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| author | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
| commit | fffcc90da91b66ee607066d410b57f34024bd1de (patch) | |
| tree | 7b8011d61013305e0bf7794a275706abd1fdb0d3 /Heat_Transfer/Chapter_3.ipynb | |
| parent | 299711403e92ffa94a643fbd960c6f879639302c (diff) | |
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adding book
Diffstat (limited to 'Heat_Transfer/Chapter_3.ipynb')
| -rwxr-xr-x | Heat_Transfer/Chapter_3.ipynb | 3026 |
1 files changed, 3026 insertions, 0 deletions
diff --git a/Heat_Transfer/Chapter_3.ipynb b/Heat_Transfer/Chapter_3.ipynb new file mode 100755 index 00000000..f90300e1 --- /dev/null +++ b/Heat_Transfer/Chapter_3.ipynb @@ -0,0 +1,3026 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Convection"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.1,Page no:3.44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Boundary layer thickness\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=10**-3 #N.s/m**2\n",
+ "#At distance y from surface\n",
+ "#ux=a+by+cy**2+dy**3\n",
+ "#At y=0,ux=0 therefore a=0\n",
+ "#i.e tao=0\n",
+ "#At edge of boundary layer,ie y=del\n",
+ "#ux=u_inf\n",
+ "#At y=o,c=0\n",
+ "#At y=del,ux=b*del+d*del**3\n",
+ "\n",
+ "#Therefore, b=-3*d*del**3\n",
+ "#d=-u_inf/(2*del**2)\n",
+ "#b=3*u_inf/(2*del)\n",
+ "\n",
+ "#For velocity profile,we have:\n",
+ "#del/x=4.64*(Nre_x)**(-1/2)\n",
+ "\n",
+ "#Evaluate N re_x\n",
+ "\n",
+ "x=75.0 #[mm]\n",
+ "x=x/1000 #[m]\n",
+ "u_inf=3.0 #[m/s]\n",
+ "rho=1000.0 #[kg/m**3] for air\n",
+ "\n",
+ "#Calculation\n",
+ "Nre_x=u_inf*rho*x/mu #Reynold number\n",
+ "#Substituting the value,we get\n",
+ "dell=x*4.64*(Nre_x**(-1.0/2.0)) #[m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Boundary layer thickness is del=\",round(dell,6),\"m or \",round(dell*1000,3),\"mm\"\n",
+ "print\"Wrong units in answer of book,m and mm are wrongly interchanged\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Boundary layer thickness is del= 0.000734 m or 0.734 mm\n",
+ "Wrong units in answer of book,m and mm are wrongly interchanged\n"
+ ]
+ }
+ ],
+ "prompt_number": 296
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.2,Page no:3.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Boundary layer thickness of plate\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=15*10**-6 #sq m /s\n",
+ "v=2 #m/s\n",
+ "L=2 #[m] length of plate\n",
+ "Nre_x=3*10**5\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "xc=Nre_x*mu/v #critical length at whihc the transition takes place\n",
+ "#Since xc is less than 2 m.Therefore the flow is laminar\n",
+ "#at any distance x,.it is calculated from\n",
+ "#del/x=4.64/(math.sqrt(NRe,x))\n",
+ "#At x=L=2 m\n",
+ "Nre_l=v*L/mu\n",
+ "del_l=4.64*L/math.sqrt(Nre_l)\n",
+ "del_l=del_l*1000 #[mm]\n",
+ "\n",
+ "#Result\n",
+ "print\"Boundary layer thickness at the trailing edge is\",round(del_l),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Boundary layer thickness at the trailing edge is 18.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.3,Page no:3.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Thickness of hydrodynamic boundary layer\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=15*10**-6 #Kinematic viscosity in [sq m /s]\n",
+ "x=0.4 #[m]\n",
+ "u_inf=3 #[m/s]\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "#At x=0.4 m,\n",
+ "Nre_x=u_inf*x/mu \n",
+ "import math\n",
+ "dell=4.64*x/math.sqrt(Nre_x) #[m]\n",
+ "dell=dell*1000 #[mm]\n",
+ "Cf_x=0.664/math.sqrt(Nre_x) \n",
+ "\n",
+ "#Result\n",
+ "print\"Since Nre,x(\",Nre_x,\")is Less than 3*10**5,..the boundary layer is laminar\"\n",
+ "print\"Thickness of boundary layer at x=\",x,\"m\",round(dell,2),\"mm\" \n",
+ "print\"Local skin friction coefficient is :\",round(Cf_x,4)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since Nre,x( 80000.0 )is Less than 3*10**5,..the boundary layer is laminar\n",
+ "Thickness of boundary layer at x= 0.4 m 6.56 mm\n",
+ "Local skin friction coefficient is : 0.0023\n"
+ ]
+ }
+ ],
+ "prompt_number": 146
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.4,Page no:3.46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Flat plate boundary layer\n",
+ "\n",
+ "import math\n",
+ "from scipy import integrate\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=1.85*10**-5 #[kg/(m.s)]\n",
+ "P=101.325 #Pressure in [kPa]\n",
+ "M_avg=29.0 #Avg molecular wt of air\n",
+ "R=8.31451 #Gas constant\n",
+ "T=300.0 #[K]\n",
+ "\n",
+ "#Calculation\n",
+ "rho=P*M_avg/(R*T) #[kg/m**3]\n",
+ "u_inf=2.0 #Viscosity in [m/s]\n",
+ "#At x=20 cm =0.2 m\n",
+ "x=0.2 #[m]\n",
+ "Nre_x=rho*u_inf*x/mu #[Reynolds number]\n",
+ "del_by_x=4.64/math.sqrt(Nre_x) #[Boundary layer]\n",
+ "dell=del_by_x*x #[m]\n",
+ "#del=del*1000 #[mm]\n",
+ "\n",
+ "#At\n",
+ "x=0.4 #[m]\n",
+ "Nre_x=(rho*u_inf*x)/mu #<3*10**5\n",
+ "#Boundary layer is laminar\n",
+ "del_by_x=4.64/math.sqrt(Nre_x) \n",
+ "del1=del_by_x*x #[m]\n",
+ "#del1=del1*1000 #[mm]\n",
+ "d=del1-dell #Del \n",
+ "def f(y):\n",
+ " m_dot=u_inf*(1.5*(y/d)-0.5*(y/d)**3)*rho\n",
+ " return(m_dot)\n",
+ "m_dot=integrate.quad(f,0,d)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Boundary layer thickness at distance 20 cm from leading edge is\",round(dell,4),\"m=\",round(dell*1000,1),\"mm\"\n",
+ "print\"Boundary layer thickness at distance 40 cm from leading edge is\",round(del1,4),\"m=\",round(del1*1000,1),\"mm\"\n",
+ "print\"Thus,Mass flow rate entering the boundary layer is\",round(m_dot[0],6),\"kg/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Boundary layer thickness at distance 20 cm from leading edge is 0.0058 m= 5.8 mm\n",
+ "Boundary layer thickness at distance 40 cm from leading edge is 0.0082 m= 8.2 mm\n",
+ "Thus,Mass flow rate entering the boundary layer is 0.003547 kg/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 152
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.5,Page no:3.47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Rate of heat removed from plate\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=3.9*10**-4 #Kinematic viscosity in sq m/s\n",
+ "k=36.4*10**-3 #Thermal conductivity in W/(m.K)\n",
+ "Npr=0.69\n",
+ "u_inf=8 #[m/s]\n",
+ "L=1 #Lenght of plate in [m]\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "Nre_l=u_inf*L/mu \n",
+ "#Since Nre_l is less than 3*10**5 ,the flow is laminar over the entire length of plate\n",
+ "Nnu=0.664*math.sqrt(Nre_l)*Npr**(1.0/3.0) #=hL/k\n",
+ "h=k*Nnu/L #w/sq m.K\n",
+ "h=3.06 #Approximation [W/sq m.K]\n",
+ "T_inf=523 #[K]\n",
+ "Tw=351 #[K]\n",
+ "W=0.3 #Width of plate [m]\n",
+ "A=W*L #Area in [sq m]\n",
+ "Q=h*A*(T_inf-Tw) # Rate of heat removal from one side in [W]\n",
+ "#from two side:\n",
+ "Q2=2*Q #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of heat removal is\",round(Q,1),\"W\"\n",
+ "print round(Q2,1),\" W heat should be removed continously from the plate\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of heat removal is 157.9 W\n",
+ "315.8 W heat should be removed continously from the plate\n"
+ ]
+ }
+ ],
+ "prompt_number": 155
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.6,Page no:3.48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat removed from plate\n",
+ "\n",
+ "#Variable declaration\n",
+ "P1=101.325 #Pressure in [kPa]\n",
+ "mu1=30.8*10**-6 #Kinematic viscosity in[sq m /s]\n",
+ "k=36.4*10**-3 #[W/(m.K)]\n",
+ "Npr=0.69\n",
+ "u_inf=8 #Velocity in [m/s]\n",
+ "Cp=1.08 #kJ/(kg.K)\n",
+ "L=1.5 #Length of plate in [m]\n",
+ "W=0.3 #Width in [m]\n",
+ "A=L*W #Area in [sq m]\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "#At constant temperature: mu1/mu2=P2/P1\n",
+ "P2=8 #[kPa]\n",
+ "mu2=mu1*P1/P2 #Kinematic viscosity at P2 in [sq m/s]\n",
+ "Nre_l=u_inf*L/mu2 #Reynold's no.\n",
+ "#Since this is less than 3*10**5\n",
+ "Nnu=0.664*math.sqrt(Nre_l)*(Npr**(1.0/3.0))\n",
+ "h=Nnu*k/L # Heat transfer coeffficient in [W/sq m.K]\n",
+ "h=round(h,1)\n",
+ "\n",
+ "T_inf=523 #[K]\n",
+ "Tw=353 #[K]\n",
+ "Q=2*h*A*(T_inf-Tw) #Heat removed from both sides in [W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of heat removed from both sides of plate is\",Q,\"W\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of heat removed from both sides of plate is 382.5 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 158
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.7,Page no:3.49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Local heat transfer coefficient\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=0.998 #kg/cubic m\n",
+ "v=20.76*10**-6 #[sq m/s]\n",
+ "Cp=1.009 #[kJ/kg.K]\n",
