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| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb | 644 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb | 426 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb | 515 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb | 632 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb | 760 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb | 802 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb | 543 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb | 610 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb | 903 | ||||
| -rw-r--r-- | Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb | 280 | ||||
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diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb new file mode 100644 index 00000000..63c6e5ec --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb @@ -0,0 +1,644 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 01 - Introduction To Heat Transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.1 - Page : 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "t1=38 # in degree C\n", + "t2=21 # in degree C\n", + "k=0.19 # unit less\n", + "x=4 #in cm\n", + "x=x*10**-2 # in meter\n", + "# Formula q=k*A*(t1-t2)/x\n", + "q_by_A=k*(t1-t2)/x \n", + "print \"The rate of heat transfer is :\",round(q_by_A,3),\"W/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate of heat transfer is : 80.75 W/m**2\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.2 - Page : 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "t_i=120 # in degree C\n", + "t_o=40 # in degree C\n", + "K=0.04 # unit less\n", + "x=0.06 #in m\n", + "Q=50 # in W\n", + "print \"Assuming steady state heat transfer in the wall\"\n", + "# Rate of heat transfer across the wall = Rate of electrical energy dissipation in the furnance\n", + "# Formula Q= K*A*(t_i-t_o)/x \n", + "A=Q*x/(K*(t_i-t_o)) \n", + "print \"Area of wall = %0.4f square meter\" %A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Assuming steady state heat transfer in the wall\n", + "Area of wall = 0.9375 square meter\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.3 - Page : 10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "t_f=30 # in degree C\n", + "t_s=400 # in degree C\n", + "d=0.04 #in m\n", + "h=20 # in W/m**2K\n", + "l=1 #in meter\n", + "A=pi*d*l#\n", + "q=h*A*(t_s-t_f) # in W\n", + "print \"Rate of heat loss = %0.3f watt\" %q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat loss = 929.911 watt\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.4 - Page : 10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "t_s=100 # in degree C\n", + "t_w=80 # in degree C\n", + "d=2*10**-3 #in m\n", + "h=3000 # in W/m**2 degree C\n", + "L=100 #in mm\n", + "L=L*10**-3 # in meter\n", + "A=pi*d*L#\n", + "# Heat loss by convection = Electric power supplied\n", + "# Formula h*A*(t_s-t_w) = Q\n", + "Q= h*A*(t_s-t_w)#\n", + "print \"Electric power supplied = %0.1f watt\" %Q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electric power supplied = 37.7 watt\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.5 - Page : 13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "A=0.6*0.9 # in square meter\n", + "x=.025 # in meter\n", + "t_s=310 # in degree C\n", + "t_f=15 # in degree C\n", + "h=22 # in W/m**2 degree C\n", + "K=45 # in W/m degree C\n", + "Q_rad=250 # in W\n", + "# Heat transfer through the plate = Convection heat loss + radiation heat loss\n", + "# Formula Q_cond = Q_conv + Q_rad\n", + "# -K*A*dt/dx = h*A*(t_s-t_f)+ Fg12*sigmaA(Ts**4-Ta64)\n", + "t_i=x*(h*A*(t_s-t_f)+Q_rad)/(K*A)+t_s \n", + "print \"The inside plate temperature = %0.2f degree C\" %t_i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The inside plate temperature = 313.86 degree C\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.6 - Page : 13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "T1=50 # in degree C\n", + "T1=T1+273 # in K\n", + "T2=20 # in degree C\n", + "T2=T2+273 # in K\n", + "d=5*10**-2 #in m\n", + "h=6.5 # in W/m**2K\n", + "l=1 #in meter\n", + "epsilon=0.8#\n", + "sigma=5.67*10**-8#\n", + "A=pi*d*l # in Square meter\n", + "q_conv = h*A*(T1-T2) # in W/m\n", + "print \"The heat loss by convection = %0.1f W/m\" %q_conv\n", + "# formula q= sigma*A*F_g12*(T1**4-T2**4) = sigma*A*epsilon*(T1**4-T2**4) (since A1<<A2, so F_g12=epsilon)\n", + "q_rad = sigma*A*epsilon*(T1**4-T2**4) # in W/m\n", + "print \"Heat loss by radiation = %0.0f W/m\" %q_rad\n", + "q_total= q_conv+q_rad#\n", + "print \"Total heat loss = %0.3f W/m\" %q_total" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat loss by convection = 30.6 W/m\n", + "Heat loss by radiation = 25 W/m\n", + "Total heat loss = 55.672 W/m\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.7 - Page : 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad\n", + "# given data\n", + "T1=1350 # in degree C\n", + "T2=50 # in degree C\n", + "L=25*10**-2 #in meter\n", + "# Formula q= -k*A*dT/dx\n", + "# or q/A= -k*dT/dx\n", + "# let q/A = q_by_A\n", + "def integrand(T):\n", + " return -0.838*(1+0.0007*T)\n", + "ans, err = quad(integrand, T1, T2)\n", + "def integrand(x):\n", + " return 1\n", + "ans2, err2 = quad(integrand, 0, L)\n", + "q_by_A=ans/ans2#\n", + "print \"Heat transfer rate per square meter through the cylinder = %0.0f watt\" %q_by_A\n", + "\n", + "# Note : Answer in the book is incorrect" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate per square meter through the cylinder = 6493 watt\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.9 - Page : 21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "K_A=0.5 # in W/m degree C\n", + "K_B=0.8 # in W/m degree C\n", + "Ti_A=600 # inside temp. of slab A in degree C\n", + "To_B=100 # outside temp. of slab B in degree C\n", + "t_A=4*10**-2 # thickness of slab A\n", + "t_B=6*10**-2 # thickness of slab B\n", + "# Heat transfer rate per square meter through the slab A\n", + "# q/A = +K_A * ( Ti_A - T) / t_A (1)\n", + "# Heat transfer rate through slab B\n", + "# q/A = +K_B * ( T - To_B) / t_B (2)\n", + "# Equating Eqns (1) and (2)\n", + "# K_A*(Ti_A - T)/t_A = K_B*(T - To_B)/t_B\n", + "T=t_A*t_B/(K_A*t_B+K_B*t_A)*(K_A*Ti_A/t_A + K_B*To_B/t_B)\n", + "print \"T, intermediate temperature of slab A and B is :\",round(T,3),\"degree C\"\n", + "#Putting the value of T in Eq(1), we get\n", + "q_by_A= K_A*( Ti_A - T) / t_A#\n", + "print \"Steady state heat transfer rate per square meter is :\",round(q_by_A,3),\"W/m**2\"\n", + "#Note : Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T, intermediate temperature of slab A and B is : 341.935 degree C\n", + "Steady state heat transfer rate per square meter is : 3225.806 W/m**2\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.10 - Page : 22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "La=3*10**-2 # in meter\n", + "Aa=1 # in m**2\n", + "ka=150 # in W/m-K\n", + "\n", + "Lb=8*10**-2 # in meter\n", + "Ab=0.5 # in m**2\n", + "kb=30 # in W/m-K\n", + "\n", + "Lc=8*10**-2 # in meter\n", + "Ac=0.5 # in m**2\n", + "kc=65 # in W/m-K\n", + "\n", + "Ld=5*10**-2 # in meter\n", + "Ad=1 # in m**2\n", + "kd=50 # in W/m-K\n", + "\n", + "T1=400 # in degree C\n", + "T2=60 # in degree C\n", + "\n", + "Ra=La/(ka*Aa)#\n", + "Rb=Lb/(kb*Ab)#\n", + "Rc=Lc/(kc*Ac)#\n", + "Rd=Ld/(kd*Ad)#\n", + "#The equivalent resistance for Rb and Rc\n", + "Re=Rb*Rc/(Rb+Rc)#\n", + "#Total Resistance\n", + "sigmaR=Ra+Re+Rd#\n", + "# heat transfer rate per square meter\n", + "q=(T1-T2)/sigmaR#\n", + "print \"Heat transfer rate per square meter = %0.2e Watt\" %q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate per square meter = 1.18e+05 Watt\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.11 - Page : 23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "k_Al=202 # in W/mK\n", + "x_Al=0.005 # in m\n", + "del_T=80 # in degree C\n", + "R_contact=0.88*10**-4 # in m**2K/W\n", + "sigmaR=x_Al/k_Al+R_contact+x_Al/k_Al # in m**2K/W\n", + "q=del_T/sigmaR # in W/m**2\n", + "#Temperature drop across the rough surface\n", + "del_T=q*R_contact #in degree C\n", + "print \"Temperature drop across the rough surface = %0.3f degree C\" %del_T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature drop across the rough surface = 51.198 degree C\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.12 - Page : 24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T1=100 # in degree C\n", + "T2=10 # in degree C\n", + "A=3*5 #in square meter\n", + "x=40*10**-2 # thickness in m**2\n", + "k=1.6 # in W/mk\n", + "h=10 # in W/m**2k\n", + "# Total resistance in heat flow path\n", + "sigmaR=x/(k*A)+1/(h*A)#\n", + "# so heat transfer rate\n", + "q=(T1-T2)/sigmaR # in Watt\n", + "q=q*10**-3 #in kW\n", + "print \"Heat transfer rate = %0.3f kW\" %q\n", + "\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate = 3.857 kW\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.13 - Page : 25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad\n", + "# given data\n", + "k='2.0+0.0005*T' # in W/m-k\n", + "A=3*5 #in square meter\n", + "T1=150 # in degree C\n", + "T2=50 # in degree C\n", + "L=20*10**-2 # thickness in m**2\n", + "# Formula q= -k*A*dt/dx\n", + "def integrand(T):\n", + " return 2.0+0.0005*T\n", + "ans, err = quad(integrand, T1, T2)\n", + "def integrand(x):\n", + " return 1\n", + "ans2, err2 = quad(integrand, 0, L)\n", + "\n", + "q=-A*ans/ans2 # in Watt\n", + "q=q*10**-3 #in kW\n", + "print \"Rate of heat transfer = %0.3f kW\" %q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat transfer = 15.375 kW\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.14 - Page : 26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T1=300 #in degree C\n", + "T2=50 #in degree C\n", + "x2=2*10**-2 # thickness of boiler wall in m\n", + "tc2=58 # thermal conductivity of wall in W/mk\n", + "x3=0.5*10**-2 # thickness of outer surface of the wall in m\n", + "tc3=116*10**-3 # thermal conductivity of outer surface of the wall in W/mk\n", + "R1=2.3*10**-3 # in k/W\n", + "R2=x2/tc2#\n", + "R3=x3/tc3#\n", + "sigmaR=R1+R2+R3 # Total Resistance\n", + "q=(T1-T2)/sigmaR#\n", + "print \"Heat transfer rate per unit area = %0.3f W/m**2\" %q\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate per unit area = 5464.687 W/m**2\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.15 - Page : 26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "Tf=80 # in degree C\n", + "I=200 # in amp\n", + "h=4000 # in W/m**2degree C\n", + "rho=70*10**-6#\n", + "L=100 # in cm\n", + "R=0.1 # in ohm\n", + "d=3 # in mm\n", + "d=d*10**-3#\n", + "As= pi*d#\n", + "#Formula I**2*R= h*As*(Tw-Tf)\n", + "Tw= I**2*R/(h*As)+Tf#\n", + "print \"Central temperature of the wire = %0.f \u00b0C\" %Tw" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Central temperature of the wire = 186 \u00b0C\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 1.16 - Page : 27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "E=500 #Absorb solar energy in W/m**2\n", + "epsilon= 0.9#\n", + "T_s= 280 # in K\n", + "T_infinite=300 # in K\n", + "h_c=20 # in W/m**2degree C\n", + "T_sky=280 # in K\n", + "sigma=5.67*10**-8#\n", + "# Formula E= h_c*(T_p-T_infinite)+epsilon*sigma*(T_P**4-T_s**4)\n", + "# On simplication T_P= 340.6-0.255*T-p**4\n", + "T_p= 315.5 # in K\n", + "print \"Equilibrium Temperature of the plate = %0.1f K\" %T_p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equilibrium Temperature of the plate = 315.5 K\n" + ] + } + ], + "prompt_number": 29 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb new file mode 100644 index 00000000..119f2bce --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb @@ -0,0 +1,426 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 - Mass Transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.1 - Page : 291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "P1=4 # in bar\n", + "P2=2 # in bar\n", + "T=25 # in degree C\n", + "Dhp=9*10**-8 # in m**2/s\n", + "S=3*10**-3 # in kg mole/m**3 bar\n", + "del_x=0.5*10**-3 # thickness in m\n", + "#(a) The molar concentration of a gas in terms of solubility\n", + "CH1=S*P1 # in kg mole/m**3\n", + "CH2=S*P2 # in kg mole/m**3\n", + "#(b) Molar diffusion flux of hydrogen through plastic memberence is given by Fick's law of diffision\n", + "#N_H= N_h/A = Dhp*(CH1-CH2)/del_x#\n", + "N_H= Dhp*(CH1-CH2)/del_x # in kg mole/s-m**2\n", + "print \"Molar diffusion flux of hydrogen through the membrane = %0.3e kg mole/s-m**2\" %N_H\n", + "#Mass_d_Flux= N_H*Molecular_Weight \n", + "Molecular_Weight=2#\n", + "Mass_d_Flux= N_H*Molecular_Weight \n", + "print \"Molar diffusion flux = %0.3e kg/s-m**2\" %Mass_d_Flux" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molar diffusion flux of hydrogen through the membrane = 1.080e-06 kg mole/s-m**2\n", + "Molar diffusion flux = 2.160e-06 kg/s-m**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.2 - Page : 292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T=25 # in degree C\n", + "T=T+273 # in K\n", + "P=1#\n", + "V1=12 #Molecular volume of H2 in cm**3/gm mole\n", + "V2=30 #Molecular volume of Air in cm**3/gm mole\n", + "M1=2 # Molecular weight of H2\n", + "M2=29 # Molecular weight of Air\n", + "#The diffusion coefficient for gases in terms of molecular volumes may be express as\n", + "D_AB= .0043*T**(3/2)/(P*(V1**(1/3)+V2**(1/3)))*(1/M1+1/M2)**(1/2)#\n", + "print \"The diffusion coefficient for gases in terms of molecular volumes = %0.3f cm**2/sec\" %D_AB" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diffusion coefficient for gases in terms of molecular volumes = 2.997 cm**2/sec\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.3 - Page : 292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T=300 # temp of gas mixture in K\n", + "D_HN2=18*10**-6 # in m**2/s at 300 K, 1 bar\n", + "T1=300 # in K\n", + "D_HO2=16*10**-6 # in m**2/s at 273 K, 1 bar\n", + "T2=273 # in K\n", + "O_2=0.2#\n", + "N_2=0.7#\n", + "H_2=0.1#\n", + "#The diffusivity at the mixture temperature and pressure are calculated as \n", + "# D1/D2 = (T1/T2)**(3/2)*(P2/P1)\n", + "D_HO2= (T/T2)**(3/2)*1/4*D_HO2#\n", + "D_HN2= (T/T1)**(3/2)*1/4*D_HN2#\n", + "#The composition of oxygen and nitrogen on a H2 free basis is \n", + "x_O= O_2/(1-H_2)#\n", + "x_N= N_2/(1-H_2)#\n", + "# The effective diffusivity for the gas mixture at given temperature and pressure is\n", + "D= 1/(x_O/D_HO2+x_N/D_HN2) # in m**2/s\n", + "print \"Effective diffusivity = %0.3e m**2/s\" %D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effective diffusivity = 4.524e-06 m**2/s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.4 - Page : 293" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "d=3 # in mm\n", + "d=d*10**-3 # in meter\n", + "T=25 # in \u00b0C\n", + "T=T+273 # in K\n", + "D= 0.4*10**-4 # in m**2/s\n", + "R= 8314#\n", + "P_A1=1 # in atm\n", + "P_A1=P_A1*10**5 # in w/m**2\n", + "P_A2=0#\n", + "C_A2=0#\n", + "x2= 15 # in meter\n", + "x1= 0#\n", + "A= pi/4*d**2#\n", + "M_A= D*A/(R*T)*(P_A1-P_A2)/(x2-x1) # in kg mole/sec\n", + "N_B= M_A#\n", + "M_B= M_A*29 # in kg/sec\n", + "print \"Value of N_B = %0.3e kg mole/sec\" %N_B\n", + "print \"Value of M_B = %0.3e kg/sec\" %M_B\n", + "# Note : The value of M_B in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of N_B = 7.608e-13 kg mole/sec\n", + "Value of M_B = 2.206e-11 kg/sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.5 - Page : 294" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "P=3 # in atm\n", + "P=P*10**5 # in N/m**2\n", + "r1=10 # in mm\n", + "r1=r1*10**-3 # in m\n", + "r2=20 # in mm\n", + "r2=r2*10**-3 # in m\n", + "R=4160 # in J/kg-K\n", + "T=303 # in K\n", + "D=3*10**-8 # in m**2/s\n", + "S=3*0.05# # Solubility of hydrogen at a pressure of 3 atm in m**3/m**3 of rubber tubing\n", + "del_x=r2-r1 # in m\n", + "L=1 # in m\n", + "Am=2*pi*L*del_x/log(r2/r1)#\n", + "#Formula P*V= m*R*T\n", + "V=S#\n", + "m=P*V/(R*T) # in kg/m**3 of rubber tubing at the inner surface of the pipe\n", + "C_A1=m#\n", + "C_A2=0#\n", + "#Diffusion flux through the cylinder is given\n", + "M=D*(C_A1-C_A2)*Am/del_x#\n", + "print \"Diffusion flux through the cylinder = %0.1e kg/sm\" %M" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusion flux through the cylinder = 9.7e-09 kg/sm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.6 - Page : 295" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "R=4160 # in J/kg-K\n", + "M=2#\n", + "D_AB=1.944*10**-8 # in m**2/s\n", + "R_H2=R/M#\n", + "S=2*0.0532# # Solubility of hydrogen at a pressure of 2 atm in cm**3/cm**3 of pipe\n", + "P=2 # in atm\n", + "P=P*1.03*10**5 # N/m**2\n", + "T=25 # in degree C\n", + "T=T+273 # in K\n", + "r1=2.5 # in mm\n", + "r1=r1*10**-3 # in m\n", + "r2=5 # in mm\n", + "r2=r2*10**-3 # in m\n", + "del_x=r2-r1 # in m\n", + "L=1 # in m\n", + "#Formula P*V= m*R*T\n", + "V=S#\n", + "m=P*V/(R*T) # in kg/m**3 of pipe\n", + "# So, Concentration of H2 at inner surface of the pipe\n", + "C_A1=0.0176 # in kg/m**3\n", + "# The resistance of diffusion of H2 away from the outer surface is negligible i.e.