14)(page no.-354), the mixed convection flow regime is \tencountered. thus we must use equation(7-77).\n", + "\t# The graetz number is calculated as \n", + "Gz = Re_f*Pr*d/L \n", + "\t# and the numerical calculation for equation(7-77) becomes\n", + "Nu = 1.75*(mu_b/mu_w)**(0.14)*(Gz+0.012*(Gz*Gr**(1.0/3.0))**(4.0/3.0))**(1.0/3.0) \n", + "\t# the average heat transfer coefficient is calculated as \n", + "h_bar = Nu*kf/d \t\t# [W/square meter degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"heat transfer coefficient is\",round(h_bar,2),\"W/square meter degree celsius\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter8.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter8.ipynb new file mode 100644 index 00000000..97eeda1a --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter8.ipynb @@ -0,0 +1,1006 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 Radiation Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total energy absorbed = 52.75 kW\n", + "total energy transmitted= 58.2 kW\n" + ] + } + ], + "source": [ + "#Example Number 8.1\n", + "# transmission and absorption in a gas plate\n", + "\n", + "# Variable declaration\n", + "\n", + "T = 2000+273 \t\t\t\t# [K] furnace temperature \n", + "L = 0.3 \t\t\t\t# [m] side length of glass plate\n", + "t1 = 0.5 \t\t\t\t# transmissivity of glass betweenb/n lambda1 \t\t\t\t\tto lambda2\n", + "lambda1 = 0.2 \t\t\t\t# [micro m] \n", + "lambda2 = 3.5 \t\t\t\t# [micro m] \n", + "E1 = 0.3 \t\t\t\t# emissivity of glass upto lambda2 \n", + "E2 = 0.9 \t\t\t\t# emissivity of glass above lambda2\n", + "t2 = 0 \t\t\t\t\t# transmissivity of glass except in the range \t\t\t\t\tof lambda1 to lambda2\n", + "\n", + "#Calculation\n", + "\n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "A = L**(2) \t\t\t\t# [square meter] area of glass plate\n", + "\t# calculating constants to use table 8-1(page no.-379-380)\n", + "K1 = lambda1*T \t\t\t\t# [micro m K]\n", + "K2 = lambda2*T \t\t\t\t# [micro m K]\n", + "\t# from table 8-1\n", + "Eb_0_lam1_by_sigmaT4 = 0 \n", + "Eb_0_lam2_by_sigmaT4 = 0.85443 \n", + "Eb = sigma*T**(4) \t\t\t# [W/square meter]\n", + "\t# total incident radiation is \n", + "\t# for 0.2 micro m to 3.5 micro m\n", + "TIR = Eb*(Eb_0_lam2_by_sigmaT4-Eb_0_lam1_by_sigmaT4)*A \t# [W]\n", + "TRT = t1*TIR \t\t\t\t# [W]\n", + "RA1 = E1*TIR \t\t\t\t# [W] for 0<lambda<3.5 micro m\n", + "RA2 = E2*(1-Eb_0_lam2_by_sigmaT4)*Eb*A # [W] for 3.5 micro m <lambda< infinity \n", + "TRA = RA1+RA2 \t\t\t\t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Total energy absorbed =\",round(TRA/1000,2),\"kW\" \n", + "print \"total energy transmitted=\",round(TRT/1000,1),\"kW\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "net radiant heat exchange between the two plates is 18.33 kW\n" + ] + } + ], + "source": [ + "#Example Number 8.2\n", + "# heat transfer between black surfaces\n", + "\n", + "# Variable declaration\n", + "\n", + "L = 1 \t\t\t\t\t# [m] length of black plate\n", + "W = 0.5 \t\t\t\t# [m] width of black plate\n", + "T1 = 1000+273 \t\t\t\t# [K] first plate temperature\n", + "T2 = 500+273 \t\t\t\t# [K] second plate temperature\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t\t# the ratios for use with figure 8-12(page no.-386) are\n", + "Y_by_D = W/W \n", + "X_by_D = L/W \n", + "\t\t# so that \n", + "F12 = 0.285 \t\t\t\t# radiation shape factor \n", + "\t\t# the heat transfer is calculated from\n", + "q = sigma*L*W*F12*(T1**(4)-T2**(4)) \n", + "print \"net radiant heat exchange between the two plates is\",round(q/1000,2),\"kW\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Shape factor between the open ends of the cylinder is 0.0768\n" + ] + } + ], + "source": [ + "#Example Number 8.3\n", + "# shape-factor algebra for open ends of cylinder\n", + "\n", + "# Variable declaration\n", + "\n", + "d1 = 0.1 \t\t\t\t# [m] diameter of first cylinder\n", + "d2 = 0.2 \t\t\t\t# [m] diameter of second cylinder\n", + "L = 0.2 \t\t\t\t# [m] length of cylinder\n", + "\t\t# we use the nomenclature of figure 8-15(page no.-388) for this \t\tproblem and designate the open ends as surfaces 3 and 4.\n", + "\t\t# we have \n", + "L_by_r2 = L/(d2/2) \n", + "r1_by_r2 = 0.5 \n", + "\t\t# so from figure 8-15 or table 8-2(page no.-389) we obtain\n", + "F21 = 0.4126 \n", + "F22 = 0.3286 \n", + "\t\t# using the reciprocity relation (equation 8-18) we have\n", + "\n", + "#Calculation\n", + "F12 = (d2/d1)*F21 \n", + "\t\t# for surface 2 we have F12+F22+F23+F24 = 1.0\n", + "\t\t# and from symmetry F23 = F24 so that\n", + "F23 = (1-F21-F22)/2 \n", + "F24 = F23 \n", + "\t\t# using reciprocity again,\n", + "import math\n", + "\n", + "A2 = math.pi*d2*L \t\t\t# [m**2]\n", + "A3 = math.pi*(d2**2-d1**2)/4 \t\t# [m**2]\n", + "F32 = A2*F23/A3 \n", + "\t\t# we observe that F11 = F33 = F44 = 0 & for surface 3 F31+F32+F34 =1.0\n", + "\t\t# so, if F31 can be determined, we can calculate the desired quantity \t\tF34. for surface 1 F12+F13+F14 = 1.0\n", + "\t\t# and from symmetry F13 = F14 so that\n", + "F13 = (1-F12)/2 \n", + "F14 = F13 \n", + "\t\t# using reciprocity gives\n", + "A1 = math.pi*d1*L \t\t\t# [square meter]\n", + "F31 = (A1/A3)*F13 \n", + "\t\t# then \n", + "F34 = 1-F31-F32 \n", + "\n", + "#Result\n", + "print \"Shape factor between the open ends of the cylinder is\",F34 " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Shape factor between the top surface and the side is 0.52\n", + "Shape factor between the side and itself is 0.398\n" + ] + } + ], + "source": [ + "#Example Number 8.4\n", + "# shape-factor algebra for truncated cone\n", + "\n", + "#Variable declaration\n", + "\n", + "d1 = 0.1 \t\t\t# [m] diameter of top of cone\n", + "d2 = 0.2 \t\t\t# [m] diameter of bottom of cone\n", + "L = 0.1 \t\t\t# [m] height of cone\n", + "\t\t#we employ figure 8-16(page no.-390) for solution of this problem and \t\ttake the nomenclature as shown, designating the top as surface 2,\n", + "\t\t# the bottom as surface 1, and the side as surface 3. thus the desired \t\tquantities are F23 and F33. we have \n", + "Z = L/(d2/2) \n", + "Y = (d1/2)/L \n", + "\t\t# thus from figure 8-16(page no.-390) \n", + "F12 = 0.12 \n", + "\t\t# from reciprcity(equatin 8-18)\n", + "\n", + "import math\n", + "A1 = math.pi*d2**(2)/4 \t\t# [square meter]\n", + "A2 = math.pi*d1**(2)/4 \t\t# [square meter]\n", + "F21 = A1*F12/A2 \n", + "\t\t#and\n", + "F22 = 0 \n", + "\t\t# so that \n", + "F23 = 1-F21 \n", + "\t\t# for surface 3 F31+F32+F33 = 1, so we must find F31 and F32 in order \t\tto evaluate F33. since F11 = 0 we have\n", + "F13 = 1-F12 \n", + "\t\t# and from reciprocity \n", + "A3 = math.pi*((d1+d2)/2)*((d1/2-d2/2)**(2)+L**(2))**(1.0/2.0) \t# [square meter]\n", + "\t\t# so from above equation\n", + "F31 = A1*F13/A3 \n", + "\t\t# a similar procedure is applies with surface 2 so that \n", + "F32 = A2*F23/A3 \n", + "\t\t# finally from above equation \n", + "F33 = 1-F32-F31 \n", + "print \"Shape factor between the top surface and the side is\",F23 \n", + "print \"Shape factor between the side and itself is\",round(F33,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of F12 is 0.424\n", + "Value of F13 is 0.262\n", + "Value of F11 is 0.313\n" + ] + } + ], + "source": [ + "#Example Number 8.5\n", + "# shape-factor algebra for cylindrical reflactor\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.6 \t\t\t\t# [m] diameter of long half-circular cylinder\n", + "L = 0.2 \t\t\t\t# [m] length of square rod\n", + "\t\t# we have given figure example 8-5(page no.-397) for solution of this \t\t\tproblem and take the nomenclature as shown, \n", + "\t\t# from symmetry we have \n", + "F21 = 0.5 \n", + "F23 = F21 \n", + "\t\t# in general, F11+F12+F13 = 1. to aid in the analysis we create the \t\t\tfictious surface 4 shown in figure example 8-5 as dashed line.\n", + "\t\t# for this surface \n", + "F41 = 1.0 \n", + "\t\t# now, all radiation leaving surface 1 will arrive either at 2 or at \t\t\t3. likewise,this radiation will arrive at the imaginary surface 4, so \t\t\tthat F41 = F12+F13 say eqn a\n", + "\t\t# from reciprocity\n", + "#Calculation\n", + "\n", + "import math\n", + "\n", + "A1 =math.pi*d/2 \t\t\t# [square meter]\n", + "A4 = L+2*math.sqrt(0.1**(2)+L**(2)) \t# [square meter]\n", + "A2 = 4*L \t\t\t\t# [square meter]\n", + "\t\t# so that \n", + "F14 = A4*F41/A1 \t\t\t# say eqn b\n", + "\t\t# we also have from reciprocity\n", + "F12 = A2*F21/A1 \t\t\t# say eqn c\n", + "\t\t# combining a,b,c, gives\n", + "F13 = F14-F12 \n", + "\t\t# finally\n", + "F11 = 1-F12-F13 \n", + "\n", + "#Result\n", + "\n", + "print \"Value of F12 is \",round(F12,3) \n", + "print \"Value of F13 is \",round(F13,3) \n", + "print \"Value of F11 is \",round(F11,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of the insulated surface is 599.4 K\n", + "Heat lost by the surface at 1000K is 8.229 kW\n" + ] + } + ], + "source": [ + "#xample Number 8.7\n", + "# surface in radiant balance\n", + "\n", + "# Variable declaration\n", + "\n", + "w = 0.5 \t\t\t# [m] width of plate \n", + "L = 0.5 \t\t\t# [m] length of plate\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t# from the data of the problem \n", + "T1 = 1000 \t\t\t# [K] temperature of first surface\n", + "T2 = 27+273 \t\t\t# [K] temperature of room\n", + "A1 = w*L \t\t\t# [square meter] area of rectangle\n", + "A2 = A1 \t\t\t# [square meter] area of rectangle\n", + "E1 = 0.6 \t\t\t# emissivity of surface 1\n", + "\t#eq(8-41) may not be used for the calculation because \tone of the heat-exchanging surfaces is not convex. The radiation \t\t\tnetwork is shown in figure example 8-7(page no.-404) where surface 3 is the \t\troom and surface 2 is the insulated surface. n\tJ2 \"floats\" in the network and is determined from the overall radiant balance.\n", + " \n", + "\t# from figure 8-14(page no.-387) the shape factors are \n", + "F12 = 0.2 \n", + "F21 = F12 \n", + "\t# because\n", + "F11 = 0 \n", + "F22 = 0 \n", + "F13 = 1-F12 \n", + "F23 = F13 \n", + "\t# the resistances are \n", + "R1 = (1-E1)/(E1*A1) \n", + "R2 = 1/(A1*F13) \n", + "R3 = 1/(A2*F23) \n", + "R4 = 1/(A1*F12) \n", + "\t# we also have\n", + "Eb1 = sigma*T1**(4) \t\t# [W/square meter]\n", + "Eb3 = sigma*T2**(4) \t\t# [W/square meter]\n", + "J3 = Eb3 \t\t\t# [W/square meter]\n", + "\t# the overall ckt is a series parallel arrangement and the heat transfer is \n", + "R_equiv = R1+(1/((1/R2)+1/(R3+R4))) \n", + "q = (Eb1-Eb3)/R_equiv \t\t # [W]\n", + "\t# this heat transfer can also be written as q = (Eb1-J1)/((1-E1)/(E1*A1))\n", + "\t# inserting the values \n", + "J1 = Eb1-q*((1-E1)/(E1*A1)) \t # [W/square meter]\n", + "\t# the value of J2 is determined from proportioning the resistances between J1 \t\tand J3, so that \n", + "\t# (J1-J2)/R4 = (J1-J3)/(R4+R2)\n", + "J2 = J1-((J1-J3)/(R4+R2))*R4 \t # [W/square meter]\n", + "Eb2 = J2\t\t\t # [W/square meter]\n", + "\t# finally, we obtain the temperature of the insulated surface as\n", + "T2 = (Eb2/sigma)**(1.0/4.0) \t # [K]\n", + "\n", + "print \"Temperature of the insulated surface is\",round(T2,1),\"K\" \n", + "print \"Heat lost by the surface at 1000K is\",round(q/1000,3),\"kW\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.8" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net radiant exchange is 798.0 W\n" + ] + } + ], + "source": [ + "#Example Number 8.8 \n", + "# open hemisphere in large room\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.3 \t\t\t# [m] diameter of hemisphere\n", + "T1 = 500+273 \t\t\t# [degree celsius] temperature of hemisphere\n", + "T2 = 30+273 \t\t\t# [degree celsius] temperature of enclosure \n", + "E = 0.4 \t\t\t# surface emissivity of hemisphere\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)] constant\n", + "\t\n", + "\t# in the given figure example 8-8(page no.-407) we take the inside of the \t\tsphere as surface 1 and the enclosure as surface 2.\n", + "\t# we also create an imaginary surface 3 covering the opening.