+ "k=0.03 #[W/m.K]\n",
+ "u_inf=3 #[m/s]\n",
+ "x=0.4 #[m]\n",
+ "w=1.5 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre_x=u_inf*x/v #Reynolds no at x=0.4 m\n",
+ "#Since this is less than 3*10**5.The flow is laminar upto x=0.4 m\n",
+ "mu=rho*v #[kg/(m.s)]\n",
+ "import math\n",
+ "Cp=1.009 #[kJ/kg.K]\n",
+ "Cp=Cp*1000 #[J/kg.K]\n",
+ "k=0.03 #W/(m.K)\n",
+ "Npr=Cp*mu/k\n",
+ "Nnu_x=0.332*(math.sqrt(Nre_x))*(Npr**(1.0/3.0))\n",
+ "hx=Nnu_x*k/x #[W/(m.K)]\n",
+ "#Average value is twice this value\n",
+ "h=round(2*hx,1) #[W/(m.K)]\n",
+ "A=x*w #Area in [sq m]\n",
+ "Tw=407 #[k]\n",
+ "T_inf=293 #[K]\n",
+ "Q=h*A*(Tw-T_inf) #[W]\n",
+ "#From both sides of the plate:\n",
+ "Q=2*Q #[W]\n",
+ "#Result\n",
+ "print\"The heat transferred from both sides of the plate is\",round(Q),\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat transferred from both sides of the plate is 1450.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 160
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.8,Page no:3.50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Width of plate\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=0.998 #[kg/cubic m]\n",
+ "v=20.76*10**-6 #[sq m/s]\n",
+ "k=0.03 #[W/m.K]\n",
+ "Npr=0.697\n",
+ "x=0.4 #[m] from leading edge of the plate\n",
+ "u_inf=3 #[m/s]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre_x=u_inf*x/v #Reynold numebr at x=0.40 m\n",
+ "#Since this is less than 3*10**5 \n",
+ "#therefore flow is laminar and \n",
+ "Nnu_x=0.332*math.sqrt(Nre_x)*(Npr**(1.0/3.0)) \n",
+ "hx=Nnu_x*k/x #[W/sq m.K]\n",
+ "#Average heat tarnsfer coefficient is twice this value\n",
+ "h=2*hx #[W/sq m.K]\n",
+ "#Given:\n",
+ "Q=1450 #[W]\n",
+ "Tw=407 #[K]\n",
+ "T_inf=293 #[K]\n",
+ "L=0.4 #[m]\n",
+ "#Q=h*w*L*(Tw-T_inf)\n",
+ "#L=Q/(h*w*(Tw-T_inf))\n",
+ "w=Q/(h*L*(Tw-T_inf)) #[m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of plate is\",round(w),\"m\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of plate is 3.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 161
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.9,Page no:3.51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat transferred in flat plate\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=17.36*10**-6 #Viscosity for air [sq m./s]\n",
+ "k=0.0275 #for air ..[W/(m.K)]\n",
+ "Cp=1.006 #[kJ/(kg.K)]\n",
+ "Npr=0.7 #for air\n",
+ "u_inf=2 #[m/s]\n",
+ "x=0.2 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre_x=u_inf*x/v #Reynolds number at x=0.2 m\n",
+ "#Since this is less than 3*10**5\n",
+ "import math\n",
+ "Nnu_x=0.332*math.sqrt(Nre_x)*(Npr**(1.0/3.0))\n",
+ "hx=Nnu_x*k/x #[W/(sq m.K]\n",
+ "#Average value of heat transfer coeff is twice this value\n",
+ "h=round(2*hx,1) #[W/sq m.K)]\n",
+ "w=1 #width in [m]\n",
+ "A=x*w #[sq m] Area of plate\n",
+ "Tw=333 #[K]\n",
+ "T_inf=300 #[K]\n",
+ "Q=h*A*(Tw-T_inf) #Heat flow in [W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat flow is :\",Q,\"W\"\n",
+ "#From both sides of plate:\n",
+ "Q=2*Q #[W]\n",
+ "print\"Heat flow from both sides of plate is\",Q,\"W\"\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat flow is : 81.18 W\n",
+ "Heat flow from both sides of plate is 162.36 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 163
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.10,Page no:3.51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Rate of heat transferred in turbulent flow\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=16.96*10**-6 #[sq m./s]\n",
+ "rho=1.128 #[kg/cubic m]\n",
+ "Npr=0.699 #Prandtl number\n",
+ "k=0.0276 #[W/m.K]\n",
+ "u_inf=15 #[m/s]\n",
+ "L=0.2 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre_l=L*u_inf/v #Reynold's number\n",
+ "import math\n",
+ "\n",
+ "#Since this is less than 3*10**5,the boundary layer is laminar over entire length\n",
+ "Nnu=0.664*math.sqrt(Nre_l)*(Npr**(1.0/3.0))\n",
+ "h=Nnu*k/L #[W/sq m.K]\n",
+ "A=L**2 #Area in [sq m]\n",
+ "Tw=293 #[K]\n",
+ "T_inf=333 #[K]\n",
+ "#Rate of heat transfer from BOTH sides is:\n",
+ "Q=2*h*A*(T_inf-Tw) #[W]\n",
+ "print\"Rate of heat transfer from both sides of plate is\",round(Q,1),\"W\\n\"\n",
+ "#ii-With turbulent boundary layer from the leading edge:\n",
+ "h=k*0.0366*(Nre_l**(0.8))*(Npr**(1.0/3.0))/L #[W/(sq m.K)]\n",
+ "#Heat transfer from both sides is :\n",
+ "Q=2*h*A*(T_inf-Tw) #[W]\n",
+ "print \"With turbulent boundary layer,\\nRate of heat transfer from both sides of the plate=\",round(Q,1)\n",
+ "print\"\\nThese calculations show that the that transfer rate is approximately doubled if boundary layer is turbulent from the leading edge \\n\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of heat transfer from both sides of plate is 109.4 W\n",
+ "\n",
+ "With turbulent boundary layer,\n",
+ "Rate of heat transfer from both sides of the plate= 226.4\n",
+ "\n",
+ "These calculations show that the that transfer rate is approximately doubled if boundary layer is turbulent from the leading edge \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.11,Page no:3.52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat transfer from plate in unit direction\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=1.906*10**-5 #[kg/(m.s)]\n",
+ "k=0.02723 #W/m.K\n",
+ "Cp=1.007 #[kJ/(kg.K)]\n",
+ "rho=1.129 #[kg/cubic m]\n",
+ "Npr=0.70\n",
+ "Mavg=29\n",
+ "u_inf=35 #[m/s]\n",
+ "L=0.75 #[m]\n",
+ "Tm=313 #[K]\n",
+ "P=101.325 #[kPa]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre_l=rho*u_inf*L/mu #Reynold's number >5*10**5\n",
+ "Nnu=0.0366*Nre_l**(0.8)*Npr**(1.0/3.0) \n",
+ "h=Nnu*k/L #[W/s m.K]\n",
+ "A=1*L #[sq m]\n",
+ "Tw=333 #[K]\n",
+ "T_inf=293 #[K]\n",
+ "Q=h*A*(Tw-T_inf) #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat transfer from the plate is\",round(Q,1),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transfer from the plate is 3179.2 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.12,Page no:3.53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat lost by sphere\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=18.23*10**-6 #sq m/s\n",
+ "k=0.02814 #[W/m.K]\n",
+ "D=0.012 #[m]\n",
+ "r=0.006 #[m]\n",
+ "u_inf=4 #[m/s]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre=D*u_inf/v #Reynold's number\n",
+ "Nnu=0.37*Nre**(0.6) \n",
+ "h=Nnu*(k/D) \n",
+ "A=4*math.pi*r**2 #Area of sphere in [sq m]\n",
+ "Tw=350 #[K]\n",
+ "T_inf=300 #[K]\n",
+ "Q=h*A*(Tw-T_inf) #Heat lost by sphere in [W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat lost by sphere is\",round(Q,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat lost by sphere is 2.21 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 170
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.13,Page no:3.53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat lost by sphere\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=15.69*10**-6 #[sq m./s]\n",
+ "k=0.02624 #[W/m.K]\n",
+ "Npr=0.708 #Prandtl number\n",
+ "mu=2.075*10**-5 #kg/m.s\n",
+ "u_inf=4 #[m/s]\n",
+ "mu_inf=1.8462*10**-5 #[m/s] velocity\n",
+ "Tw=350 #[K]\n",
+ "T_inf=300 #[K]\n",
+ "D=0.012 #[m]\n",
+ "r=D/2 #Radius in [m]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre=u_inf*D/v #Reynold's numbe\n",
+ "Nnu=2+(0.4*Nre**(1.0/2.0)+0.06*Nre**(2.0/3.0))*Npr**(0.4)*(mu_inf/mu)**(1.0/4.0)\n",
+ "h=Nnu*k/D #[W/sq m.K]\n",
+ "import math\n",
+ "A=4*math.pi*r**2 #Area in [sq m]\n",
+ "Q=h*A*(Tw-T_inf) \n",
+ "\n",
+ "#Result\n",
+ "print\"\\n Heat lost by the sphere is\",round(Q,3),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Heat lost by the sphere is 1.554 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 171
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.14,Page no:3.54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Percent power lost in bulb\n",
+ "#Variable declaration\n",
+ "v=2.08*10**-5 #[sq m/s]\n",
+ "k=0.03 #W/(m.K)\n",
+ "Npr=0.697 #Prandtl number\n",
+ "D=0.06 #[m]\n",
+ "u_inf=0.3 #[m/s]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre=D*u_inf/v #Reynolds number\n",
+ "#Average nusselt number is given by:\n",
+ "Nnu=0.37*(Nre**0.6) \n",
+ "h=Nnu*k/D #W/sq m.K\n",
+ "Tw=400 #[K]\n",
+ "T_inf=300 #[K]\n",
+ "D=0.06 #[m]\n",
+ "r=0.03 #[m]\n",
+ "import math\n",
+ "A=4*math.pi*r**2 #Area in [sq m]\n",
+ "Q=h*A*(Tw-T_inf) #[W]\n",
+ "per=Q*100/100 #Percent of heat lost by forced convection\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat transfer rate is\",round(Q,2),\"W,And percentage of power lost by convection is:\",round(per,2),\"%\" \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transfer rate is 12.1 W,And percentage of power lost by convection is: 12.1 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 172