\n", + "C_A2=0#\n", + "Am=2*pi*L*del_x/log(r2/r1)#\n", + "# Loss of H2 by diffusion \n", + "M_A= D_AB*(C_A1-C_A2)*Am/del_x#\n", + "print \"Loss of H2 by diffusion = %0.3e kg/s\" %M_A\n", + "\n", + "\n", + "#Note: In the book , they put wrong value of C_A1 to calculate M_A, so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of H2 by diffusion = 3.101e-09 kg/s\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.7 - Page : 296" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "from math import log\n", + "# given data\n", + "Px1= 0.14 # in bar\n", + "Px2= 0#\n", + "P=1.013 # in bar\n", + "Py1=P-Px1# # in bar\n", + "Py2=P-Px2# # in bar\n", + "D=8.5*10**-6 # in m**2/s\n", + "d=5 # diameter in meter\n", + "L=1 # in mm\n", + "L=L*10**-3 #in meter\n", + "M=78 # molecular weight\n", + "Am_x= 1/4*pi*d**2*M#\n", + "R=8314#\n", + "del_x=3 # thickness in mm\n", + "del_x=del_x*10**-3 # in m\n", + "T=20 # in degree C\n", + "T=T+273 # in K\n", + "P=P*10**5 # in N/m**2\n", + "m_x= D*Am_x*P*log(Py2/Py1)/(R*T*del_x)#\n", + "# The mass of the benzene to be evaporated\n", + "mass= 1/4*pi*d**2*L#\n", + "density=880 # in kg/m**3\n", + "m_b= mass*density#\n", + "toh=m_b/m_x # in sec\n", + "print \"Time taken for the entire organic compound to evaporate = %0.3f seconds\" %toh\n", + "\n", + "\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken for the entire organic compound to evaporate = 643.788 seconds\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.8 - Page : 297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "A=0.5 # in m**2\n", + "Pi=2.2 # in bar\n", + "Pi=Pi*10**5 # in N/m**2\n", + "Pf=2.18 # in bar\n", + "Pf=Pf*10**5 # in N/m**2\n", + "T=300 # in K\n", + "S=0.072 # in m**3\n", + "V=0.028 # in m**3\n", + "L=10 # in mm\n", + "L=L*10**-3 # in meter\n", + "R=287#\n", + "# Diffusivity of air in rubber D\n", + "# Initial mass of air in the tube\n", + "mi= Pi*V/(R*T) # in kg\n", + "#final mass of air in the tube\n", + "mf= Pf*V/(R*T) # in kg\n", + "# Mass of air escaped\n", + "ma = mi-mf #in kg\n", + "# Formula Na = ma/A = mass of air escaped / Time elapsed * area\n", + "A=6*24*3600*0.5#\n", + "Na = ma/A #in kg/sm**2\n", + "# Solubility of air should be calculated at mean temperature\n", + "S_meanTemperature=(2.2+2.18)/2 # in bar\n", + "#Solubility of air at the mean inside Pressure is \n", + "S=S*S_meanTemperature # in m**3/m**3 of rubber\n", + "V1=S#\n", + "V2=0.072#\n", + "T1=T#\n", + "T2=T#\n", + "P1=2.19*10**5 # in N/m**2\n", + "P2=1*10**5 # in N/m**2\n", + "# The corresponding mass concentration at the inner and outer surface of the tube, from gas equation are calculated as\n", + "Ca1= P1*V1/(R*T1) # in kg/m**3\n", + "Ca2= P2*V2/(R*T2) # in kg/m**3\n", + "# The diffusion flux rate of air through the rubber is given by\n", + "# Na = ma/A = D*(Ca1-Ca2)/del_x, here\n", + "del_x=L#\n", + "D=Na*del_x/(Ca1-Ca2)#\n", + "print \"Diffusivity of air in rubber = %0.3e m**2/s\" %D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusivity of air in rubber = 7.905e-11 m**2/s\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb new file mode 100644 index 00000000..b5e164ff --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb @@ -0,0 +1,515 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 02 - General Heat Conduction Equation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.5 - Page : 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import log, pi\n", + "# given data\n", + "r1=5 # in cm\n", + "r2=5+4 # in cm\n", + "r3= 9+2.5 # in cm\n", + "k1=0.0701 # in W/mK\n", + "k2=0.1 # in W/mK\n", + "L=20 # in m\n", + "T1=234.36 # in degree C\n", + "T3=24 # in degree C\n", + "sigmaR= (log(r2/r1)/(2*pi*k1*L) + log(r3/r2)/(2*pi*k2*L))#\n", + "\n", + "# Part (i)\n", + "q=(T1-T3)/sigmaR # in watt\n", + "print \"Heat transfer rate = %0.3f watt\" %q\n", + "\n", + "# Part(ii)\n", + "# Formula q= (T1-T2)/(log(r2/r1)/(2*pi*k1*L))\n", + "T2 =T1- (q*(log(r2/r1)/(2*pi*k1*L)))#\n", + "print \"Interface temperature of insulation = %0.3f degree\" %T2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate = 2439.473 watt\n", + "Interface temperature of insulation = 71.585 degree\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.6 - Page : 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from numpy import pi\n", + "# given data\n", + "k_brick=0.93 # in W/mK\n", + "k_insulation=0.12 # in W/mK\n", + "k_wood=0.175 # in W/mK\n", + "k_Al=204 # in W/mK\n", + "k1=k_brick#\n", + "k2=k_insulation#\n", + "k3=k_wood#\n", + "T1=200 # in degree C\n", + "T4=10 # in degree C\n", + "x1=10*10**-2 # in m\n", + "x2=25*10**-2 # in m\n", + "x3=1*10**-2 # in m\n", + "A=0.1 # in m**2\n", + "sigmaR= x1/(k1*A)+x2/(k2*A)+x3/(k3*A)#\n", + "q1=(T1-T4)/sigmaR#\n", + "print \"Heat transfer rate without rivet = %0.2f Watt\" %q1\n", + "\n", + "# Heat transfer rate with rivet\n", + "d=3*10**-2 # in meter\n", + "x=x1+x2+x3#\n", + "k_rivet=k_Al#\n", + "A_rivet=pi*d**2/4 # in m**2\n", + "R_rivet= x/(k_rivet*A_rivet)#\n", + "A_eff=A-A_rivet # in m**2\n", + "sigmaRw= 1/A_eff*(x1/k1+x2/k2+x3/k3) # in k/W\n", + "R_eq= R_rivet*sigmaRw/(R_rivet+sigmaRw) # in k/W\n", + "q2=(T1-T4)/R_eq # in watt\n", + "print \"Heat transfer rate with rivet = %0.3f Watt\" %q2\n", + "percentIncrease=(q2-q1)*100/q1 # percent increase in heat flow due to rivet\n", + "print \"Percentage increase in heat flow due to rivet = %0.f %%\" %math.ceil(percentIncrease)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate without rivet = 8.45 Watt\n", + "Heat transfer rate with rivet = 84.497 Watt\n", + "Percentage increase in heat flow due to rivet = 900 %\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.7 - Page : 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "k_cu=384 # in W/mK\n", + "k_s=1.75 # in W/mK\n", + "k1=k_cu#\n", + "k2=k_s#\n", + "hi=221 # in W/m**2K\n", + "ho=3605 # in W/m**2K\n", + "Ti=100 # in degree C\n", + "To=125 # in degree C\n", + "r1=0.2 # in m\n", + "r2=0.02+0.006 # in m\n", + "r3=0.026+0.003 # in m\n", + "ri=0.02 # in m\n", + "L=1 # in m\n", + "# Part(i)\n", + "Ao= 2*pi*r3*L#\n", + "Ai= 2*pi*r1*L#\n", + "# Formula Uo= 1/Ao*sigmaR\n", + "Uo= 1/(r3/(ri*hi) + r3/k1*log(r2/r1) + r3/k2*log(r3/r2) + 1/ho) # in w/m**2K\n", + "print \"Overall heat transfer coefficient based on outer area = %0.3f W/m**2K\" %Uo\n", + "\n", + "#Part(ii)\n", + "del_T= To-Ti#\n", + "q=Uo*Ao*del_T#\n", + "print \"Water to air heat transfer rate = %0.3f W/m\" %q\n", + "\n", + "#Part (iii)\n", + "# Formula q= T/(log(r3/r2)/(2*pi*k*L)) , where T=T2-T3 and k=k_s\n", + "k=k_s#\n", + "T= q*log(r3/r2)/(2*pi*k*L)#\n", + "print \"Temperature drop across the scale deposited = %.3f degree C\" %T\n", + "\n", + "# Note: In Part (i), they put wrong value of r2 and r1 in log(r2/r1) to calculate the value of Uo. So there is some difference in answer of coding and book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Overall heat transfer coefficient based on outer area = 117.730 W/m**2K\n", + "Water to air heat transfer rate = 536.298 W/m\n", + "Temperature drop across the scale deposited = 5.326 degree C\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.8 - Page : 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# given data\n", + "k=0.175 # in W/mK\n", + "h_infinite=9.3 # in W/m**2K\n", + "T_infinite=30 # in degree C\n", + "T_s=70 # in degree C\n", + "d=10*10**-3 # in m\n", + "r=d/2#\n", + "L=1 # in m\n", + "rc=k/h_infinite # in m\n", + "CriticalThickness = rc-r # in meter\n", + "CriticalThickness=CriticalThickness*10**3#\n", + "print \"Critical thickness = %0.1f mm\" %CriticalThickness\n", + "\n", + "q1=2*pi*r*L*h_infinite*(T_s-T_infinite) # in W/m\n", + "q2= (T_s-T_infinite)/(log(rc/r)/(2*pi*k*L)+1/(2*pi*rc*h_infinite)) # in W/m\n", + "PerIncHeatDiss= (q2-q1)*100/q1#\n", + "print \"Percentage increase in heat dissipation rate = %0.3f %%\" %PerIncHeatDiss\n", + "#Also q1=I1**2*R with bare cable\n", + "# q2=I2**2*R with insulated cable\n", + "I2_by_I1 = sqrt(q2/q1)#\n", + "# ( I2-I1 ) / I1 = (I2_by_I1 -1) / 1\n", + "# Percentage increase in current carrying capacity\n", + "PerIncCurrent = (I2_by_I1 -1) / 1 *100#\n", + "print \"Increase in current carrying capacity = %0.f %%\" %math.floor(PerIncCurrent)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical thickness = 13.8 mm\n", + "Percentage increase in heat dissipation rate = 61.845 %\n", + "Increase in current carrying capacity = 27 %\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.9 - Page : 59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "k_in=0.3 # in W/mK\n", + "k_gw=0.038 # in W/mK\n", + "ro=1.5 # in cm\n", + "ho=12 # in W/m**2 degree C\n", + "rc=k_in/ho # in m\n", + "rc=rc*10**2 # in cm\n", + "print \"Critical radius = %0.1f cm\" %rc\n", + "if ro<rc:\n", + " print \"Since radius of insulation (\",round(ro,3),\"cm) is less than critical radius of insulation (\",round(rc,3),\"cm)\"\n", + " print \"so heat transfer rate will increase by adding thsi insulation and hence it is not effective\"\n", + "ro=ro*10**-2 # in meter\n", + "# For effective insulation\n", + "# ro>=rc\n", + "# Kin/ho<= ro\n", + "roho=ro*ho # in W/mK\n", + "# Kin<= ro*ho\n", + "print \"Maximum value of thermal conductivity = %0.2f W/mK\" %roho" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical radius = 2.5 cm\n", + "Since radius of insulation ( 1.5 cm) is less than critical radius of insulation ( 2.5 cm)\n", + "so heat transfer rate will increase by adding thsi insulation and hence it is not effective\n", + "Maximum value of thermal conductivity = 0.18 W/mK\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.10 - Page : 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "d=1.2*10**-3 # in m\n", + "r=d/2 # in m\n", + "rc=1.8*10**-3 # in m\n", + "T1=100 # in degree C\n", + "T_infinite=30 # in degree C\n", + "k=0.3 # in W/mK\n", + "h=10 # in W/m**2K\n", + "L=1 # in m\n", + "ke=5.1*10**7#\n", + "q=(T1-T_infinite)/(log(rc/r)/(2*pi*k)+1/(2*pi*rc*h)) # in W/m\n", + "# Volume of wire for one meter length\n", + "vol= pi*r**2*L # in m**3\n", + "print \"In steady state heat transfer process, the heat produced by the wire is dissipated to surrounding.\" \n", + "# Heat produced per unit volume of the wire\n", + "HeatProduced= q/vol # in w/m**2\n", + "# Formula HeatProduced= I**2*R = I**2/ke\n", + "I=sqrt(HeatProduced*ke) # in amp/m**2\n", + "# Area of wire\n", + "A= pi*r**2#\n", + "# so current carrying capacity of the given wire\n", + "Current= I*A#\n", + "print \"The current carried by the copper wire = %0.3f amphere\" %Current" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In steady state heat transfer process, the heat produced by the wire is dissipated to surrounding.\n", + "The current carried by the copper wire = 20.698 amphere\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.11 - Page : 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "d_i=.1 # inner dia in m\n", + "r_i=d_i/2 # in m\n", + "Ti=473 # in K\n", + "T_infinite=293 # in K\n", + "k=1 # in W/mK\n", + "h=8 # in W/m**2K\n", + "rc=k/h # in m\n", + "print \"Critical radius = %0.3f meter\" %rc\n", + "#when\n", + "ro=rc#\n", + "q_by_L= (Ti-T_infinite)/(log(rc/r_i)/(2*pi*k)+1/(2*pi*rc*h)) # in W/m\n", + "print \"Heat loss per meter length of pipe = %0.3f W/m\" %q_by_L\n", + "\n", + "# Note: To calculate the value of q_by_L the calculation is wrong in the book so answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical radius = 0.125 meter\n", + "Heat loss per meter length of pipe = 590.189 W/m\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.12 - Page : 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "r1=100*10**-3 # in m\n", + "r2=200*10**-3 # in m\n", + "q1=1.16*10**5 # in W/m**2\n", + "t2=30 # in degree C\n", + "k=50 # in W/mK\n", + "L=1 # in m\n", + "# Total heat passing through the cylinder q\n", + "#q=q1*2*pi*r1*L (1)\n", + "# and heat conducted through the cylinder\n", + "# q= 2*pi*k*L(t1-t2)/log(r2/r1) (2)\n", + "# From (1) and (2)\n", + "t1= t2+ q1*2*pi*r1*L*log(r2/r1)/(2*pi*k*L) # in degree C\n", + "print \"Temperature of inner surface = %0.2f degree C\" %t1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature of inner surface = 190.81 degree C\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.13 - Page : 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "d1=1*10**-3 # in m\n", + "d2=3*10**-3 # in m\n", + "r1=d1/2#\n", + "r2=d2/2#\n", + "kp=384 # in W/mK\n", + "kw=0.35 # in W/mK\n", + "rho=1.96*10**-8 # in Wm\n", + "t_s=95 # in degree C\n", + "t_infinite=40 # in degree C\n", + "h=8.75 # in W/m**2K\n", + "q_by_L= (t_s-t_infinite)/(log(r2/r1)/(2*pi*kp)+1/(2*pi*r2*h))#\n", + "# Also q_by_L = I**2*R/L = I**2*rho/(pi/4*d**2)\n", + "I= sqrt(q_by_L*(pi/4*d1**2)/rho) # in amp\n", + "print \"The maximum steady state current = %0.3f amphere\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum steady state current = 13.481 amphere\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 2.16 - Page : 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log, sqrt\n", + "from math import pi\n", + "from __future__ import division\n", + "# given data\n", + "d1=10*10**-3 # in mm\n", + "r1=d1/2#\n", + "K=0.2 # in W/mK\n", + "T_max=177 # in degree C\n", + "T_infinite=27 # in degree C\n", + "ho=10 # in W/m**2K\n", + "R=10# # in W/m\n", + "rc=K/ho # in m\n", + "x=rc-r1 # in m\n", + "q_by_L= (T_max-T_infinite)/(log(rc/r1)/(2*pi*K)+1/(2*pi*ho*rc))#\n", + "# Also q_by_L = I**2*R\n", + "I= sqrt(q_by_L/R) # in amp\n", + "print \"The maximum possible current = %0.3f amphere\" %I\n", + "\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum possible current = 2.811 amphere\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb new file mode 100644 index 00000000..870158aa --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb @@ -0,0 +1,632 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 - Fins Heat Transfer From Extended Surface" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.1 - Page : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from numpy import pi\n", + "from math import sqrt\n", + "# given data\n", + "d=20 # in mm\n", + "d=d*10**-3 #in m\n", + "h=5 # in W/m**2K\n", + "T_0=100 # in degree C\n", + "T_infinite=20 # in degree C\n", + "K=15 # in W/m-K\n", + "#(i)Temperature distribution equation\n", + "rho=pi*d # in m\n", + "A=pi*d**2/4 #in square meter\n", + "m=sqrt(h*rho/(K*A))#\n", + "print \"(i) Temperature distribution equation is = theta/theta_0 = (T-T_infinite)/(T_0-T_infinite) = e**- %0.2f*x\" %m\n", + "#(ii)Heat loss from the rod\n", + "t_0=100 # in degree C\n", + "t_infinite=20 # in degree C\n", + "q=sqrt(K*A*h*rho)*(t_0-t_infinite)#\n", + "print \"(ii) Heat loss from the road is :\",round(q,3),\"watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Temperature distribution equation is = theta/theta_0 = (T-T_infinite)/(T_0-T_infinite) = e**- 8.16*x\n", + "(ii) Heat loss from the road is : 3.078 watt\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.2 - Page : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log, pi\n", + "from __future__ import division\n", + "# given data\n", + "d=3 # in cm\n", + "d=d*10**-2 #in m\n", + "h=20 # in W/m**2K\n", + "T1=140 # in degree C\n", + "T2=100 # in degree C\n", + "L=15*10**-2 # in meter\n", + "T_infinite=30 # in degree C\n", + "T_0=140 # in degree C\n", + "# Let at\n", + "x=0#T_0=T1#\n", + "x=15 #in cm\n", + "x=x*10**-2 # in m\n", + "T=100 # in degree C\n", + "rho=pi*d#\n", + "A=pi*d**2/4#\n", + "# Formula (T-T_infinite)/(T_0-T_infinite) = %e**-m*x\n", + "m=log((T_0-T_infinite)/(T-T_infinite))/x#\n", + "# Formula m=sqrt(h*rho/(k*A))\n", + "k=h*rho/(m**2*A)#\n", + "print \"Thermal conductivity of the rod material = %0.3f W/m-k\" %k" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thermal conductivity of the rod material = 293.699 W/m-k\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.3 - Page : 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tanh\n", + "# given data\n", + "t=1 # in mm\n", + "t=t*10**-3 # in meter\n", + "L= 10 # in mm\n", + "L= L*10**-3 # in meter\n", + "k= 380 # W/mK\n", + "To= 230 # in \u00b0C\n", + "T_inf= 30 # in \u00b0C\n", + "h= 40 # in W/m**2K\n", + "B= 1 # in meter\n", + "Ac= B*t # in m**2\n", + "rho= 2*(B+t)#\n", + "m= sqrt(h*rho/(k*Ac))#\n", + "# Part(a)\n", + "nita= tanh(m*L)/(m*L)*100 # fin efficiency in %\n", + "print \"Fin efficiency = %0.3f %%\" %nita\n", + "\n", + "# Part(b)\n", + "N=1000/9+1 # number of fin\n", + "Af= N*rho*L # in square meter\n", + "A1= 1 # plate area in m**2\n", + "A2= N*1*1*10**-3 # Area where fins are attached in square meter\n", + "Au= A1-A2 # in square meter\n", + "q_T= N*sqrt(h*rho*k*Ac)*(To-T_inf)*tanh(m*L)+Au*h*(To-T_inf) # in W/m**2\n", + "print \"Total heat transfer per square meter of plane wall surface = %0.3f kW/m**2\" %(q_T*10**-3)\n", + "\n", + "# Part(c)\n", + "A=1*1 # in m**2\n", + "q= h*A*(To-T_inf) # in W/m**2\n", + "print \"Heat transfer if there were no fins attached = %0.0f kW/m**2\" %(q*10**-3)\n", + "\n", + "# Note : Answer of part(b) in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fin efficiency = 99.303 %\n", + "Total heat transfer per square meter of plane wall surface = 24.934 kW/m**2\n", + "Heat transfer if there were no fins attached = 8 kW/m**2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.4 - Page : 87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, tanh\n", + "# given data\n", + "w=5*10**-2 # in meter\n", + "L=1 # in meter\n", + "t=2.5*10**-2 # in meter\n", + "h=47 # in W/m**2K\n", + "k=16.3 # in W/mK (for 18.8 steel)\n", + "T_0=100 # in degree C\n", + "T_infinite=20 # in degree C\n", + "Ac=w*t # in square meter\n", + "rho=2*(w+t)#\n", + "m=sqrt(h*rho/(k*Ac))#\n", + "q_fin=k*Ac*m*(T_0-T_infinite)*((tanh(m*L)+h/(k*m) )/(1+h/(m*k)*tanh(m*L)))\n", + "print \"The heat lost by the fin of one meter length is :\",round(q_fin,3),\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat lost by the fin of one meter length is : 30.32 W\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.5 - Page : 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "# given data\n", + "w=1 # in meter\n", + "L=2.5*10**-2 # in meter\n", + "t=0.8*10**-3 # in meter\n", + "l=1 # in meter\n", + "T_0=150 # in degree C\n", + "T_infinite=40 # in degree C\n", + "h=20 # in W/m**2K\n", + "k=65 # in W/mK (for 18.8 steel)\n", + "Ac=w*t#\n", + "d=5*10**-2 # Cylinder dia in meter\n", + "rho=2*(w+t)#\n", + "rho=floor(rho)#\n", + "\n", + "m=sqrt(h*rho/(k*Ac))#\n", + "mL=m*L#\n", + "# heat transfer rate from 12 fins\n", + "q_fin=12*k*Ac*m*(T_0-T_infinite)*((tanh(m*L)+h/(k*m) )/(1+h/(m*k)*tanh(m*L)))#\n", + "print \"Heat transfer rate from 12 fins is :\",round(q_fin,3),\"watt\"\n", + "Au=pi*d*l-12*w*t#\n", + "qu=h*Au*(T_0-T_infinite)#\n", + "print \"Now heat transfer from unfinned surface area is :\",round(qu,3),\"watt\"\n", + "q=q_fin+qu#\n", + "print \"Total head transfer rate from the cylinder is :\",round(q,3),\"watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate from 12 fins is : 1155.94 watt\n", + "Now heat transfer from unfinned surface area is : 324.455 watt\n", + "Total head transfer rate from the cylinder is : 1480.395 watt\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.6 - Page : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sinh\n", + "# given data\n", + "T_0=100 # in degree C\n", + "T_infinite=30 # in degree C\n", + "T_L=100 # in degree C\n", + "d=6*10**-3 # copper rod dia in meter\n", + "L=50*10**-2 # developed length in meter\n", + "Ac=pi*d**2/4 # in square meter\n", + "rho=pi*d # in meter\n", + "h=30 # in W/m**2K\n", + "k=330 # in W/mK \n", + "m=sqrt(h*rho/(k*Ac))#\n", + "#(i) Temperature distribution equation for the fin\n", + "# (T-T_infinite)/(T_0-T_infinite)=([(T_L-T_infinite)/(T_0-T_infinite)]*sinh(m*x)+sinh(m*(L-x)))/sinh(m*L)\n", + "#Temperature at\n", + "x=0.25 # in m\n", + "T= (((T_L-T_infinite)/(T_0-T_infinite))*sinh(m*x)+sinh(m*(L-x)))/sinh(m*L)*(T_0-T_infinite)+T_infinite#\n", + "print \"(i) Temperature at the centre of the rod is :\",round(T,1),\"degree C\"\n", + "print \"(ii) Heat transfer rate from the fin- This is equivalent to two fins of length 25 cm long with insulated tip\"\n", + "L=25*10**-2 # in meter\n", + "q=2*sqrt(h*rho*k*Ac)*(T_0-T_infinite)*tanh(m*L)#\n", + "print \"Heat transfer by the rod is :\",round(q,2),\"watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Temperature at the centre of the rod is : 49.6 degree C\n", + "(ii) Heat transfer rate from the fin- This is equivalent to two fins of length 25 cm long with insulated tip\n", + "Heat transfer by the rod is : 9.76 watt\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.7 - Page : 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cosh\n", + "# given data\n", + "T_0=100 # in degree C\n", + "T_infinite=25 # in degree C\n", + "d=5*10**-2 # in meter\n", + "L=15*10**-2 # in meter\n", + "h=8 # in W/m**2K\n", + "k=20 # in W/mK \n", + "rho=pi*d # in meter\n", + "Ac=pi*d**2/4 # in square meter\n", + "m=sqrt(h*rho/(k*Ac))#\n", + "#(ii) Temperature at free end i.e. at \n", + "x=L\n", + "# Formula (T_L-T_infinite)/(T_0-T_infinite)= 1/(cosh(m*L)+h/(k*m)*sinh(m*L) )\n", + "T_L=(1/(cosh(m*L)+h/(k*m)*sinh(m*L) ))*(T_0-T_infinite)+T_infinite#\n", + "print \"(ii) Temperature at free end is :\",round(T_L,3),\"degree C\"\n", + "\n", + "#(iii) Heat flow out the source means heat transfer from the fin\n", + "q_f=sqrt(h*rho*k*Ac)*(T_0-T_infinite)*((h/(k*m)+tanh(m*L))/(1+h*tanh(m*L)/(k*m)))\n", + "print \"(iii) Heat flow out the source :\",round(q_f,3),\"watt\"\n", + "\n", + "# (iv) Heat flow rate at free end\n", + "q_L=h*Ac*(T_L-T_infinite)\n", + "print \"(iv) Heat flow rate at free end is :\",round(q_L,3),\"watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(ii) Temperature at free end is : 76.739 degree C\n", + "(iii) Heat flow out the source : 12.089 watt\n", + "(iv) Heat flow rate at free end is : 0.813 watt\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.8 - Page : 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "import math\n", + "from math import sqrt, tanh\n", + "# given data\n", + "T_0=150 # in degree C\n", + "T_infinite=40 # in degree C\n", + "w=1 # in m\n", + "t=0.75*10**-3 # in m\n", + "d=5*10**-2 # in meter\n", + "L=25*10**-3 # in meter\n", + "k=75 # in W/mK \n", + "h=23.3 # in W/m**2K\n", + "N=12 # numbers of fins\n", + "Ac=w*t #in square meter\n", + "rho=2*(w+t) # in meter\n", + "delta=Ac/rho#\n", + "L_c=L+delta#\n", + "ML_c=L_c*sqrt(h*rho/(k*Ac))\n", + "q_fin= N*sqrt(h*rho*k*Ac)*(T_0-T_infinite)*tanh(ML_c)#\n", + "q_fin=math.floor(q_fin)#\n", + "A_0=pi*d*w-12*Ac\n", + "q_unfin= h*A_0*(T_0-T_infinite)#\n", + "q_total=q_fin+q_unfin#\n", + "print \"Rate of heat transfer is :\",round(q_total,1),\"watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat transfer is : 1711.5 watt\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.9 - Page : 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# given data\n", + "L=0.06 # in meter\n", + "A=4.64*10**-4 # in m**2\n", + "rho=0.12 # in m\n", + "h=442 # in W/m**2\n", + "T_0=773 # in K\n", + "T_infinite=1143 # in K\n", + "K=23.2 # in W/mK\n", + "m=sqrt(h*rho/(K*A))#\n", + "q=sqrt(h*rho*K*A)*(T_0-T_infinite)*tanh(m*L)#\n", + "print \"Heat transfer rate is :\",round(q,3),\"watt\"\n", + "\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is : -279.457 watt\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.10 - Page : 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from math import cosh\n", + "# given data\n", + "L=0.12 # in meter\n", + "t=.15*10**-2 # thickness in m\n", + "K=55.5 # in W/mK\n", + "h=23.5 # in W/mK\n", + "T_L=357 # in K\n", + "T_0=313 # in K\n", + "\n", + "# Formula m=sqrt(h*rho/(K*A)) and rho=pi*d and A=pi*d*t, putting value of rho and A\n", + "m=sqrt(h/(K*t))#\n", + "mL=m*L#\n", + "mL=math.floor(mL)#\n", + "# Formula (T_L-T_infinite)/(T_0-T_infinite)= 1/cosh(m*L)\n", + "T_infinite=(T_L-T_0/cosh(mL))/(1-1/cosh(mL))#\n", + "T_infinite=math.ceil(T_infinite)#\n", + "measurement_error=T_infinite-T_L#\n", + "print \"Measurement Error is :\",round(measurement_error,0),\"K\"\n", + "\n", + "# Note: In the book, Unit of answer is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Measurement Error is : 16.0 K\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.11 - Page : 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# given data\n", + "k=20 # in W/mK\n", + "T_L=150 # in degree C\n", + "T_0=70 # in degree C\n", + "L=12*10**-2 # in meter\n", + "h=80 # in W/m**2K\n", + "t=3*10**-3 # in m\n", + "# Formula m=sqrt(h*rho/(K*A)) and rho=pi*d and A=pi*d*t, putting value of rho and A\n", + "m=sqrt(h/(k*t))#\n", + "# Formula (T_L-T_infinite)/(T_0-T_infinite)= 1/cosh(m*L)\n", + "T_infinite=(T_L-T_0/cosh(m*L))/(1-1/cosh(m*L))#\n", + "PercentageError=(T_infinite-T_L)*100/T_infinite#\n", + "print \"Percentage Error is :\",round(PercentageError,2),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage Error is : 1.35 %\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.12 - Page : 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import acosh\n", + "# given data\n", + "k=30 # in W/mK\n", + "h=100 # in W/m**2K\n", + "T_infinite=300 # in degree C\n", + "d=2*10**-2 # in m\n", + "t=1*10**-3 # in m\n", + "err=1 # in % of applied temperature difference\n", + "# Formula m=sqrt(h*rho/(K*A)) and rho=pi*d and A=pi*d*t, putting value of rho and A\n", + "m=sqrt(h/(k*t))#\n", + "\n", + "# From (T_L-T_infinite)/(T_0-T_infinite)= 1/100 = 1/cosh(m*L)\n", + "L=acosh(100)/m # in meter\n", + "L=L*10**3 # in mm\n", + "print \"Minimum length os pocket is :\",round(L,2),\"mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum length os pocket is : 91.77 mm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 3.13 - Page : 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "k=32 # in W/m**2 degree C\n", + "h=14.8 # in W/m**2 degree C\n", + "t_o=480 # in degree C\n", + "t_i=55 # in degree C\n", + "t_a=20 # in degree C\n", + "d=2.5*10**-2 # in m\n", + "rho=pi*d # in m\n", + "Ac=pi*d**2/4 # in m**2\n", + "m=sqrt(h*rho/(k*Ac))#\n", + "# From (t-t_a)/(t_o-t_a) = cosh(m(L-x))/cosh(m*L)\n", + "L=acosh((t_o-t_a)/(t_i-t_a))/m# # at x=L,t=t_i\n", + "print \"Length of shaft specified between the motor and the pump is :\",round(L,4),\"meter\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of shaft specified between the motor and the pump is : 0.3798 meter\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb new file mode 100644 index 00000000..9a0797ed --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb @@ -0,0 +1,760 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 - Transient (Unsteady State) Heat Conduction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.1 - Page : 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "L=1 # in m\n", + "rho=1600 # in kg/m**3\n", + "k=40 # in w/mK\n", + "Cp=4*10**3 # in J/kgK\n", + "a=900 # in degree C\n", + "b=-300 # in degree C/m\n", + "c=-50 # in degree C/m**2\n", + "Qg=1*10**3 # in kW/m**2\n", + "A=10 # area in m**2\n", + "#t=a+b*x+c*x**2 at any instant, so\n", + "# dtBYdx= b+2*c*x\n", + "# d2tBYdx2 = 2*c, then\n", + "\n", + "# Part(a)\n", + "#q1= -k*A*dtBYdx , at\n", + "x=0#\n", + "q1= -k*A*(b+2*c*x) # in w\n", + "#q2= -k*A*dtBYdx , at\n", + "x=L#\n", + "q2= -k*A*(b+2*c*x) # in w\n", + "E_stored= (q1-q2)+Qg*A*L # in watt\n", + "print \"The rate of change of energy storage = %0.1e watt\" %E_stored\n", + "\n", + "# Part(b)\n", + "alpha= k/(rho*Cp) # in m**2s\n", + "d2tBYdx2 = 2*c#\n", + "dtBYdtoh= alpha*(d2tBYdx2+Qg/k ) # in degree C/sec\n", + "print \"Rate of change of temperature = %0.3e degree C/sec\" %dtBYdtoh\n", + "print \"Since dt by dx is independent of x. Hence time rate of charge of temperature throughout wall will remain same.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate of change of energy storage = -3.0e+04 watt\n", + "Rate of change of temperature = -4.688e-04 degree C/sec\n", + "Since dt by dx is independent of x. Hence time rate of charge of temperature throughout wall will remain same.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.2 - Page : 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "from numpy import log, pi\n", + "# given data\n", + "k=40 # in W/mK\n", + "rho=7800 # in kg/m**3\n", + "C=450 # in J/kgK\n", + "d=20*10**-3 # in m\n", + "r=d/2#\n", + "t_i=400 # in degree C\n", + "t=85 # in degree C\n", + "t_infinite=25 # in degree C\n", + "h=80 # in W/m**2K\n", + "#l_s=V/A = (4/3*pi*r**3)/(4*pi*r**2) = r/3\n", + "l_s=r/3 # in m\n", + "Bi= h*l_s/k#\n", + "# since Biot number is less than 0.1, hence lumped heat capacity system analysis can be applied\n", + "\n", + "# Part(a)\n", + "# Formula (t-t_infinite)/(t_i-t_infinite)= %e**(-h*A*toh/(rho*V*C)) = %e**(-h*toh/(rho*l_s*C))\n", + "toh= -log((t-t_infinite)/(t_i-t_infinite))*(rho*l_s*C)/h # in sec\n", + "print \"The time require to cool the sphere = %0.3f sec\" %toh\n", + "\n", + "# Part(b)\n", + "# dtBYdtoh = h*A*(t_i-t_infinite)/(rho*V*C) = h*(t_i-t_infinite)/(rho*l_s*C)\n", + "dtBYdtoh = h*(t_i-t_infinite)/(rho*l_s*C) # in degree C/sec\n", + "print \"Initial rate of cooling = %0.3f degree C/sec\" %dtBYdtoh\n", + "\n", + "# Part(c)\n", + "A=4*pi*r**2#\n", + "toh=60#\n", + "q_in= h*A*(t_i-t_infinite)*exp(-h*toh/(rho*l_s*C)) # in watt\n", + "print \"Instantaneous heat transfer rate = %0.3f watt\" %q_in\n", + "\n", + "# Part(d) Total energy transferred during first one minute\n", + "V=4/3*pi*r**3#\n", + "TotalEnergy = rho*C*V*(t_i-t_infinite)*(1-exp(-h*toh/(rho*C*l_s)))#\n", + "print \"Total energy transferred during first one minute = %0.3f watt\" %TotalEnergy\n", + "\n", + "# Note: Answer of first and last part in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time require to cool the sphere = 268.015 sec\n", + "Initial rate of cooling = 2.564 degree C/sec\n", + "Instantaneous heat transfer rate = 25.013 watt\n", + "Total energy transferred during first one minute = 1855.401 watt\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.3 - Page : 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# given data\n", + "k=40 # in W/mK\n", + "rho=8200 # in kg/m**3\n", + "C=400 # in J/kgK\n", + "D=6*10**-3 # in m\n", + "R=D/2#\n", + "t_i=30 # in degree C\n", + "t_infinite1=400 # for 10 sec in degree C\n", + "t_infinite2=20 # for 10 sec in degree C\n", + "h=50 # in W/m**2K\n", + "\n", + "# Part(a)\n", + "#l_s= V/A = R/3\n", + "l_s= R/3 # in m\n", + "#toh= rho*V*C/(h*A) = rho*C*l_s/h\n", + "toh= rho*C*l_s/h # in sec\n", + "print \"Time constance = %0.1f sec\" %toh\n", + "\n", + "# Part (b)\n", + "Bi= h*l_s/k#\n", + "# since Bi < 0.1 , hence lumped heat capacity analysis is valid. Now , temperature attained by junction in 10 seconds when exposed to hot air at 400 degree C\n", + "toh=10 # in sec\n", + "# (t-t_infinite1)/(t_i-t_infinite1)= %e**(-h*A*toh/(rho*V*C)) = %e**(-h*toh/(rho*l_s*C))\n", + "t= exp(-h*toh/(rho*l_s*C))*(t_i-t_infinite1)+t_infinite1 # in degree C\n", + "\n", + "print \"The junction is taken out from hot air stream and placed in stream of still air 20 degree C.\"\n", + "print \"The initial temperature in this case = \",round(t,3)\n", + "t_i=t#\n", + "toh=20 # in sec\n", + "t= exp(-h*toh/(rho*l_s*C))*(t_i-t_infinite2)+t_infinite2 # in degree C\n", + "print \"The temperature attained by junction = %0.3f degree C\" %t\n", + "\n", + "# Note: In the last, calculation to find the value of t is wrong so Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time constance = 65.6 sec\n", + "The junction is taken out from hot air stream and placed in stream of still air 20 degree C.\n", + "The initial temperature in this case = 82.314\n", + "The temperature attained by junction = 65.939 degree C\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.4 - Page : 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "k=8 # in W/mK\n", + "alpha=4*10**-6 # in m**2/s\n", + "h=50 # in W/m**2K\n", + "D=6*10**-3 # in m\n", + "R=D/2#\n", + "T=0.5 # where T = (t-t_infinite)/(t_i-t_infinite)\n", + "#l_s= V/A = R/3\n", + "l_s= R/2 # in m\n", + "Bi= h*l_s/k#\n", + "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n", + "# toh= rho*V*C/(h*A) = rho*C*l_s/h = k*l_s/(h*alpha)\n", + "toh= k*l_s/(h*alpha) # in seconds\n", + "print \"Time constant = %0.f seconds\" %toh\n", + "# It is given that (t-t_infinite)/(t_i-t_infinite) = 0.5 = %e**(-h*A*c /(rho*V*C)) = %e**(-h*c/(rho*l_s*C)) = %e**(-h*alpha*c/(l_s))\n", + "# or (t-t_infinite)/(t_i-t_infinite) = %e**(-h*alpha*c/(l_s)#\n", + "c= -log(T)*l_s/(h*alpha) # in sec\n", + "print \"The time required to temperature change to reach half of its initial value = %0.1f seconds\" %c" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time constant = 60 seconds\n", + "The time required to temperature change to reach half of its initial value = 5.2 seconds\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.5 - Page : 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "#t=450-500*x+100*x**2+150*x**3 at any instant, so\n", + "# dtBYdx= -500+200*x+450*x**2\n", + "L=0.5 # thickness of the wall in meter\n", + "k=10 # in W/mK\n", + "# Rate of heating entering in the wall per unit area, at\n", + "x=0#\n", + "#q1= -k*dtBYdx\n", + "q1= -k*(-500+200*x+450*x**2) # in W/m**2\n", + "# Rate of heat going out of the wall per unit area , at\n", + "x=L#\n", + "q2= -k*(-500+200*x+450*x**2) # in W/m**2\n", + "E=q1-q2 # in W/m**2\n", + "print \"Heat energy stored per unit area = %0.0f W/m**2\" %E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat energy stored per unit area = 2125 W/m**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.6 - Page : 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "k=385 # in W/mK\n", + "h=100 # in W/m**2K\n", + "delta =2*10**-3 # thickness of plate in meter\n", + "A=25*25 # area of plate in square meter\n", + "rho=8800 # kg/m**3\n", + "C=400 # J/kg-K\n", + "# l_s= V/A= L*B*delta/(2*L*B) = delta/2\n", + "l_s= delta/2 # in meter\n", + "Bi= h*l_s/k#\n", + "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n", + "\n", + "# Part(i)\n", + "# toh= rho*V*C/(h*A) = rho*C*l_s/h\n", + "toh= rho*C*l_s/h # in second\n", + "print \"Time constant = %0.1f seconds\" %toh\n", + "\n", + "# Part(ii)\n", + "t_i=400 # in degree C\n", + "t=40 # in degree C\n", + "t_infinite=25 # in degree C\n", + "# (t-t_infinite)/(t_i-t_infinite) = %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) \n", + "toh= -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h # in sec\n", + "print \"The time required for the plate to reach the temperature of 40 degree C = %0.1f seconds\" %toh" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time constant = 35.2 seconds\n", + "The time required for the plate to reach the temperature of 40 degree C = 113.3 seconds\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.7 - Page : 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "k=380 # in W/mK\n", + "delta =6*10**-2 # thickness of plate in meter\n", + "rho=8800 # kg/m**3\n", + "C=400 # J/kg-K\n", + "# l_s= V/A = delta/2\n", + "l_s= delta/2 # in meter\n", + "t=80 # in degree C\n", + "t_i=200 # in degree C\n", + "t_inf=30 # in degree C\n", + "hw= 75 # in W/m**2K\n", + "ha= 10 # in W/m**2K\n", + "\n", + "# Part(i)\n", + "# ha*A*(t-t_inf_a)+ hw*A*(t-t_inf_w) = -rho*V*C*dtBYdtho, since t_ini_a = t_inf_w = t_inf = 30 degree C\n", + "# (ha+hw)*A*(t-t_inf)= -rho*V*C*dtBYdtho\n", + "# (ha+hw)/(rho*C*V)*A*dtoh = -dt/(t-t_inf)\n", + "# integrate('(ha+hw)/(rho*V*C)*A','toh',0,toh) = integrate('1/(t-t_inf)','t',t_i,t)\n", + "toh= -rho*l_s*C/(ha+hw)*log((t-t_inf)/(t_i-t_inf))#\n", + "print \"Time required to cool plate to 80 degree C is :\",round(toh,1),\"seconds =\",round(toh/60,2),\"minutes\"\n", + "\n", + "# Part (ii)\n", + "t= -rho*l_s*C/(2*ha)*log((t-t_inf)/(t_i-t_inf))#\n", + "print \"Time required to cool plate in only air is :\",round(t,1),\"seconds =\",round(t/60,2),\"minutes\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required to cool plate to 80 degree C is : 1520.4 seconds = 25.34 minutes\n", + "Time required to cool plate in only air is : 6461.5 seconds = 107.69 minutes\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.8 - Page : 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "k=45 # in W/m degree C\n", + "d =0.1 # in meter\n", + "l =0.30 # in meter\n", + "t=800 # in degree C\n", + "t_i=100 # in degree C\n", + "t_infinite=1200 # in degree C\n", + "h= 120 # in W/m**2 degree C\n", + "alpha=1.2*10**-5 # in meter\n", + "rhoC= k/alpha#\n", + "V=pi/4*d**2*l # in m**3\n", + "A= pi*d*l + 2*pi/4*d**2 # in m**2\n", + "# l_s= V/A = (pi/4*d**2*l)/(pi*d*l + 2*pi/4*d**2) = d*l/(4*l+2*d**2)\n", + "l_s = d*l/(4*l+2*d**2)#\n", + "Bi= h*l_s/k#\n", + "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n", + "# (t-t_infinite)/(t_i-t_infinite) = %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) = %e**(-h*toh/(rhoC*l_s))\n", + "toh = -log((t-t_infinite)/(t_i-t_infinite))*rhoC*l_s/h # in sec\n", + "\n", + "# So, the velocity of ingot passing through the furnace\n", + "FurnaceLength = 8*100 # in cm\n", + "time = toh#\n", + "Velocity = FurnaceLength/time # in cm/sec\n", + "print \"Maximum speed = %0.4f cm/sec\" %Velocity" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum speed = 1.0291 cm/sec\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.9 - Page : 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "rho=8500 # in kg/m**3\n", + "C=400 # J/kgK\n", + "toh=1 # in sec\n", + "h= 400 # in W/m**2 degree C\n", + "t=198 # in degree C\n", + "t_i=25 # in degree C\n", + "t_infinite=200 # in degree C\n", + "\n", + "# Part (1)\n", + "# toh =rho*V*C/(h*A) = rho*C*l_s/h\n", + "l_s= toh*h/(rho*C)#\n", + "# l_s = V/A = r/3 \n", + "r=3*l_s # in m\n", + "r=r*10**3 # in mm\n", + "d=2*r # in m\n", + "print \"Junction diameter needed for the thermocouple = %0.3f mili miter\" %d\n", + "\n", + "# Part(ii)\n", + "# toh= -rho*V*C/(h*A)*log((t-t_infinite)/(t_i-t_infinite)) \n", + "toh = -toh*log((t-t_infinite)/(t_i-t_infinite))#\n", + "print \"Time required for the thermocouple junction to reach 198 degree C = %0.3f seconds\" %toh" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Junction diameter needed for the thermocouple = 0.706 mili miter\n", + "Time required for the thermocouple junction to reach 198 degree C = 4.472 seconds\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.10 - Page : 118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "L=40*10**-2 # in m\n", + "k=1.5 # in W/mK\n", + "A=4 # in square meter\n", + "alpha=1.65*10**-3 # in m**2/h\n", + "#T = 50-40*x+10*x**2+20*x**3-15*x**4 , so\n", + "# dtBYdx= -40+20*x+60*x**2-60*x**3\n", + "# d2tBYdx2 = 20+120*x-180*x**2\n", + "\n", + "# Part (a) Heat entering the slab\n", + "#q1= -k*A*dtBYdx , at\n", + "x=0#\n", + "qi= -k*A*(-40+20*x+60*x**2-60*x**3) # in w\n", + "print \"Heat entering the slab = %0.0f watt\" %qi\n", + "# Heat leaving the slab\n", + "#ql= -k*A*dtBYdx , at\n", + "x=L#\n", + "ql= -k*A*(-40+20*x+60*x**2-60*x**3) # in w\n", + "print \"Heat leaving the slab = %0.2f watt\" %ql\n", + "\n", + "# Part (b) Rate of heat storage\n", + "RateOfHeatStorage = qi-ql # in watt\n", + "print \"Rate of heat storage = %0.2f watt\" %RateOfHeatStorage\n", + "\n", + "# Part (c) Rate of temperature change\n", + "# d2tBYdx2 = 1/alpha*dtBYdtoh\n", + "# dtBYdtoh= alpha*d2tBYdx2, at\n", + "x=0#\n", + "dtBYdtoh = alpha*(20+120*x-180*x**2) # in degree C/h\n", + "print \"The rate of temperature change at entering the slab = %0.3f degree C/h\" %dtBYdtoh\n", + "# dtBYdtoh= alpha*d2tBYdx2, at\n", + "x=L\n", + "dtBYdtoh = alpha*(20+120*x-180*x**2) # in degree C/h\n", + "print \"The rate of temperature change at leaving the slab = %0.3f degree C/h\" %dtBYdtoh\n", + "\n", + "# Part (d) for the rate of heating or cooling to be maximum\n", + "# dBYdx of dtBYdtoh = 0\n", + "# dBYdx of (alpha*d2tBYdx2) =0\n", + "# d3tBYdx3 = 0\n", + "x=120/360 # in meter\n", + "print \"The point where rate of heating or cooling is maximum = %0.3f meter\" %x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat entering the slab = 240 watt\n", + "Heat leaving the slab = 157.44 watt\n", + "Rate of heat storage = 82.56 watt\n", + "The rate of temperature change at entering the slab = 0.033 degree C/h\n", + "The rate of temperature change at leaving the slab = 0.065 degree C/h\n", + "The point where rate of heating or cooling is maximum = 0.333 meter\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.11 - Page : 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "k=40 # in W/m degree C\n", + "d =12*10**-3 # in meter\n", + "t=127 # in degree C\n", + "t_i=877 # in degree C\n", + "t_infinite=52 # in degree C\n", + "h= 20 # in W/m**2 degree C\n", + "rho=7800 # in W/m**2K\n", + "C=600 # in J/kg K\n", + "r=d/2 # in meter\n", + "#l_s = V/A = r/3\n", + "l_s = r/3#\n", + "Bi= h*l_s/k#\n", + "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n", + "# (t-t_infinite)/(t_i-t_infinite) = %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) = %e**(-h*toh/(rho*C*l_s))\n", + "toh = -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h # in sec\n", + "print \"Time required for cooling process =\",round(toh,3),\"seconds =\",round(toh/60,2),\"minutes\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required for cooling process = 1122.215 seconds = 18.7 minutes\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.12 - Page : 127" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "D=10*10**-2 # in m\n", + "b=D/2#\n", + "h= 100 # in W/m**2 degree C\n", + "T_o=418 # in degree C\n", + "T_i=30 # in degree C\n", + "T_infinite=1000 # in degree C\n", + "\n", + "print \" (A) For copper cylinder \"\n", + "k=350 # in W/mK\n", + "alpha=114*10**-7 # in m**2/s\n", + "Bi= h*b/k#\n", + "theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)#\n", + "Fo=18.8#\n", + "# Formula Fo= alpha*t/b**2\n", + "t=Fo*b**2/alpha#\n", + "print \"Time required to reach for the cylinder centreline temperature 418 degree C =\",round(t,3),\"seconds =\",round(t/3600,3),\"hours\"\n", + "\n", + "# (2) Temperature at the radius of 4 cm\n", + "theta_0_t = 0.985#\n", + "# Formula theta_0_t = (T-T_infinite)/(T_o-T_infinite)\n", + "T= theta_0_t*(T_o-T_infinite)+T_infinite # in degree C\n", + "print \"Temperature at the radius of 4 cm = %0.2f degree C\" %T \n", + "print \"It has very less temperature gradients over 4 cm radius\"\n", + "\n", + "print \" (B) For asbestos cylinder \"\n", + "k=0.11 # in W/mK\n", + "alpha=0.28*10**-7 # in m**2/s\n", + "Bi= h*b/k#\n", + "theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)#\n", + "Fo=0.21#\n", + "# Formula Fo= alpha*t/b**2\n", + "t=Fo*b**2/alpha#\n", + "print \"Time required to reach for the cylinder centreline temperature 418 degree C =\",round(t,3),\"seconds =\",round(t/3600,1),\"hours\"\n", + "\n", + "# (2) Temperature at the radius of 4 cm\n", + "theta_x_t = 0.286#\n", + "# Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)\n", + "T= theta_x_t*(T_o-T_infinite)+T_infinite # in degree C\n", + "print \"Temperature at the radius of 4 cm = %0.3f degree C\" %T \n", + "print \"It has large temperature gradients\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " (A) For copper cylinder \n", + "Time required to reach for the cylinder centreline temperature 418 degree C = 4122.807 seconds = 1.145 hours\n", + "Temperature at the radius of 4 cm = 426.73 degree C\n", + "It has very less temperature gradients over 4 cm radius\n", + " (B) For asbestos cylinder \n", + "Time required to reach for the cylinder centreline temperature 418 degree C = 18750.0 seconds = 5.2 hours\n", + "Temperature at the radius of 4 cm = 833.548 degree C\n", + "It has large temperature gradients\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 4.13 - Page : 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "D=5*10**-2 # in m\n", + "b=D/2#\n", + "h= 500 # in W/m**2 degree C\n", + "k=60 # in W/m**2K\n", + "rho=7850 # in kg/m**3\n", + "C=460 # in J/kg\n", + "alpha=1.6*10**-5 # in m**2/s\n", + "T_i=225 # in degree C\n", + "T_infinite=25 # in degree C\n", + "t=2 # in minute\n", + "\n", + "# Part(i)\n", + "Bi= h*b/k#\n", + "Fo= alpha*t/b**2#\n", + "theta_0_t = 0.18#\n", + "# Formula theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)\n", + "T_o= theta_0_t*(T_i-T_infinite)+T_infinite # in degree C\n", + "print \"Centreline Temperature of the sphere after 2 minutes of exposure = %0.f degree C \" %T_o\n", + "\n", + "# Part(2)\n", + "depth= 10*10**-3 # in meter\n", + "r=b-depth # in meter\n", + "rBYb=r/b#\n", + "theta_x_t = 0.95#\n", + "# Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)\n", + "T= theta_x_t*(T_o-T_infinite)+T_infinite # in degree C\n", + "print \"The Temperature at the depth of 1 cm from the surface after 2 minutes = %0.1f degree C\" %T\n", + "\n", + "# Part (3)\n", + "BiSquareFo= Bi**2*Fo#\n", + "QbyQo= 0.8 # in kJ\n", + "A=4/3*pi*b**3#\n", + "Qo= rho*A*C*(T_i-T_infinite) # in J\n", + "Qo=Qo*10**-3 # in kJ\n", + "# The heat transffered during 2 minute, \n", + "Q= Qo*QbyQo # in kJ\n", + "print \"The heat transffered during 2 minutes = %0.3f kJ\" %Q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Centreline Temperature of the sphere after 2 minutes of exposure = 61 degree C \n", + "The Temperature at the depth of 1 cm from the surface after 2 minutes = 59.2 degree C\n", + "The heat transffered during 2 minutes = 37.814 kJ\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb new file mode 100644 index 00000000..d27ae294 --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb @@ -0,0 +1,802 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 - Forced Convection Heat Transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.1 - Page : 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# given data\n", + "rho=1.14 # in kg/m**3\n", + "k=2.73*10**-2 # in W/mK\n", + "Cp=1.005 # in kg/kgK\n", + "v= 16*10**-6 # in m**2/s\n", + "Pr=0.67#\n", + "# Other data given in the problem are\n", + "V=2 # in m/s\n", + "w=20*10**-2 # in m\n", + "t_infinite= 10 # in degree C\n", + "t_s=65 # in degree C\n", + "x=0.25 # in m from leading edge\n", + "# Re= rho*Vx/miu = V*x/v\n", + "Re= V*x/v#\n", + "#Since Re<5*10**5 , hence the flow is a laminar flow\n", + "#(a) Boundary layer thickness\n", + "delta= 5*x/(sqrt(Re)) # in m\n", + "delta=delta*10**2 # in cm\n", + "print \"Boundary layer thickness = %0.3f cm\" %delta\n", + "\n", + "#(b) Thermal boundary layer thickness\n", + "delta_t= delta/Pr**(1/3) # in cm\n", + "print \"Thermal boundary layer thickness = %0.3f cm\" %delta_t\n", + "\n", + "#(c) Local friction coefficient\n", + "Cfx= 0.664/sqrt(Re)#\n", + "print \"Local friction coefficient = %0.3f\" %Cfx\n", + "Cf=2*Cfx#\n", + "print \"Average friction coefficient %0.3f\" %Cf\n", + "\n", + "#(d) Total drag force\n", + "A=.25*.2 # in m**2\n", + "toh_o=Cf*(rho*V**2/2)#\n", + "F=toh_o*A#\n", + "print \"Total drag force = %0.3f N\" %F\n", + "\n", + "#(e) \n", + "# Formula Nux= hx*x/k = 0.332*Re**(1/2)*Pr**(1/3)\n", + "hx= 0.332*k/x*Re**(1/2)*Pr**(1/3) # in W/m**2K\n", + "print \"Local heat transfer coefficient = %0.3f W/m**2K\" %hx\n", + "h=2*hx#\n", + "print \"Average heat transfer coefficient = %0.3f W/m**2K\" %h\n", + "#(f)\n", + "q=h*A*(t_s-t_infinite)#\n", + "print \"Rate of heat transfer = %0.3f W/m**2K\" %q\n", + "\n", + "#Note: In the book, they calculated wrong value of Re so all the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boundary layer thickness = 0.707 cm\n", + "Thermal boundary layer thickness = 0.808 cm\n", + "Local friction coefficient = 0.004\n", + "Average friction coefficient 0.008\n", + "Total drag force = 0.001 N\n", + "Local heat transfer coefficient = 5.608 W/m**2K\n", + "Average heat transfer coefficient = 11.216 W/m**2K\n", + "Rate of heat transfer = 30.844 W/m**2K\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.2 - Page : 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "rho=998 # in kg/m**3\n", + "k=.648 # in W/mK\n", + "v= 0.556*10**-6 # in m**2/s\n", + "Pr=3.54#\n", + "V=2 # in m/s\n", + "t_infinite= 10 # in degree C\n", + "t_s=90 # in degree C\n", + "Re=5*10**5#\n", + "A=1*1 # in m**2\n", + "# Re= rho*Vx/miu = V*x/v\n", + "x=Re*v/V # in m\n", + "print \"Length of the plate = %0.3f m\" %x\n", + "# Nu = h*x/k =Pr**(1/3)*(0.037*Re**0.8-872)\n", + "x=1#\n", + "Re= V*x/v#\n", + "h= Pr**(1/3)*(0.037*Re**0.8-873)*k/x # in W/m**2\n", + "q=h*A*(t_s-t_infinite)#\n", + "print \"Heat transfer from entire plate = %0.f kW\" %int(q*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of the plate = 0.139 m\n", + "Heat transfer from entire plate = 444 kW\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.3 - Page : 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "rho=1.06 # in kg/m**3\n", + "K=.0289#\n", + "v= 18.97*10**-6 # in m**2/s\n", + "Pr=0.696#\n", + "V=2.2 # in m/s\n", + "L=0.9 # in m\n", + "B=0.45 # in m\n", + "t_infinite= 30 # in degree C\n", + "t_s=90 # in degree C\n", + "#(a) For first half of the plate\n", + "x=L/2 # in m\n", + "Re=V*x/v#\n", + "# Nu = h*x/K = 0.664*Re**(1/2)*Pr**(1/3)\n", + "h= 0.664*Re**(1/2)*Pr**(1/3)*K/x # in W/m**2 degree C\n", + "A=x*B#\n", + "Q1=h*A*(t_s-t_infinite) # in watt\n", + "print \"Heat transfer rate from first half of the plate = %0.f watt\" %Q1\n", + "\n", + "#(b) Heat transfer from entire plate\n", + "x=L # in m\n", + "Re=V*x/v#\n", + "# Nu = h*x/K = 0.664*Re**(1/2)*Pr**(1/3)\n", + "h= 0.664*Re**(1/2)*Pr**(1/3)*K/x # in W/m**2 degree C\n", + "A=L*B#\n", + "Q2=h*A*(t_s-t_infinite) # in watt\n", + "print \"Heat transfer rate from entire plate = %0.3f watt\" %Q2\n", + "\n", + "#(c) From next half of the plate\n", + "Q3= Q2-Q1#\n", + "print \"Heat transfer rate from next half of the plate %0.3f\" %Q3" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate from first half of the plate = 105 watt\n", + "Heat transfer rate from entire plate = 148.342 watt\n", + "Heat transfer rate from next half of the plate 43.448\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.4 - Page : 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "from numpy import pi\n", + "# given data\n", + "rho=985 # in kg/m**3\n", + "k=.654 # in W/mK\n", + "Cp=4.18 # in kgJ/kgK\n", + "Cp=Cp*10**3 # in J/kgK\n", + "v= 0.517*10**-6 # in m**2/s\n", + "Pr=3.26#\n", + "V=1.2 # in m/s\n", + "t_s=85 # in degree C\n", + "t_i=40 # in degree C\n", + "t_o=70 # in degree C\n", + "Ax=15*35 # in mm\n", + "P=15+35#\n", + "de=4*Ax/(2*P) # in mm\n", + "de=de*10**-3 # in m\n", + "Re=V*de/v#\n", + "# Formula Nu= h*de/k = 0.023Re**0.8*Pr**0.4\n", + "h=0.023*Re**0.8*Pr**0.4*k/de # in W/m**2K\n", + "m=pi*de**2*V*rho/4#\n", + "d=de#\n", + "L=m*Cp*log((t_s-t_i)/(t_s-t_o))/(pi*d*h)#\n", + "print \"The length of tube = %0.3f meter\" %L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The length of tube = 4.407 meter\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.5 - Page : 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "k=.026 # in W/mK\n", + "v= 16.8*10**-6 # in m**2/s\n", + "miu=2*10**-5 # in kg/ms\n", + "Pr=0.708#\n", + "V=15 # in m/s\n", + "x=2 # in m\n", + "A=2*1 # in m**2\n", + "Re=V*x/v#\n", + "del_t=40-10 # in degree C\n", + "# since Re > 3 *10**5, hence turbulent flow at x=2 m length of laminar flow region is x_L then\n", + "Re_1=3*10**5#\n", + "# Re_1 = 3*10**5 = V*x_L/v\n", + "x_L= Re_1*v/V#\n", + "\n", + "# Part (a)\n", + "#Nu= h*x_L/k = 0.664*Re_1**(1/2)*Pr**(1/3)#\n", + "h= 0.664*Re_1**(1/2)*Pr**(1/3)*k/x_L # in W/m**2\n", + "print \"The average heat transfer coefficient over the laminar boundary layer = %0.3f W/m**2 \" %h\n", + "\n", + "# Part(b)\n", + "#Nu= h*x/k = (0.037*Re**0.8-872)*Pr**(1/3)#\n", + "h= (0.037*Re**0.8-872)*Pr**(1/3)*k/x # in W/m**2\n", + "print \"The average heat transfer coefficient over entire plate = %0.3f W/m**2\" %h\n", + "\n", + "# Part (c)\n", + "q=h*A*del_t#\n", + "print \"Total heat transfer rate = %0.3f watt\" %q\n", + "\n", + "# Note: Calculation of the part(a) in this book is wrong, so answer of the part(a) in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The average heat transfer coefficient over the laminar boundary layer = 25.083 W/m**2 \n", + "The average heat transfer coefficient over entire plate = 32.910 W/m**2\n", + "Total heat transfer rate = 1974.607 watt\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.6 - Page : 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "rho=997 # in kg/m**3\n", + "k=0.608 # in W/mK\n", + "Cp= 4180 # in J/kg K\n", + "miu=910*10**-6 # in Ns/m**2\n", + "d=30*10**-3 # in m\n", + "m=0.02 # in kg/s\n", + "t_o=30 # in degree C\n", + "t_i=20 # in degree C\n", + "Re= 4*m/(pi*d*miu)#\n", + "q_desh=12*10**3 # in W/m**2\n", + "# since Re < 2300, flow is laminar one\n", + "\n", + "# Part(a)\n", + "# Nu = h*d/k = 4.36\n", + "h=4.36*k/d#\n", + "print \"Heat transfer coefficient = %0.3f W/m**2K\" %h\n", + "\n", + "# Part (b)\n", + "L=m*Cp*(t_o-t_i)/(q_desh*pi*d)#\n", + "print \"Length of pipe = %0.3f meter\" %L\n", + "\n", + "# Part(c)\n", + "# q_desh= h*(t_infinite-t_o)\n", + "t_infinite = q_desh/h+t_o#\n", + "print \"The inner tube surface temperature at the outlet = %0.3f degree C\" %t_infinite\n", + "\n", + "# Part(d)\n", + "f=64/Re#\n", + "print \"Friction Factor = %0.3f \" %f\n", + "\n", + "# Part(e)\n", + "V=4*m/(pi*d**2*rho) # in m/s ( because m= rho*V*A , m= rho*V*pi*d**2/4 )\n", + "del_P= f*L*rho*V**2/(d*2) # in N/m**2\n", + "print \"The pressure drop in the pipe = %0.3f N/m**2\" %del_P\n", + "\n", + "# Note: In part(b) value of L is miss printed actual value is .739 m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient = 88.363 W/m**2K\n", + "Length of pipe = 0.739 meter\n", + "The inner tube surface temperature at the outlet = 165.804 degree C\n", + "Friction Factor = 0.069 \n", + "The pressure drop in the pipe = 0.679 N/m**2\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.7 - Page : 159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "from math import log\n", + "# given data\n", + "rho=977.3 # in kg/m**3\n", + "kf=0.665 # in W/mK\n", + "Cp= 4186 # in J/kg K\n", + "miu=4.01*10**-4 # in kg/m-s\n", + "Pr=2.524#\n", + "d=0.02 # in m\n", + "m=0.5 # in kg/s\n", + "t_o=70 # in degree C\n", + "t_i=20 # in degree C\n", + "t_s=100 # in degree C\n", + "Re= 4*m/(pi*d*miu)#\n", + "\n", + "# Since Re > 2300, flow is turbulent flow. Then Nusselt Number\n", + "# Nu = h*d/k = 0.023*Re**0.8*Pr**0.4\n", + "h=0.023*Re**0.8*Pr**0.4*kf/d # in W/m**2\n", + "print \"Average heat transfer coefficient = %0.3f W/m**2\" %h\n", + "L=m*Cp*log((t_s-t_i)/(t_s-t_o))/(pi*d*h) # in meter\n", + "print \"Length of tube = %0.3f meter\" %L\n", + "# Note: Calculation of Re is wrong so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average heat transfer coefficient = 9207.036 W/m**2\n", + "Length of tube = 3.549 meter\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.8 - Page : 160" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "rho=977 # in kg/m**3\n", + "k=0.608 # in W/mK\n", + "Cp= 4180 # in J/kg K\n", + "miu=910*10**-6 # in poise\n", + "d=0.02 # in m\n", + "m=0.02 # in kg/s\n", + "t_o=40 # in degree C\n", + "t_i=10 # in degree C\n", + "q_desh= 20*10**3 # in W/m**2\n", + "\n", + "# Part (a)\n", + "Re= 4*m/(pi*d*miu)#\n", + "print \"Reynold number = %0.3f\" %Re\n", + "\n", + "# Part(b)\n", + "# Nu = h*d/k = 4.364\n", + "h=4.364*k/d#\n", + "print \"Heat transfer coefficient = %0.2f W/m**2K\" %h\n", + "\n", + "# Part (c)\n", + "# q= q_desh*A = m*Cp*(t_o-t_i)\n", + "# q_desh *( pi*d*l) = m*Cp*(t_o-t_i)\n", + "l=m*Cp*(t_o-t_i)/(q_desh*pi*d)#\n", + "print \"Length of pipe = %0.3f meter\" %l" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reynold number = 1399.164\n", + "Heat transfer coefficient = 132.67 W/m**2K\n", + "Length of pipe = 1.996 meter\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.9 - Page : 161" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "rho=7.7*10**3 # in kg/m**3\n", + "k=12 # in W/mK\n", + "Cp= 130 # in J/kg degree C\n", + "Pr=0.011#\n", + "delta=8*10**-8 # in m**2/s\n", + "d=0.06 # in m\n", + "m=4 # in kg/s\n", + "t_i=200 # in degree C\n", + "del_t=25 # in degree C\n", + "miu=rho*delta#\n", + "Re= 4*m/(pi*d*miu)#\n", + "# From correlation Nu =h*d/k = 4.82+0.0185*Pe**0.827\n", + "Pe=Re*Pr#\n", + "h=(4.82+0.0185*Pe**0.827)*k/d # in W/m**2K\n", + "# Length of tube required by doing every balance\n", + "# m*Cp*del_t = h*A*(t_s-t_b) = h*(pi*d*l)*(t_s-t_b) # its given (t_s-t_b) = 40 degree C\n", + "l= m*Cp*del_t/(h*(pi*d)*40) # in meter\n", + "print \"Length of tube = %0.3f meter\" %l" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of tube = 0.678 meter\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.10 - Page : 162" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "d=0.058 # in m\n", + "t_infinite=30 # in degree C\n", + "t_s=155 # in degree C\n", + "V=52 # in m/s\n", + "T_f=(t_s+t_infinite)/2 # in degree C\n", + "T_f=T_f+273 # in K\n", + "# Fluid properties at 92.5 degree C and 1 atm\n", + "miu= 2.145*10**-5 # in kg/ms\n", + "Pr=0.696#\n", + "P=1.0132*10**5#\n", + "R=287#\n", + "k=0.0312 # in W/mK\n", + "rho=P/(R*T_f) # in kg/m**3\n", + "Re=rho*V*d/miu#\n", + "C=0.0266#\n", + "n=0.805#\n", + "# Nu = h*d/k = C*(Re)**n*Pr**(1/3)\n", + "h=C*(Re)**n*Pr**(1/3)*k/d # in W/m**2K\n", + "#So, heat transfer rate per unit length from cylinder\n", + "q_by_L= h*(pi*d)*(t_s-t_infinite) # in W/m\n", + "print \"Heat transfer rate per unit length from cylinder = %0.3f W/m\" %q_by_L\n", + "# Note: Calculation of q_by_L in the book is wrong , so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate per unit length from cylinder = 3914.183 W/m\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.11 - Page : 163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "delta=15.68*10**-6 # in m**2/s\n", + "t_infinite=25+273 # in K\n", + "t_s=80+273 # in K\n", + "t_infinite=25+273 # in K\n", + "k=0.02625 # in W/m degree C\n", + "Pr=0.708#\n", + "miu_infinite=1.846*10**-5 #in kg/ms\n", + "miu_s= 2.076*10**-5 # in kg/ms\n", + "d=10*10**-3 # in m\n", + "V=5 # in m/s\n", + "A=4*pi*(d/2)**2#\n", + "Re=V*d/delta#\n", + "Nu= 2+ (0.4*Re**(1/2)+0.06*Re**(2/3))*Pr**0.4*(miu_infinite/miu_s)**(1/4)#\n", + "# Nu = h*d/k\n", + "h=Nu*k/d # in W/m**2K\n", + "# heat transfer rate\n", + "q=h*A*(t_s-t_infinite) # in watt\n", + "print \"Heat transfer rate = %0.3f watt\" %q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate = 1.456 watt\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.12 - Page : 164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# given data\n", + "Cp=4179 # in J/kg-K\n", + "rho= 997 # in kg/m**3\n", + "V=2 # in m/s\n", + "miu= 855*10**-6 # in Ns/m**2\n", + "Pr=5.83#\n", + "k=0.613#\n", + "Do=6 #outer dia in cm\n", + "Di=4 #inner dia in cm\n", + "# de= 4*A/P = 4*pi/4*(Do**2-Di**2)/(pi*(Do+Di))\n", + "# or\n", + "de= Do-Di # in cm\n", + "de=de*10**-2 # in m\n", + "Re= rho*V*de/miu#\n", + "# Since Re > 2300, hence flow is turbulent. Hence using Dittus Boelter equation\n", + "# Nu= 0.023*Re**0.8*Pr**0.4 =h*de/k\n", + "h= 0.023*Re**0.8*Pr**0.4*k/de # in W/m**2K\n", + "print \"Heat transfer coefficient = %0.f W/m**2K\" %math.floor(h)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient = 7752 W/m**2K\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.13 - Page : 165" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "Cp=0.138 # in KJ/kg-K\n", + "m=8.33 # in kg/sec\n", + "Pr=0.0238#\n", + "k=8.7 # in W/mk\n", + "d=1.5*10**-2 # in m\n", + "miu=1.5*10**-3 # in kg/ms\n", + "Re=4*m/(pi*miu*d)#\n", + "Pe=Re*Pr#\n", + "# Nu = h*d/k = 7+0.025*Pe**0.8\n", + "h= (7+0.025*Pe**0.8)*k/d # in W/m**2 degree C\n", + "print \"Heat transfer coefficient = %0.3f W/m**2 degree C\" %h" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient = 29255.771 W/m**2 degree C\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.14 - Page : 166" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "rho=887 # in kg/m**3\n", + "Pr=0.026#\n", + "k=25.6 # in W/mk\n", + "d=2.5*10**-2 # in m\n", + "miu=0.58*10**-3 # in kg/ms\n", + "V=3 # in m/s\n", + "Re=rho*V*d/(miu)#\n", + "Pe=Re*Pr#\n", + "Nu = 4.8+0.015*Pe**0.85*Pr**0.08\n", + "h= Nu*k/d # in W/m**2 degree C\n", + "print \"Heat transfer coefficient = %0.3f W/m**2 degree C\" %h\n", + "#Note: There is some difference in coding and book answer because they did not take aqurate calculation" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient = 15217.633 W/m**2 degree C\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 5.15 - Page : 166" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "delta=38.1*10**-6 # in m**2/s\n", + "Pr=501#\n", + "Prs=98#\n", + "K=0.138 # in W/mk\n", + "T_infinite=353 # in K\n", + "T_s=423 # in K\n", + "V=2 # in m/s\n", + "d=12.5*2*10**-3 # in m\n", + "Re=V*d/delta#\n", + "n=0.36# for Pr >= 10\n", + "C=0.26 # for Re between 10**3 and 2*10**5\n", + "m=0.6 # for Re between 10**3 and 2*10**5\n", + "Nu= C*Re**m*Pr**n*(Pr/Prs)**(1/4)#\n", + "h= Nu*K/d # in W/m**2 degree C\n", + "A=pi*25*10**-3#\n", + "del_t=T_s-T_infinite#\n", + "# Formula q=h*A*del_t\n", + "q_by_L = h*A*del_t#\n", + "print \"Initial rate of heat loss per meter length of cylinder = %0.3f\" %q_by_L\n", + "# Note: calculation in the book is wrong so answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Initial rate of heat loss per meter length of cylinder = 8260.795\n" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb new file mode 100644 index 00000000..e5eca739 --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb @@ -0,0 +1,543 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 - Free Convection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.1 - Page : 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "# (i) when\n", + "x=.3 # in m\n", + "T_s=100 # in degree C\n", + "T_infinite=30 # in degree C\n", + "T_f=(T_s+T_infinite)/2 # in degree C\n", + "T_f=T_f+273 # in K\n", + "Bita=1/T_f#\n", + "# Other fluid properties at film temperature\n", + "Pr=0.703#\n", + "K=0.0301 # in W/mK\n", + "T=1.8*10**-5 # in m**2/s\n", + "g=9.81#\n", + "del_T=T_s-T_infinite#\n", + "Gr=(g*Bita*del_T*x**3)/T**2#\n", + "Ra=Gr*Pr#\n", + "print \"Rayleigh Number is = %0.2e\" %Ra\n", + "#Since Ra<10**9, hence flow is laminar, then correlation for vertical plate in laminar flow\n", + "# Formula Nu=0.59*Ra**(1/4)=h*x/K\n", + "h=0.59*Ra**(1/4)*K/x # in W/m**2K\n", + "A=2*.3*.5#\n", + "q1=h*A*(T_s-T_infinite)#\n", + "print \"Heat transfer rate from the plate, when the vertical height is 0.3 m :\",round(q1,3),\"W\"\n", + "\n", + "#(ii) when\n", + "x=0.5 # in m\n", + "Gr=(g*Bita*del_T*x**3)/T**2#\n", + "Ra=Gr*Pr#\n", + "# Formula Nu=0.59*Ra**(1/4)=h*x/K\n", + "h=0.59*Ra**(1/4)*K/x # in W/m**2K\n", + "q2=h*A*(T_s-T_infinite)#\n", + "print \"Heat transfer rate from the plate, when the vertical height is 0.5 m :\",round(q2,3),\"W\"\n", + "PercentageDecrease=(q1-q2)/q1*100#\n", + "print \"Percentage decreases in heat transfer rate when x=0.5 m as compared to when x=0.3 m is :\",round(PercentageDecrease,3),\"%\"\n", + "\n", + "#Note : In the book ,In part (b), calculation of getting the value of h is wrong " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rayleigh Number is = 1.19e+08\n", + "Heat transfer rate from the plate, when the vertical height is 0.3 m : 129.844 W\n", + "Heat transfer rate from the plate, when the vertical height is 0.5 m : 114.277 W\n", + "Percentage decreases in heat transfer rate when x=0.5 m as compared to when x=0.3 m is : 11.989 %\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.2 - Page : 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "Pr=0.694#\n", + "K=0.0296 # in W/mK\n", + "rho=1.029 # in kg/m**3\n", + "miu=20.6*10**-6 # in poise\n", + "x=.2 # in m\n", + "T_s=110 # in degree C\n", + "T_infinite=30 # in degree C\n", + "T_f=(T_s+T_infinite)/2 # in degree C\n", + "T_f=T_f+273 # in K\n", + "Bita=1/T_f#\n", + "g=9.81#\n", + "del_T=T_s-T_infinite#\n", + "Gr=(rho**2*g*Bita*del_T*x**3)/miu**2#\n", + "Ra=Gr*Pr#\n", + "#since Rayleigh number is less than 10**10, hence\n", + "Nu=0.68*Pr**(1/2)*Gr**(1/4)/((.952+Pr)**(1/4))#\n", + "h=Nu*K/x#\n", + "A=2*0.2*1#\n", + "q=h*A*(T_s-T_infinite)#\n", + "print \"Heat transfer rate is :\",round(q,1),\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is : 194.7 W\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.3 - Page : 179" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "d=7.5*10**-2 # in m\n", + "x=2 # in m\n", + "T_s=70 # in degree C\n", + "T_infinite=10 # in degree C\n", + "del_T=T_s-T_infinite#\n", + "g=9.81#\n", + "calculation=4.5*10**10# # value of g*Bita*rho**2*C_p/(miu*k)\n", + "K=2.75*10**-2 # in W/mK\n", + "# g*Bita*rho**2*C_p/(miu*k) = g*Bita*rho**2/miu**2 * miu*C_p/k = (g*Bita*del_T*x**3/T**2 * miu*C_p/k)/(del_T*x**3)\n", + "GrxPr= calculation*del_T*x**3# # value of Gr*Pr\n", + "Nu= 0.13*(GrxPr)**(1/3)#\n", + "# Formula Nu = h*x/k\n", + "h= Nu*K/x # in W/m**2K\n", + "A=2*pi*d#\n", + "q=h*A*(del_T) # in W\n", + "q=q*60*60 # in J/h\n", + "print \"Heat transfer rate = %0.3e J/h\" %q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate = 5.067e+06 J/h\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.4 - Page : 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "m=15 # in kg\n", + "C_p=420 # in J/kg K\n", + "T_s=200 # in degree C\n", + "T_infinite=30 # in degree C\n", + "T_f=(T_s+T_infinite)/2 # in degree C\n", + "T_f=T_f+273 # in K\n", + "Pr=0.688#\n", + "K=0.0321 # in W/mK\n", + "delta=23.18*10**-6 # in m**2/s\n", + "Bita=1/T_f#\n", + "g=9.81#\n", + "x=0.3 # in m\n", + "del_T=T_s-T_infinite#\n", + "Gr=(g*Bita*del_T*x**3)/delta**2#\n", + "Ra=Gr*Pr#\n", + "#Since Ra<10**9, hence it is laminar flow using the relation\n", + "# Formula Nu=0.59*Ra**(1/4)=h*x/K\n", + "h=0.59*Ra**(1/4)*K/x # in W/m**2K\n", + "print \"(i) Heat transfer coefficient is :\",round(h,3),\"W/m**2K\"\n", + "\n", + "# (b) Initial rate of cooling\n", + "# Formula h*A*(T_s-T_infinite) = m*C_p*dt_by_toh\n", + "A=2*0.3*0.5#\n", + "dt_by_toh = h*A*(T_s-T_infinite)/(m*C_p) # in degree C/sec\n", + "dt_by_toh=dt_by_toh*60 # in degree C /min\n", + "print \"(ii) Initial rate of cooling of the plate is :\",round(dt_by_toh,3),\"degreeC per min\"\n", + "\n", + "#(c) Time taken by plate to cool from 200 degree C to 50 degree C\n", + "T_i=200 # in degree C\n", + "T=50 # in degree C\n", + "# Formula (T-T_infinite)/(T_i-T_infinite)= %e**(-h*A*toh/(m*C_p))#\n", + "toh= -log((T-T_infinite)/(T_i-T_infinite))*m*C_p/(h*A) # in sec\n", + "toh=toh/60 # in min\n", + "print \"(iii) Time required to cool plate from 200 degree C to 50 degree C is :\",round(toh,2),\"minutes\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Heat transfer coefficient is : 6.97 W/m**2K\n", + "(ii) Initial rate of cooling of the plate is : 3.385 degreeC per min\n", + "(iii) Time required to cool plate from 200 degree C to 50 degree C is : 107.46 minutes\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.5 - Page : 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "rho=0.8 # in kg/m**3#\n", + "C_p=1.01 # in KJ/kg K\n", + "Pr=0.684#\n", + "d=15*10**-2 # diameter in meter\n", + "K=0.035 # in W/mK\n", + "delta=2.78*10**-5 # in m**2/s\n", + "g=9.81#\n", + "x=2 # in m\n", + "T_s=250 # in degree C\n", + "T_infinite=30 # in degree C\n", + "T_f=(T_s+T_infinite)/2 # in degree C\n", + "T_f=T_f+273 # in K\n", + "Bita=1/T_f#\n", + "del_T=T_s-T_infinite#\n", + "#Heat loss from vertical part by free convection\n", + "Gr=(g*Bita*del_T*x**3)/delta**2#\n", + "Ra=Gr*Pr#\n", + "#Since Ra>10**9, hence turbulent flow\n", + "# Formula Nu= h*x/K =0.13*Ra**(1/3)\n", + "h=0.13*Ra**(1/3)*K/x # in W/m**2K\n", + "A=2*pi*d#\n", + "q1=h*A*del_T # w\n", + "q1=q1*10**-3 # in kW\n", + "print \"Heat loss from vertical part is :\",round(q1,3),\"kW\"\n", + "#Heat loss for Horizontal part\n", + "# here\n", + "x=d#\n", + "Gr=(g*Bita*del_T*x**3)/delta**2#\n", + "Ra=Gr*Pr#\n", + "#Since Ra<10**9, hence laminar fluid flow\n", + "# Formula Nu= h*x/K =0.53*Ra**(1/4)\n", + "h=0.53*Ra**(1/4)*K/x # in W/m**2K\n", + "A=pi*d*8#\n", + "q2=h*A*del_T # w\n", + "q2=q2*10**-3 # in kW\n", + "print \"Heat loss for horizontal part is :\",round(q2,3),\"kW\"\n", + "\n", + "#Heat loss by radiation\n", + "sigma=5.67*10**-8#\n", + "epsilon=0.65 # emissivity of steel\n", + "A=pi*d*10#\n", + "T_s=T_s+273 # in K\n", + "T_infinite=T_infinite+273 # in K\n", + "q3=sigma*A*epsilon*(T_s**4-T_infinite**4) # in w\n", + "q3=q3*10**-3 # in kW\n", + "print \"Heat loss by radiation is :\",round(q3,3),\"kW\"\n", + "#Total heat loss\n", + "theta=q1+q2+q3#\n", + "print \"Total heat loss is :\",round(theta,3),\"kW\"\n", + "\n", + "\n", + "#Note : value of q3 and theta in the book is wrong so answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss from vertical part is : 1.572 kW\n", + "Heat loss for horizontal part is : 6.447 kW\n", + "Heat loss by radiation is : 11.53 kW\n", + "Total heat loss is : 19.549 kW\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.6 - Page : 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "rho=1.205 # in kg/m**3#\n", + "C_p=1006 # in J/kg K\n", + "Pr=0.71#\n", + "K=0.0256 # in W/mK\n", + "delta=1.506*10**-5 # in m**2/s\n", + "T_s=35 # in degree C\n", + "T_infinite=5 # in degree C\n", + "T_f=(T_s+T_infinite)/2 # in degree C\n", + "T_f=T_f+273 # in K\n", + "Bita=1/T_f#\n", + "del_T=T_s-T_infinite#\n", + "g=9.81#\n", + "# Formula 1/x= 1/Lh + 1/Lv\n", + "Lh=50 # in cm\n", + "Lv=50 # in cm\n", + "x=Lh*Lv/(Lh+Lv) # in cm\n", + "x=x*10**-2 # in m\n", + "# Formula Gr=(g*Bita*del_T*x**3)/delta**2#\n", + "Gr=(g*Bita*del_T*x**3)/delta**2#\n", + "Ra=Gr*Pr#\n", + "# Formula Nu= h*x/K =0.53*Ra**(1/4)\n", + "h=0.53*Ra**(1/4)*K/x # in W/m**2K\n", + "A=2*(0.5+0.5)#\n", + "q=h*A*del_T # w\n", + "print \"Heat loss per meter length of pipe is :\",round(q,3),\"watt\"\n", + "\n", + "# Note: In the book, value of h is wrong due to place miss value of x, so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss per meter length of pipe is : 272.624 watt\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.7 - Page : 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad\n", + "# given data\n", + "L=3 # in m\n", + "delta=0#\n", + "hx='10*x**(-1/4)'\n", + "# (a) Average heat transfer coefficient\n", + "def integrand(x):\n", + " return 10*x**(-1/4)\n", + "\n", + "ans, err = quad(integrand, delta, L)\n", + "\n", + "h=1/L*ans\n", + "print \"(a) Average heat transfer coefficient is :\",round(h,2),\"W/m**2K\"\n", + "\n", + "# (b) Heat transfer rate\n", + "A=3*.3 # in m**2\n", + "Tp=170 # plate temp. in degree C\n", + "Tg=30 # gas temp. in degree C\n", + "del_T=Tp-Tg#\n", + "q=h*A*del_T # in W\n", + "print \"(b) Heat transfer rate is :\",round(q,3),\"W\"\n", + "\n", + "# (c) \n", + "x=2 # in m\n", + "qx_by_A= 10*x**(-1/4)*(Tp-Tg)#\n", + "print \"Local heat flux 2 m from the leading edge is :\",round(qx_by_A,1),\"W/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) Average heat transfer coefficient is : 10.13 W/m**2K\n", + "(b) Heat transfer rate is : 1276.524 W\n", + "Local heat flux 2 m from the leading edge is : 1177.3 W/m**2\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.8 - Page : 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "Pr=0.712#\n", + "K=0.026 # in W/mK\n", + "delta=1.57*10**-5 # in m**2/s\n", + "T_s=320 # in K\n", + "T_infinite=280 # in K\n", + "del_T=T_s-T_infinite#\n", + "T_f=(T_s+T_infinite)/2 # in K\n", + "Bita=1/T_f#\n", + "d1=20 # in cm\n", + "d2=30 # in cm\n", + "x=(d2-d1)/2 # in cm\n", + "x=x*10**-2 # in m\n", + "g=9.81#\n", + "Gr=(g*Bita*del_T*x**3)/delta**2#\n", + "Ra=Gr*Pr#\n", + "# Formula Nu= h*x/K =0.228*Ra**(0.226)\n", + "h=0.228*Ra**(0.226)*K/x # in W/m**2K\n", + "A=pi*(d1*10**-2)**2#\n", + "q=h*A*del_T # w\n", + "print \"Heat transfer rate is :\",round(q,1),\"watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is : 11.4 watt\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 6.9 - Page : 186" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "K=0.0278 # in W/mK\n", + "rho=1.092 # in kg/m**3\n", + "miu=19.57*10**-6 # in kg/ms\n", + "Cp=1007 # in kg/kg degree C\n", + "epsilon=0.9#\n", + "sigma=5.67*10**-8#\n", + "d=75+2*25 # in mm\n", + "d=d*10**-3 # in meter\n", + "T_s=80 # in degree C\n", + "T_infinite=20 # in degree C\n", + "T_f=(T_s+T_infinite)/2 # in degree C\n", + "T_f=T_f+273 # in K\n", + "Bita=1/T_f#\n", + "g=9.81#\n", + "del_T=T_s-T_infinite#\n", + "Pr=miu*Cp/K#\n", + "Gr=(rho**2*g*Bita*del_T*d**3)/miu**2#\n", + "# Formula Nu= h*d/K = 0.53*(Gr*Pr)**(1/4)#\n", + "h= 0.53*(Gr*Pr)**(1/4)*K/d#\n", + "#(a) Heat loss from 6 m length of pipe\n", + "A=pi*d*6#\n", + "Q_conv=h*A*del_T#\n", + "Q_rad=epsilon*sigma*A*((T_s+273)**4-(T_infinite+273)**4)#\n", + "#total heat transfer rate\n", + "Q=Q_conv+Q_rad#\n", + "print \"Total heat transfer rate = %0.f W\" %Q\n", + "\n", + "# (b) Overall heat transfer coefficient\n", + "# Formula Q=U*A*del_T\n", + "U=Q/(A*del_T)#\n", + "print \"Overall heat transfer coefficient is :\",round(U,3),\"W/m**2 degree C\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total heat transfer rate = 1863 W\n", + "Overall heat transfer coefficient is : 13.178 W/m**2 degree C\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb new file mode 100644 index 00000000..1a92384e --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb @@ -0,0 +1,610 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 - Radiation Heat Transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.1 - Page : 212" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import exp\n", + "# given data\n", + "lamda=2*10**-6 # in m\n", + "C1=0.374*10**-15#\n", + "T=2000+273 # in K'\n", + "C2=1.4388*10**-2#\n", + "\n", + "#(a)\n", + "# Formula Eb_lamda= (C1*lamda**-5)/(exp(C2/(lamda*T))-1)\n", + "Eb_lamda= (C1*lamda**-5)/(exp(C2/(lamda*T))-1)#\n", + "print \"Monochromatic emissive power at 2 micro wavelength = %0.2e W/m**2\" %Eb_lamda\n", + "\n", + "#(b)\n", + "# Formula lamda_max * T =2898 # in micro m K\n", + "lamda_max= 2898/T # in micro m\n", + "print \"Wave-length at which the emission is maximum = %0.3f micro m\" %lamda_max\n", + "\n", + "#(c)\n", + "Elamdab_max=1.285*10**-5*T**5 # in W/m**2-m\n", + "print \"Maximum emissive power = %0.4e W/m**2-m\" %Elamdab_max\n", + "\n", + "#(d)\n", + "sigma=5.67*10**-8#\n", + "E=sigma*T**4#\n", + "print \"Total emissive power = %0.3e W/m**2\" %E\n", + "\n", + "#Note: Answer of part (a) in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Monochromatic emissive power at 2 micro wavelength = 5.15e+11 W/m**2\n", + "Wave-length at which the emission is maximum = 1.275 micro m\n", + "Maximum emissive power = 7.7965e+11 W/m**2-m\n", + "Total emissive power = 1.513e+06 W/m**2\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.2 - Page : 213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "lamda=2*10**-6 # in m\n", + "C1=0.374*10**-15#\n", + "T=2000+273 # in K'\n", + "C2=1.4388*10**-2#\n", + "\n", + "epsilon=0.3#\n", + "sigma=5.67*10**-8#\n", + "T1=300 # in K\n", + "T2=200 # in K\n", + "del_T=T1-T2#\n", + "h=12 # in W/m**2 degree C\n", + "d=4*10**-2 # diameter in m\n", + "l=1 # in m\n", + "A=pi*d*l#\n", + "# Heat transfer rate by radiation,\n", + "q_r= epsilon*sigma*A*(T1**4-T2**4) # in W\n", + "# Heat transfer rate by convection,\n", + "q_c=h*A*del_T # in W\n", + "# Total heat transfer,\n", + "q=q_r+q_c#\n", + "# Formula q=U*A*del_T\n", + "U=q/(A*del_T) # Overall heat tranfer coefficient\n", + "print \"Overall heat tranfer coefficient = %0.3f W/m**2 degree C\" %U\n", + "#Note: Value of q_c is wrong in the book, so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Overall heat tranfer coefficient = 13.106 W/m**2 degree C\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.3 - Page : 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# given data\n", + "epsilon=0.5#\n", + "T1=1200 # in K\n", + "T2=300 # in K\n", + "#(a) Heat transfer rate between the two plates is \n", + "# Formula Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))\n", + "epsilon1=epsilon#\n", + "epsilon2=epsilon#\n", + "A1byA2=1#\n", + "Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#\n", + "# Formula q12= sigma*A*Fg12*(T1**4-T2**4) \n", + "sigma=5.67*10**-8#\n", + "q12byA=sigma*Fg12*(T1**4-T2**4) # in W/m**2\n", + "print \"Heat transfer rate between the two plates = %0.4e W/m**2\" %q12byA\n", + "\n", + "#(b)\n", + "epsilon3=.05#\n", + "Fg13=1/(1/epsilon1+(1/epsilon3-1)*A1byA2)#\n", + "Fg32=1/(1/epsilon3+(1/epsilon2-1)*A1byA2)#\n", + "# q13=q32\n", + "# sigma*A*Fg13*(T1**4-T3**4) = sigma*A*Fg32*(T3**4-T2**4) \n", + "T3= ((T1**4+T2**4)/2)**(1/4)#\n", + "T3=math.floor(T3)#\n", + "q13byA=sigma*Fg13*(T1**4-T3**4) # in W/m**2\n", + "print \"Heat transfer rate if a radiation shield with an emissivity of 0.05 on both sides is placed \"\n", + "print \"between the two plates = %0.3e W/m**2 \" %q13byA" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate between the two plates = 3.9038e+04 W/m**2\n", + "Heat transfer rate if a radiation shield with an emissivity of 0.05 on both sides is placed \n", + "between the two plates = 2.789e+03 W/m**2 \n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.4 - Page : 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T1=800+273 # in K\n", + "A= 5*6 # in square meter\n", + "epsilon=0.45#\n", + "sigma=5.67*10**-8#\n", + "q=epsilon*sigma*A*T1**4 #in watt\n", + "print \"Energy emitted by a grey surface = %0.5e watt\" %q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy emitted by a grey surface = 1.01465e+06 watt\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.5 - Page : 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# given data\n", + "A=5 # in m**2\n", + "intensity=660 # in W/m**2\n", + "alpha=6/11#\n", + "rho=6/22#\n", + "toh=6/33#\n", + "energy_absorbed= intensity*alpha*A # in watt\n", + "print \"Energy absorbed = %0.f watt\" %energy_absorbed\n", + "energy_transmitted=intensity*rho*A #in watt\n", + "print \"Energy transmitted = %0.f watt\" %energy_transmitted\n", + "energy_emitted= intensity*toh*A # in watt\n", + "print \"Energy emitted = %0.f watt\" %energy_emitted" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy absorbed = 1800 watt\n", + "Energy transmitted = 900 watt\n", + "Energy emitted = 600 watt\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.5 - Page : 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T1=200+273 # in K\n", + "T2=100+273 # in K\n", + "A= 1*2 # in square meter\n", + "sigma=5.67*10**-8#\n", + "x_D= 1/4#\n", + "y_D= 1/2#\n", + "Fg12= 0.033#\n", + "q12= Fg12*sigma*A*(T1**4-T2**4) # in watt\n", + "print \"The net heat exchange between two surfaces = %0.2f watt\" %q12" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The net heat exchange between two surfaces = 114.88 watt\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.6 - Page : 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "d=20*10**-2 #diameter of pipe in m\n", + "l=1 # length of pipe in m\n", + "s=30*10**-2 # side of duct in m\n", + "A1=pi*d*l # area of pipe in m**2\n", + "A2=4*s*s # area of duct in m**2\n", + "epsilon1=0.8#\n", + "epsilon2=0.9#\n", + "T1=200+273 # in K\n", + "T2=20+273 # in K\n", + "# Formula Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))\n", + "Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))#\n", + "# Heat transfer rate between pipe and duct\n", + "sigma=5.67*10**-8#\n", + "q12=sigma*Fg12*A1*(T1**4-T2**4) # in W\n", + "print \"Heat transfer rate between pipe and duct = %0.3f W\" %q12\n", + "#Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate between pipe and duct = 1053.148 W\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.10 - Page : 220" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "D=150*10**-3 # in m\n", + "H=400*10**-3 # in m\n", + "T1=500 # in K\n", + "epsilon=0.7#\n", + "# Formula F11=(4*H)/(4*H+D)\n", + "F11=(4*H)/(4*H+D)#\n", + "sigma=5.67*10**-8#\n", + "A1=pi*D*H#\n", + "q=sigma*A1*epsilon*T1**4*((1-F11)/(1-F11*(1-epsilon)))#\n", + "print \"Heat Heat loss for cavity = %0.3f W\" %q\n", + "#Note: Answer in the book is not accutate" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat Heat loss for cavity = 55.227 W\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.11 - Page : 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "epsilon1=.04#\n", + "epsilon2=epsilon1#\n", + "T1=-153+273 # in K\n", + "T2=27+273 # in K\n", + "h_fg=209 # in kJ/kg\n", + "h_fg=h_fg*10**3 # in J/kg\n", + "d1=20*10**-2 # in m\n", + "d2=30*10**-2 # in m\n", + "A1=d1**2 # in square meter\n", + "A2=d2**2 # in square meter\n", + "A=4*pi*(d2-d1)**2#\n", + "Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))#\n", + "sigma=5.67*10**-8# \n", + "q12=sigma*A*Fg12*(T1**4-T2**4) # in W\n", + "print \"Net radiant heat transfer rate = %0.3f watt\" %q12\n", + "print \"Negative sign indicates that heat flows into the sphere\"\n", + "q12=-q12 #\n", + "m=q12*60/h_fg #\n", + "print \"Rate of evaporation per minutes = %0.3e kg/min\" %m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net radiant heat transfer rate = -1.577 watt\n", + "Negative sign indicates that heat flows into the sphere\n", + "Rate of evaporation per minutes = 4.526e-04 kg/min\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.12 - Page : 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T1=500 # in K\n", + "T2=300 # in K\n", + "sigma=5.67*10**-8#\n", + "A=2 # surface area of each plate in m**2\n", + "#(a) If the plates are perfectly black\n", + "F12=1#\n", + "q12=sigma*A*F12*(T1**4-T2**4)#\n", + "print \"Radiation heat transfer between two black parellel plates = %0.4e watt\" %q12\n", + "\n", + "#(b) If the plates are gray surface\n", + "#in this case\n", + "F12=1#\n", + "#A1=A2, so\n", + "A1byA2=1\n", + "epsilon1=.4#\n", + "epsilon2=epsilon1#\n", + "#Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#\n", + "Fg12=1/((1-epsilon1)/epsilon1 + 1/F12 + ((1-epsilon2)/epsilon2)*A1byA2)#\n", + "q12=sigma*A*Fg12*(T1**4-T2**4) # in W\n", + "print \"Heat transfer rate = %0.3e watt\" %q12" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radiation heat transfer between two black parellel plates = 6.1690e+03 watt\n", + "Heat transfer rate = 1.542e+03 watt\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.13 Page No - 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "T1=800 # in K\n", + "T3=200 # in K\n", + "sigma=5.67*10**-8#\n", + "d1=20*10**-2 # in m\n", + "d2=30*10**-2 # in m\n", + "d3=40*10**-2 # in m\n", + "A1=4*pi*(d1/2)**2 # in m**2\n", + "A2=4*pi*(d2/2)**2 # in m**2\n", + "A3=4*pi*(d3/2)**2 # in m**2\n", + "epsilon1=0.2#\n", + "epsilon2=epsilon1\n", + "epsilon3=epsilon1\n", + "Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1/A2)#\n", + "Fg23=1/(1/epsilon2+(1/epsilon3-1)*A2/A3)#\n", + "# Under steady state condition \n", + "# q12 = q23\n", + "# A1*Fg12*sigma*(T1**4-T2**4) = A2*Fg23*sigma*(T2**4-T3**4)\n", + "T2 = ((A2*Fg23*T3**4/(A1*Fg12)+T1**4)/(A2*Fg23/(A1*Fg12) + 1))**(1/4)\n", + "print \"Steady state temperature of the intermediate sphere = %0.f K\" %T2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Steady state temperature of the intermediate sphere = 604 K\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.14 Page No : 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "T1=400 # in K\n", + "T2=500 # in K\n", + "T3=1200 # in K\n", + "alpha1=0.70#\n", + "alpha2=0.6#\n", + "alpha3=0.4#\n", + "# First part\n", + "sigma=5.67*10**-8#\n", + "qa=alpha3*sigma*T3**4#\n", + "print \"The rate of energy absorption = %0.2e W/m**2\" %qa\n", + "# Second part\n", + "qa=alpha1*sigma*T1**4#\n", + "print \"The rate of emission of radiation energy = %0.2f W/m**2\" %qa\n", + "# Note : Answer of the first part in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate of energy absorption = 4.70e+04 W/m**2\n", + "The rate of emission of radiation energy = 1016.06 W/m**2\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 7.15 Page No : 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "d1=100 # in mm\n", + "d1=d1*10**-3 # in m\n", + "d2=100+10*2 # in mm\n", + "d2=d2*10**-3 # in m\n", + "l=1 # in m\n", + "A1byA2=d1**2/d2**2#\n", + "A1=pi*d1*l # in m**2\n", + "sigma=5.67*10**-8#\n", + "T1=120+273 # in K\n", + "T2=35+273 # in K\n", + "epsilon1=.8#\n", + "epsilon2=.1#\n", + "Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#\n", + "# Radiant heat transfer from the tube\n", + "q=A1*Fg12*sigma*(T1**4-T2**4)\n", + "print \" Radiant heat transfer from the tube = %0.3f W/m\" %q\n", + "#Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Radiant heat transfer from the tube = 35.282 W/m\n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb new file mode 100644 index 00000000..dbf7d5a0 --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb @@ -0,0 +1,903 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 - Heat Exchanger" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.1 - Page : 248" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import log\n", + "# given data\n", + "t_hi=80 # in degree C\n", + "t_ci=30 # in degree C\n", + "t_ho=40 # in degree C\n", + "Mh=0.278 # in kg/s\n", + "Mc=0.278 # in kg/s\n", + "Cph=2.09# # in kJ/kg degree C\n", + "Cpc=4.18 # in kJ/kg degree C\n", + "U=24 # in W/m**2 degree C\n", + "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n", + "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci # in degree C\n", + "del_t1=t_hi-t_co #in degree C\n", + "del_t2=t_ho-t_ci #in degree C\n", + "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n", + "Cph=Cph*10**3 # in J/kg degree C\n", + "q=Mh*Cph*(t_hi-t_ho)#\n", + "#Formula q=U*A*del_tm\n", + "A=q/(U*del_tm) # in m**2\n", + "print \"Surface area of heat exchange = %0.1f square meter\" %A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface area of heat exchange = 53.2 square meter\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.2 - Page : 249" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "t_hi=160 # in degree C\n", + "t_ci=25 # in degree C\n", + "t_ho=60 # in degree C\n", + "Mh=2 # in kg/s\n", + "Mc=2 # in kg/s\n", + "Cph=2.035# # in kJ/kg degree C\n", + "Cpc=4.187 # in kJ/kg degree C\n", + "U=250 # in W/m**2 K\n", + "d=0.5 # in m\n", + "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n", + "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci # in degree C\n", + "del_t1=t_hi-t_co #in degree C\n", + "del_t2=t_ho-t_ci #in degree C\n", + "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n", + "Cph=Cph*10**3 # in J/kg degree C\n", + "q=Mh*Cph*(t_hi-t_ho)#\n", + "#Formula q=U*pi*d*l*del_tm\n", + "l=q/(U*pi*d*del_tm)#\n", + "print \"Length of the heat exchanger = %0.2f meter\" %l" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of the heat exchanger = 18.22 meter\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.3 - Page : 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "from __future__ import division\n", + "# given data\n", + "t_hi=110 # in degree C\n", + "t_ci=35 # in degree C\n", + "t_co=75 # in degree C\n", + "Mh=2.5 # in kg/s\n", + "Mc=1 # in kg/s\n", + "Cph=1.9# # in kJ/kg K\n", + "Cpc=4.18 # in kJ/kg K\n", + "U=300 # in W/m**2 K\n", + "\n", + "# Energy balance Mc*Cpc*(t_co-t_ci) = Mh*Cph*(t_hi-t_ho)\n", + "t_ho=t_hi- Mc*Cpc*(t_co-t_ci)/(Mh*Cph) # in degree C\n", + "del_t1=t_hi-t_co #in degree C\n", + "del_t2=t_ho-t_ci #in degree C\n", + "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n", + "Cph=Cph*10**3 # in J/kg degree C\n", + "q=Mh*Cph*(t_hi-t_ho)#\n", + "#Formula q=U*A*del_tm\n", + "A=q/(U*del_tm)#\n", + "print \"Area of the heat exchanger = %0.2f square meter\" %A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Area of the heat exchanger = 14.92 square meter\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.4 - Page : 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "Fi=0.00014 # in m**2 degree C/W\n", + "hi=2000 # in W/m**2degree C\n", + "Fo=0.00015 # in m**2 degree C/W\n", + "ho=1000 # in W/m**2degree C\n", + "di=3*10**-2 # in m\n", + "do=4*10**-2 #in m\n", + "ro=do/2#\n", + "ri=di/2#\n", + "k=53 # in W/m degree C\n", + "Uo=1/(do/di*1/hi+ do/(2*k)*log(ro/ri) + 1/ho + do*Fi/di + Fo)#\n", + "print \"The overall heat transfer coefficient = %0.3f W/m**2 degree C\" %Uo\n", + "# Note : Answer in the book is not accurate" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The overall heat transfer coefficient = 473.509 W/m**2 degree C\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.5 - Page : 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# given data\n", + "V=0.15 # in m/s\n", + "di=2.5*10**-2 # in m\n", + "\n", + "delta=0.364*10**-6 # in m**2/s\n", + "k=0.668 # in W/m degree C\n", + "Pr=2.22#\n", + "\n", + "Re=V*di/delta#\n", + "# Formula Nu= hi*di/k = 0.023*Re**0.8*Pr**0.3\n", + "hi=0.023*Re**0.8*Pr**0.3*k/di # in W/m**2 degree C\n", + "\n", + "# Now, Reynold number for flow of air across the tube\n", + "delta=18.22*10**-6 # in m**2/s\n", + "k=0.0281 # in W/m degree C\n", + "Pr=0.703#\n", + "d=2.5*10**-2 # in m\n", + "u=10 # in m/s\n", + "Re=u*d/delta#\n", + "Re=math.floor(Re)#\n", + "#The Nusselt number for this case\n", + "Nu=(0.04*Re**0.5+ 0.006*Re**(2/3))*Pr**0.4\n", + "# Formula Nu= ho*do/k\n", + "do=di#\n", + "ho=Nu*k/do # in W/m**2 degree C\n", + "print \"Heat transfer coefficient = %0.3f W/m**2 degree C\" %ho\n", + "U=1/(1/hi+1/ho)#\n", + "print \"The overall heat transfer coefficient neglecting the wall resistance = %0.3f W/m**2 degree C\" %U\n", + "l=1 # in m\n", + "Ti=90 # in degree C\n", + "To=10 # in degree C\n", + "q=U*pi*d*l*(Ti-To) #\n", + "print \"Heat loss per meter length of the tube = %0.3f W/m\" %q\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient = 7.931 W/m**2 degree C\n", + "The overall heat transfer coefficient neglecting the wall resistance = 7.882 W/m**2 degree C\n", + "Heat loss per meter length of the tube = 49.523 W/m\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.6 - Page : 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# given data\n", + "t_hi=83 # in degree C\n", + "t_ho=45 # in degree C\n", + "t_ci=25 # in degree C\n", + "Mh=5 # in kg/min\n", + "Mc=9 # in kg/min\n", + "Cph=4.18# # in kJ/kg K\n", + "Cpc=2.85 # in kJ/kg K\n", + "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci) \n", + "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci # in degree C\n", + "print \"t_co = %0.2f degree C\" %t_co\n", + "if(t_co>t_ho) :\n", + " print \"since t_co > t_ho, hence counter flow arrangment will be suitable\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t_co = 55.96 degree C\n", + "since t_co > t_ho, hence counter flow arrangment will be suitable\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.7 - Page : 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "# (a) For parallel flow arrangment\n", + "del_t1=60-10 # in degree C\n", + "del_t2=40-30 # in degree C\n", + "del_tm=(del_t1-del_t2)/log(del_t1/del_t2) # in degree C\n", + "q=100*10**3 # in W\n", + "U=75 # in W/m**2 degree C\n", + "# Formula q=U*A*del_tm#\n", + "A=q/(U*del_tm)#\n", + "print \"Area for paraller flow arrangment = %0.1f square meter\" %A\n", + "# (b) For counter flow heat exchange\n", + "del_t1=60-30 # in degree C\n", + "del_t2=40-10 # in degree C\n", + "# In this case\n", + "del_tm=(del_t1+del_t2)/2 # in degree C\n", + "A=q/(U*del_tm)#\n", + "print \"Area For counter flow heat exchange = %0.2f square meter\" %A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Area for paraller flow arrangment = 53.6 square meter\n", + "Area For counter flow heat exchange = 44.44 square meter\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.8 - Page : 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "Cp=4180 # in J/kg degree C\n", + "miu=0.86*10**-3 # in kg/m-s\n", + "Pr=60#\n", + "k=0.60 # in W/m degree C\n", + "h_fg=2372400 # in W\n", + "ho=6000 # in W/m**2 degree C\n", + "di=2*10**-2 # in m\n", + "d_o=3*10**-2 # in m\n", + "t_co=35 # in degree C\n", + "t_ci=15 # in degree C\n", + "M=0.9#\n", + "Re=4*M/(pi*di*miu)#\n", + "# since Re > 2300, hence flow inside tube is a turbulent flow.\n", + "# Hence Nu= hi*di/k = 0.023*Re**0.8*Pr**0.4\n", + "hi= 0.023*Re**0.8*Pr**0.4*k/di#\n", + "Uo= 1/(1/10213.6*(d_o/di)+1/ho)#\n", + "del_t1=50-15 # in degree C\n", + "del_t2=50-35 # in degree C\n", + "del_tm=(del_t1-del_t2)/log(del_t1/del_t2) # in degree C\n", + "# Formula q= Uo*pi*d_i*L*del_tm = M*Cp*(t_co-t_ci)\n", + "L= M*Cp*(t_co-t_ci)/(Uo*pi*d_o*del_tm)#\n", + "print \"Length of tube = %0.3f meter\" %L\n", + "q=M*Cp*(t_co-t_ci) # in watt\n", + "m=q/h_fg#\n", + "print \"Rate of condensation = %0.3f kg/sec\" %m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of tube = 10.604 meter\n", + "Rate of condensation = 0.032 kg/sec\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.9 - Page : 257" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "Cph=3850# # in J/kg degree C\n", + "t_hi=100 # in degree C\n", + "t_ci=20 # in degree C\n", + "t_ho=50 # in degree C\n", + "Mh=8 # in kg/s\n", + "Mc=10 # in kg/s\n", + "Cpc=4.18*10**3 # in J/kg degree C\n", + "U=400 # in W/m**2 degree C\n", + "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n", + "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci # in degree C\n", + "# Heat load\n", + "q=Mh*Cph*(t_hi-t_ho) # in W\n", + "# (a) Parallel flow\n", + "del_t1=90 # in degree C\n", + "del_t2=3.16 # in degree C\n", + "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n", + "A=q/(U*del_tm)#\n", + "print \"Surface area for parallel flow = %0.3f meter square\" %A\n", + "# (b) Counter flow heat exchanger\n", + "del_t1=53.16 # in degree C\n", + "del_t2=40 # in degree C\n", + "del_tm_counter= (del_t1-del_t2)/log(del_t1/del_t2)#\n", + "A=q/(U*del_tm_counter)#\n", + "print \"Surface area for counter flow heat exchanger = %0.1f meter square\" %A\n", + "#(c) One shell pass and two tube pass.\n", + "#here\n", + "t1=10 # in degree C\n", + "t2=46.84 # in degree C\n", + "T1=100 # in degree C\n", + "T2=50 # in degree C\n", + "P=(t2-t1)/(T1-t1)#\n", + "R=(T1-T2)/(t2-t1)#\n", + "F=0.88#\n", + "del_tm=F*del_tm_counter # in degree C\n", + "A=q/(U*del_tm)#\n", + "print \"Surface area for one shell pass and two tube pass = %0.2f meter square\" %A\n", + "# (d) For cross flow, correction factor \n", + "F=0.9#\n", + "del_tm=F*del_tm_counter#\n", + "A=q/(U*del_tm)#\n", + "print \"Surface area for cross flow = %0.3f meter square\" %A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface area for parallel flow = 148.486 meter square\n", + "Surface area for counter flow heat exchanger = 83.2 meter square\n", + "Surface area for one shell pass and two tube pass = 94.56 meter square\n", + "Surface area for cross flow = 92.456 meter square\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.10 Page No : 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# given data\n", + "Cpc=4.18*10**3 # in J/kg degree C\n", + "Mc=1 # in kg/s\n", + "Mh=2.4 # in kg/s\n", + "Cph=2050# # in J/kg degree C\n", + "t_hi=100 # in degree C\n", + "t_ci=20 # in degree C\n", + "C_c=Mc*Cpc # in W/degree C\n", + "C_h=Mh*Cph # in W/degree C\n", + "U=300 # in W/m**2 degree C\n", + "A=10 # in m**2\n", + "C_min=C_c#\n", + "C_max=C_h#\n", + "N= A*U/C_min#\n", + "C=C_min/C_max#\n", + "# Effectiveness for counter flow heat exchanger\n", + "epsilon= (1-exp(-N*(1-C)))/(1-C*exp(-N*(1-C)))#\n", + "# Total heat transfer\n", + "q=epsilon*C_min*(t_hi-t_ci) # in watt\n", + "print \"Total heat transfer = %0.3f kW\" %(q*10**-3) \n", + "t_co=t_ci+epsilon*C*(t_hi-t_ci) #\n", + "print \"Exit temperature of water = %0.3f degree C\" %t_co" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total heat transfer = 144.170 kW\n", + "Exit temperature of water = 49.303 degree C\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.11 Page No : 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "t_hi=135 # in degree C\n", + "t_ci=20 # in degree C\n", + "t_ho=65 # in degree C\n", + "t_co=50 # in degree C\n", + "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n", + "# C =C_min/C_max = Mh*Cph/( Mc*Cpc)\n", + "C= (t_co-t_ci)/(t_hi-t_ho)#\n", + "epsilon=(t_hi-t_ho)/(t_hi-t_ci)#\n", + "# Also epsilon = epsilon_parallel = (1-exp(-NTU*(1+C)))/(1+C)\n", + "NTU= -log(1-epsilon*(1+C))/(1+C)#\n", + "# if the existing heat exchanger is to be used as counter flow mode, its NTU will not change, i.e.\n", + "epsilon_c= (1-exp(-NTU*(1-C)))/((1-C*exp(-NTU*(1-C))))#\n", + "# Exit temperature\n", + "# (i) Hot fluid\n", + "t_ho=t_hi-epsilon_c*(t_hi-t_ci) # in degree C\n", + "print \"Exit temperature for hot fluid = %0.3f degree C\" %t_ho\n", + "\n", + "#(ii) Cold fluid\n", + "t_co= t_ci+epsilon_c*C*(t_hi-t_ci)#\n", + "print \"Exit temperature for cold fluid = %0.1f degree C\" %t_co\n", + "\n", + "# (iii) # If the parallel flow heat exchanger is too long, then body fluid will have common outlet temperature (t)\n", + "# From MCp_h*(t_hi-t) = MCp_c*(t-t_ci)\n", + "\n", + "t=(C*t_hi+t_ci)/(1+C)#\n", + "print \"The minimum temperature to which the oil may be cooled by increasing the tube length with parallel flow operation,\"\n", + "print \"in degree C = %0.1f\" %t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Exit temperature for hot fluid = 55.909 degree C\n", + "Exit temperature for cold fluid = 53.9 degree C\n", + "The minimum temperature to which the oil may be cooled by increasing the tube length with parallel flow operation,\n", + "in degree C = 54.5\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.12 - Page : 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "t_hi=78 # in degree C\n", + "t_ci=23 # in degree C\n", + "t_ho=65 # in degree C\n", + "t_co=36 # in degree C\n", + "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n", + "# C =C_min/C_max = Mh*Cph/( Mc*Cpc)\n", + "C= (t_co-t_ci)/(t_hi-t_ho)#\n", + "epsilon=(t_hi-t_ho)/(t_hi-t_ci)#\n", + "# Formula epsilon = (1-exp(-N*(1+C)))/(1+C)\n", + "N= -log(1-epsilon*(1+C))/(1+C)#\n", + "# When flow rates of both fluids are doubled , the deat capacity ratio will not change, i.e.\n", + "# C=1\n", + "# MCp_new =2* MCp_old\n", + "# N=U*A/C_min=N/2\n", + "N=N/2#\n", + "epsilon=(1-exp(-N*(1+C)))/(1+C)#\n", + "# exit temperature\n", + "t_ho=t_hi-epsilon*(t_hi-t_ci) # in degree C\n", + "t_co= t_ci+epsilon*(t_hi-t_ci)#\n", + "print \"Exit temperature in degree C is : \",round(t_ho,2),\"and\",round(t_co,2)\n", + "# Note: Answer in the book is wrong due to put wrong value of t_ci in second last line" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Exit temperature in degree C is : 70.47 and 30.53\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.13 - Page : 262" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "t_hi=125 # in degree C\n", + "t_ci=22 # in degree C\n", + "Mh=21 # in kg/s\n", + "Mc=5 # in kg/s\n", + "C_ph=2100 # in J/kg K\n", + "C_pc=4100 # in J/kg K\n", + "Ch=Mh*C_ph # in Js/kg\n", + "Cc=Mc*C_pc # in Js/kg\n", + "C_min=Cc # in Js/kg\n", + "C_max=Ch # in Js/kg\n", + "U=325 # in W/m**2 K\n", + "d=2.2*10**-2 # in m\n", + "l=5 # in m\n", + "total_tube=195 # number of total tubes\n", + "A=pi*d*l*total_tube\n", + "NTU=U*A/C_min#\n", + "C=C_min/C_max#\n", + "epsilon = (1-exp(-NTU*(1-C)))/(1-C*exp(-NTU*(1-C)))#\n", + "t_co= t_ci+epsilon*(t_hi-t_ci)#\n", + "t_ho= t_hi-epsilon*Cc/Ch*(t_hi-t_ci)#\n", + "print \"Exit temperature in degree C :\",round(t_co,1),\"and\",round(t_ho,2)\n", + "# Total heat transfer\n", + "q=epsilon*C_min*(t_hi-t_ci)#\n", + "print \"Total heat transfer = %0.3f kW\" %(q*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Exit temperature in degree C : 82.8 and 96.73\n", + "Total heat transfer = 1246.589 kW\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.14 - Page : 263" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# given data\n", + "t_hi=94 # in degree C\n", + "t_ci=15 # in degree C\n", + "Mw=0.36 # in kg/s\n", + "Mo=0.153 # in kg/s\n", + "C_po=2*10**3 # in J/kg K\n", + "C_pw=4.186*10**3 # in J/kg K\n", + "U=10.75*10**2 # in W/m**2 K\n", + "A=1 # in m**2\n", + "Ch=Mo*C_po # in kW/K\n", + "Cc=Mw*C_pw # in kW/K\n", + "C_min=Ch # in W/K\n", + "C_max=Cc # in W/K\n", + "C=C_min/C_max#\n", + "NTU=U*A/C_min#\n", + "# Effectiveness\n", + "N=NTU#\n", + "epsilon = (1-exp(-N*(1-C)))/(1-C*exp(-N*(1-C)))#\n", + "mCp_min=C_min#\n", + "q_max= mCp_min*(t_hi-t_ci) # in W\n", + "q_actual= epsilon*q_max # in W\n", + "print \"Total heat transfer = %0.3f watt\" %q_actual\n", + "# Outlet temp. of water\n", + "t_co= q_actual/Cc+t_ci # in degree C\n", + "print \"Outlet temperature of water = %0.3f degree C\" %t_co\n", + "# Outlet temp. of oil\n", + "t_ho=t_hi-q_actual/Ch #in degree C\n", + "print \"Outlet temperature of oil = %0.3f degree C\" %t_ho\n", + "#Note: Evaluation of Cc and Ch in the book is wrong so the Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total heat transfer = 22987.488 watt\n", + "Outlet temperature of water = 30.254 degree C\n", + "Outlet temperature of oil = 18.877 degree C\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.15 - Page : 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "U=1800 # in W/m**2 degree C\n", + "h_fg=2200*10**3 # in J/kg\n", + "t_ci=20 # in degree C\n", + "t_co=90 # in degree C\n", + "del_t1=120-20 # in degree C\n", + "del_t2=120-90 # in degree C\n", + "del_tm=(del_t1-del_t2)/log(del_t1/del_t2) # in degree C\n", + "Mc=1000/3600 # in kg/s\n", + "Cc=4180 # in kg/s\n", + "# Rate of heat transfer\n", + "q=Mc*Cc*(t_co-t_ci) # in watt\n", + "# Formula q=U*A*del_tm\n", + "A=q/(U*del_tm)#\n", + "print \"Surface area = %0.3f square meter\" %A\n", + "#Rate of condensation of steam\n", + "ms=q/h_fg # in kg/sec\n", + "print \"Rate of condensation of steam = %0.4f kg/sec\" %ms" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface area = 0.777 square meter\n", + "Rate of condensation of steam = 0.0369 kg/sec\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.16 - Page : 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# given data\n", + "Mh=10000/3600 # in kg/sec\n", + "Mc=8000/3600 # in kg/sec\n", + "Cph=2095 # in J/kg K\n", + "Cpc=4180 # in J/kg K\n", + "t_hi=80 # in degree C\n", + "t_ci=25 # in degree C\n", + "t_ho=50 # in degree C\n", + "U=300 # in W/m**2 K\n", + "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n", + "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci # in degree C\n", + "del_t1=t_hi-t_co #in degree C\n", + "del_t2=t_ho-t_ci #in degree C\n", + "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n", + "q=Mh*Cph*(t_hi-t_ho)#\n", + "#Formula q=U*A*del_tm\n", + "A=q/(U*del_tm) # in m**2\n", + "print \"Surface area of heat exchange = %0.2f square meter\" %A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface area of heat exchange = 19.23 square meter\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 8.17 - Page : 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "from numpy import pi\n", + "# given data\n", + "ho=5000 # in W/m**2 degree C\n", + "rho=988.1 # in kg/m**3\n", + "K=0.6474#\n", + "D=555*10**-9 # in m**2/s\n", + "Pr=3.54#\n", + "n=100#\n", + "d_i=2.5*10**-2 # in m\n", + "ri=d_i/2#\n", + "d_o=2.9*10**-2 # in m\n", + "ro=d_o/2#\n", + "Cp=4174 # in J/kg degree C\n", + "Mc=8.333 # in kg/s\n", + "Mw=Mc#\n", + "t_c1=30 # in degree C\n", + "t_c2=70 # in degree C\n", + "t_n1=100 # in degree C\n", + "t_n2=t_n1 # in degree C\n", + "R_fi=0.0002 # in m**2 degree C/W (In the book, there is miss print in this line,they took here R_fi = .002)\n", + "# Heat gain by water\n", + "Q=Mc*Cp*(t_c2-t_c1)#\n", + "# Also Q= U*A*del_tm\n", + "del_t1=t_n1-t_c1 #in degree C\n", + "del_t2=t_n2-t_c2 #in degree C\n", + "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)# \n", + "# Mw= 1/4*pi*d_i**2*V*rho*N, here\n", + "N=n#\n", + "V=4*Mw/(pi*d_i**2*rho*N)#\n", + "# Formula Re=V*d_i/v, here\n", + "v=D#\n", + "Re=V*d_i/v#\n", + "# Formula Nu= hi*d_i/K = 0.023*Re**0.8*Pr**0.33\n", + "hi= 0.023*Re**0.8*Pr**0.33*K/d_i#\n", + "# Formula 1/Vi= 1/hi + R_fi + ri/ro*1/ho\n", + "Vi= 1/(1/hi + R_fi + ri/ro*1/ho) # in W/m**2 degree C\n", + "#Formula Q = Vi*(N*pi*d_i*L)*del_tm\n", + "L= Q/(Vi*(N*pi*d_i)*del_tm)#\n", + "print \"Length of the tube bundle = %0.1f m\" %L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of the tube bundle = 4.6 m\n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb new file mode 100644 index 00000000..a03c60e2 --- /dev/null +++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb @@ -0,0 +1,280 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 - Condensation And Boiling" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 9.1 Page No : 277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "from __future__ import division\n", + "# given data\n", + "h_fg=2256*10**3 # in J/kg\n", + "rho=970 # in kg/m**3\n", + "rho_v=0.596 # in kg/m**3\n", + "k=0.66 # in W/mK\n", + "miu=3.7*10**-4 # in kg/m-s\n", + "T_sat=100 # in degree C\n", + "T_s=40 # in degree C\n", + "L=1.5 # in m\n", + "d=0.09 # in m\n", + "g=9.81#\n", + "# heat transfer coefficient\n", + "#h_bar = 1.13*[ rho*g*(rho-rho_v)*h_fg*k**3/(miu*L*(T_sat-T_s))]**(1/4) # in W/m**2k\n", + "h_bar= 1.13*(rho*g*(rho-rho_v)*h_fg*k**3/(miu*L*(T_sat-T_s)))**(1/4)#\n", + "# heat transfer rate\n", + "q=h_bar*pi*d*L*(T_sat-T_s) # in watt\n", + "print \"Heat transfer rate = %0.3f kW\" %(q*10**-3)\n", + "#rate of condensation\n", + "m=q/h_fg # in kg/s\n", + "print \"Rate of condenstion = %0.4f kg/s\" %m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate = 105.276 kW\n", + "Rate of condenstion = 0.0467 kg/s\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 9.2 Page No : 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "from __future__ import division\n", + "# given data\n", + "h_fg=2310*10**3 # in J/kg\n", + "rho=980 # in kg/m**3\n", + "k=0.67 # in W/mK\n", + "Cp=4.18#\n", + "delta=.41*10**-6 # in m**2/s\n", + "miu=rho*delta#\n", + "T_sat=70 # in degree C\n", + "T_s=55 # in degree C\n", + "L=1 # in m\n", + "d=0.03 # in m\n", + "g=9.81#\n", + "N=5#\n", + "# (a) for Horizontal tube\n", + "h_bar = 0.725*(rho**2*g*h_fg*k**3/(N*miu*d*(T_sat-T_s)))**(1/4) # in W/m**2k\n", + "# heat transfer rate\n", + "q=h_bar*pi*d*L*N**2*(T_sat-T_s) # in watt\n", + "print \"Heat transfer rate for horizontal tube = %0.3f kW\" %(q*10**-3)\n", + "#rate of condensation\n", + "m=q/h_fg # in kg/s\n", + "print \"Rate of condenstion = %0.3f kg/s\" %m\n", + "\n", + "# (b) For Vertical tube\n", + "h_bar = 1.13*(rho**2*g*h_fg*k**3/(miu*L*(T_sat-T_s)))**(1/4) # in W/m**2k\n", + "# heat transfer rate\n", + "q=h_bar*pi*d*L*N**2*(T_sat-T_s) # in watt\n", + "print \"Heat transfer rate for vertical tube = %0.3f kW\" %(q*10**-3)\n", + "#rate of condensation\n", + "m=q/h_fg # in kg/s\n", + "print \"Rate of condenstion = %0.3f kg/s\" %m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate for horizontal tube = 236.364 kW\n", + "Rate of condenstion = 0.102 kg/s\n", + "Heat transfer rate for vertical tube = 229.269 kW\n", + "Rate of condenstion = 0.099 kg/s\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 9.3 Page No : 280" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# given data\n", + "h_fg=2392*10**3 # in J/kg\n", + "rho=993 # in kg/m**3\n", + "k=0.63 # in W/mK\n", + "miu=728*10**-6 # in kJ/m-s\n", + "N=10#\n", + "T_sat=45.7 # in degree C\n", + "T_s=25 # in degree C\n", + "d=4*10**-3 # in m\n", + "g=9.81#\n", + "h_bar = 0.725*(rho**2*g*h_fg*k**3/(N*miu*d*(T_sat-T_s)))**(1/4) # in W/m**2k\n", + "m=300/(60*60)#\n", + "# Formula m=q/h_fg\n", + "q=m*h_fg#\n", + "print \"Heat transfer rate = %0.3f kW\" %(q*10**-3)\n", + "# Formula q=h_bar*pi*d*L*N**2*(T_sat-T_s)\n", + "L=q/(h_bar*pi*d*N**2*(T_sat-T_s))#\n", + "print \"Length of tube = %0.3f m\" %L\n", + "\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate = 199.333 kW\n", + "Length of tube = 1.068 m\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 9.4 Page No : 280" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin\n", + "from numpy import pi\n", + "# given data\n", + "h_fg=2400*10**3 # in J/kg\n", + "rho=993 # in kg/m**3\n", + "rho_v=0.0563 # in kg/m**3\n", + "t_mf=(40+30)/2 # in degree C\n", + "k=0.625 # in W/mK\n", + "miu=728*10**-6 # in kJ/m-s\n", + "x=0.25#\n", + "T_sat=40 # in degree C\n", + "T_s=30 # in degree C\n", + "g=9.81#\n", + "\n", + "# (a) Thickness of condensate film\n", + "delta=(4*k*(T_sat-T_s)*miu*x/(rho*(rho-rho_v)*g*h_fg))**(1/4) # in meter\n", + "print \"Thickness of condensate film = %0.5f mm\" %(delta*10**3)\n", + "\n", + "#(b) Local value of heat transfer coefficient\n", + "hx=k/delta # in W/m**2\n", + "L=0.5 # in m\n", + "hm=4/3*(L/x)**(1/4)*hx#\n", + "print \"Average heat transfer coefficient = %0.1f W/m**2\" %hm\n", + "# The heat transfer rate\n", + "A=0.5*0.5 # in m**2\n", + "q=hm*A*(T_sat-T_s) # in watt\n", + "print \"The heat transfer rate = %0.2f kW\" %(q*10**-3)\n", + "\n", + "# (c) \n", + "theta=45 # in degree\n", + "h_vertical=hm#\n", + "h_inclined=h_vertical*(sin(theta*pi/180))**(1/4)#\n", + "print \"Average heat transfer coefficient when plate is inclined at 45 degree = %0.1f W/m**2K\" %h_inclined" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of condensate film = 0.11832 mm\n", + "Average heat transfer coefficient = 8375.5 W/m**2\n", + "The heat transfer rate = 20.94 kW\n", + "Average heat transfer coefficient when plate is inclined at 45 degree = 7680.4 W/m**2K\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 9.5 Page No : 282" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given correlataion\n", + "#h_A=5.56*(det_T)**3\n", + "#h_P=h_A*(rho/rho_a)**0.4\n", + "del_T=25 # in degree C\n", + "h_A=5.56*(del_T)**3 # in W/m**2K\n", + "print \"The heat transfer coefficient = %0.3f kW/m**2K\" %(h_A*10**-3)\n", + "# and at 20 bar\n", + "rho=20#\n", + "rho_a=1#\n", + "h_P=h_A*(rho/rho_a)**0.4 # in W/m**2\n", + "print \"Value of h_P = %0.1f kW/m**2\" %(h_P*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat transfer coefficient = 86.875 kW/m**2K\n", + "Value of h_P = 287.9 kW/m**2\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
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