\n", + "\t# then the heat transfer is given by\n", + "#Calculation\n", + "Eb1 = sigma*T1**(4) \t\t# [W/square meter]\n", + "Eb2 = sigma*T2**(4) \t\t# [W/square meter]\n", + "\n", + "import math\n", + "A1 = 2*math.pi*(d/2)**(2) \t# [square meter] area of surface 1\n", + "\t# calculating the surface resistance \n", + "R1 = (1-E)/(E*A1) \n", + "\t# since A2 tends to 0 so R2 also tends to 0\n", + "R2 = 0 \n", + "\t# recognize that all of the radiation leaving surface 1 which \twill \teventually arrive at enclosure 2 will also hit the imaginary surface \t\t\t3(F12=F13). we also recognize that A1*F13 = A3*F31. but \n", + "F31 = 1.0 \n", + "A3 = math.pi*(d/2)**(2) \t# [square meter]\n", + "F13 = (A3/A1)*F31 \n", + "F12 = F13 \n", + "\t# then calculating space resistance \n", + "R3 = 1/(A1*F12) \n", + "\t# we can claculate heat transfer by inserting the quantities in eq (8-40):\n", + "q = (Eb1-Eb2)/(R1+R2+R3) \t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Net radiant exchange is\",round(q),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.9" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For emissivity of 0.2 the value of effective emissivity is 0.467\n", + " For emissivity of 0.5 the value of effective emissivity is 0.738\n", + "For emissivity of 0.8 the value of effective emissivity is 0.907\n" + ] + } + ], + "source": [ + "#Example Number 8.9\n", + "# effective emissivity of finned surface\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# for unit depth in the z-dimension we have \n", + "A1 = 10 \t\t\t# [square meter]\n", + "A2 = 5 \t\t\t\t# [square meter]\n", + "A3 = 60 \t\t\t# [square meter]\n", + "\t# the apparent emissivity of the open cavity area A1 is given by \tequation(8-47) as \n", + "\t# Ea1 = E*A3/[A1+E*(A3-A1)]\n", + "\t# for const surface emissivity the emitted energy from the total area A1+A2 is\n", + "\t# e1 = Ea1*A1+E*A2*Eb\n", + "\t# and the energy emitted per unit area for that total area is \n", + "\t# e_t = [(Ea1*A1+E*A2)/(A1+A2)]*Eb\n", + "\t# the coeff of Eb is the effective emissivity, E_eff of the combination of the \tsurface and open cavity. inserting \n", + "\t# above equations gives the following values\n", + "\n", + "\n", + "\n", + "#Calculation & Results\n", + "\n", + "\t# for E = 0.2\n", + "\n", + "E = 0.2 \n", + "Ea1 = E*A3/(A1+E*(A3-A1)) \n", + "E_eff = ((Ea1*A1+E*A2)/(A1+A2)) \n", + "\n", + "print \"For emissivity of 0.2 the value of effective emissivity is\",round(E_eff,3) \n", + "\n", + "\t# for E = 0.5\n", + "\n", + "E = 0.5 \n", + "Ea1 = E*A3/(A1+E*(A3-A1)) \n", + "E_eff = ((Ea1*A1+E*A2)/(A1+A2))\n", + " \n", + "print \" For emissivity of 0.5 the value of effective emissivity is \",round(E_eff,3) \n", + "\n", + "\t# for E = 0.8\n", + "\n", + "E = 0.8 \n", + "Ea1 = E*A3/(A1+E*(A3-A1)) \n", + "E_eff = ((Ea1*A1+E*A2)/(A1+A2)) \n", + "\n", + "print \"For emissivity of 0.8 the value of effective emissivity is \",round(E_eff,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.10" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat tranfer is reduced by 93.2 percent\n" + ] + } + ], + "source": [ + "#Example Number 8.10\n", + "# heat transfer reduction with parallel plate shield\n", + "\n", + "# Variable declaration\n", + "\n", + "E1 = 0.3 \t\t\t# emissivity of first plane\n", + "E2 = 0.8 \t\t\t# emissivity of second plane\n", + "E3 = 0.04 \t\t\t# emissivity of shield\n", + "\n", + "#Calculation\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t\t# the heat transfer without the shield is given by \n", + "\t\t# q_by_A = sigma*(T1**4-T2**4)/((1/E1)+(1/E2)-1) = \t\t0.279*sigma*(T1**4-T2**4)\n", + "\t\t# T1 is temp of 1st plane & T2 is temperature of second plane\n", + "\t\t# the radiation network for the problem with the shield in place is \t\t\tshown in figure (8-32) (page no.-410). \n", + "\t\t# the resistances are \n", + "R1 = (1-E1)/E1 \n", + "R2 = (1-E2)/E2 \n", + "R3 = (1-E3)/E3 \n", + "\t\t# the total resistance with the shield is \n", + "R = R1+R2+R3 \n", + "\t\t# and the heat transfer is \n", + "\t\t# q_by_A = sigma*(T1**4-T2**4)/R = 0.01902*sigma*(T1**4-T2**4)\n", + "\n", + "#Result\n", + "\n", + "print \"Heat tranfer is reduced by\",round((((0.279-0.01902)/0.279)*100),1),\"percent\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.11" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of the outer cylinder is 716.0 K\n", + "Total heat lost by inner cylinder is 2826.0 W\n" + ] + } + ], + "source": [ + "#Example Number 8.11\n", + "# open cylindrical shield in large room\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# two concentric cylinders of example(8.3) have \n", + "T1 = 1000 \t\t\t# [K] \n", + "E1 = 0.8 \n", + "E2 = 0.2 \n", + "T3 = 300 \t\t\t# [K] room temperature \n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t# refer to figure example 8-11(page no.-413) for radiation network\n", + "\t# the room is designed as surface 3 and J3 = Eb3, because the room is very \t\tlarge,(i.e.its surface is very small) \n", + "\t# in this problem we must consider the inside and outside of surface 2 and \t\tthus have subscripts i and o to designate the respective quantities. \n", + "\t# the shape factor can be obtained from example 8-3 as\n", + "F12 = 0.8253 \n", + "F13 = 0.1747 \n", + "F23i = 0.2588 \n", + "F23o = 1.0 \n", + "\t# also\n", + "\n", + "#Calculations\n", + "import math\n", + "A1 = math.pi*0.1*0.2 \t\t# [square meter] area of first cylinder\n", + "A2 = math.pi*0.2*0.2 \t\t# [square meter] area of second cylinder\n", + "Eb1 = sigma*T1**4 \t\t# [W/square meter]\n", + "Eb3 = sigma*T3**4 \t\t# [W/square meter]\n", + "\t# the resistances may be calculated as \n", + "R1 = (1-E1)/(E1*A1) \n", + "R2 = (1-E2)/(E2*A2) \n", + "R3 = 1/(A1*F12) \n", + "R4 = 1/(A2*F23i) \n", + "R5 = 1/(A2*F23o) \n", + "R6 = 1/(A1*F13) \n", + "\t# the network could be solved as a series-parallel circuit to obtain the heat \t\ttransfer, butwe will need the radiosities anyway, so we setup three nodal\t\tequations to solve for J1,J2i, and J2o.\n", + "\t# we sum the currents into each node and set them equal to zero:\n", + "\t# node J1: (Eb1-J1)/R1+(Eb3-J3)/R6+(J2i-J1)/R3 = 0\n", + "\t# node J2i: (J1-J2i)/R3+(Eb3-J2i)/R4+(J2o-J2i)/(2*R2) = 0\n", + "\t# node J2o: (Eb3-J2o)/R5+(J2i-J2o)/(2*R2) = 0\n", + "\t# these equations can be solved by matrix method and the solution is \n", + "J1 = 49732 \t\t\t# [W/square meter]\n", + "J2i = 26444 \t\t\t# [W/square meter]\n", + "J2o = 3346 \t\t\t# [W/square meter]\n", + "\t# the heat transfer is then calculated from\n", + "q = (Eb1-J1)/((1-E1)/(E1*A1)) \t# [W]\n", + "\t# from the network we see that\n", + "Eb2 = (J2i+J2o)/2\t\t # [W/square meter]\n", + "\t# and \n", + "T2 = (Eb2/sigma)**(1.0/4.0) \t# [K]\n", + "\t# if the outer cylinder had not been in place acting as a \"shield\" the heat \t\tloss from cylinder 1 could have been calculated from equation(8-43a) as \n", + "q1 = E1*A1*(Eb1-Eb3)\t\t # [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Temperature of the outer cylinder is\",round(T2),\"K\" \n", + "print \"Total heat lost by inner cylinder is\",round(q1),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.12" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat-transfer rate between the two planes is 5621.0 W/square meter\n", + "Temperature of the gas is 592.4 K\n", + "Ratio of heat-transfer with presence of gas to without presence of gas is 0.97\n" + ] + } + ], + "source": [ + "#Example Number 8.12\n", + "# network for gas radiation between parallel plates\n", + "\n", + "# Variable declaration\n", + "\n", + "T1 = 800 \t\t\t\t# [K] temperature of first plate \n", + "E1 = 0.3 \t\t\t\t# emissivity\n", + "T2 = 400 \t\t\t\t# [K] temperature of second plate\n", + "E2 = 0.7 \t\t\t\t# emissivity\n", + "Eg = 0.2 \t\t\t\t# emissivity of gray gas\n", + "tg = 0.8 \t\t\t\t# transmissivity of gray gas \n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "\t# the network shown in figure 8-39(page no.-419) applies to this problem. all \t\tthe shape factors are unity for large planes and the various resistors can be \t\tcomputed on a unit area basis as \n", + "\n", + "#Calculation\n", + "F12 = 1 \n", + "F1g = 1 \n", + "F2g = F1g \n", + "R1 = (1-E1)/E1 \n", + "R2 = (1-E2)/E2 \n", + "R3 = 1/(F12*(1-Eg)) \n", + "R4 = 1/(F1g*Eg) \n", + "R5 = 1/(F2g*Eg) \n", + "Eb1 = sigma*T1**(4) \t\t\t# [W/square meter]\n", + "Eb2 = sigma*T2**(4) \t\t\t# [W/square meter]\n", + "\n", + "\t# the equivalent resistance of the center \"triangle\" is \n", + "\n", + "R = 1/((1/R3)+(1/(R4+R5))) \n", + "\n", + "\t# the total heat transfer is then \n", + "\n", + "q_by_A = (Eb1-Eb2)/(R1+R2+R) \t\t# [W/square meter]\n", + "\n", + "\t# heat transfer would be given by equation (8-42):\n", + "\n", + "q_by_A1 = (Eb1-Eb2)/((1/E1)+(1/E2)-1) \t# [W/square meter]\n", + "\n", + "\t# the radiosities may be computed from q_by_A = (Eb1-J1)*(E1/(1-E1)) = \t(J2-Eb2)*(E2/(1-E2))\n", + "\n", + "J1 = Eb1-q_by_A*((1-E1)/E1) \t\t# [W/square meter]\n", + "J2 = Eb2+q_by_A*((1-E2)/E2) \t\t# [W/square meter]\n", + "\n", + "\t# for the network Ebg is just the mean of these values\n", + "\n", + "Ebg = (J1+J2)/2 \t\t\t# [W/square meter]\n", + "\n", + "\t# so that the temperature of the gas is\n", + "Tg = (Ebg/sigma)**(1.0/4.0) \t\t# [K]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat-transfer rate between the two planes is\",round(q_by_A),\"W/square meter\" \n", + "print \"Temperature of the gas is\",round(Tg,1),\"K\" \n", + "print \"Ratio of heat-transfer with presence of gas to without presence of gas is\",round(q_by_A/q_by_A1,2) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.13" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Apparent emissivity of covered opening is 0.6269\n", + "If there were no cover present, the value of Ea(apparent emissivity) would be 0.8571\n" + ] + } + ], + "source": [ + "#Example Number 8.13\n", + "# cavity with transparent cover \n", + "\n", + "#Variable declaration\n", + "\n", + "E1 = 0.5 \t\t\t# emissivity of rectangular cavity\n", + "t2 = 0.5 \t\t\t# transmissivity\n", + "rho2 = 0.1 \t\t\t# reflectivity\n", + "E2 = 0.4 \t\t\t# emissivity\n", + "\t# from example 8-9 we have\n", + "\t# per unit depth in the z direction we have \n", + "\n", + "#Calculation\n", + "A1 = 60\n", + "A2 = 10.0 \n", + "\t# we may evaluate K from equation(8-96a)\n", + "K = E1/(t2+(E2/2))\n", + "\n", + "\n", + "\t# the value of Ea is then computed from equation (8-96) as \n", + "Ea = (t2+(E2/2))*K/((A2/A1)*(1-E1)+K) \n", + "print \"Apparent emissivity of covered opening is \",round(Ea,4) \n", + "\t# when no cover present, the value of Ea would be given by eq (8-47) as\n", + "Ea1 = E1*A1/(A2+E1*(A1-A2)) \n", + "\n", + "#Result\n", + "print \"If there were no cover present, the value of Ea(apparent emissivity) would be \",round(Ea1,4) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.14" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radiation lost through the quartz window to a room temperature of 30 degree celsius is 112171.0 W/square meter\n", + "With no windows at all, the heat transfer would be 148397.0 W/square meter\n" + ] + } + ], + "source": [ + "#Example Number 8.14\n", + "# Transmitting and reflecting system for furnace opening\n", + "\n", + "# Variable declaration\n", + "\t\n", + "T1 = 1000+273 \t\t\t# [K] temperature of furnace\n", + "lamda = 4.0 \t\t\t# [micro meter]\n", + "\t\t#for 0 < lamda < 4 micro meter\n", + "t1 = 0.9 \n", + "E1 = 0.1 \n", + "rho1 = 0 \n", + "\t\t#for 4 micro meter < lamda < infinity \n", + "t2 = 0 \n", + "E2 = 0.8 \n", + "rho2 = 0.2 \n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "T3 = 30+273\t\t \t# [K] room temperature\n", + "\t\t# because the room is large it may be treated as a blackbody.\n", + "\t\t# analyze the problem by calculating the heat transfer for each \t\t\twavelength band and then adding them together to obtain the total. the \t\tnetwork for each band is a modification of figure 8-57 \n", + "A1 = 1.0 \t\t\t# [square meter]\n", + "A2 = 1.0 \t\t\t# [square meter]\n", + "A3 = 1.0 \t\t\t# [square meter]\n", + "F12 = 1.0 \n", + "F13 = 1.0 \n", + "F32 = 1.0 \n", + "\t\t# the total emissive powers are \n", + "\n", + "#Calculation\n", + "\n", + "Eb1 = sigma*T1**(4) \t\t# [W/square meter]\n", + "Eb3 = sigma*T3**(4) \t\t# [W/square meter]\n", + "\t\t# to determine the fraction of radiation in each wavelength band\n", + "lamba_into_T1 = lamda*T1 \t# [micro meter K]\n", + "lamba_into_T3 = lamda*T3 \t# [micro meter K]\n", + "\t\t# consulting table 8-1, we find \n", + "Eb1_0_to_4 = 0.6450*Eb1 \t# [W/square meter]\n", + "Eb3_0_to_4 = 0.00235*Eb3 \t# [W/square meter]\n", + "Eb1_4_to_inf = (1-0.6450)*Eb1 \t# [W/square meter]\n", + "Eb3_4_to_inf = (1-0.00235)*Eb3 \t# [W/square meter]\n", + "\t\t# apply these numbers to the network for the two wavelengths bands, \t\t\twith unit areas.\n", + "\n", + "\t\t# 0 < lamda < 4 micro meter band:\n", + "R1 = 1/(F13*t1) \n", + "R2 = 1/(F32*(1-t1)) \n", + "R3 = 1/(F12*(1-t1)) \n", + "R4 = rho1/(E1*(1-t1)) \n", + "\t\t# the net heat transfer from the network is then \n", + "R_equiv_1 = 1/(1/R1+1/(R2+R3+R4)) \n", + "q1 = (Eb1_0_to_4-Eb3_0_to_4)/R_equiv_1 # [W/square meter]\n", + "\n", + "\t\t# 4 micro meter < lamda < infinity band:\n", + "R2 = 1/(F32*(1-t2)) \n", + "R3 = 1/(F12*(1-t2)) \n", + "R4 = rho2/(E2*(1-t2)) \n", + "\t\t# the net heat transfer from the network is then \n", + "\t\t# R1 is infinity\n", + "R_equiv_2 = R2+R3+R4*2 \n", + "q2 = (Eb1_4_to_inf-Eb3_4_to_inf)/R_equiv_2 # [W/square meter]\n", + "\n", + "\t\t# the total heat loss is then \n", + "q_total = q1+q2 \t\t\t# [W/square meter]\n", + "\t\t# with no windows at all, the heat transfer would have been the \t\t\tdifference in blackbody emissive powers,\n", + "Q = Eb1-Eb3 \t\t\t\t# [W/square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Radiation lost through the quartz window to a room temperature of 30 degree celsius is\",round(q_total),\"W/square meter\" \n", + "\n", + "print \"With no windows at all, the heat transfer would be\",round(Q),\"W/square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.20" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with :\n", + "White paint is 39.5 degree celsius\n", + "Flat black lacquer is 104.8 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 8.20\n", + "# solar-environment equilibrium temperatures \n", + "\n", + "# Variable declaration\n", + "\n", + "q_by_A_sun = 700 \t\t\t# [W/m**(2)] solar flux\n", + "T_surr = 25+273 \t\t\t# [K] surrounding temperature\n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "\t# at radiation equilibrium the netenergy absorbed from sun must equal the \tlong-wavelength radiation exchange with the surroundings,or\n", + "\t# (q_by_A_sun)*alpha_sun = alpha_low_temp*sigma*(T**4-T_surr**4) (a)\n", + "\n", + "\t# case (a) for white paint\n", + "\n", + "\t# for white paint we obtain from table 8-4\n", + "\n", + "#Calculation\n", + "\n", + "alpha_sun = 0.12 \n", + "alpha_low_temp = 0.9 \n", + "\t# so that equation (a) becomes\n", + "T = ((q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr**(4))**(1.0/4.0)\t # [K]\n", + "\n", + "\n", + "#Result\n", + "\n", + "print\"Radiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with :\"\n", + "print \"White paint is\",round(T-273,1),\"degree celsius\" \n", + "\n", + "\t# case (b) for flat black lacquer we obtain\n", + "\n", + "alpha_sun = 0.96 \n", + "alpha_low_temp = 0.95 \n", + "\t# so that equation (a) becomes\n", + "T = ((q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr**(4))**(1.0/4.0) \t# [K]\n", + "\n", + "print \"Flat black lacquer is\",round(T-273,1),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.23" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the true air temperature is 28.6 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 8.23\n", + "# temperature measurement error caused by radiation \n", + "\n", + "# Variable declaration\n", + "\n", + "E = 0.9 \t\t\t\t# emissivity of mercury-in-glass thermometer \n", + "Tt = 20+273 \t\t\t\t# [K] temperature indicated by thermometer \n", + "Ts = 5+273 \t\t\t\t# [K] temperature of walls\n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K^(4)]\n", + "h = 8.3 \t\t\t\t# [W/sq m] heat transfer coefficient for \t\t\t\t\tthermometer\n", + "\t# we employ equation(8-113) for the solution: h*(Tinf-Tt) =sigma*E*(Tt^4-Ts^4)\n", + "\t# inserting the values in above equation \n", + "\n", + "Tinf = sigma*E*(Tt**4-Ts**4)/h+Tt \t# [K]\n", + "print\"the true air temperature is\",round(Tinf-273,1),\"degree celsius\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb new file mode 100644 index 00000000..977f0e4b --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb @@ -0,0 +1,308 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 Condensation and Boiling Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of reynolds no. is 49.7 so the laminar assumption was correct\n", + "The heat transfer is 2368.0 w\n", + "Total mass flow condensate is 3.78 kg/h\n" + ] + } + ], + "source": [ + "#Example Number 9.1\n", + "# condensation on vertical plate\n", + "\n", + "#Variable declaration\n", + "\n", + "\t# we have to check the reynolds no. to that film is laminar or turbulent\n", + "Tf = (100+98)/2 \t\t\t# [degree celsius]\n", + "Tw = 98 \t\t\t\t# [degree celsius]\n", + "RHOf=960 \t\t\t\t# [kg/cubic meter] \n", + "MUf=2.82*10**(-4) \t\t\t# [kg/m s]\n", + "Kf=0.68 \t\t\t\t# [W/m degree celsius]\n", + "g=9.81 \t\t\t\t\t# [m/s**(2)]\n", + "L=0.3 \t\t\t\t\t# [m]\n", + "\t# RHOf(RHOf-RHOv)~RHOf**(2)\n", + "\t# let us assume laminar film condensate \n", + "Tsat=100 \t\t\t\t# [degree celsius]\n", + "Tg=100 \t\t\t\t\t# [degree celsius]\n", + "Hfg=2255*10**(3) \t\t\t# [J/kg]\n", + "\n", + "#Calculation\n", + "\n", + "hbar=0.943*((RHOf**(2)*g*Hfg*Kf**(3)/(L*MUf*(Tg-Tw)))**(0.25)) \t# [W/sq m deg celsius]\n", + "h=hbar \t\t\t\t\t# [W/square meter degree celsius]\n", + "\t# checking reynolds no. with equation(9-17)\n", + "Ref=4*h*L*(Tsat-Tw)/(Hfg*MUf) \n", + "\n", + "print \"Value of reynolds no. is\",round(Ref,1),\"so the laminar assumption was correct\" \n", + "\t# the heat transfer is now calculated from \n", + "A=0.3*0.3 \t\t\t\t# [square meter]\n", + "q=hbar*A*(Tsat-Tw) \t\t\t# [W]\n", + "mdot=q/Hfg \t\t\t\t# [kg/h]\n", + "\n", + "#Result\n", + "\n", + "print \"The heat transfer is\",round(q),\"w\" \n", + "mdot=mdot*3600 \t\t\t\t# [kg/h]\n", + "print \"Total mass flow condensate is\",round(mdot,2),\"kg/h\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total surface area is 3.99 square meter/m\n", + "Heat transfer is 100.1 kW/m\n", + "Total mass flow of condensate is 159.8 kg/h\n" + ] + } + ], + "source": [ + "#Example Number 9.2\n", + "# condensation on tube tank\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# the condensate properties are obtained from previous example\n", + "\t# replacing L by n*d\n", + "Tw=98 \t\t\t\t# [degree celsius]\n", + "RHOf=960 \t\t\t# [kg/cubic meter] \n", + "MUf=2.82*10**(-4) \t\t# [kg/m s]\n", + "Kf=0.68 \t\t\t# [W/m degree celsius]\n", + "g=9.81 \t\t\t\t# [m/s^(2)]\n", + "Tsat=100 \t\t\t# [degree celsius]\n", + "Tg=100 \t\t\t\t# [degree celsius]\n", + "Hfg=2255*10**(3) \t\t# [J/kg]\n", + "d=0.0127 \t\t\t# [m]\n", + "n=10 \n", + "\n", + "#Calculation\n", + "\n", + "hbar=0.725*((RHOf**(2)*g*Hfg*Kf**(3)/(n*d*MUf*(Tg-Tw)))**(0.25)) # [W/sq m deg C]\n", + "\t# total surface area is \n", + "n=100 \n", + "Al=n*22*d/7 \t\t\t# [square meter]\n", + "print \"Total surface area is\",round(Al,2),\"square meter/m\" \n", + "\t# so the heat transfer is \n", + "Ql=hbar*Al*(Tg-Tw)\t\t # [W]\n", + "\n", + "#Result \n", + "\n", + "print \"Heat transfer is\",round(Ql/1000,2),\"kW/m\" \n", + "\t# total mass flow of condensate is then \n", + "mdotl=Ql/Hfg \t\t\t# [kg/h]\n", + "mdotl=mdotl*3600\t\t# [kg/h]\n", + "print \"Total mass flow of condensate is\",round(mdotl,1),\"kg/h\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer in a 1.0 m length of tube is 2810.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 9.4\n", + "# Flow boiling\n", + "# Variable declaration\n", + "\n", + "p =0.5066\t\t\t# [MPa] pressure of water \n", + "d = 0.0254 \t\t\t# [m] diameter of tube \n", + "Tw = 10.0 \t\t\t# [degree celsius]\n", + "\t\t\t\t# for calculation we use equation (9-45), noting that \n", + "dT = 10.0 \t\t\t# [degree celsius]\n", + "\t\t\t\t# the heat transfer coefficient is calculated as \n", + "\n", + "import math\n", + "h = 2.54*Tw**(3)*math.exp(p/1.551) \t# [W/square meter degree celsius]\n", + "\n", + "\t\t# the surface area for a 1-m length of tube is \n", + "L = 1 \t\t\t\t# [m]\n", + "import math\n", + "A = math.pi*d*L \t\t# [square meter]\n", + "\n", + "\t\t# so the heat transfer is \n", + "q = h*A*dT \t\t\t# [W/m]\n", + "print \"Heat transfer in a 1.0 m length of tube is\",round(q),\"W/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flux obtained is 22.8 kW/square meter\n", + "If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is 23.0 percent\n" + ] + } + ], + "source": [ + "#Example Number 9.5\n", + "# water boiling in a pan \n", + "\n", + "# Variable declaration\n", + "\n", + "p = 0.101\t\t\t\t# [MPa] pressure of water \n", + "dT_x = 8 \t\t\t\t# [degree celsius]\n", + "p1 = 0.17 \t\t\t\t# [MPa] given operating pressure\n", + "\t# we will use the simplified relation of table 9-13(page no.-506) for the \t\testimates.we do not know the value of q_by_A and so must choose one of the two \trelation for a horizontal surface from the table\n", + "\t# we anticipate nucleate boiling, so choose\n", + "h = 5.56*dT_x**(3) \t\t\t# [W/square meter degree celsius]\n", + "\t\t\t\t\t# and the heat flux is \n", + "q_by_A = h*dT_x \t\t\t# [W/square meter]\n", + "\t# for operation as a pressure cooker we obtain the value of h from \t\tequation(9-44)\n", + "\n", + "#Calculation\n", + "\n", + "hp = h*(p1/p)**(0.4) \t\t\t# [W/square meter degree celsius]\n", + "\t# the corresponding heat flux is \n", + "q_by_A1 = hp*dT_x \t\t\t# [W/square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat flux obtained is\",round(q_by_A/1000,1),\"kW/square meter\" \n", + "per_inc = 100*(q_by_A1-q_by_A)/q_by_A \n", + "\n", + "print \"If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is\",round(per_inc),\"percent\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thus the heat transfers 14.0 times the heat of a pure copper rod with a substantial temperature gradient\n" + ] + } + ], + "source": [ + "#Example Number 9.6\n", + "# heat-flux comparisons \n", + "\n", + "# Variable declaration\n", + "\n", + "Tw = 200.0 \t\t\t\t# [degree celsius] water temperature \n", + "L = 0.08 \t\t\t\t# [m] length of solid copper bar\n", + "dT = 100.0 \t\t\t\t# [degree C] temp differential in copper bar\n", + "\t#using the data of table 9-4(page no.-508)\n", + "\t# the heat flux per unit area is expressed as q_by_A = -k*del_T/dx\n", + "\t# from table A-2(page no.-) the thermal conductivity of copper is \n", + "k = 374.0 \t\t\t\t# [W/m degree celsius]\n", + "\n", + "#Calculation\n", + "\n", + "q_by_A = -k*(-dT)/L \t\t\t# [W/square meter]\n", + "\n", + "\t# from table 9-4(page no.-508) the typical axial heat flux for a water heat \t\tflux for a water heat pipe is \n", + "q_by_A_axial = 0.67 \t\t\t# [kW/csquare meter]\n", + "q_by_A = q_by_A/(1000*10**(4.0)) \t# [kW/csquare meter]\n", + "time=q_by_A_axial/q_by_A\n", + "\n", + "#Result\n", + "\n", + "print \"Thus the heat transfers \",round(time),\"times the heat of a pure copper rod with a substantial temperature gradient\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_in_SI_units_by_Holman/README.txt b/Heat_Transfer_in_SI_units_by_Holman/README.txt new file mode 100644 index 00000000..948e81e2 --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/README.txt @@ -0,0 +1,10 @@ +Contributed By: Deepak Shakya +Course: btech +College/Institute/Organization: DCRUST +Department/Designation: Chemical Engg +Book Title: Heat Transfer in SI units +Author: Holman +Publisher: Tata McGraw - Hill Education, New Delhi +Year of publication: 2002 +Isbn: 0-07-0634513 +Edition: 9
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100644 index 00000000..662bf254 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter1.ipynb @@ -0,0 +1,558 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Advanced Operational Amplifier Principles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.1,Page 6" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "open output voltage is 0.5 V\n", + "resistance lower loaded is 333.333 ohm\n", + "loaded output voltage is 0.25 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=1000.0;\n", + "R2=1000.0;\n", + "Rl=500.0#load resistance\n", + "V=1.0#input voltage\n", + "\n", + "#calculation\n", + "Vo=(R2/(R1+R2))*V;\n", + "Rll=1/((1/R2)+(1/Rl))#lower loaded resistance\n", + "Vol=(Rll/(R2+Rll))*V;\n", + "\n", + "#result\n", + "print \"open output voltage is\",round(Vo,3),\"V\"\n", + "print \"resistance lower loaded is\",round(Rll,3),\"ohm\"\n", + "print \"loaded output voltage is\",round(Vol,3),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.2,Page 11" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistance is 1.01 Kohm\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=100000.0#resistance\n", + "Acl=100.0#amplifier gain\n", + "\n", + "#calculation\n", + "Ri=Rf/(Acl-1);\n", + "\n", + "#result\n", + "print \"input resistance is\",round(Ri/1000,2), \"Kohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.3,Page 17" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current through Ri1 is 178.571 microAmp\n", + "current through Ri2 is 31.915 microAmp\n", + "current through Ri2 is 31.915 microAmp\n", + "current through Rf is 210.486 microAmp\n", + "voltage dropped is 2.105 V\n", + "output voltage 1 is -2.105 V\n", + "output voltage is 2.105 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vni=0.0#non inverting voltage\n", + "Vinv=0.0;#inverting voltage\n", + "Vri1=1.0;\n", + "Vri2=15.0;\n", + "Ri1=5600.0#resistance\n", + "Ri2=470000.0;\n", + "Rf=10000.0#load resistance\n", + "\n", + "#calculation\n", + "Ir1=Vri1/Ri1;\n", + "Ir2=Vri2/Ri2;\n", + "Irf=(Vri1/Ri1)+(Vri2/Ri2);\n", + "Vr=Irf*Rf;\n", + "Vo1=-Vr;\n", + "Vo=Irf*Rf;\n", + "\n", + "#result\n", + "print \"current through Ri1 is\",round(Ir1*1e6,3), \"microAmp\"\n", + "print \"current through Ri2 is\",round(Ir2*1e6,3), \"microAmp\"\n", + "print \"current through Ri2 is\",round(Ir2*1e6,3),\"microAmp\"\n", + "print \"current through Rf is\",round(Irf*1e6,3), \"microAmp\"\n", + "print \"voltage dropped is\",round(Vr,3), \"V\"\n", + "print \"output voltage 1 is\",round(Vo1,3), \"V\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.4,Page 25" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inverting voltage is 4.955 V\n", + "non inverting voltage is 4.955 V\n", + "current through Rf2 is 42.698 microA\n", + "current through Ri2 is 42.698 microA\n", + "voltage dropped is 4.056 V\n", + "output voltage is 884.897 mV\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ri1=950.00;#ohm\n", + "Ri2=1050.00;\n", + "Rf1=105000.00;#resistance\n", + "Rf2=95000.00;\n", + "Vin=5.00;#voltage\n", + "\n", + "#calculation\n", + "Vinv=(Rf1/(Rf1+Ri1))*Vin;\n", + "Vni=Vinv;\n", + "Irf2=(Vin-Vinv)/Ri2;\n", + "Iri2=Irf2;\n", + "Vrf2=Irf2*Rf2;\n", + "Vo=Vinv-Vrf2-.014;\n", + "\n", + "#result\n", + "print \"inverting voltage is\",round(Vinv,3), \"V\"\n", + "print \"non inverting voltage is\",round (Vni,3), \"V\"\n", + "print \"current through Rf2 is\",round(Irf2*1e6,3), \"microA\"\n", + "print \"current through Ri2 is\",round(Iri2*1e6,3), \"microA\"\n", + "print \"voltage dropped is\",round(Vrf2,3), \"V\"\n", + "print \"output voltage is\",round(Vo*1000,3), \"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.5,Page 27" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 272.222 microA\n", + "input resistor current is 500.0 microA\n", + "feedback resistor current is 227.778 microAmp\n", + "resistor voltage is 227.778 mV\n", + "1st output voltage is 2.222 V\n", + "input resistor current is 327.778 microA\n", + "input resistor current is 827.778 microA\n", + "feedback resistor voltage is 7.45 V\n", + "2nd output voltage is 10.0 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vniu1=2.45;#V\n", + "Vniu2=2.55;#V\n", + "Vinvu1=2.45;\n", + "Vinvu2=2.55;\n", + "Ri1=9000.0;#ohm\n", + "Ri2=1000.0;#ohm\n", + "Rf1=1000.0;\n", + "Rf2=9000.0;\n", + "Rg=200.0;#load resistance\n", + "\n", + "#calculation\n", + "Iri1=Vniu1/Ri1;\n", + "Irg=(Vniu2-Vniu1)/Rg;\n", + "Irf1=Irg-Iri1;\n", + "Vrf1=Irf1*Rf1;\n", + "Vou1=Vniu1-Vrf1;\n", + "Iri2=(Vniu2-Vou1)/Ri2;\n", + "Irf2=Iri2+Irg;\n", + "Vrf2=Irf2*Rf2#feedback resistor voltage\n", + "Vo=Vrf2+Vniu2;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Iri1*1e6,3), \"microA\"\n", + "print \"input resistor current is\",round(Irg*1e6,3), \"microA\"\n", + "print \"feedback resistor current is\",round(Irf1*1e6,3), \"microAmp\"\n", + "print \"resistor voltage is\",round(Vrf1*1000,3), \"mV\"\n", + "print \"1st output voltage is\",round(Vou1,3), \"V\"\n", + "print \"input resistor current is\",round(Iri2*1e6,3), \"microA\"\n", + "print \"input resistor current is\",round(Irf2*1e6,3),\"microA\"\n", + "print \"feedback resistor voltage is\",round(Vrf2,3), \"V\"\n", + "print \"2nd output voltage is\",round(Vo,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.6.a,Page 29" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 128.0 microA\n", + "feedback resistor current is 128.0 microA\n", + "feedback resistor voltage is 5.018 V\n", + "output resistor voltage is 5.018 V\n", + "output voltage is 3.818 V\n", + "load current is 0.5 A\n", + "load power is 2.5 W\n", + "power dissipated in LM317 is 5.0 W\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vniu1=0;#V\n", + "Vinvu2=0;#V\n", + "Vref=2.56;\n", + "Rl=10000.0;#ohm\n", + "Rf=39200.0;#ohm\n", + "Ro=10.0;#resistance\n", + "Vdc1=5.0;\n", + "Vdc2=15.0;\n", + "Idc=0.5;#current\n", + "\n", + "#calculation\n", + "Iu1=(Vref/Rl)*.5;\n", + "Irf=Iu1;\n", + "Vrf=Irf*Rf;\n", + "Vout=Vrf+Vinvu2;\n", + "Eo=Vout-1.2;\n", + "Iload=Vdc1/Ro;\n", + "Pload=Vdc1**2/Ro;\n", + "Plm317=(Vdc2-Vdc1)*Idc;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Iu1*1e6,3), \"microA\"\n", + "print \"feedback resistor current is\",round(Irf*1e6,3), \"microA\"\n", + "print \"feedback resistor voltage is\",round(Vrf,3), \"V\"\n", + "print \"output resistor voltage is\",round(Vout,3), \"V\"\n", + "print \"output voltage is\",round(Eo,3), \"V\"\n", + "print \"load current is\",round(Iload,3), \"A\"\n", + "print \"load power is\",round(Pload,3), \"W\"\n", + "print \"power dissipated in LM317 is\",round(Plm317,3), \"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.6.b,Page 31" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 360.36 microamp\n", + "inverting voltage 1 & 2 is 396.396 mV\n", + "current across Rs is 3.964 A\n", + "emitter voltage is 8.324 V\n", + "output voltage is 10.124 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=4;#V\n", + "Vs=1.8;#V\n", + "Rf=10000.0;#ohm\n", + "Ri=1100.0;#ohm\n", + "Rl=2.0;#ohm\n", + "Rs=0.1;#ohm\n", + "\n", + "#calculation\n", + "Irf=Vin/(Rf+Ri);\n", + "Vni=Irf*Ri;\n", + "Ir=Vni/Rs;\n", + "Ve=Ir*(Rl+Rs);\n", + "Vo=Ve+Vs;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Irf*1e6,3),\"microamp\"\n", + "print \"inverting voltage 1 & 2 is\",round(Vni*1000,3), \"mV\"\n", + "print \"current across Rs is\",round(Ir,3), \"A\"\n", + "print \"emitter voltage is\",round(Ve,3), \"V\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.7,Page 36" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 9.899 V\n", + "power delivered is 12.25 W\n", + "load voltage is 28.284 V\n" + ] + } + ], + "source": [ + "#finding voltage and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vs=18.0;#V\n", + "Rl=8.0;#load resistance\n", + "Pll=100.0;#power\n", + "\n", + "#calculation\n", + "Vlp=Vs-4;\n", + "Vlr=Vlp/(2**(.5));\n", + "Pl=(Vlr**2)/Rl;\n", + "Vl=(Pll*Rl)**(.5);\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vlr,3), \"V\"\n", + "print \"power delivered is\",round(Pl,3), \"W\"\n", + "print \"load voltage is\",round(Vl,3), \"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.9,Page 44" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 37.34 V\n", + "V+ is 45.34 V ;V- is 29.34 V\n" + ] + } + ], + "source": [ + "#finding output volatage and range \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "import numpy as np\n", + "Vp=6.0;#V\n", + "Ra=10.0;#Kohm\n", + "Rb=1800.0;#ohm\n", + "V=8.0;\n", + "#solving for Ir & Vo\n", + "a=np.array([[1.0,-124.6e-6],[7800.0,-1.0]])\n", + "b=np.array([134.6e-6,0.0])\n", + "\n", + "#calculation\n", + "x=np.linalg.solve(a,b);\n", + "Vo=x[1];\n", + "Va=Vo+V;\n", + "Vb=Vo-V;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vo,2), \"V\"\n", + "print \"V+ is\",round(Va,2), \"V ;V- is\",round(Vb,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.11,Page 50" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output current is 4.091 mA\n", + "output voltage is 45.409 V\n", + "gain output voltage 1 is 13.356 V\n", + "gain output voltage 2 is 0.38 V\n" + ] + } + ], + "source": [ + "#finding output voltage and gain output voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=4.5;\n", + "R1=1100.0;\n", + "R2=10000.0;\n", + "\n", + "G1=3.4#gain 1\n", + "G2=120.0#gain 2\n", + "\n", + "#calculation\n", + "Ir=Vin/R1;\n", + "Vo=Ir*(R1+R2);\n", + "Vuo1=Vo/G1;\n", + "Vuo2=Vo/G2;\n", + "\n", + "#result\n", + "print \"output current is\",round(Ir*1000,3),\"mA\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"\n", + "print \"gain output voltage 1 is\",round(Vuo1,3), \"V\"\n", + "print \"gain output voltage 2 is\",round(Vuo2,2),\"V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter2.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter2.ipynb new file mode 100644 index 00000000..9d88fe28 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter2.ipynb @@ -0,0 +1,65 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Power Electronics Circuit Layout" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.9,Page 83" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load current is 3.75 A\n", + "wiring resistance is 26.67 mohm\n", + "resistance per inch is 1666.67 microohm/inch\n" + ] + } + ], + "source": [ + "#finding an appropriate wire gauge\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=15.0;#voltage\n", + "R=4.0;#resistance\n", + "Vl=.1;\n", + "D=8.0;#duty cycle\n", + "\n", + "#calculation\n", + "Il=V/R;\n", + "Rw=Vl/Il#wiring resistance\n", + "Ri=Rw/(2*D);\n", + "\n", + "#result\n", + "print \"load current is\",round(Il,2), \"A\"\n", + "print \"wiring resistance is\",round(Rw*1000,2), \"mohm\"\n", + "print \"resistance per inch is\",round(Ri*1e6,2), \"microohm/inch\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter3.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter3.ipynb new file mode 100644 index 00000000..4129f088 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter3.ipynb @@ -0,0 +1,481 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Power Parameter Calculations" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.1,Page 109" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ramp current is 450.0 kAt/s\n", + "current at 5 micro sec is 2.25 A\n" + ] + } + ], + "source": [ + "#finding ramp current and current at 5 micro sec\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=3.0;\n", + "f=150000.0;\n", + "t=5.0e-6;\n", + "\n", + "#calculation\n", + "T=1/f;\n", + "It=Ip/T;\n", + "I5=It*t;\n", + "\n", + "#result\n", + "print \"ramp current is\",round(It/1000,3), \"kAt/s\"\n", + "print \"current at 5 micro sec is\",round(I5,3), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.2,Page 110" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current in time 0<=t<800ns is 3.575 A\n", + "current in time 800ns<=t<2 microsec is 0.0 A\n", + "current in time 400ns is 1.85 A\n", + "current in time 1 microsec is 0.0 A\n" + ] + } + ], + "source": [ + "#finding current at different time\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=2.0;\n", + "f=500000.0;\n", + "Ir=.3;\n", + "Cd=.4#duty cycle\n", + "t1=4.0e-7;\n", + "t2=1.0e-6;\n", + "I1=0;\n", + "\n", + "#calculation\n", + "T=1/f;\n", + "Im=Ip-Ir;\n", + "I4=(Ip-Im)*t1/(Cd*T)+Im;\n", + "It=(Ip-Im)*t/(Cd*T)+Im;\n", + "It1=0\n", + "\n", + "#resilt\n", + "print \"current in time 0<=t<800ns is\",round(It,3),\"A\"\n", + "print \"current in time 800ns<=t<2 microsec is\",round(It1,2), \"A\"\n", + "print \"current in time 400ns is\",round(I4,2), \"A\"\n", + "print \"current in time 1 microsec is\",round(I1,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.3,Page 115" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 54.02 V\n" + ] + } + ], + "source": [ + "#finding average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vr=120;\n", + "\n", + "#calculation\n", + "V=(Vr*2**.5)/pi;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(V,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.4,Page 119" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average current is 0.98 A\n" + ] + } + ], + "source": [ + "#finding average current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "f=100000.0;\n", + "Cd=.35#duty cycle\n", + "Ip=3.0;\n", + "Ir=.4;\n", + "\n", + "#calculation\n", + "Im=Ip-Ir;\n", + "T=1/f;\n", + "I=Cd*((Ip-Im)/2+Im)\n", + "\n", + "#result\n", + "print \"average current is\",round(I,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.5,Page 124" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 8.87 V\n" + ] + } + ], + "source": [ + "#finding rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=15.0;\n", + "Cd=.35;\n", + "f=100000.0;\n", + "\n", + "#calculation\n", + "V=Vp*Cd**.5;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(V,2), \"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.6,Page 127" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms current is 1.73 A\n" + ] + } + ], + "source": [ + "#finding rms current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=3.0;\n", + "f=100000.0;\n", + "\n", + "#calculation\n", + "I=Ip/3**.5;\n", + "\n", + "#result\n", + "print \"rms current is\",round(I,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.7,Page 133" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 85.0 V\n" + ] + } + ], + "source": [ + "#finding rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=170.0;\n", + "f=60.0;\n", + "\n", + "#calculation\n", + "Vr=Vp/2;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vr,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.8,Page 140" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power required is 2.42 hp\n", + "Pick a 5HP motor\n", + "current required is 18.84 amp\n" + ] + } + ], + "source": [ + "#finding current and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "M=1000.0;\n", + "H=40.0;\n", + "T=30.0;\n", + "E1=.9;\n", + "E2=.5;\n", + "V=220.0;\n", + "P1=5.0;\n", + "\n", + "#calculation\n", + "W=M*H;\n", + "P=(W)/(T*550);\n", + "Pe=P1/E1;\n", + "I=(Pe*746)/V;\n", + "\n", + "#result\n", + "print \"power required is\",round(P,2), \"hp\"\n", + "print('Pick a 5HP motor')\n", + "print \"current required is\",round(I,2), \"amp\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.9,Page 145" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered to the load is 6.36 Watt\n", + "power provided by each supply is 7.23 Watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=1.0;\n", + "Ri=1100.0;\n", + "Rf=10000.0;\n", + "Rl=8.0;\n", + "Vs=18.0;\n", + "\n", + "#calculation\n", + "Ir=Vin/Ri;\n", + "Vl=Ir*(Ri+Rf);\n", + "Ip=Vl/Rl;\n", + "Pl=(Vl*Ip)/2;\n", + "Ps=(Vs*Ip)/pi;\n", + "\n", + "#result\n", + "print \"power delivered to the load is\",round(Pl,2),\"Watt\"\n", + "print \"power provided by each supply is\",round(Ps,2), \"Watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.10,Page 149" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 141.67 Watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=170.0;\n", + "R=51.0;\n", + "\n", + "#calculation\n", + "I=V/R;\n", + "P=(V*I)/4;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(P,2), \"Watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.11,Page 151" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power dissipated is 7.05 watt\n", + "power dissipated when transistor resistance is 0.2 hm is 0.35 watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=7.2;\n", + "Rq=.2;\n", + "Rl=4;\n", + "D=.6;\n", + "\n", + "#calculation\n", + "Ip=V/(Rq+Rl);\n", + "Vl=Ip*Rl;\n", + "P=D*Vl*Ip;\n", + "Vq=Ip*Rq;\n", + "Pq=D*Vq*Ip;\n", + "\n", + "#result\n", + "print \"power dissipated is\",round(P,2), \"watt\"\n", + "print \"power dissipated when transistor resistance is 0.2 hm is\",round(Pq,2), \"watt\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter4.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter4.ipynb new file mode 100644 index 00000000..a01cb417 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter4.ipynb @@ -0,0 +1,733 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Linear Power Amplifier Integrated Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.1,Page 162" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 500.0 mV\n" + ] + } + ], + "source": [ + "#finding voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=1;\n", + "Ri=10;\n", + "Vi=0;\n", + "Ip=500;\n", + "\n", + "#calculation\n", + "Vrf=Ip*Rf;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vrf,2), \"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.2,Page 165" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency of OPA548 is 67.26 KHz\n", + "slew rate of OPA548 is 1.12 Mhz\n", + "the OPA548 can be used\n" + ] + } + ], + "source": [ + "#finding frequency\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=300.0;\n", + "P=35.0;\n", + "R=8.0;\n", + "S=10000.0;\n", + "fh=20.0;\n", + "\n", + "#calculation\n", + "Vl=(P*R)**.5;\n", + "Vp=Vl*2**.5;\n", + "Il=Vl/R;\n", + "f=S/(2*pi*Vp);\n", + "Ao=Vl/Vi;\n", + "G=Ao*fh;\n", + "\n", + "#result\n", + "print \"frequency of OPA548 is\",round(f,2), \"KHz\"\n", + "print \"slew rate of OPA548 is\",round(G,2), \"Mhz\"\n", + "print('the OPA548 can be used')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.3,Page 168" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 3.5 watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=10.0;\n", + "V=12.0;\n", + "Vl=5.0;\n", + "\n", + "#calculation\n", + "Pl=Vl**2/Rl;\n", + "I=Vl/Rl;\n", + "Ps=V*I;\n", + "Pic=Ps-Pl;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(Pic,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.4,Page 170" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vload Iload Pload Ps Pic\n", + "0.0 0.0 0.0 0.0 0.0\n", + "0.2 0.0 0.0 0.24 0.24\n", + "0.4 0.0 0.02 0.48 0.46\n", + "0.6 0.1 0.04 0.72 0.68\n", + "0.8 0.1 0.06 0.96 0.9\n", + "4.8 0.5 2.3 5.76 3.46\n", + "5.0 0.5 2.5 6.0 3.5\n", + "5.2 0.5 2.7 6.24 3.54\n", + "5.4 0.5 2.92 6.48 3.56\n", + "5.6 0.6 3.14 6.72 3.58\n", + "5.8 0.6 3.36 6.96 3.6\n", + "6.0 0.6 3.6 7.2 3.6\n", + "6.2 0.6 3.84 7.44 3.6\n", + "6.4 0.6 4.1 7.68 3.58\n", + "11.4 1.1 13.0 13.68 0.68\n", + "11.6 1.2 13.46 13.92 0.46\n", + "11.8 1.2 13.92 14.16 0.24\n", + "12.0 1.2 14.4 14.4 0.0\n" + ] + }, + { + "data": { + "image/png": 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E8BnQNcHzXcifoFaUd4E9RRzjl5FOJbZ3rw2De+IJaNrU62hEyq4uXWzNsC5d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\nREREREREREREREREREREJBn/DwWIrS4GjkBZAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f414b298f90>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Vload vs Pic graph\n", + "\n", + "#initialisation of variable\n", + "%matplotlib inline\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "Vload=[0.0, 0.2, 0.4, 0.6, 0.8, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8, 6.0, 6.2, 6.4, 11.4, 11.6, 11.8, 12.0];\n", + "Iload=[0.0, 0.0, 0.0, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.6, 0.6, 0.6, 0.6, 0.6, 1.1, 1.2, 1.2, 1.2];\n", + "Pload=[0.00, 0.00, 0.02, 0.04, 0.06, 2.30, 2.50, 2.70, 2.92, 3.14, 3.36, 3.60, 3.84, 4.10, 13.00, 13.46, 13.92, 14.40];\n", + "Ps=[0.00, 0.24, 0.48, 0.72, 0.96, 5.76, 6.00, 6.24, 6.48, 6.72, 6.96, 7.20, 7.44, 7.68, 13.68, 13.92, 14.16, 14.40];\n", + "Pic=[0.00, 0.24, 0.46, 0.68, 0.90, 3.46, 3.50, 3.54, 3.56, 3.58, 3.60, 3.60, 3.60, 3.58, 0.68, 0.46, 0.24, 0.00];\n", + "\n", + "#result\n", + "print('Vload Iload Pload Ps Pic');\n", + "for i in range(0,18):\n", + " print Vload[i],\" \",Iload[i],\" \",Pload[i],\" \",Ps[i],\" \", Pic[i]\n", + " \n", + "plt.plot(Vload,Pic);\n", + "plt.xlabel('load voltage (V)')\n", + "plt.ylabel('IC Power(W)')\n", + "plt.title('load voltage vs IC Power')\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.5,Page 173" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IC power is 2.57 W\n", + "total power is 3.82 W\n", + "dc supply current is 159.155 mA\n", + "power delivered is 1.25 watt\n" + ] + } + ], + "source": [ + "#finding different power and current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=12.0;\n", + "Vp=5.0;\n", + "R=10.0;\n", + "\n", + "#calculation\n", + "Ip=Vp/R;\n", + "Il=Ip/2**.5;\n", + "Pl=(Vp*Ip)/2;\n", + "Id=Ip/pi;\n", + "Pt=2*V*Ip/pi;\n", + "Pic=Pt-Pl;\n", + "\n", + "#result\n", + "print \"IC power is\",round(Pic,2), \"W\"\n", + "print \"total power is\",round(Pt,2), \"W\"\n", + "print \"dc supply current is\",round(Id*1000,3), \"mA\"\n", + "print \"power delivered is\",round(Pl,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.6,Page 179" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thermal resistance is 24.61 C/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ts=40.0;\n", + "P=2.92;\n", + "Qj=2.5;\n", + "Qc=2.0;\n", + "Tj=125.0;\n", + "\n", + "#calculation\n", + "Qs=(Tj-Ts)/P-Qj-Qc;\n", + "\n", + "#result\n", + "print \"thermal resistance is\",round(Qs,2),\"C/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.7,Page 180" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vload Iload Pload Ps Pic\n", + "9.4 0.94 4.42 14.36 9.94\n", + "9.6 0.96 4.61 14.67 10.06\n", + "10.0 power delivered by IC in watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=24.0;\n", + "R=10.0;\n", + "Qs=4.0;\n", + "Tj=125.0;\n", + "Ta=40.0;\n", + "Qj=2.5;\n", + "Qc=2.0;\n", + "Vload=[9.4, 9.6];\n", + "Iload=[.94, .96];\n", + "Pload=[4.42, 4.61];\n", + "Ps=[14.36, 14.67];\n", + "Pic=[9.94, 10.06];\n", + "\n", + "#calculation\n", + "P=(Tj-Ta)/(Qj+Qc+Qs);\n", + "\n", + "#result\n", + "print('Vload Iload Pload Ps Pic');\n", + "for i in range(0,2):\n", + " print Vload[i],\" \",Iload[i],\" \",Pload[i],\" \",Ps[i],\" \", Pic[i]\n", + "print round(P,2),\"power delivered by IC in watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.8,Page 182" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "gain is 23.0\n", + "limit current is 4.01 A\n", + "output voltage is 46.0 V\n", + "maximum output voltage is 32.0 V\n" + ] + } + ], + "source": [ + "#finding current and voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=22.0;\n", + "Ri=1.0;\n", + "Rs=15.0;\n", + "I=4.75;\n", + "Rc=4.0;\n", + "Vp=2.0;\n", + "Rl=8.0;\n", + "Im=4.0;\n", + "\n", + "#calculation\n", + "Av=1+(Rf/Ri);\n", + "Il=(Rs*I)/(Rc+13.75);\n", + "Vo=Vp*Av;\n", + "V=Im*Rl;\n", + "\n", + "#result\n", + "print \"gain is\",round(Av,2)\n", + "print \"limit current is\",round(Il,2), \"A\"\n", + "print \"output voltage is\",round(Vo,2), \"V\"\n", + "print \"maximum output voltage is\",round(V,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.9,Page 185" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "loudness ofsound is 108.06 dB\n" + ] + } + ], + "source": [ + "#finding loudness\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "D=8.0;\n", + "d=1.0;\n", + "I=90.0;\n", + "\n", + "#calculation\n", + "Is=20*log(d/D,10);\n", + "Ir=I-Is;\n", + "\n", + "#result\n", + "print \"loudness ofsound is\",round(Ir,2), \"dB\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.10,Page 186" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "19.95 power provided in watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "D=1.0;\n", + "I1=108.0;\n", + "I2=95.0;\n", + "P=1.0;\n", + "\n", + "#calculation\n", + "I=I1-I2;\n", + "Pr=P*10**(I/10);\n", + "\n", + "#result\n", + "print \"power provided is\",round(Pr,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.11,Page 188" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 12.65 V\n", + "gain is 10.28\n" + ] + } + ], + "source": [ + "#finding output voltage and gain\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "P=20;\n", + "R=8;\n", + "Vi=1.23;\n", + "\n", + "#calculation\n", + "V=(P*R)**.5;\n", + "G=V/Vi;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(V,2), \"V\"\n", + "print \"gain is\",round(G,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.12,Page 191" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistor b/w pins 1&8 is 600.0 ohm\n", + "thus pick a 620 ohm resistor\n", + "capacitor b/w pins 1&8 is 22.46 microF\n", + "thus pick a 27 microF capacitor\n" + ] + } + ], + "source": [ + "#finding resistor and capacitor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "G=40.0;\n", + "f=80.0;\n", + "R1=15000.0;\n", + "R2=150.0;\n", + "\n", + "#calculation\n", + "R=2*(R1/G)-R2;\n", + "R11=620;\n", + "C=1/(2*pi*f*R11/7);\n", + "\n", + "#result\n", + "print \"resistor b/w pins 1&8 is\",round(R,2),\"ohm\"\n", + "print('thus pick a 620 ohm resistor')\n", + "print \"capacitor b/w pins 1&8 is\",round(C*1e6,2), \"microF\"\n", + "print('thus pick a 27 microF capacitor')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.13,Page 193" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power dissipated is 140.0 mW\n", + "thermal resistance is 628.93 degree C/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "V=5.0#voltage\n", + "Tm=150.0#temperature\n", + "Ta=50.0#temperature\n", + "Qa=107.0;\n", + "Qc=37.0;\n", + "Ps=299.0;\n", + "\n", + "#calculation\n", + "Vd=V/2;\n", + "Vm=V-1;\n", + "Vp=Vm-Vd;\n", + "Vr=Vp/2**.5;\n", + "Pl=1000*Vr**2/R;\n", + "Pl=140;\n", + "Pic=Ps-Pl;\n", + "Q=(Tm-Ta)/Pic;\n", + "\n", + "#result\n", + "print \"power dissipated is\",round(Pl,2), \"mW\"\n", + "print \"thermal resistance is\",round(Q*1000,2),\"degree C/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.14,Page 197" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power deliverd is 562.5 mwatt\n" + ] + } + ], + "source": [ + "#finding power delivered\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "V=5.0#voltage\n", + "\n", + "#calculation\n", + "Vl=V-1;\n", + "Vp=Vl-1;\n", + "Vr=Vp/2**.5;\n", + "P=Vr**2/R;\n", + "\n", + "#result\n", + "print \"power deliverd is\",round(P*1000,2), \"mwatt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.15,Page 201" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 85.56 watt\n", + "thermal resistance is 1.4 degreeC/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "Ts=35.0#temperature\n", + "Ta=150.0#temperature\n", + "Vm=42.0#voltage\n", + "\n", + "#calcuation\n", + "Vp=Vm-5;\n", + "Vr=Vp/2**.5;\n", + "Pm=Vr**2/R;\n", + "P=45;\n", + "Qs=(Ta-Ts)/P-1.2;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(Pm,2), \"watt\"\n", + "print \"thermal resistance is\",round(round(Qs*10)/10,2), \"degreeC/W\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter5.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter5.ipynb new file mode 100644 index 00000000..58305b46 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter5.ipynb @@ -0,0 +1,647 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Discrete Linear Power Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.1,Page 215" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Id=0 from 0 to 2 so not shown in the graph\n" + ] + }, + { + "data": { + "image/png": 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8oVgLIhWJQSTrtWwJ77xjE+KOO852j7vuOsjLcx2ZZBrVGETS0IoVMGCA7Rg3diy0auU6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\nIiIiIiIiIiIiIiIiIiIiknn+HwC9g+9EmLPsAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7fb0a512a8d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Id vs Vgs\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "%matplotlib inline\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "Vth=3.6;\n", + "Vgs=4;#voltage\n", + "#volt change beyond 3.6 causes a major increase in Id as it is cut off voltage\n", + "\n", + "#result\n", + "print('Id=0 from 0 to 2 so not shown in the graph')\n", + "x=np.linspace(2,3.6,300);\n", + "y=(-2.5*(x-3.6))**.5;\n", + "plt.plot(x,y)\n", + "plt.xlabel('Vgs(volts)');\n", + "plt.ylabel('Id(amps)');\n", + "plt.title('Id vs Vgs');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.2,Page 217" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "drain current is 3.8 A\n", + "Vth=4V is assumed\n" + ] + } + ], + "source": [ + "#finding drain current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=4.5;#voltage\n", + "T=25;#degreeC\n", + "Id=3.8;\n", + "\n", + "#result\n", + "print \"drain current is\",round(Id,2), \"A\"\n", + "print('Vth=4V is assumed')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.3,Page 219" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "MOSFET is IRF530N\n", + "lower limit of Vth is -4.0 V\n", + "upper limit of Vth is -2.0 V\n", + "current is 2.3 A\n" + ] + } + ], + "source": [ + "#finding drain current of IRF530N\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vgs=-5;#voltage\n", + "Vthl=-4;\n", + "Vthu=-2;\n", + "Id=2.3;#current\n", + "\n", + "#result\n", + "print('MOSFET is IRF530N')\n", + "print \"lower limit of Vth is\",round(Vthl,2), \"V\"\n", + "print \"upper limit of Vth is\",round(Vthu,2), \"V\"\n", + "print \"current is\",round(Id,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example5.5,Page 225" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 1.51 kohm\n", + "load voltage is 36.68 V\n", + "Pq is 40.02 watt\n", + "Ps is 82.0 watt\n", + "Pl is 41.97 watt\n" + ] + } + ], + "source": [ + "#finding Pq,Pl,Ps,resistance,load voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=22.0;#resistance\n", + "Vg=3.6;#voltage\n", + "Vd=56.0;\n", + "G=.98;#gain\n", + "Vi=40.0;\n", + "Rl=8.0;#load resistance\n", + "Vp=36.5;\n", + "\n", + "#calculation\n", + "Vr=Vd-Vg;\n", + "Ir=Vr/R1;\n", + "R2=Vg/Ir;\n", + "Va=(R1/(R1+R2))*Vi;\n", + "Vl=G*Va;\n", + "Il=Vp/Rl;\n", + "Pl=Vp*4.6/4;\n", + "Ps=Vd*4.6/pi;\n", + "Pq=Ps-Pl;\n", + "\n", + "#result\n", + "print \"resistance is\",round(R2,2), \"kohm\"\n", + "print \"load voltage is\",round(Vl,2), \"V\"\n", + "print \"Pq is\",round(Pq,2), \"watt\"\n", + "print \"Ps is\",round(Ps,2), \"watt\"\n", + "print \"Pl is\",round(Pl,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.6,Page 232" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current is 2.45 mA\n", + "resistance2 is 814.815 ohm\n", + "pick R2=R3=820ohm R1=R4=22 kohm\n" + ] + } + ], + "source": [ + "#finding resistance and current \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=22.0;#resistance\n", + "V1=56.0;#voltage\n", + "V2=2.0;#voltage\n", + "\n", + "#calculation\n", + "I=(V1-V2)/R1;\n", + "R2=V2/I;\n", + "\n", + "#result\n", + "print \"current is\",round(I,2), \"mA\"\n", + "print \"resistance2 is\",round(R2*1000,3), \"ohm\"\n", + "print('pick R2=R3=820ohm R1=R4=22 kohm')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.7,Page 234" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load voltage is 10.01 V\n" + ] + } + ], + "source": [ + "#finding load voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=350.0;#voltage\n", + "f=100.0;#frequency\n", + "Rf=10000.0;#resistance\n", + "Ri=520.0;\n", + "\n", + "#calculation\n", + "Vp=(1+(Rf/Ri))*Vi*2**.5;\n", + "\n", + "#result\n", + "print \"load voltage is\",round(Vp/1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.8,Page 238" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load rms voltage is 20.0 V\n", + "resistance is 54.04 kohm\n", + "current is 1.18 mA\n", + "load current is 4.41 A\n", + "supply power is 39.3 watt\n", + "load power is 38.9 W\n", + "power is 19.552 W\n", + "thermal resistance is 3.01 degreC/W\n" + ] + } + ], + "source": [ + "#designing amplifier\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=50.0;#power\n", + "Z=4.7#impedence\n", + "R=4.0;#resistance\n", + "Ta=40.0;#degreeC\n", + "Tj=140.0;#degreeC\n", + "Vd=28.0;\n", + "R2=22.0;\n", + "\n", + "#calculation\n", + "Vr=(P*R)**.5;\n", + "Vp=Vr*2**.5;\n", + "Av=-Vr/1.23;\n", + "Rf=-Av*Z;\n", + "I=(Vd-2)/R2;\n", + "Vm=.63*Vd;\n", + "Ip=Vm/R;\n", + "Ps=Vd*Ip/pi;\n", + "Pl=Ip**2/2*R;\n", + "Pq=round(Ps)-Pl/2;\n", + "Qs=(Tj-Ta)/Pq-2.1;\n", + "\n", + "#result\n", + "print \"load rms voltage is\",round(Vp,2), \"V\"\n", + "print \"resistance is\",round(Rf,2), \"kohm\"\n", + "print \"current is\",round(I,2), \"mA\"\n", + "print \"load current is\",round(Ip,2), \"A\"\n", + "print \"supply power is\",round(Ps,2), \"watt\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"power is\",round(Pq,3), \"W\"\n", + "print \"thermal resistance is\",round(Qs,2), \"degreC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.9,Page 243" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 129.37 mV\n", + "load current is 32.34 mA\n" + ] + } + ], + "source": [ + "#finding load current,output voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=7.5e-3;#voltage\n", + "Ib=800e-9;#current\n", + "R=53.9e3;#resistance\n", + "\n", + "#calculation\n", + "Vo=11.5*Vi+Ib*R;\n", + "Id=Vo/4;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vo*1000,2), \"mV\"\n", + "print \"load current is\",round(Id*1000,2), \"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.10,Page 249" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 0.31 ohm\n", + "thus pick a .33ohm rsistance\n", + "voltage is 0.55 V\n", + "power is 0.23 W\n", + "thermal resistance is 8.1 degreeC/W\n" + ] + } + ], + "source": [ + "#finding resistance,voltage,power \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "G=6.4;#A/V\n", + "I=5.0;#current\n", + "Pq=9.8;#W\n", + "Tj=140.0;\n", + "Ta=40.0;\n", + "R1=.33;\n", + "\n", + "#calculation\n", + "R=2/G;\n", + "Im=I/3;\n", + "Vr=Im*R1;\n", + "P=Vr*Im/4;\n", + "Qs=(Tj-Ta)/Pq-2.1;\n", + "\n", + "#result\n", + "print \"resistance is\",round(R,2), \"ohm\"\n", + "print('thus pick a .33ohm rsistance')\n", + "print \"voltage is\",round(Vr,2), \"V\"\n", + "print \"power is\",round(P,2), \"W\"\n", + "print \"thermal resistance is\",round(Qs,2), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.11,Page 251" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "limit level current is 8.49 A\n" + ] + } + ], + "source": [ + "#finding limit current \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=200;#power\n", + "R=8;#ohm\n", + "\n", + "#calculation\n", + "Il=(P/R)**.5*2**.5;\n", + "Ilm=1.2*Il;\n", + "\n", + "#result\n", + "print \"limit level current is\",round(Ilm,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.12,Page 253" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 0.1 ohm\n", + "power is 1.8 W\n", + "MOSFET power is 84.0 W\n", + "temperature is 468.4 degreeC\n" + ] + } + ], + "source": [ + "#finding resistance,power,temperature \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=6;#current\n", + "V=.6;#voltage\n", + "D=.5;#duty cycle\n", + "T=40;#temperature\n", + "\n", + "#calculation\n", + "Rs=V/I;\n", + "Pr=D*V*I;\n", + "Vp=28;\n", + "Pm=D*Vp*I;\n", + "Tj=T+Pm*5.1;\n", + "\n", + "#result\n", + "print \"resistance is\",round(Rs,2), \"ohm\"\n", + "print \"power is\",round(Pr,2), \"W\"\n", + "print \"MOSFET power is\",round(Pm,2), \"W\"\n", + "print \"temperature is\",round(Tj,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.13,Page 255" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum safe temperature is 89.05 degreeC\n" + ] + } + ], + "source": [ + "#finding maximum safe temperature \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "T=130;#temperature\n", + "P=19.5;#power\n", + "\n", + "#calculation\n", + "Ts=T-P*2.1;\n", + "\n", + "#result\n", + "print \"maximum safe temperature is\",round(Ts,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.14,Page 257" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "reactance is 8.86 ohm\n", + "voltage across resistor is 10.03 V\n", + "-48 is the angle of the voltage in degrees\n", + "power dissipated by load is 12.5 watts\n", + "current across the resistance is 1.77 A\n", + "power supply is 15.8 W\n", + "power dissipated by transistor is 9.55 watts\n" + ] + } + ], + "source": [ + "#finding 3 powers and current across resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=15.0;#voltage\n", + "f=300.0;#frequency\n", + "L=4.7;#inductance\n", + "Vdc=28;#V\n", + "Pr=12.5;\n", + "\n", + "#calculation\n", + "Xl=2*pi*f*L;\n", + "Zload=sqrt(8**2+8.9**2);#magnitude of Zload\n", + "Vload=15.0;#msgnitude of Vload\n", + "Vr=Vload*8/Zload;\n", + "I=Vr/8*sqrt(2);\n", + "Psupply=Vdc*I/pi;\n", + "Pq=Psupply-Pr/2;\n", + "\n", + "#result\n", + "print \"reactance is\",round(Xl/1000,2), \"ohm\"\n", + "print \"voltage across resistor is\",round(Vr,2), \"V\"\n", + "print \"-48 is the angle of the voltage in degrees\";\n", + "print \"power dissipated by load is\",round(Pr,2), \"watts\"\n", + "print \"current across the resistance is\",round(I,2), \"A\"\n", + "print \"power supply is\",round(Psupply,2), \"W\"\n", + "print \"power dissipated by transistor is\",round(Pq,2), \"watts\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter6.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter6.ipynb new file mode 100644 index 00000000..0c7858a4 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter6.ipynb @@ -0,0 +1,528 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Power Switches" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.1,Page 274" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current is 1.81 A\n", + "load power is 49.32 W\n", + "transistor power is 1.45 W\n", + "least value of base current is 90.67 mA\n", + "max value of base resistance is 4.85 ohm\n", + "thus pick Rb=33ohm\n" + ] + } + ], + "source": [ + "#finding Ic,Pload,Pq,Ibase,Rbase\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vs=28.0;#V\n", + "Vi=5.0;#V\n", + "Rl=15.0;#ohm\n", + "Vc=.8;#V\n", + "b=20.0;\n", + "\n", + "#calculation\n", + "Ic=(Vs-Vc)/Rl;\n", + "Pl=Ic**2*Rl;\n", + "Pq=Ic*Vc;\n", + "Ib=Ic/b*1000;\n", + "Rb=(Vi-.6)/Ib;\n", + "\n", + "#result\n", + "print \"collector current is\",round(Ic,2), \"A\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"transistor power is\",round(Pq,2), \"W\"\n", + "print \"least value of base current is\",round(Ib,2), \"mA\"\n", + "print \"max value of base resistance is\",round(Rb*100,2), \"ohm\"\n", + "print ('thus pick Rb=33ohm')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.4,Page 282" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load resistance is 554.0 ohm\n", + "thus pick Rl=560ohm\n", + "max value of Rb is 3.0 kohm\n", + "thus pick Rb=2.2kohm\n", + "load current is 49.46 mA\n", + "load power is 685.08 mW\n", + "power delivered is 7.42 mW\n" + ] + } + ], + "source": [ + "#finding Pload,Pq,Iload,resistances\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vd=28.0;#V\n", + "f=100.0;#frequency\n", + "I=50.0;#current\n", + "Rl1=560.0;\n", + "Vp=2.4;\n", + "Ib=500.0;#microAmp\n", + "D=.5;#duty cycle\n", + "\n", + "#calculation\n", + "Rl=(Vd-.3)/I;\n", + "Rb=(Vp-.9)/Ib;\n", + "Vl=Vd-.3;\n", + "Ip=Vl/Rl1;\n", + "Pl=D*Vl*Ip;\n", + "Pq=D*Ip*.3;\n", + "\n", + "#result\n", + "print \"load resistance is\",round(Rl*1000), \"ohm\"\n", + "print('thus pick Rl=560ohm')\n", + "print \"max value of Rb is\",round(Rb*1000,2),\"kohm\"\n", + "print('thus pick Rb=2.2kohm')\n", + "print \"load current is\",round(Ip*1000,2), \"mA\"\n", + "print \"load power is\",round(Pl*1000,2), \"mW\"\n", + "print \"power delivered is\",round(Pq*1000,2), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.5,Page 286" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time of rise is 788.48 ns\n", + "time of rise is 4.65 microsec\n" + ] + } + ], + "source": [ + "#finding time of rise\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "C=640.0;#capacitor\n", + "R1=560.0;#load resistance\n", + "R2=3.3;#kohm\n", + "\n", + "#calculation\n", + "t1=2.2*R1*C;\n", + "t2=2.2*R2*C;\n", + "\n", + "#result\n", + "print \"time of rise is\",round(t1/1000,2), \"ns\"\n", + "print \"time of rise is\",round(t2/1000,2), \"microsec\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.6,Page 287" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 682.5 ohm\n", + "pick up resistance=680 ohm\n", + "rise time is 957.44 ns\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f4e9da3f2d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Rpick up,time of rise\n", + "\n", + "#initialisation of variable\n", + "%matplotlib inline\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "Vol=0.7;\n", + "Iol=40.0/1000;#current\n", + "Rpullup1=680.0;\n", + "C=640.0;\n", + "Epullup=28.0;\n", + "#for plotting\n", + "x=[0, .1, 1.9, 4.1, 5, 5.1, 5.3, 5.6, 6.0, 9.3];\n", + "y=[27.8, .1, .1, .1, .1, 5, 13.5, 21.0, 27.0, 27.8];\n", + "\n", + "#calculation\n", + "Rpullup=(Epullup-Vol)/Iol;\n", + "trise=2.2*Rpullup1*C;\n", + "plt.plot(x,y,'r');\n", + "plt.xlabel ('time(mus)')\n", + "plt.ylabel ('Vout')\n", + "plt.title ('Vout vs time')\n", + "\n", + "#result\n", + "print \"resistance is\",round(Rpullup,2), \"ohm\"\n", + "print('pick up resistance=680 ohm');\n", + "print \"rise time is\",round(trise/1000,2), \"ns\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.7,Page 289" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "worst case resistance is 0.286 ohm\n", + "load current is 4.45 A\n", + "load voltage is 26.73 V\n", + "load power is 47.62 W\n", + "drop voltage is 1.27 V\n", + "power is 2.27 W\n", + "temperature is 182.6 deg.C\n" + ] + } + ], + "source": [ + "#finding worst case resistance and power characteristics\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=.11;#resistance\n", + "Vd=28.0;#voltage\n", + "R2=6.0;#ohm\n", + "D=.4;#duty cycle\n", + "Q=62.0;\n", + "\n", + "#calculation\n", + "Ro=2.6*R1;\n", + "Ip=Vd/(R2+Ro);\n", + "Vl=Ip*R2;\n", + "Pl=D*Vl*Ip;\n", + "Vq=Ip*Ro;\n", + "Pq=D*Vq*Ip;\n", + "T=40+round(Pq*10)/10*Q;\n", + "\n", + "#result\n", + "print \"worst case resistance is\",round(Ro,3), \"ohm\"\n", + "print \"load current is\",round(Ip,2),\"A\"\n", + "print \"load voltage is\",round(Vl,2), \"V\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"drop voltage is\",round(Vq,2), \"V\"\n", + "print \"power is\",round(Pq*10,2)/10, \"W\"\n", + "print \"temperature is\",round(T,2), \"deg.C\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.8,Page 292" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage is 150.0 V\n" + ] + } + ], + "source": [ + "#finding voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "L=10.0;#inductor\n", + "I=4.5;#current\n", + "t=300.0#time\n", + "\n", + "#calculation\n", + "V=L*I/t;\n", + "\n", + "#result\n", + "print \"voltage is\",round(V*1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.9,Page 298" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1+R2 is 682.5 ohm\n", + "pick R1=330ohm & R2=360ohm as they divide Vd setting 8V<Vg<18V\n", + "node voltage for V1 is 28.0 V\n", + "node voltage for V2 is 0.7 V\n", + "gate voltage is 15.31 V\n", + "gate & source diff is -12.69 V\n", + "load voltage is 26.73 V\n", + "load current is 2.23 A\n", + "load power is 47.63 W\n", + "Pq is 2.26 W\n", + "thermal resistance is 44.92 degreeC/W\n" + ] + } + ], + "source": [ + "#finding resistance and power characteristics\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=12.0;#load resistance\n", + "V1=.8;#voltage\n", + "V2=2.4;#voltage\n", + "D=.8;#duty cycle\n", + "Tj=150.0;#degreeC\n", + "Ta=40.0;#degreeC\n", + "Vd=28.0;\n", + "Vo=.7;\n", + "I=40.0;#mA;\n", + "R1=330;\n", + "R2=360;\n", + "Vn1=28;\n", + "Vn2=.7;\n", + "\n", + "#calculation\n", + "k=(Vd-Vo)/I;\n", + "Vg=R2*Vd/(R1+R2)+Vn2;\n", + "Vgs=Vg-Vd;\n", + "Vl=Vd*Rl/(Rl+.57);\n", + "Il=Vl/Rl;\n", + "Pl=D*Vl*Il;\n", + "Vq=Il*.57;\n", + "Pq=D*Vq*Il;\n", + "Q=(Tj-Ta)/Pq-3.7;\n", + "\n", + "#result\n", + "print \"R1+R2 is\",round(k*1000,2), \"ohm\"\n", + "print('pick R1=330ohm & R2=360ohm as they divide Vd setting 8V<Vg<18V')\n", + "print \"node voltage for V1 is\",round(Vn1,2),\"V\"\n", + "print \"node voltage for V2 is\",round(Vn2,2), \"V\"\n", + "print \"gate voltage is\",round(Vg,2), \"V\"\n", + "print \"gate & source diff is\",round(Vgs,2), \"V\"\n", + "print \"load voltage is\",round(Vl,3), \"V\"\n", + "print \"load current is\",round(Il,2), \"A\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"Pq is\",round(Pq,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,2), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.10,Page 305" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time is 1.1 micro s\n" + ] + } + ], + "source": [ + "#finding time\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=40.0;#current\n", + "Q=44.0;#nC\n", + "\n", + "#calculation\n", + "t=Q/I;\n", + "\n", + "#result\n", + "print \"time is\",round(t,2), \"micro s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.11,Page 313" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current is 1.8 A\n", + "load voltage is 26.97 V\n", + "power is 40.85 W\n", + "high side voltage is 0.67 V\n", + "high side power is 1.03 W\n", + "low side voltage is 0.36 V\n", + "low side power is 0.55 W\n", + "IC power is 1.56 W\n", + "thermal resistance is 55.49 degreeC/W\n" + ] + } + ], + "source": [ + "#finding different voltages and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=15.0;#load resistance\n", + "D=.85;#duty cycle\n", + "Ts=60.0;#degreeC\n", + "Vd=28.0;#voltage\n", + "R1=.375;\n", + "R2=.2;\n", + "\n", + "#calculation\n", + "I=Vd/(R1+R2+Rl);\n", + "Vl1=I*Rl;\n", + "P=D*Vl*I;\n", + "Vh=I*R1;\n", + "Ph=D*Vh*I;\n", + "Vl=I*R2;\n", + "Pl=D*Vl*I;\n", + "Pic=Ph+Pl;\n", + "Pic=1.56;\n", + "Tj=150;\n", + "Ta=60;\n", + "Q=(Tj-Ta)/Pic-2.2;\n", + "\n", + "#result\n", + "print \"current is\",round(I,2), \"A\"\n", + "print \"load voltage is\",round(Vl1,2), \"V\"\n", + "print \"power is\",round(P,2), \"W\"\n", + "print \"high side voltage is\",round(Vh,2), \"V\"\n", + "print \"high side power is\",round(Ph,2), \"W\"\n", + "print \"low side voltage is\",round(Vl,2), \"V\"\n", + "print \"low side power is\",round(Pl,2), \"W\"\n", + "print \"IC power is\",round(Pic,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,2), \"degreeC/W\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter7.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter7.ipynb new file mode 100644 index 00000000..45af7d98 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter7.ipynb @@ -0,0 +1,505 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: Switiching Power Supplies" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.1,Page 326" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 0.25\n", + "average voltage is 3.0 V\n" + ] + } + ], + "source": [ + "#finding duty cycle and average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "T=20.0;#time\n", + "Vp=12.0;#voltage\n", + "t=5.0;\n", + "\n", + "#calculation\n", + "D=t/T;\n", + "Vd=(D*Vp);\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D,3)\n", + "print \"average voltage is\",round(Vd,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.2,Page 238" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 42.0 %\n", + "time is 10.0 microsec\n", + "on time is 4.167 microsec\n", + "ripple current is 133.636 mA\n", + "load current is 500.0 mA\n", + "peak inductor current is 566.818 mA\n" + ] + } + ], + "source": [ + "#finding on time ripple,load,peak inductor current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vd=12.0;#voltage\n", + "Vl=5.0;#load voltage\n", + "Rl=10.0;#load resistance\n", + "f=100.0;#frequency\n", + "L=220.0;#inductor\n", + "\n", + "#calculation\n", + "D=Vl/Vd;\n", + "T=1/f;\n", + "t=D*T;\n", + "Vr=Vd-Vl;\n", + "I=Vr*round(t*10000)/10/L;\n", + "Il=Vl/Rl;\n", + "Ip=Il+I/2;\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D*100), \"%\"\n", + "print \"time is\",round(T*1000,3), \"microsec\"\n", + "print \"on time is\",round(t*10000,2)/10, \"microsec\"\n", + "print \"ripple current is\",round(I*1000,3),\"mA\"\n", + "print \"load current is\",round(Il*1000,3), \"mA\"\n", + "print \"peak inductor current is\",round(Ip*1000,3), \"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.3,Page 335" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms current is 325.01 mA\n", + "by trapezium method\n", + "rms current is 324.04 mA\n", + "by rectangle method\n", + "\n", + " rectangle method gives good result than trapezium method\n" + ] + } + ], + "source": [ + "#finding rms current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Id=500.0;#load current\n", + "i=134;#mA\n", + "D=.42;#duty cycle\n", + "\n", + "#calculation\n", + "Ip=Id+i/2;\n", + "Im=Id-i/2;\n", + "I1=((D/3)*(Ip**2+Im*Ip+Im**2))**.5;\n", + "I2=D**.5*Id;\n", + "\n", + "#result\n", + "print \"rms current is\",round(I1,2), \"mA\"\n", + "print('by trapezium method')\n", + "print \"rms current is\",round(I2,2), \"mA\"\n", + "print('by rectangle method')\n", + "print '\\n rectangle method gives good result than trapezium method'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.4,Page 336" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage is 0.3 V\n", + "dissipated power is 63.0 mW\n" + ] + } + ], + "source": [ + "#finding voltage and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=.3;#voltage\n", + "I=500.0;#current\n", + "D=.42;#duty cycle\n", + "T=150.0;#temperature\n", + "R=.6;#ohm\n", + "\n", + "#calculation\n", + "Vq=I*R;\n", + "Pq=D*Vq*I;\n", + "\n", + "#result\n", + "print \"voltage is\",round(Vq/1000,2), \"V\"\n", + "print \"dissipated power is\",round(Pq/1000,2), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.5,Page 341" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 42.0 %\n", + "time period is 6.667 microsec\n", + "on time is 2.778 microsec\n", + "load current is 500.0 mA\n", + "ripple current is 125.0 mA\n", + "inductor voltage is 7.0 V\n", + "inductor is 155.556 microH\n", + "inductor current is 562.5 mA\n", + "minimum capacitor current is 250.0 mA\n", + "minimum capacitor voltage is 18.0 V\n", + "Rf/Ri is 3.07\n", + "power of LM2595 is 0.33 W\n", + "thermal resistance is 210.998 degreeC/W\n" + ] + } + ], + "source": [ + "#finding all componenets\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R=10.0;#resistance\n", + "V1=5.0;#V\n", + "V2=12.0;#V\n", + "Ta=80.0;#degreeC\n", + "Tb=150.0;\n", + "f=150.0;#frequency\n", + "\n", + "#calculation\n", + "D=V1/V2;\n", + "T=1/f;\n", + "t=D*T;\n", + "Id=V1/R;\n", + "i=.25*Id;\n", + "Vl=V2-V1;\n", + "L=Vl*t/i;\n", + "Ip=Id+i/2;\n", + "Ic=Id/2;\n", + "Vc=1.5*V2;\n", + "K=V1/1.23-1;\n", + "P=.01*V2+D*Id*1;\n", + "Q=(Tb-Ta)/P-2.2;\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D*100), \"%\"\n", + "print \"time period is\",round(T*1000,3), \"microsec\"\n", + "print \"on time is\",round(t*1000,3), \"microsec\"\n", + "print \"load current is\",round(Id*1000,3), \"mA\"\n", + "print \"ripple current is\",round(i*1000,3), \"mA\"\n", + "print \"inductor voltage is\",round(Vl,2), \"V\"\n", + "print \"inductor is\",round(L*1000,3), \"microH\"\n", + "print \"inductor current is\",round(Ip*1000,2), \"mA\"\n", + "print \"minimum capacitor current is\",round(Ic*1000,2), \"mA\"\n", + "print \"minimum capacitor voltage is\",round(Vc,3), \"V\"\n", + "print \"Rf/Ri is\",round(K,2)\n", + "print \"power of LM2595 is\",round(P,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,3), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.6,Page 349" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load power is 15.4 W\n", + "supply power is 17.11 W\n", + "dc current is 1.4 A\n", + "inductor current is 1.57 A\n", + "duty cycle is 0.45\n", + "inductor is 154.29 microH\n", + "transistor power is 352.8 mW\n", + "diode power is 385.0 mW\n", + "capacitor is 157.5 microF\n" + ] + } + ], + "source": [ + "#finding different power,inductor current,inductor value\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0#V\n", + "V2=22.0;#V\n", + "I=.7;#A\n", + "f=100.0;#kHz\n", + "R=.4;#ohm\n", + "Vd=.5;\n", + "\n", + "#calculation\n", + "Pl=V2*I;\n", + "Ps=Pl/.9;\n", + "Id=round(Ps/V1*10)/10;\n", + "i=.25*Id;\n", + "Ip=Id+i/2;\n", + "D=round((1-V1/V2)*100)/100;\n", + "t=D/f;\n", + "L=V1*t/i;\n", + "Vp=Id*R;\n", + "Pq=D*Vp*Id;\n", + "Pd=(1-D)*.5*Id;\n", + "C=Id*t/2/20;\n", + "\n", + "#result\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"supply power is\",round(Ps,2), \"W\"\n", + "print \"dc current is\",round(Id,2), \"A\"\n", + "print \"inductor current is\",round(Ip,2), \"A\"\n", + "print \"duty cycle is\",round(D,2)\n", + "print \"inductor is\",round(L*1000,2), \"microH\"\n", + "print \"transistor power is\",round(Pq*1000,2), \"mW\"\n", + "print \"diode power is\",round(Pd*100,2)*10, \"mW\"\n", + "print \"capacitor is\",round(C*1e6,2), \"microF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.7,Page 355" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rf/Ri is 16.886\n", + "pick Rf=22; Ri=1.3;\n", + "rms current is 1.4 A\n", + "switch power is 132.3 mW\n", + "IC power is 151.2 mW\n", + "total power is 283.5 mW\n", + "IC temperature is 98.43 degreeC\n" + ] + } + ], + "source": [ + "#finding feedback resistor,power,current and temperature\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0;#V\n", + "V2=22.0;#V\n", + "I=.7;#A\n", + "Ta=80.0;#degreeC\n", + "Ps=17.1#supply power\n", + "\n", + "#calculation\n", + "K=V2/1.23-1;\n", + "Id=round(Ps/V1*10)/10;\n", + "D=round((1-(V1/V2))*100)/100;\n", + "Ir=D**.5*Id;\n", + "Ps=Ir**2*.15;\n", + "Pi=D*Id*V1/50;\n", + "P=Ps+Pi;\n", + "T=Ta+P*65;\n", + "\n", + "#result\n", + "print \"Rf/Ri is\",round(K,3)\n", + "print('pick Rf=22; Ri=1.3;')\n", + "print \"rms current is\",round(Id,2), \"A\"\n", + "print \"switch power is\",round(Ps*1000,2), \"mW\"\n", + "print \"IC power is\",round(Pi*1000,2), \"mW\"\n", + "print \"total power is\",round(P*1000,2), \"mW\"\n", + "print \"IC temperature is\",round(T,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.8,Page 359" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum voltage is 18.25 V\n", + "diode voltage is 20.0 V\n", + "duty cycle is 0.34\n", + "power delivered is 5.0 W\n", + "average current is 466.67 mA\n", + "mid primary current is 1.37 A\n", + "rms current is 800.33 mA\n", + "ramp current is 480.0 mA\n", + "maximum transistor current is 1.61 A\n", + "minimum transistor current is 1.13 A\n", + "diode peak current is 2.02 A\n", + "secondary rms current is 1.23 A\n", + "capacitor is 170.0 microF\n" + ] + } + ], + "source": [ + "#designing circuit and finding circuit parameter\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0;#V\n", + "V2=5.0;#V\n", + "Il=1.0;#load current\n", + "T=10.0;#microsec\n", + "K=1.25;#Npri/Nsec\n", + "L=85.0;#microH\n", + "\n", + "#calculation\n", + "Vq=V1+V2*K;\n", + "Vd=V1*K+V2;\n", + "D=round((K*V2)*100/(V1+K*V2))/100;\n", + "Po=V2*Il;\n", + "Pi=round(Po/.09)/10;\n", + "Id=Pi/V1;\n", + "Im=Id/D;\n", + "Ir=(Im*D**.5);\n", + "i=V1*D*T/L;\n", + "IM=Im+.24;\n", + "Imin=Im-.24;\n", + "Ip=K*IM;\n", + "Imid=Il/(1-D);\n", + "Irms=Imid*(1-D)**.5;\n", + "C=D*Il*T/20;\n", + "\n", + "#result\n", + "print \"maximum voltage is\",round(Vq,2), \"V\"\n", + "print \"diode voltage is\",round(Vd,2), \"V\"\n", + "print \"duty cycle is\",round(D,2)\n", + "print \"power delivered is\",round(Po,2), \"W\"\n", + "print \"average current is\",round(Id*1000,2), \"mA\"\n", + "print \"mid primary current is\",round(Im,2), \"A\"\n", + "print \"rms current is\",round(Ir*1000,2),\"mA\"\n", + "print \"ramp current is\",round(i*1000,2), \"mA\"\n", + "print \"maximum transistor current is\",round(IM,2),\"A\"\n", + "print \"minimum transistor current is\",round(Imin,2),\"A\"\n", + "print \"diode peak current is\",round(Ip,2), \"A\"\n", + "print \"secondary rms current is\",round(Irms,2),\"A\"\n", + "print \"capacitor is\",round(C*1000,2), \"microF\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter8.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter8.ipynb new file mode 100644 index 00000000..7cf6b405 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter8.ipynb @@ -0,0 +1,326 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Thyristors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.3,Page 397" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inductance is 6.22 microH\n", + "load impedence at angle 90 degree is 0.00195 ohm\n" + ] + } + ], + "source": [ + "#finding inductance,load impedence\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "f=50.0;#hertz\n", + "R=80.0;#load resistance\n", + "K=50.0;#di/dt\n", + "\n", + "#calculation\n", + "L=V*2**.5/K;\n", + "Z=2*pi*f*L;\n", + "\n", + "#result\n", + "print \"inductance is\",round(L,2),\"microH\"\n", + "print \"load impedence at angle 90 degree is\",round(Z*1e-6,5), \"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.4,Page 400" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "minimum value of capacitor is 0.067 micfoF\n", + "\n", + "choose C=.1 micoF\n", + "capacitor impedence at angle -90degree is 31.83 ohm\n", + "Load current in mA at an angle 90 degrees is 6.91\n", + "Potential drop in V at an angle 90 degrees is 0.55\n", + "Power dissipated is 3 mW\n" + ] + } + ], + "source": [ + "#finding capacitor,current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "f=50.0;#hertz\n", + "R=80.0;#load resistance\n", + "K=75.0;#dv/dt\n", + "Vd=400.0;#DRM voltage\n", + "\n", + "\n", + "#calculation\n", + "C=Vd/R/K;\n", + "C1=.1;\n", + "Z=1/(2*pi*f*C1);\n", + "Iload=V/1000/(-Z*cos(180*pi/180)+R*round(cos(90*pi/180)));\n", + "Vload=Iload/1000*R;\n", + "P=Vload*Iload;\n", + "\n", + "#result\n", + "print \"minimum value of capacitor is\",round(C,3), \"micfoF\"\n", + "print('\\nchoose C=.1 micoF')\n", + "print \"capacitor impedence at angle -90degree is\",round(Z*1000,2), \"ohm\"\n", + "print \"Load current in mA at an angle 90 degrees is\",round(Iload,2)\n", + "print \"Potential drop in V at an angle 90 degrees is\",round(Vload,2)\n", + "print \"Power dissipated is\",int(P), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.5,Page 402" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "snubbing resistnce is 7.39 ohm\n" + ] + } + ], + "source": [ + "#finding snubbing resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220;#line voltage\n", + "f=50;#hertz\n", + "R=80;#load resistance\n", + "I=46;#TSM current\n", + "\n", + "#calculation\n", + "Rs=V*2**.5/(I-V*2**.5/R);\n", + "\n", + "#result\n", + "print \"snubbing resistnce is\",round(Rs,2), \"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.6,Page 414" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "line period is 16.67 ms\n", + "half-cycle time is 8.333 ms\n", + "no. of cycles is 10.0\n", + "voltage for t1 is 54.0 V\n", + "power for t1 is 291.6 W\n", + "voltage for t2 is 100.0 V\n", + "voltage for t2 is 1000.0 W\n" + ] + } + ], + "source": [ + "#finding voltage , power and cycles\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R=10.0;#load\n", + "V=120.0;#rms voltage\n", + "f=60.0;#hertz\n", + "T=83.3;#ms\n", + "t1=15;#ms\n", + "t2=55;#ms\n", + "\n", + "#calculation\n", + "Tl=1/f;\n", + "Th=Tl/2;\n", + "C=round(T/Th/100)*100;\n", + "D1=.2;\n", + "V1=round(V*D1**.5);\n", + "P1=V1**2/R;\n", + "D2=.7;\n", + "V2=round(V*D2**.5);\n", + "P2=V2**2/R;\n", + "\n", + "#result\n", + "print \"line period is\",round(Tl*1000,2), \"ms\"\n", + "print \"half-cycle time is\",round(Th*1000,3), \"ms\"\n", + "print \"no. of cycles is\",C/1000\n", + "print \"voltage for t1 is\",round(V1,3), \"V\"\n", + "print \"power for t1 is\",round(P1,3), \"W\"\n", + "print \"voltage for t2 is\",round(V2,3), \"V\"\n", + "print \"voltage for t2 is\",round(P2,3), \"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.8,Page 427" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 42.0 V\n", + "dc voltage is 41.0 V\n", + "\n", + "the markers indicae Vp=163V Vave=41\n", + "full-wave rms voltage is 108.0 V\n", + "rms voltage is 108.0 V\n", + "\n", + "the markers indicate Vp=169V ;Vave=106V\n" + ] + } + ], + "source": [ + "#finding dc volatge,average voltage,rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=120.0;#line voltage\n", + "A=60.0;#degree\n", + "D=0.35;\n", + "\n", + "#calculation\n", + "Va=D*V;\n", + "Vd=V*2**.5*(cos(A*pi/180)+1)/2/pi;\n", + "Vr=.9*V;\n", + "Vrms=V*(2**.5)*(.5*(pi-1.047)+.25*sin(2*A*pi/180))**.5/pi**.5;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(Va,3), \"V\"\n", + "print \"dc voltage is\",round(Vd), \"V\"\n", + "print('\\nthe markers indicae Vp=163V Vave=41')\n", + "print \"full-wave rms voltage is\",round(Vr), \"V\"\n", + "print \"rms voltage is\",round(Vrms), \"V\"\n", + "print('\\nthe markers indicate Vp=169V ;Vave=106V')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.9,Page 430" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 141.18 V\n", + "double checked value of rms voltage is 141.18 V\n" + ] + } + ], + "source": [ + "#finding rms voltage and double checked rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "P=1.3;#kW\n", + "R=15.0;#ohm\n", + "\n", + "#calculation\n", + "Vr=round((P*1000*R)**.5);\n", + "D=Vr/V;\n", + "Vr=V*2**.5*(.5*(pi-1.710)+sin(196*pi/180)/4)**.5/pi**.5;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vr,2), \"V\"\n", + "print \"double checked value of rms voltage is\",round(Vr,2), \"V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter9.ipynb b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter9.ipynb new file mode 100644 index 00000000..d50d131a --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/Chapter9.ipynb @@ -0,0 +1,405 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Power Conversation and Motor Drive Operations" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.1,Page 457" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "peak voltage is 37.6 V\n", + "load voltage is 35.7 V\n", + "ripple voltage is 3.96 V\n", + "approx. load voltage is 35.62 V\n" + ] + } + ], + "source": [ + "#finding peak,load,ripple voltages\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=28.0;#V\n", + "C=4700.0;#microF\n", + "R=16.0;#load\n", + "f=120.0;#hertz\n", + "\n", + "#calculation\n", + "Vp=V*2**.5-2;\n", + "Vd=0.95*Vp;\n", + "Id=Vd/R;\n", + "v=Id/f/C;\n", + "#approximation\n", + "Vd1=Vp-v*1e6/2;\n", + "\n", + "#result\n", + "print \"peak voltage is\",round(Vp,2), \"V\"\n", + "print \"load voltage is\",round(Vd,1), \"V\"\n", + "print \"ripple voltage is\",round(v*1e6,2), \"V\"\n", + "print \"approx. load voltage is\",round(Vd1,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2,Page 459" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zth is 1.0 + 1.0 in ohm\n", + "inductor is 2.65 mH\n" + ] + } + ], + "source": [ + "#finding inductor,Zth\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=120.0;#pri voltage\n", + "V2=28.0;#sec voltage\n", + "I=2.0;#pri current\n", + "f=60.0;#Hz\n", + "Vth=28.8;#open voltage\n", + "V3=12.1;#pri-short voltage\n", + "Is=2.0;#short current at 45 degree\n", + "\n", + "#calculation\n", + "Zi=(V2*V3)/V1/Is*cos(45*pi/180);\n", + "Zj=(V2*V3)/V1/Is*sin(45*pi/180);\n", + "L=Zi/(2*pi*f);\n", + "\n", + "#result\n", + "print'Zth is',round(Zi),'+',round(Zj),'in ohm'\n", + "print \"inductor is\",round(L*1000,2), \"mH\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.4,Page 463" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power factor is 0.32\n" + ] + } + ], + "source": [ + "#finding power factor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I1=1.8;#current\n", + "R=16.0;#resistance\n", + "I2=5.7;#A\n", + "V=28.8;#Voltage\n", + "\n", + "#calculation\n", + "P=I1**2*R;\n", + "S=I2*V;\n", + "Pf=P/S;\n", + "\n", + "#result\n", + "print \"power factor is\",round(Pf,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.5, Page 468" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "aparrent power is 8.14 kVA\n", + "dissipated power is 7.84 kW\n", + "power factor is 0.96\n" + ] + } + ], + "source": [ + "#finding power factor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=22.6;#current\n", + "I2=28.00;\n", + "V=120.0;#Voltage\n", + "V2=280.0;\n", + "\n", + "#calculation\n", + "Pt=3*I*V;\n", + "Pl=I2*V2;\n", + "Pf=Pl/Pt;\n", + "\n", + "#result\n", + "print \"aparrent power is\",round(Pt/1000,2),\"kVA\"\n", + "print \"dissipated power is\",round(Pl/1000,2),\"kW\"\n", + "print \"power factor is\",round(Pl/Pt,2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.6,Page 474" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ratio is 0.72\n", + "firing angle is 58 degrees\n", + "dc voltage is 148.85 V\n", + "time delay is 2.69 ms\n" + ] + } + ], + "source": [ + "#finding firing angle, time delay,Vd\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=208.0;#voltage\n", + "R=100.0;#load\n", + "Vd=150.0;#V\n", + "\n", + "#calculation\n", + "r=Vd/V;\n", + "a=58;#degree\n", + "Vd=3*2**.5*208*(cos(pi/3+a*pi/180)-cos(2*pi/3+a*pi/180))/pi;\n", + "t=a*16.7/360;\n", + "\n", + "#result\n", + "print \"ratio is\",round(r,2)\n", + "print('firing angle is 58 degrees');\n", + "print \"dc voltage is\",round(Vd,2), \"V\"\n", + "print \"time delay is\",round(t,2), \"ms\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.7,Page 480" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max. current is 41.67 A\n", + "dissipated power is 8.68 W\n" + ] + } + ], + "source": [ + "#finding maximum current and power dissipated\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=150.0;#power\n", + "V=8.0;#voltage\n", + "R=.01;#resistance\n", + "D=.5;#duty cycle\n", + "\n", + "#calculation\n", + "I=P/.9/D/V;\n", + "Ir=I*D**.5;\n", + "Pq=Ir**2*R;\n", + "\n", + "#result\n", + "print \"max. current is\",round(I,2), \"A\"\n", + "print \"dissipated power is\",round(Pq,2),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.8,Page 489" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "pwm fundamental frequency is 30.72 kHz\n", + "output voltage is 9.46 V\n" + ] + } + ], + "source": [ + "#finding fundamental frequency and output voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "f1=60.0;#frequency\n", + "V=150.0;#voltage\n", + "f2=31.0;#kHz\n", + "\n", + "#calculation\n", + "f3=f1*4;\n", + "Vo=V*10**(-4.2);\n", + "\n", + "#result\n", + "print \"pwm fundamental frequency is\",round(f3*2**7/1000,2), \"kHz\"\n", + "print \"output voltage is\",round(Vo*1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.9,Page 491" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 127.32 V\n", + "\n", + "Va-d @ 200Vin=4.2V\n", + "\n", + "\n", + "pick R1=47kohm\n", + "current through dividers is 2.62 mA\n", + "R2 is 1.6 kohm\n", + "capacitor is 27.01 microF\n" + ] + } + ], + "source": [ + "#finding resistances,capacitor,average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=120.0;#load voltage\n", + "f=60.0;#Hz\n", + "Vp=200.0;#V\n", + "Vd=5.0;#V\n", + "\n", + "\n", + "#calculation\n", + "Vdc=2*Vp/pi;\n", + "Va=4.2;\n", + "R1=47.0;\n", + "I=(Vdc-Va)/R1;\n", + "R2=Va/I;\n", + "K=1.0/(1/R1+1/R2)# R1 \\\\ R2\n", + "C=1.0/2/pi/3.8/K;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(Vdc,2), \"V\"\n", + "print('\\nVa-d @ 200Vin=4.2V')\n", + "print('\\n\\npick R1=47kohm')\n", + "print \"current through dividers is\",round(I,2), \"mA\"\n", + "print \"R2 is\",round(R2,2), \"kohm\"\n", + "print \"capacitor is\",round(C*1000,2), \"microF\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/README.txt b/Power_Electronics_Principles_and_Applications_by_Jacob/README.txt new file mode 100644 index 00000000..274ef8b6 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/README.txt @@ -0,0 +1,10 @@ +Contributed By: Amit Kumar Das +Course: btech +College/Institute/Organization: ISM Dhanbad +Department/Designation: Mining Machinery Engineering +Book Title: Power Electronics Principles and Applications +Author: Jacob +Publisher: Cengage Learning, New Delhi +Year of publication: 2009 +Isbn: 9780766823327 +Edition: 4
\ No newline at end of file diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/4.png b/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/4.png Binary files differnew file mode 100644 index 00000000..5199042e --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/4.png diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/5.png b/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/5.png Binary files differnew file mode 100644 index 00000000..76933e63 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/5.png diff --git a/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/6.png b/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/6.png Binary files differnew file mode 100644 index 00000000..607dd0a1 --- /dev/null +++ b/Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/6.png diff --git a/sample_notebooks/AviralYadav/Chapter5.ipynb b/sample_notebooks/AviralYadav/Chapter5.ipynb new file mode 100644 index 00000000..5780e6b5 --- /dev/null +++ b/sample_notebooks/AviralYadav/Chapter5.ipynb @@ -0,0 +1,146 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5:INTRODUCTION TO THERMAL POWER PLANT" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.1:pg-271" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "The coal required per hour,W= 50.95 tons/hr\n" + ] + } + ], + "source": [ + "# Example 5_1\n", + " \n", + "# Given data\n", + "P=100.0;# Plant capacity in Mw\n", + "CV=25600.0;# Calorific value in kJ/kg\n", + "n_th=30.0;# The thermal efficiency of the plant in %\n", + "n_eg=92.0;# Electrical generation efficiency in %\n", + "\n", + "# Calculation\n", + "# Mechanical energy available=W*CV*(n_th/100) in kJ/hr\n", + "# Electrical energy available=W*CV*(n_th/100)*(n_eg/100) in kJ/hr\n", + "q_e=P*10**3*3600;# Heat equivalent in kJ/hr\n", + "W=(q_e/(CV*(n_th/100)*(n_eg/100)));# The coal required per hour in kg/hr\n", + "W=(W/1000);# The coal required per hour in tons/hr\n", + "print \"\\nThe coal required per hour,W=\",round(W,2),\" tons/hr\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.2:pg-273" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " The capacity of the power plant,P= 62.05 MW\n" + ] + } + ], + "source": [ + "# Example 5_2\n", + " \n", + "# Given data\n", + "CV=28900.0;#kJ/kg\n", + "n_b=83.0;# The boiler efficiency in %\n", + "n_t=32.0;# The turbine efficiency in %\n", + "n_g=97.0;# The generator efficiency in %\n", + "W=30.0;# The coal consumption of the station in tons/hr\n", + "\n", + "# Calculation\n", + "P=((W*1000*CV)*(n_b/100)*(n_t/100)*(n_g/100))/(3600*1000);# The capacity of the power plant in MW\n", + "print \"\\n The capacity of the power plant,P=\",round(P,2),\" MW\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.3:pg-273" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volume of gas required per hour,V= 450.0 m**3/hr\n" + ] + } + ], + "source": [ + "#Example 5_3\n", + "\n", + "# Given values\n", + "P=100.0;# Power in kW\n", + "CV=4000.0;# Calorific value in kJ/m**3\n", + "n_o=0.20;# Over all efficiency of the plant\n", + "\n", + "# Calculation\n", + "V=(3600*P)/(CV*n_o);# m**3/hr\n", + "print \"The volume of gas required per hour,V=\",round(V,2),\" m**3/hr\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |