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.15,Page no:3.55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "#Heat lost by cylinder \n",
+ "u_inf=50 #velocity in [m/s]\n",
+ "mu=2.14*10**-5 #[kg/(m.s)]\n",
+ "rho=0.966 #[kg/cubic m]\n",
+ "k=0.0312 #[W/(m.K)]\n",
+ "Npr=0.695 #Prandtl number\n",
+ "D=0.05 #Diameter in [m]\n",
+ "Nre=D*u_inf*rho/mu #Reynold's number\n",
+ "Nnu=0.0266*Nre**0.805*Npr**(1.0/3.0) \n",
+ "h=round(Nnu*k/D,1) #W/sq m.K\n",
+ "Tw=423 #[K]\n",
+ "T_inf=308 #[K]\n",
+ "import math\n",
+ "#Heat loss per unit length is :\n",
+ "Q_by_l=h*math.pi*D*(Tw-T_inf) #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat lost per unit length of cylinder is\",round(Q_by_l),\"W\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat lost per unit length of cylinder is 3102.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.16,Page no:3.55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat transfer in tube\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=20.92*10**-6 #sq m/s\n",
+ "k=3.0*10.0**-2 #W/(m.K)\n",
+ "Npr=0.7\n",
+ "u_inf=25.0 #[m/s]\n",
+ "d=50.0 #[mm]\n",
+ "d=d/1000 #[m]\n",
+ "Nre=u_inf*d/v #Reynold's number\n",
+ "Tw=397.0 #[K]\n",
+ "T_inf=303.0 #[K]\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "#Case 1: Circular tube\n",
+ "Nnu=0.0266*Nre**(0.805)*Npr**(1.0/3.0) \n",
+ "h=Nnu*k/d #[W/sq m.K]\n",
+ "A=math.pi*d #Area in [sq m]\n",
+ "Q=h*A*(Tw-T_inf) #[W]\n",
+ "Q_by_l1=h*math.pi*d*(Tw-T_inf) #[W/m]\n",
+ "\n",
+ "#Case 2:Square tube\n",
+ "A=50.0*50.0 #Area in [sq mm]\n",
+ "P=2.0*(50.0+50.0) #Perimeter [mm]\n",
+ "l=4.0*A/P #[mm]\n",
+ "l=l/1000 #[m]\n",
+ "Nnu=0.102*(Nre**0.675)*(Npr**(1.0/3.0))\n",
+ "h=Nnu*k/d #W/(sq m.K)\n",
+ "A=4*l*l #[sq m]\n",
+ "\n",
+ "Q=h*A*(Tw-T_inf)\n",
+ "Q_by_l2=Q/l #[W/m]\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"Rate of heat flow from the circular pipe is\",round(Q_by_l1,1),\"W/m\" \n",
+ "print\"Rate of heat flow from the square pipe=\",round(Q_by_l2,1),\"W/m\"\n",
+ "print\"Hence rate of heat flow from square pipe is more than that from circular pipe\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of heat flow from the circular pipe is 1464.2 W/m\n",
+ "Rate of heat flow from the square pipe= 1711.2 W/m\n",
+ "Hence rate of heat flow from square pipe is more than that from circular pipe\n"
+ ]
+ }
+ ],
+ "prompt_number": 186
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Example no:3.17,Page no:3.63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Heat transfer coefficient\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=0.8 #Viscosity of flowing fluid [N.s/sq m]\n",
+ "rho=1.1 #Density of flowinf fluid [g/cubic cm]\n",
+ "rho=rho*1000 #Density in [kg/cubic m]\n",
+ "Cp=1.26 #Specific heat [kJ/kg.K]\n",
+ "Cp=Cp*10**3 # in[J/(kg.K)]\n",
+ "k=0.384 #[W/(m.K)]\n",
+ "mu_w=1.0 #Viscosity at wall temperature [N.s/sq m]\n",
+ "L=5.0 #[m]\n",
+ "vfr=300.0 #Volumetric flow rate in [cubic cm/s]\n",
+ "vfr=vfr*10.0**-6 #[cubic m/s]\n",
+ "mfr=vfr*rho #Mass flow rate of flowinf fluid [kg/s]\n",
+ "Di=20.0 #Inside diameter in[mm]\n",
+ "Di=Di/1000 #[m]\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Area=(math.pi/4)*Di**2 #Area of cross-section [sq m]\n",
+ "u=vfr/Area #Veloctiy in [m/s]\n",
+ "Nre=Di*u*rho/mu #Reynold's number\n",
+ "#As reynold's number is less than 2100,he flow is laminar\n",
+ "Npr=Cp*mu/k #Prandtl number\n",
+ "Nnu=1.86*(Nre*Npr*Di/L)**(1.0/3.0)*(mu/mu_w)**(0.14)\n",
+ "hi=Nnu*k/Di #inside heat transfer coefficient [W/sq m.K]\n",
+ "\n",
+ "#Result\n",
+ "print\"Inside heat transfer coefficient is\",round(hi),\"W/(sq m.K)\"\n",
+ "#Note:\n",
+ "print\"NOTE:The answer given in book..ie 1225 is wrong.please redo the calculation of last line manually to check\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inside heat transfer coefficient is 225.0 W/(sq m.K)\n",
+ "NOTE:The answer given in book..ie 1225 is wrong.please redo the calculation of last line manually to check\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.18,Page no:3.64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat transfer coefficient in heated tube\n",
+ "\n",
+ "#Variable declaration\n",
+ "m=5500.0 #Mass flow rate in [kg/h]\n",
+ "m=m/3600.0 #[kg/s]\n",
+ "rho=1.07 #Density of fluid in [g/cm**3]\n",
+ "rho=rho*1000 #[kg/m**3]\n",
+ "vfr=m/rho #Volumetric flow rate in [m**3/s]\n",
+ "Di=40.0 #Diameter of tube [mm]\n",
+ "Di=Di/1000 #[m]\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "A=(math.pi/4)*Di**2 #Area of cross-section in [sq m]\n",
+ "u=vfr/A #Velocity of flowing fluid [m/s]\n",
+ "rho=1070.0 #Density in [kg/m**3]\n",
+ "mu=0.004 #Viscosity in [kg/m.s]\n",
+ "Nre=Di*u*rho/mu\n",
+ "Nre=12198.0 #Approx\n",
+ "#Since this reynold's number is less than 10000,the flow is turbulent\n",
+ "Cp=2.72 #Specific heat in [kJ/kg.K]\n",
+ "Cp=Cp*10**3 #Specific heat in [J/kg.K]\n",
+ "k=0.256 #thermal conductivity in [W/m.K]\n",
+ "Npr=Cp*mu/k #Prandtl number\n",
+ "Nnu=0.023*(Nre**0.8)*(Npr**0.4) #Nusselt number\n",
+ "hi=k*Nnu/Di #Inside heat transfer coefficient in [W/m**2.K]\n",
+ "\n",
+ "#Result\n",
+ "print\"Inside heat transfer coefficient is \",round(hi,1),\"W/sq m.K\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inside heat transfer coefficient is 1225.4 W/sq m.K\n"
+ ]
+ }
+ ],
+ "prompt_number": 189
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.19,Page no:3.66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#h of water flowing in tube\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=984.1 #Density of water [kg/m**3]\n",
+ "Cp=4187.0 #Specific heat in [J/kg.K]\n",
+ "mu=485.0*10**-6 #Viscosity at 331 K[Pa.s]\n",
+ "k=0.657 #[W/(m.K)]\n",
+ "mu_w=920.0*10**-6 #Viscosity at 297 K [Pa.s]\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "D=16.0 #Diameter in [mm]\n",
+ "D=D/1000 #Diameter in [m]\n",
+ "u=3.0 #Velocity in [m/s]\n",
+ "rho=984.1 #[kg/m**3]\n",
+ "Nre=D*u*rho/mu #Reynolds number\n",
+ "Nre=round(Nre)\n",
+ "Npr=Cp*mu/k #Prandtl number\n",
+ "\n",
+ "#Dittus-Boelter equation (i)\n",
+ "Nnu=0.023*(Nre**0.8)*(Npr**0.3) #nusselt number\n",
+ "h=k*Nnu/D #Heat transfer coefficient [W/m**2.K]\n",
+ "\n",
+ "#Result\n",
+ "print\"ANSWER-(i) \\nBy Dittus-Boelter equation we get h=\",round(h,1),\"W/sq m.K\"\n",
+ "#sieder-tate equation (ii)\n",
+ "Nnu=0.023*(Nre**0.8)*(Npr**(1.0/3.0))*((mu/mu_w)**0.14) #Nusselt number\n",
+ "h=k*Nnu/D #Heat transfer coefficient in [W/sq m.K]\n",
+ "print\"Answer-(ii)\\nBy Sieder-Tate equation we get h=\",round(h,2),\"W/sq m.K\"\n",
+ "print\"\\nNOTE:Calculation mistake in book in part 2 ie sieder tate eqn\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ANSWER-(i) \n",
+ "By Dittus-Boelter equation we get h= 12972.6 W/sq m.K\n",
+ "Answer-(ii)\n",
+ "By Sieder-Tate equation we get h= 12315.04 W/sq m.K\n",
+ "\n",
+ "NOTE:Calculation mistake in book in part 2 ie sieder tate eqn\n"
+ ]
+ }
+ ],
+ "prompt_number": 191
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.20,Page no:3.67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Overall heat transfer coefficient\n",
+ "\n",
+ "#Variable declaration\n",
+ "m_dot=2250 #Mass flow arte in [kg/h]\n",
+ "Cp=3.35 #Specific heat in [kJ/(kg.K)]\n",
+ "dT=316-288.5 #Temperature drop for oil [K]\n",
+ "Q=Cp*m_dot*dT #Rate of heat transfer in [kJ/h]\n",
+ "Q=round(Q*1000/3600) #[J/s] or[W]\n",
+ "Di=0.04 #Inside diameter [m]\n",
+ "Do=0.048 #Outside diamter in [m]\n",
+ "hi=4070 #for steam [W/sq m.K]\n",
+ "ho=18.26 #For oil [W/sq m.K]\n",
+ "Rdo=0.123 #[sq m.K/W]\n",
+ "Rdi=0.215 #[sq m.K/W]\n",
+ "\n",
+ "#Calculation\n",
+ "Uo=1.0/(1.0/ho+Do/(hi*Di)+Rdo+Rdi*(Do/Di)) #[W/m**2.K]\n",
+ "Uo=2.3\n",
+ "import math\n",
+ "dT1=373-288.5 #[K]\n",
+ "dT2=373-316 #[K]\n",
+ "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n",
+ "Ao=Q/(Uo*dTm) #Heat transfer area in [m**2]\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat transfer area is:\",round(Ao,1),\"m**2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transfer area is: 358.4 m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 192
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.21,Page no:3.68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Number of tubes in exchanger\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k_tube=111.65 #[W/m.K]\n",
+ "W=4500.0 #[kg/h]\n",
+ "rho=995.7 #[kg/sq m]\n",
+ "Cp=4.174 #[kJ/(kg.K)]\n",
+ "k=0.617 #[W/(m.K)]\n",
+ "v=0.659*10**-6 #Kinematic viscosity [sq m/s]\n",
+ "m_dot=1720.0 #kg/h\n",
+ "T1=293.0 #Initial temperature in [K]\n",
+ "T2=318.0 #Final temperature in [K]\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "dT=T2-T1 #[K]\n",
+ "Q=m_dot*Cp*dT #Heat transfer rate in [kJ/h]\n",
+ "Q=Q*1000.0/3600.0 #[J/s] or [W]\n",
+ "Di=0.0225 #[m]\n",
+ "u=1.2 #[m/s]\n",
+ "#Nre=Di*u*rho/mu or\n",
+ "Nre=Di*u/v #Reynolds number\n",
+ "#As Nre is greater than 10000,Dittus Boelter equation is applicable\n",
+ "Cp=Cp*10**3 #J/(kg.K)\n",
+ "mu=v*rho #[kg/(m.s)]\n",
+ "Npr=Cp*mu/k #Prandtl number\n",
+ "#Dittus-Boelter equation for heating is \n",
+ "Nnu=0.023*(Nre**0.8)*(Npr**0.4)\n",
+ "hi=k*Nnu/Di #Heat transfer coefficient [W/(sq m.K)]\n",
+ "Do=0.025 #[m]\n",
+ "Dw=(Do-Di)/math.log(Do/Di) #math.log mean diameter in [m]\n",
+ "ho=4650.0 #[W/sq m.K]\n",
+ "k=111.65 #[W/m.K]\n",
+ "xw=(Do-Di)/2 #[m]\n",
+ "Uo=1.0/(1.0/ho+Do/(hi*Di)+xw*Do/(k*Dw)) #Overall heat transfer coefficient in W/(m**2.K)\n",
+ "T_steam=373.0 #Temperature of condensing steam in [K]\n",
+ "dT1=T_steam-T1+10 #[K]\n",
+ "dT2=T_steam-T2+10 #[K]\n",
+ "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n",
+ "Ao=Q/(Uo*dTm)#Area in [m**2]\n",
+ "L=4.0 #length of tube [m]\n",
+ "n=Ao/(math.pi*Do*L) #number of tubes\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"No. of tubes required=\",round(n) \n",
+ "print\"\\nNOTE: there is an error in book in calculation of dT1 and dT2,\\n373-293 is written as 90,instead of 80...similarly in dT2,\\nSo,in compliance with the book,10 is added to both of them\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of tubes required= 1.0\n",
+ "\n",
+ "NOTE: there is an error in book in calculation of dT1 and dT2,\n",
+ "373-293 is written as 90,instead of 80...similarly in dT2,\n",
+ "So,in compliance with the book,10 is added to both of them\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.22,Page no:3.71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Convective film coefficient\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "m_dot=25000 #massflow rate of water [kg/h]\n",
+ "rho=992.2 #[kg/m**3]\n",
+ "k=0.634 #[W/m.K]\n",
+ "vfr=m_dot/rho #[m**3/h]\n",
+ "Npr=4.31 #Prandtl numberl\n",
+ "Di=50 #[mm]\n",
+ "Di=0.05 #[m]\n",
+ "dT=10 #[K] as the wall is at a temperature of 10 K above the bulk temperature\n",
+ "\n",
+ "#Calculation\n",
+ "u=round((vfr/3600)/(math.pi*(Di/2)**2),2) #Velocity of water in [m/s]\n",
+ "#Nre=Di*u*rho/mu=Di*u/v as v=mu/rho\n",
+ "v=0.659*10**-6 #[m**2/s]\n",
+ "Nre=Di*u/v #Reynolds number\n",
+ "#As it is less than 10000,the flow is in the turbulent region for heat transfer and Dittus Boelter eqn is used\n",
+ "Nnu=0.023*(Nre**0.8)*(Npr**0.4) #Nusselt number\n",
+ "hi=Nnu*k/Di #Heat transfer coefficiet in [W/sq m.K]\n",
+ "q_by_l=hi*math.pi*Di*dT #Heat transfer per unit length[kW/m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Average value of convective film coefficient is hi=\",round(hi),\" W/sq m.K\"\n",
+ "print\"Heat transferred per unit length is Q/L=\",round(q_by_l/1000,1),\"kW/m\"\n",
+ " \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average value of convective film coefficient is hi= 11584.0 W/sq m.K\n",
+ "Heat transferred per unit length is Q/L= 18.2 kW/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 196
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.23,Page no:3.72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Length of tube\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "vfr=1200.0 #Water flow rate in [l/h]\n",
+ "rho=0.98 #Density of water in g/[cubic cm]\n",
+ "m_dot=vfr*rho #Mass flow rate of water [kg/h]\n",
+ "m_dot2=m_dot/3600.0 #[kg/s]\n",
+ "Cp=4.187*10**3 #[J/kg.K]\n",
+ "Di=0.025 #Diameter in [m]\n",
+ "mu=0.0006 #[kg/(m.s)]\n",
+ "\n",
+ "#Calculation\n",
+ "Ai=math.pi*((Di/2)**2) #Area of cross-section in [m**2]\n",
+ "Nre=(Di/mu)*(m_dot2/Ai) #Reynolds number\n",
+ "k=0.63 #for metal wall in [W/(m.K)]\n",
+ "Npr=Cp*mu/k #Prandtl number\n",
+ "#Since Nre>10000\n",
+ "#therefore ,Dittus boelter eqn for heating is \n",
+ "Nnu=0.023*(Nre**(0.8))*(Npr**(0.4))\n",
+ "ho=5800.0 #Film heat coefficientW/(m**2.K)\n",
+ "hi=Nnu*k/Di #Heat transfer coeffcient in [W/(sq m.K)]\n",
+ "Do=0.028 #[m]\n",
+ "Di=0.025 #[m]\n",
+ "xw=(Do-Di)/2 #[m]\n",
+ "Dw=(Do-Di)/math.log(Do/Di) #[m]\n",
+ "k=50.0 #for metal wall in [W/(m.K)]\n",
+ "Uo=1.0/(1.0/ho+Do/(hi*Di)+xw*Do/(k*Dw)) #in [W/sq m.K]\n",
+ "dT=343.0-303.0 #[K]\n",
+ "dT1=393.0-303.0 #[K]\n",
+ "dT2=393.0-343.0 #[K]\n",
+ "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n",
+ "Cp=Cp/1000.0 #[in [kJ/kg.K]]\n",
+ "Q=m_dot*Cp*dT #Rate of heat transfer in [kJ/h]\n",
+ "Q=Q*1000.0/3600.0 #[J/s] or [W]\n",
+ "Ao=Q/(Uo*dTm) #Heat transfer area in [sq m]\n",
+ "#Also,..Ao=math.pi*Do*L ..implies that\n",
+ "L=Ao/(math.pi*Do) #[m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Length of tube required is\",round(L),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of tube required is 5.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.24,Page no:3.73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Cooling coil\n",
+ "#1.For initial conditions:\n",
+ "import math\n",
+ "from scipy.optimize import fsolve\n",
+ "\n",
+ "#Variable declaration\n",
+ "T=360 #[K]\n",
+ "T1=280 #[K]\n",
+ "T2=320 #[K]\n",
+ "dT1=T-T1 #[K]\n",
+ "dT2=T-T2 #[K]\n",
+ "\n",
+ "#Calculation\n",
+ "#Q1=m1_dot*Cp1*(T2-T1)\n",
+ "Cp1=4.187 #Heat capacity \n",
+ "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n",
+ "m1_by_UA=dTlm/(Cp1*(T2-T1)) \n",
+ "#For final conditions :\n",
+ "#m2_dot=m1_dot\n",
+ "#U2=U1\n",
+ "#A2=5*A1\n",
+ "def f(t):\n",
+ " x=m1_by_UA*Cp1*(t-T1)-5*((dT1-(T-t))/math.log(dT1/(T-t)))\n",
+ " return(x)\n",
+ "T=fsolve(f,350.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"Outlet temperature of water is\",T[0],\"K\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Outlet temperature of water is 357.5 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 197
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.25,Page no:3.74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Outlet temperature of water\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "#Variable declaration\n",
+ "mo_dot=60.0 #Mass flow rate of oilin [g/s]\n",
+ "mo_dot=6.0*10**-2 #[kg/s]\n",
+ "Cpo=2.0 #Specific heat of oil in [kJ/(kg.K)]\n",
+ "T1=420.0 #[K]\n",
+ "T2=320.0 #[K]\n",
+ "\n",
+ "#Calculation\n",
+ "Q=mo_dot*Cpo*(T1-T2) #Rate of heat flow in [kJ/s]\n",
+ "mw_dot=mo_dot #Mass flow rate of water #kg/s\n",
+ "t1=290.0 #[K]\n",
+ "Cpw=4.18 #[kJ/(kg.K)]\n",
+ "#For finding outlet temperature of water\n",
+ "t2=t1+Q/(mw_dot*Cpw) #[K]\n",
+ "dT1=T1-t2 #[K]\n",
+ "dT2=T2-t1 #[K]\n",
+ "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n",
+ "ho=1.6 #Oil side heat transfer coefficient in [kW/(sq m.K)]\n",
+ "hi=3.6 #Water side heat transfer coeff in [kW/(sq m.K)]\n",
+ "#Overall heat transfer coefficient is:\n",
+ "U=1.0/(1.0/ho+1.0/hi) #[kW/(m**2.K)]\n",
+ "A=Q/(U*dTm) #[sq m]\n",
+ "Do=25.0 #[mm]\n",
+ "Do=Do/1000 #[m]\n",
+ "L=A/(math.pi*Do) #Length of tube in [m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Outlet temperature of water is:\",round(t2),\"K\" \n",
+ "print\"Area of heat transfer required is\",round(A,2),\"m^2\"\n",
+ "print\"Length of tube required is\",round(L,2),\"m\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Outlet temperature of water is: 338.0 K\n",
+ "Area of heat transfer required is 0.21 m^2\n",
+ "Length of tube required is 2.66 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 199
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.26,Page no:3.76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Inside heat transfer coefficient\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=0.14 # for oil[W/m.K]\n",
+ "Cp=2.1 # for oil [kJ/kg.K]\n",
+ "Cp=Cp*10**3 #J/kg.K\n",
+ "mu=154 #[mN.s/sq m]\n",
+ "mu_w=87 #(mn.s/sq m)\n",
+ "L=1.5 #[m]\n",
+ "m_dot=0.5 #Mass flow rate of oil[kg/s]\n",
+ "Di=0.019 #Diameter of tube [m]\n",
+ "mean_T=319 #Mean temperature of oil [K]\n",
+ "\n",
+ "#Calculation\n",
+ "mu=mu*10**-3 #[N.s/sq m] or [kg/(m.s)]\n",
+ "A=math.pi*(Di/2)**2 #[sq m]\n",
+ "G=m_dot/A #Mass velocity in [kg/sq m.s]\n",
+ "Nre=Di*G/mu #Reynolds number\n",
+ "#As Nre<2100,the flow is laminar\n",
+ "mu_w=mu_w*10**-3 #[N.s/sq m] or kg/(m.s)\n",
+ "#The sieder tate equation is \n",
+ "hi=(k*(2.0*((m_dot*Cp)/(k*L))**(1.0/3.0)*(mu/mu_w)**(0.14)))/Di #Heat transfer coeff in [W/sq m.K]\n",
+ "\n",
+ "#Result\n",
+ "print\"The inside heat transfer coefficient is\",round(hi,2),\"W/(m**2.K) \" \n",
+ "print\"NOTE:Calculation mistake in last line.ie in the calculation of hi in book,please perform the calculation manually to check the answer\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The inside heat transfer coefficient is 272.97 W/(m**2.K) \n",
+ "NOTE:Calculation mistake in last line.ie in the calculation of hi in book,please perform the calculation manually to check the answer\n"
+ ]
+ }
+ ],
+ "prompt_number": 200
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.27,Page no:3.77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Film heat transfer coefficient\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "m_dot=0.217 #Water flow rate in [kg/s]\n",
+ "Do=19.0 #Outside diameter in [mm]\n",
+ "rho=1000.0 #Density\n",
+ "t=1.6 #Wall thickness in [mm]\n",
+ "Di=Do-2*t #i.d of tube in [mm]\n",
+ "Di=Di/1000.0 #[m]\n",
+ "Do=Do/1000.0 #[m]\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "Ai=math.pi*(Di/2)**2 #Cross-sectional area in sq m\n",
+ "u=m_dot/(rho*Ai) #Water velocity through tube [m/s]\n",
+ "u=1.12 #approx in book\n",
+ "Di=0.0157 #apprx in book\n",
+ "T1=301.0 #Inlet temperature of water in [K]\n",
+ "T2=315.0 #Outlet temperature of water in [K]\n",
+ "T=(T1+T2)/2 #[K]\n",
+ "hi=(1063.0*(1+0.00293*T)*(u**0.8))/(Di**0.20) #Inside heat transfer coefficient W/(sq m.K)\n",
+ "hi=5084.0 #Approximation\n",
+ "hio=hi*(Di/Do) #Inside heat transfer coeff based on outside diameter in W/(sq m.K)\n",
+ "\n",
+ "#Result\n",
+ "print\"Based on outside temperature,Inside heat transfer coefficient is\",round(hio),\"W/(m**2.K) or \",round(hio/1000,1),\"kW/(m**2.K)\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Based on outside temperature,Inside heat transfer coefficient is 4201.0 W/(m**2.K) or 4.2 kW/(m**2.K)\n"
+ ]
+ }
+ ],
+ "prompt_number": 202
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.28,Page no:3.77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Area of exchanger\n",
+ "\n",
+ "#Variable declaration\n",
+ "mair_dot=0.90 #[kg/s]\n",
+ "T1=283.0 #[K]\n",
+ "T2=366.0 #[K]\n",
+ "dT=(T1+T2)/2 #[K]\n",
+ "Di=12.0 #[mm]\n",
+ "Di=Di/1000.0 #[m]\n",
+ "G=19.9 #[kg/(sq m.s)]\n",
+ "mu=0.0198 #[mN.s/(sq m)]\n",
+ "mu=mu*10**-3 #[N.s/sq m] or [kg/(m.s)]\n",
+ "\n",
+ "#Calculation\n",
+ "Nre=Di*G/mu #Reynolds number\n",
+ "#It is greater than 10**4\n",
+ "k=0.029 #W/(m.K)\n",
+ "Cp=1.0 #[kJ/kg.K]\n",
+ "Cp1=Cp*10**3 #[J/kg.K]\n",
+ "Npr=Cp1*mu/k #Parndtl number\n",
+ "#Dittus-Boelter equation is\n",
+ "hi=0.023*(Nre**0.8)*(Npr**0.4)*k/Di #[W/sq m.K]\n",
+ "ho=232.0 #W/sq m.K\n",
+ "U=1.0/(1.0/hi+1.0/ho) #Overall heat transfer coefficient [W/m**2.K]\n",
+ "Q=mair_dot*Cp*(T2-T1) #kJ/s\n",
+ "Q=Q*10**3 #[J/s] or [W]\n",
+ "T=700.0 #[K]\n",
+ "dT1=T-T2 #[K]\n",
+ "dT2=T2-T1 #[K]\n",
+ "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n",
+ "#Q=U*A*dTm\n",
+ "A=Q/(U*dTm) #Area in sq m\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat transfer area of equipment is\",round(A,2),\"m^2\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transfer area of equipment is 6.5 m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 203
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.29,Page no:3.82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Natural and forced convection\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=18.41*10**-6 #[sq m./s]\n",
+ "k=28.15*10**-3 #[W/m.K]\n",
+ "Npr=0.7 #Prandtl number\n",
+ "Beta=3.077*10**-3 #K**-1\n",
+ "g=9.81 #m/s**2\n",
+ "Tw=350 #[K]\n",
+ "T_inf=300 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "L=0.3 #[m]\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "#1.Free Convection\n",
+ "Ngr=(g*Beta*dT*L**3)/(v**2) #Grashof number\n",
+ "Npr=0.7 #Prandtl number\n",
+ "Nnu=0.59*(Ngr*Npr)**(1.0/4.0) #Nusselt number\n",
+ "h=Nnu*k/L #Average heat transfer coefficient [W/sq m K]\n",
+ "\n",
+ "#2.Forced Convestion\n",
+ "u_inf=4 #[m/s]\n",
+ "Nre_l=u_inf*L/v\n",
+ "Nnu=0.664*(Nre_l**(1.0/2.0))*(Npr**(1.0/3.0)) #Nusselt number\n",
+ "h1=Nnu*k/L #[W/sq m.K]\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"In free convection,heat transfer coeff,h=\",round(h,1),\"W/(m^2.K)\"\n",
+ "print\"In forced convection,heat transfer coeff,h=\",round(h1,2),\"W/(m^2.K)\"\n",
+ "print\"From above it is clear that heat transfer coefficient in forced convection is much larger than that in free convection\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "In free convection,heat transfer coeff,h= 5.3 W/(m^2.K)\n",
+ "In forced convection,heat transfer coeff,h= 14.12 W/(m^2.K)\n",
+ "From above it is clear that heat transfer coefficient in forced convection is much larger than that in free convection\n"
+ ]
+ }
+ ],
+ "prompt_number": 206
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.30,Page no:3.83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Natural convection\n",
+ "\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=0.02685 #W/(m.K)\n",
+ "v=16.5*10**-6 #kg/(m.s)\n",
+ "Npr=0.7 #Prandtl number\n",
+ "Beta=3.25*10**-3 #K**-1\n",
+ "g=9.81 #m/(s**2)\n",
+ "Tw=333 #[k]\n",
+ "T_inf=283 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "L=4 #Length/height of plate [m]\n",
+ "\n",
+ "#Calculation\n",
+ "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashoff number\n",
+ "#Let const=Ngr*Npr\n",
+ "const=Ngr*Npr\n",
+ "#Sice it is >10**9\n",
+ "Nnu=0.10*(const**(1.0/3.0)) #Nusselt number\n",
+ "h=round(Nnu*k/L,1) #W/(sq m.K)\n",
+ "W=7 #width in [m]\n",
+ "A=L*W #Area of heat transfer in [sq m]\n",
+ "Q=h*A*dT #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat transferred is\",Q,\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transferred is 6020.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 208
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.31,Page no:3.84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Free convection in vertical pipe\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=18.97*10**-6 #m**2/s\n",
+ "k=28.96*10**-3 #W/(m.K)\n",
+ "Npr=0.696\n",
+ "D=100.0 #Outer diameter [mm]\n",
+ "D=D/1000 #[m]\n",
+ "Tf=333.0 #Film temperature in [K]\n",
+ "Tw=373.0 #[K]\n",
+ "T_inf=293.0 #[K]\n",
+ "\n",
+ "#Calculation\n",
+ "dT=Tw-T_inf #[K]\n",
+ "Beta=1.0/Tf #[K**-1]\n",
+ "g=9.81 #[m/s**2]\n",
+ "L=3.0 #Length of pipe [m]\n",
+ "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashof number\n",
+ "Nra=Ngr*Npr\n",
+ "Nnu=0.10*(Ngr*Npr)**(1.0/3.0) #nusselt number for vertical cylinder\n",
+ "h=Nnu*k/L #W/(sq m.K)\n",
+ "Q_by_l=h*math.pi*D*dT #Heat loss per metre length [W/m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Hence,Heat loss per metre length is\",round(Q_by_l,2),\"W/m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hence,Heat loss per metre length is 120.68 W/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 210
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.32,Page no:3.84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat loss per unit length\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=0.630 #W/(m.K\n",
+ "Beta=3.04*10**-4 #K**-1\n",
+ "rho=1000.0 #kg/m**3\n",
+ "mu=8.0*10**-4 #[kg/(m.s)]\n",
+ "Cp=4.187 #kJ/(kg.K)\n",
+ "g=9.81 #[m/(s**2)]\n",
+ "Tw=313.0 #[K]\n",
+ "T_inf=298.0 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "D=20.0 #[mm]\n",
+ "D=D/1000 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "Ngr=9.81*(rho**2)*Beta*dT*(D**3)/(mu**2) #Grashoff number\n",
+ "Cp1=Cp*1000.0 #[J/kg.K]\n",
+ "Npr=Cp1*mu/k #Prandtl number\n",
+ "#Average nusselt number is\n",
+ "Nnu=0.53*(Ngr*Npr)**(1.0/4.0)\n",
+ "h=Nnu*k/D #[W/ sqm.K]\n",
+ "Q_by_l=h*math.pi*D*dT #Heat loss per unit length [W/m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat loss per unit length of the heater is\",round(Q_by_l,1),\"W/m(APPROX)\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat loss per unit length of the heater is 653.4 W/m(APPROX)\n"
+ ]
+ }
+ ],
+ "prompt_number": 213
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.33,Page no:3.85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Free convection in pipe\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=0.03406 #[W/(m/K)]\n",
+ "Beta=2.47*10**-3 #K**-1\n",
+ "Npr=0.687 #Prandtl number\n",
+ "v=26.54*10**-6 #m**2/s\n",
+ "g=9.81 #[m/s**2]\n",
+ "Tw=523.0 #[K]\n",
+ "T_inf=288.0 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "D=0.3048 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "Ngr=(g*Beta*dT*(D**3))/(v**2) #Grashof number\n",
+ "Nra=Ngr*Npr \n",
+ "#For Nra less than 10**9,we have for horizontal cylinder\n",
+ "Nnu=0.53*(Nra**(1.0/4.0)) #Nusselt number\n",
+ "h=Nnu*k/D #[W/sq m.K]\n",
+ "Q_by_l=h*math.pi*D*dT #W/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat loss of heat transfer per meter lengh is\",round(Q_by_l,1),\"W/m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat loss of heat transfer per meter lengh is 1492.4 W/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 214
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.34,Page no:3.86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Free convection in plate\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=960.63 #Density in [kg/m**3]\n",
+ "Cp=4.216*10**3 #Specific heat in [J/(kg.K)]\n",
+ "D=16.0 #Diameter in [cm]\n",
+ "D=D/100 #[m]\n",
+ "k=0.68 #Thermal conductivity in [W/m.K]\n",
+ "\n",
+ "#Calculation\n",
+ "A=(math.pi*(D/2)**2)\n",
+ "L=A/(math.pi*D) #Length=A/P in [m]\n",
+ "Beta=0.75*10**-3 #[K**-1]\n",
+ "alpha=1.68*10**-7 #[m**2/s]\n",
+ "g=9.81 #[m/s**2]\n",
+ "Tw=403.0 #[K]\n",
+ "T_inf=343.0 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "v=0.294*10**-6 #[m**2/s]\n",
+ "Nra=(g*Beta*(L**3)*dT)/(v*alpha) \n",
+ "#1.For Top surface\n",
+ "Nnu=0.15*(Nra)**(1.0/3.0) #Nusselt number\n",
+ "ht=Nnu*k/L #Heat transfer coeff for top surface in W/(m**2.K)\n",
+ "ht=round(ht)\n",
+ "#2.For bottom surface\n",
+ "Nnu=0.27*Nra**(1.0/4.0) #Nusselt number\n",
+ "hb=Nnu*k/L #[W/sq m.K]\n",
+ "hb=round(hb)\n",
+ "Q=(ht+hb)*A*dT #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"The rate of heat input is\",round(Q,1),\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate of heat input is 3410.4 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 215
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.35,Page no:3.87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat transfer from disc\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=2.0*10**-5 #[m**2/s]\n",
+ "Npr=0.7 #Prandtl number\n",
+ "k=0.03 #[W/m.K]\n",
+ "D=0.25 #Diameter in [m]\n",
+ "L=0.90*D #Characteristic length,let [m]\n",
+ "T1=298.0 #[K]\n",
+ "T2=403.0 #[K]\n",
+ "dT=T2-T1 #[K]\n",
+ "\n",
+ "#Calculation\n",
+ "Tf=(T1+T2)/2 #[K]\n",
+ "Beta=1.0/Tf #[K**-1]\n",
+ "A=math.pi*(D/2)**2 #Area in[sq m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "\n",
+ "#Case 1: Hot surface facing up\n",
+ "Ngr=g*Beta*dT*(L**3)/(v**2) #Grashoff number\n",
+ "Nnu=0.15*((Ngr*Npr)**(1.0/3.0)) #Nusselt number\n",
+ "print \"Nnu in the book is wrongly calculated as 66.80,\\nActually it is:58.22\"\n",
+ "h=Nnu*k/L #[W/sq m.K]\n",
+ "Q=h*A*dT #[W]\n",
+ "\n",
+ "#Case 2:For hot surface facing down\n",
+ "Nnu=0.27*(Ngr*Npr)**(1.0/4.0) #Grashof Number\n",
+ "h=Nnu*k/L #[W/sqm.K]\n",
+ "Q1=h*A*dT #[W]\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nHeat transferred when hot surface is facing up is\",round(Q,2),\"W\"\n",
+ "print\"NOTE:Taking into consideration the correct value of Nnu\\n\"\n",
+ "print\"Heat transferred when hot surface is facing down is\",round(Q1,2),\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nnu in the book is wrongly calculated as 66.80,\n",
+ "Actually it is:58.22\n",
+ "\n",
+ "Heat transferred when hot surface is facing up is 40.04 W\n",
+ "NOTE:Taking into consideration the correct value of Nnu\n",
+ "\n",
+ "Heat transferred when hot surface is facing down is 16.23 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 234
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.36,Page no:3.88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Rate of heat input to plate\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=960 #[kg/m**3]\n",
+ "Beta=0.75*10**-3 #[K**-1]\n",
+ "k=0.68 #[W/m.K]\n",
+ "alpha=1.68*10**-7 #[m**2/s]\n",
+ "v=2.94*10**-7 #[m**2/s]\n",
+ "Cp=4.216 #[kJ/kg.K]\n",
+ "Tw=403 #[K]\n",
+ "T_inf=343 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "g=9.81 #[m/s**2]\n",
+ "l=0.8 #[m]\n",
+ "W=0.08 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "A=l*W #Area in [m**2]\n",
+ "P=2*(0.8+0.08) #Perimeter in [m]\n",
+ "L=A/P #Characteristic dimension/length,L in [m]\n",
+ "Nra=g*Beta*L**3*dT/(v*alpha) \n",
+ "#(i) for natural convection,heat transfer from top/upper surface heated \n",
+ "Nnu=0.15*(Nra**(1.0/3.0)) #Nusselt number\n",
+ "ht=Nnu*k/L #[W/m**2.K]\n",
+ "ht=2115.3 #Approximation in book,If done manually then answer diff\n",
+ "#(ii)For the bottom/lower surface of the heated plate\n",
+ "Nnu=0.27*(Nra**(1.0/4.0)) #Nusselt number\n",
+ "hb=Nnu*k/L #[W/(m**2.K)]\n",
+ "hb=round(hb)\n",
+ "#Rate of heat input is equal to rate of heat dissipation from the upper and lower surfaces of the plate\n",
+ "Q=(ht+hb)*A*(Tw-T_inf) #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of heat input is equal to heat dissipation =\",round(Q,1),\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of heat input is equal to heat dissipation = 10914.4 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 235
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.37,Page no:3.89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Two cases in disc\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=0.03 #W/(m.K)\n",
+ "Npr=0.697 #Prandtl number\n",
+ "v=2.076*10**-6 #m**2/s\n",
+ "Beta=0.002915 #K**-1\n",
+ "D=25.0 #[Diameter in cm]\n",
+ "D=D/100 #[m]\n",
+ "Tf=343.0 #Film temperature in [K]\n",
+ "\n",
+ "#Calculation\n",
+ "A=math.pi*(D/2)**2 #Area in [m**2]\n",
+ "P=math.pi*D #Perimeter [m]\n",
+ "T1=293.0 #[K]\n",
+ "T2=393.0 #[K]\n",
+ "g=9.81 #[m/s**2]\n",
+ "#Case (i) HOT SURFACE FACING UPWARD\n",
+ "L=A/P #Characteristic length in [m]\n",
+ "Beta=1.0/Tf #[K**-1]\n",
+ "dT=T2-T1 #[K]\n",
+ "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashoff number\n",
+ "Nra=Ngr*Npr \n",
+ "Nnu=0.15*(Nra**(1.0/3.0)) #Nusselt number\n",
+ "h=Nnu*k/L #[W/m**2.K]\n",
+ "Q1=h*A*dT #[W]\n",
+ "\n",
+ "#Case-(ii) HOT FACE FACING DOWNWARD\n",
+ "Nnu=0.27*(Nra**(1.0/4.0)) #Nusselt number\n",
+ "h=Nnu*k/L #W/(m**2.K)\n",
+ "Q2=h*A*dT #[W]\n",
+ "\n",
+ "#Case-(iii)-For disc vertical \n",
+ "L=0.25 #Characteristic length[m] \n",
+ "D=L #dia[m]\n",
+ "A=math.pi*((D/2)**2) #[sq m]\n",
+ "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashoff number\n",
+ "Npr=0.697\n",
+ "Nra=Ngr*Npr \n",
+ "Nnu=0.10*(Nra**(1.0/3.0)) #Nusselt number\n",
+ "h=Nnu*k/D #[W/(m**2.K)]\n",
+ "Q3=h*A*dT #[W]\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat transferred when disc is horizontal with hot surface facing upward is\",round(Q1,1),\"W\"\n",
+ "print\"Heat transferred when disc is horizontal with hot surface facing downward is\",round(Q2,1),\"W\" \n",
+ "print\"For vertical disc,heat transferred is\",round(Q3),\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transferred when disc is horizontal with hot surface facing upward is 170.8 W\n",
+ "Heat transferred when disc is horizontal with hot surface facing downward is 65.6 W\n",
+ "For vertical disc,heat transferred is 114.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 239
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.38,Page no:3.91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Total heat loss in a pipe\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=23.13*10**-6 #[m**2/s]\n",
+ "k=0.0321 #[W/m.K]\n",
+ "Beta=2.68*10**-3 #[K**-1]\n",
+ "Tw=443.0 #[K]\n",
+ "T_inf=303.0 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "g=9.81 #[m/s**2]\n",
+ "Npr=0.688 #Prandtl number\n",
+ "D=100.0 #Diameter [mm]\n",
+ "D=D/1000 #Diameter [m]\n",
+ "\n",
+ "#Calculation\n",
+ "Nra=(g*Beta*dT*(D**3)*Npr)/(v**2)\n",
+ "Nnu=0.53*(Nra**(1.0/4.0)) #Nusselt number\n",
+ "h=Nnu*k/D #[W/(m**2.K)]\n",
+ "h=7.93 #Approximation\n",
+ "e=0.90 #Emissivity\n",
+ "sigma=5.67*10**-8 \n",
+ "#Q=Q_conv+Q_rad #Total heat loss\n",
+ "#for total heat loss per meter length\n",
+ "Q_by_l=h*math.pi*D*dT+sigma*e*math.pi*D*(Tw**4-T_inf**4) #[W/m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Total heat loss per metre length of pipe is\",round(Q_by_l,1),\"W/m\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total heat loss per metre length of pipe is 831.1 W/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 240
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.39,Page no:3.91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat loss by free convection\n",
+ "import math\n",
+ "\n",
+ "#Result\n",
+ "k=0.035 #[W/(m.K)]\n",
+ "Npr=0.684 #Prandtl number\n",
+ "Beta=2.42*10**-3 #[K**-1]\n",
+ "v=27.8*10**-6 #[m**2/s]\n",
+ "Tw=533.0 #[K]\n",
+ "T_inf=363.0 #[K]\n",
+ "dT=Tw-T_inf #[K]\n",
+ "D=0.01 #[m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "\n",
+ "#Calculation\n",
+ "Nra=(g*Beta*dT*(D**3))/(v**2)\n",
+ "#For this <10**5,we have for sphere\n",
+ "A=4*math.pi*(D/2)**2 #Area of sphere in [m**2]\n",
+ "Nnu=(2+0.43*Nra**(1.0/4.0))#Nusslet number\n",
+ "h=Nnu*k/D #W/(m**2.K)\n",
+ "Q=h*A*dT #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of heat loss is\",round(Q,2),\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of heat loss is 1.06 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 241
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.40,Page no:3.92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat loss from cube\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=17.95*10**-6 #[m**2/s]\n",
+ "dT=353.0-293.0 #[K]\n",
+ "k=0.0283 #[W/m.K]\n",
+ "g=9.81 #[m/s**2]\n",
+ "Npr=0.698 #Prandtl number\n",
+ "Cp=1005.0 #J/(kg.K)\n",
+ "Tf=323.0 #Film temperature in [K]\n",
+ "Beta=1.0/Tf #[K**-1]\n",
+ "l=1.0 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "Nra=(g*Beta*dT*(l**3)*Npr)/(v**2)\n",
+ "#In textbook result of above statement is wrongly calculated,So\n",
+ "Nra=3.95*10**8\n",
+ "#For Nra <10**9,for a vertical plate,the average nusselt number is\n",
+ "Nnu=0.59*Nra**(1.0/4.0) #Nusselt number\n",
+ "h=round(Nnu*k/l,2) #[W/m**2.K]\n",
+ "A=l**2 #Area [m**2]\n",
+ "#Heat loss form 4 vertical faces of 1m*1m is \n",
+ "Q1=4.0*(h*A*dT) #[W]\n",
+ "#For top surface \n",
+ "P=4.0*l #Perimeter in [m]\n",
+ "L=A/P #[m]\n",
+ "Nra=(Npr*g*Beta*dT*(L**3))/(v**2)\n",
+ "Nnu=0.15*Nra**(1.0/3.0) #Nusselt number\n",
+ "h=round(Nnu*k/L,1) #[W/m**2.K]\n",
+ "Q2=h*A*dT #[W]\n",
+ "Q_total=Q1+Q2 #Total heat loss[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Total heat loss is\",Q_total,\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "-"
+ }
+ },
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total heat loss is 966.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 246
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.41,Page no:3.93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Plate exposed to heat\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=0.910 #Density in [kg/m**3]\n",
+ "Cp=1.009*1000 #[J/kg.K]\n",
+ "k=0.0331 #[W/m.K]\n",
+ "mu=22.65*10**-6 #[N.s/m**2]\n",
+ "\n",
+ "#Calculation\n",
+ "#Let a=smaller side\n",
+ "#b=bigger side\n",
+ "#Qa=ha*A*dT\n",
+ "#Qb=hb*A*dT\n",
+ "#Qa=1.14*Qb\n",
+ "#Given a*b=15*10**-4\n",
+ "#On solving we get:\n",
+ "a=0.03 #[m]\n",
+ "b=0.05 #[m]\n",
+ "A=a*b #Area in [sq m]\n",
+ "Tf=388 #[K]\n",
+ "Beta=1.0/Tf #[K**-1]\n",
+ "T1=303 #[K]\n",
+ "T2=473 #[K]\n",
+ "dT=T2-T1 #[K]\n",
+ "v=mu/rho \n",
+ "g=9.81 #m/s**2[acceleration due to gravity ]\n",
+ "hb=0.59*(((g*Beta*dT*(b**3))/(v**2))*Cp*mu/k)**(1.0/4.0)*(k/b) #[W/sq m.K]\n",
+ "Qb=hb*A*(dT) #[W]\n",
+ "\n",
+ "Qa=1.14*Qb #[W]\n",
+ "\n",
+ "#Result\n",
+ "print\"Dimensions of the plate are\",a,\"x\",b,\"m=\",a*100,\"x\",b*100,\"cm\"\n",
+ "print\"Heat transfer when the bigger side held vertical is\",round(Qb,2),\"W\"\n",
+ "print\"Heat transfer when the small side held vertical is\",round(Qa,2),\"W\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dimensions of the plate are 0.03 x 0.05 m= 3.0 x 5.0 cm\n",
+ "Heat transfer when the bigger side held vertical is 2.77 W\n",
+ "Heat transfer when the small side held vertical is 3.16 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 249
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.42,Page no:3.99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Nucleate poolboiling\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Ts=373.0 #[K]\n",
+ "rho_l=957.9 #rho*l[kg/m**3]\n",
+ "Cpl=4217.0 #[J/kg.K]\n",
+ "mu_l=27.9*10**-5 #[kg/(m.s)]\n",
+ "rho_v=0.5955 #[kg/m**3]\n",
+ "Csf=0.013\n",
+ "sigma=5.89*10**-2 #[N/m]\n",
+ "Nprl=1.76\n",
+ "lamda=2257.0 #[kJ/kg]\n",
+ "lamda=lamda*1000 #in [J/kg]\n",
+ "n=1 #for water\n",
+ "m_dot=30.0 #Mass flow rate [kg/h]\n",
+ "\n",
+ "#Calculation\n",
+ "m_dot=m_dot/3600 #[kg/s]\n",
+ "D=30.0 #Diameter of pan [cm]\n",
+ "D=D/100 #[m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "A=math.pi*(D/2)**2 #Area in [sq m]\n",
+ "Q_by_A=m_dot*lamda/A #[W/sq m]\n",
+ "#For nucleate boiling point we have:\n",
+ "dT=(lamda/Cpl)*Csf*(((Q_by_A)/(mu_l*lamda))*math.sqrt(sigma/(g*(rho_l-rho_v))))**(1.0/3.0)*(Nprl**n) #[K]\n",
+ "Tw=Ts+dT #[K]\n",
+ "\n",
+ "#Result\n",
+ "print\"Temperature of the bottom surface of the pan is\",round(Tw,1),\"W/(m^2)\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature of the bottom surface of the pan is 385.5 W/(m^2)\n"
+ ]
+ }
+ ],
+ "prompt_number": 251
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.43,Page no:3.100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Peak Heat flux\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "lamda=2257.0 #[kJ/kg]\n",
+ "lamda=lamda*1000 #in [J/kg]\n",
+ "rho_l=957.9 #rho*l[kg/m**3]\n",
+ "rho_v=0.5955 #[kg/m**3]\n",
+ "sigma=5.89*10**-2 #[N/m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "\n",
+ "#Calculation\n",
+ "#Peak heat flux is given by\n",
+ "Q_by_A_max=(math.pi/24)*(lamda*rho_v**0.5*(sigma*g*(rho_l-rho_v))**(1.0/4.0)) #W/m**2\n",
+ "Q_by_A_max=Q_by_A_max/(10**6) #MW/(sq m)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Peak heat flux is\",round(Q_by_A_max,3),\"MW/sq m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak heat flux is 1.106 MW/sq m\n"
+ ]
+ }
+ ],
+ "prompt_number": 252
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.44,Page no:3.100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Stable film pool boiling\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho_l=957.9 #[kg/m**3]\n",
+ "lamda=2257.0 #[kJ/kg]\n",
+ "lamda=lamda*10**3 #[J/kg]\n",
+ "rho_v=31.54 #[kg/m**3]\n",
+ "Cpv=4.64 #[kJ/kg.K]\n",
+ "Cpv=Cpv*10**3 #[J/kg.K]\n",
+ "kv=58.3*10**-3#[W/(m.K)]\n",
+ "g=9.81 #[m/s**2]\n",
+ "mu_v=18.6*10**-6 #[kg/(m.s)]\n",
+ "e=1.0 #Emissivity\n",
+ "sigma=5.67*10**-8 \n",
+ "Ts=373.0 #[K]\n",
+ "Tw=628.0 #[K]\n",
+ "\n",
+ "#Calculation\n",
+ "dT=Tw-Ts #[K]\n",
+ "D=1.6*10**-3 #[m]\n",
+ "T=(Tw+Ts)/2 #[K]\n",
+ "\n",
+ "hc=0.62*((kv**3)*rho_v*(rho_l-rho_v)*g*(lamda+0.40*Cpv*dT)/(D*mu_v*dT))**(1.0/4.0) #Convective heat transfer coeff [W/sq m.K]\n",
+ "hr=e*sigma*(Tw**4-Ts**4)/(Tw-Ts) #Radiation heat transfer coeff in [W/sq m.K]\n",
+ "h=hc+(3.0/4.0)*hr #Total heat transfer coefficient W/(sq m.K)\n",
+ "Q_by_l=h*math.pi*D*dT #Heat dissipation rate per unit length in [kW/m]\n",
+ "Q_by_l=Q_by_l/1000 #[kW/m]\n",
+ "#Result\n",
+ "print\"Stable film boiling point heat transfer coefficient is\",round(h,1),\"W/(sq m.K)\"\n",
+ "print\"Heat dissipated per unit length of the heater is\",round(Q_by_l,1),\"kW/m\"\n",
+ "\n",
+ "print\"\\nNOTE:In textbook,value of hc is wrongly calculated as 1311.4,Actually it is 1318.9,\"\n",
+ "print\"So,there is a difference in final values of 'h'\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stable film boiling point heat transfer coefficient is 1340.9 W/(sq m.K)\n",
+ "Heat dissipated per unit length of the heater is 1.7 kW/m\n",
+ "\n",
+ "NOTE:In textbook,value of hc is wrongly calculated as 1311.4,Actually it is 1318.9,\n",
+ "So,there is a difference in final values of 'h'\n"
+ ]
+ }
+ ],
+ "prompt_number": 262
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.45,Page no:3.102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Heat transfer in tube\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "dT=10 #[K]\n",
+ "P=506.625 #[kPa]\n",
+ "P=P/10**3 #[Mpa]\n",
+ "D=25.4 #Diameter [mm]\n",
+ "D=D/1000 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "h=2.54*(dT**3)*(math.exp(P/1.551)) #[W/sq m.K]\n",
+ "#Q=h*math.pi*D*L*dT\n",
+ "#Heat transfer rate per meter length of tube is \n",
+ "Q_by_l=h*math.pi*D*dT #[W/m]\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of heat transfer per 1m length of tube is\",round(Q_by_l),\"W/m\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of heat transfer per 1m length of tube is 2810.0 W/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 136
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.46,Page no:3.102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Nucleat boiling and heat flux\n",
+ "\n",
+ "#Variable declaration\n",
+ "dT=8.0 #[K]\n",
+ "P=0.17 #[Mpa]\n",
+ "P=P*1000 #[kPa]\n",
+ "h1=2847.0 #[W/(sq m.K)]\n",
+ "P1=101.325 #[kPa]\n",
+ "h=5.56*(dT**3) #[W/sq m.K]\n",
+ "\n",
+ "#Calculation\n",
+ "Q_by_A=h*dT #[W/sq m]\n",
+ "hp=h*(P/P1)**(0.4) #[W/sq m.K]\n",
+ "#Corresponding heat flux is :\n",
+ "Q_by_A1=hp*dT #[W/sq m]\n",
+ "per=(Q_by_A1-Q_by_A)*100.0/Q_by_A #Percent increase in heat flux\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat flux when pressure is 101.325 kPa is\",Q_by_A,\"W/m^2(APPROX)\"\n",
+ "print\"Percent increase in heat flux is\",round(per),\"%\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat flux when pressure is 101.325 kPa is 22773.76 W/m^2(APPROX)\n",
+ "Percent increase in heat flux is 23.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 265
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.47,Page no:3.110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Dry steam condensate\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "mu=306*10**-6 #[N.s/m**2]\n",
+ "k=0.668 #[W/m.K]\n",
+ "rho=974.0 #[kg/m**3]\n",
+ "lamda=2225.0 #[kJ/kg]\n",
+ "lamda=lamda*10**3 #[J/kg.K]\n",
+ "g=9.81 #[m/s**2]\n",
+ "Ts=373.0 #[K]\n",
+ "Tw=357.0 #[K]\n",
+ "dT=Ts-Tw #[K]\n",
+ "Do=25.0 #[mm]\n",
+ "Do=Do/1000 #[m]\n",
+ "\n",
+ "#Calculation\n",
+ "h=0.725*((rho**2*g*lamda*k**3)/(mu*Do*dT))**(1.0/4.0) #[W/sq m.K]\n",
+ "Q_by_l=h*math.pi*Do*dT #[W/m]\n",
+ "m_dot_byl=(Q_by_l/lamda) #[kg/s]\n",
+ "m_dot_byl=m_dot_byl*3600 #[kg/h]\n",
+ "\n",
+ "#Result\n",
+ "print\"Mean heat transfer coefficient is\",round(h),\"W/(m^2.K)\" \n",
+ "print\"Heat transfer per unit length is\",round(Q_by_l),\"W/m\" \n",
+ "print\"Condensate rate per unit length is\",round(m_dot_byl,1),\"kg/h\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mean heat transfer coefficient is 10864.0 W/(m^2.K)\n",
+ "Heat transfer per unit length is 13653.0 W/m\n",
+ "Condensate rate per unit length is 22.1 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 269
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.48,Page no:3.111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Laminar Condensate film\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=960.0 #[kh/m**3]\n",
+ "mu=2.82*10**-4 #[kg/(m.s)]\n",
+ "k=0.68 #[W/(m.K)]\n",
+ "lamda=2255.0 #[kJ/kg]\n",
+ "lamda=lamda*10**3 #[J/kg]\n",
+ "Ts=373.0 #Saturation temperature of steam [K]\n",
+ "Tw=371.0 #[K]\n",
+ "dT=Ts-Tw #[K]\n",
+ "L=0.3 #Dimension [m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "\n",
+ "#Calculation\n",
+ "h=0.943*(rho**2*g*lamda*k**3/(L*mu*dT))**(1.0/4.0) #W/sq m.K\n",
+ "A=L**2 #[sq m] \n",
+ "Q=h*A*(Ts-Tw) #[W]=[J/s]\n",
+ "m_dot=Q/lamda #Condensate rate[kg/s]\n",
+ "m_dot=m_dot*3600.0 #[kg/h]\n",
+ "\n",
+ "#Result\n",
+ "print\"Average heat transfer coefficient is\",round(h),\" W/(m^2.K)(APPROX)\"\n",
+ "print\"Heat transfer rate is\",round(Q),\"J/kg\"\n",
+ "print\"Steam condensate rate per hour is\",round(m_dot,2),\"kg/h\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average heat transfer coefficient is 13156.0 W/(m^2.K)(APPROX)\n",
+ "Heat transfer rate is 2368.0 J/kg\n",
+ "Steam condensate rate per hour is 3.78 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 276
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.49,Page no:3.112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Saturated vapour condensate in array\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=1174.0 #[kg/m**3]\n",
+ "k=0.069 #[W/(m.K)]\n",
+ "mu=2.5*10**-4 #[N.s/m**2]\n",
+ "lamda=132*10**3 #[J/kg]\n",
+ "g=9.81 #[m/s**2]\n",
+ "Ts=323.0 #[K]\n",
+ "Tw=313.0 #[K]\n",
+ "dT=Ts-Tw #[K]\n",
+ "\n",
+ "#Calculation\n",
+ "#For square array,n=4\n",
+ "n=4.0 #number of tubes\n",
+ "Do=12.0 #[mm]\n",
+ "Do=Do/1000 #[m]\n",
+ "h=0.725*(rho**2*lamda*g*k**3/(n*Do*mu*dT))**(1.0/4.0) #W/(sq m.K) \n",
+ "#For heat transfer area calcualtion,n=16\n",
+ "A=n*math.pi*Do #[sq m]\n",
+ "A=0.603\n",
+ "Q=h*A*dT#[W/m]\n",
+ "m_dot=round(Q/lamda,3) #[kg/s]\n",
+ "\n",
+ "m_dot=m_dot*3600 #[kg/h]\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of condensation per unit length is\",m_dot,\"kg/h\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of condensation per unit length is 176.4 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 279
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.50,Page no:3.113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Mass rate of steam condensation\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=960.0 #[kg/m**3]\n",
+ "k=0.68 #[W/m.K]\n",
+ "mu=282.0*10**-6 #[kg/(m.s)]\n",
+ "Tw=371.0 #Tube wall temperature [K]\n",
+ "Ts=373.0 #Saturation temperature in [K]\n",
+ "dT=Ts-Tw #[K]\n",
+ "lamda=2256.9 #[kJ/kg]\n",
+ "lamda=lamda*10**3 #[J/kg]\n",
+ "\n",
+ "#Calculation\n",
+ "#For a square array with 100tubes,n=10\n",
+ "Do=0.0125 #[m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "n=10.0\n",
+ "h=0.725*(((rho**2)*g*lamda*(k**3)/(mu*n*Do*dT))**(1.0/4.0)) #W/(sq m.K)\n",
+ "\n",
+ "L=1.0 #[m]\n",
+ "#n=100\n",
+ "n=100.0 \n",
+ "A=n*math.pi*Do*L #[m**2/m length]\n",
+ "Q=h*A*dT #Heat transfer rate in [W/m]\n",
+ "ms_dot=Q/lamda #[kg/s]\n",
+ "ms_dot=ms_dot*3600.0 #[kg/h]\n",
+ "\n",
+ "#Result\n",
+ "print\"Mass rate of steam condensation is\",round(ms_dot),\"kg/h\" \n",
+ "print\"NOTE:ERROR in Solution in book.Do is wrongly taken as 0.012 in lines 17 and 22 of the book,Also A is wrongly calculated\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass rate of steam condensation is 158.0 kg/h\n",
+ "NOTE:ERROR in Solution in book.Do is wrongly taken as 0.012 in lines 17 and 22 of the book,Also A is wrongly calculated\n"
+ ]
+ }
+ ],
+ "prompt_number": 286
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.51,Page no:3.114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Saturated tube condensate in a wall\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=975.0 #[kg/m**3]\n",
+ "k=0.871 #[W/m.K]\n",
+ "dT=10.0 #[K]\n",
+ "mu=380.5*10**-6 #[N.s/m**2]\n",
+ "lamda=2300.0 #[kJ/kg]\n",
+ "lamda=lamda*1000 # Latent heat of condensation [J/kg]\n",
+ "Do=100.0 #Outer diameter [mm]\n",
+ "Do=Do/1000 #[m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "\n",
+ "#Calculation\n",
+ "#for horizontal tube\n",
+ "h1=0.725*((rho**2*lamda*g*k**3)/(mu*Do*dT))**(1.0/4.0) #Average heat transfer coefficient\n",
+ "#for vertical tube\n",
+ "#h2=0.943*((rho**2*lambda*g*k**3)/(mu*L*dT))**(1/4) #Average heat transfer coefficient\n",
+ "h2=h1 #For vertical tube\n",
+ "#implies that\n",
+ "L=(0.943*((rho**2*lamda*g*k**3)**(1.0/4.0))/(h1*((mu*dT)**(1.0/4.0))))**4 #[m]\n",
+ "L=0.29 #Approximate in book\n",
+ "h=0.943*((rho**2*lamda*g*k**3)/(mu*L*dT))**(1.0/4.0) #[W/(sq m.K)]\n",
+ "A=math.pi*Do*L #Area in [m**2]\n",
+ "Q=h*A*dT #Heat transfer rate [W]\n",
+ "mc_dot=Q/lamda #[Rate of condensation]in [kg/s]\n",
+ "mc_dot=mc_dot*3600 #[kg/h]\n",
+ "\n",
+ "#Result\n",
+ "print\"Tube length is\",L,\"m\"\n",
+ "print\"Rate of condensation per hour is\",round(mc_dot,2),\"kg/h\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Tube length is 0.29 m\n",
+ "Rate of condensation per hour is 14.32 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 288
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.52,Page no:3.115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Condensation rate\n",
+ "\n",
+ "#Variable declaration\n",
+ "m1_dot=50.0 # For horizontal position[kg/h]\n",
+ "Do=10.0 #[mm]\n",
+ "Do=Do/1000 #[m]\n",
+ "L=1.0 #[m]\n",
+ "#For 100 tubes n=10\n",
+ "n=10.0 \n",
+ "\n",
+ "#Calculation\n",
+ "#We know that\n",
+ "#m_dot=Q/lambda=h*A*dT/lambda\n",
+ "#m_dot is proportional to h\n",
+ "#m1_dot prop to h1\n",
+ "#m2_dot propn to h2\n",
+ "#m1_dot/m2_dot=h1/h2\n",
+ "#or :\n",
+ "m2_dot=m1_dot/((0.725/0.943)*(L/(n*Do))**(1.0/4.0)) #[kg/h]\n",
+ "\n",
+ "#Result\n",
+ "print\"For vertical position,Rate of condensation is\",round(m2_dot,2),\"kg/h\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For vertical position,Rate of condensation is 36.57 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 289
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.53,Page no:3.116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Condensation on vertical plate\n",
+ "rho=975 #[kg/m**3]\n",
+ "k=0.671 #[W/(m.K)]\n",
+ "mu=3.8*10**-4 #[N.s/m**2]\n",
+ "dT=10 #[K]\n",
+ "lamda=2300*10**3 #[J/kg]\n",
+ "L=1 #[m]\n",
+ "g=9.81 #[m/s**2]\n",
+ "\n",
+ "#Calculation\n",
+ "ha=0.943*((rho**2*lamda*g*k**3)/(mu*L*dT))**(1.0/4.0) #W/(sq m.K) #[W/sq m.K]\n",
+ "#Local heat transfer coefficient\n",
+ "#at x=0.5 #[m]\n",
+ "x=0.5 #[m]\n",
+ "h=((rho**2*lamda*g*k**3)/(4*mu*dT*x))**(1.0/4.0) #[W/sq m.K]\n",
+ "delta=((4*mu*dT*k*x)/(lamda*rho**2*g))**(1.0/4.0) #[m]\n",
+ "delta=delta*10**3 #[mm]\n",
+ "\n",
+ "#Result\n",
+ "print\"(i)- Average heat transfer coefficient is\",round(ha),\" W/(m**2.K)\"\n",
+ "print\"(ii)-Local heat transfer coefficient at 0.5 m height is\",round(h),\"W/(m^2.K)\"\n",
+ "print\"(iii)-Film thickness is\",round(delta,3),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)- Average heat transfer coefficient is 6060.0 W/(m**2.K)\n",
+ "(ii)-Local heat transfer coefficient at 0.5 m height is 5404.0 W/(m^2.K)\n",
+ "(iii)-Film thickness is 0.124 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 295
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |