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diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb new file mode 100644 index 00000000..85b7eec5 --- /dev/null +++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb @@ -0,0 +1,1176 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 07:Principles of forced convection" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.1:pg-296" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\n", + "First we check from reynolds no. that the flow is laminar or tubulent\n", + "Reynold number is\n", + "Re= 20000.0\n", + "which is less than critical reynolds number,So the flow is laminar.\n", + "The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)\n", + "NuL= 93.9037805416\n", + "Heat flux in W/(m**2*K) is\n", + "h= 2.72320963571\n", + "The rate of heat transfer per unit width in W is\n", + "Q= 408.481445356\n" + ] + } + ], + "source": [ + " \n", + "import math \n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\"\n", + "#Engine oil at temprature,Tinf=60°C with a velocity of Uinf=1m/s flows over plate of length(L)=5m whose temprature(Tw)=30°C\n", + "Tw=30;\n", + "L=5;\n", + "Uinf=1;\n", + "Tinf=60;\n", + "#The properties at a film temprature of 45°C are as follows density(rho=870kg/m**3),Prandtl number(Pr=2850),conductivity(k=0.145W/(m*°C)),kinematic viscosity(nu=250*10**-6m**2/s).\n", + "rho=870;\n", + "Pr=2850;\n", + "k=0.145;\n", + "nu=250*10**-6;\n", + "print\"First we check from reynolds no. that the flow is laminar or tubulent\"\n", + "#Reynolds number is given by Re=(Uinf*L)/nu\n", + "print\"Reynold number is\"\n", + "Re=(Uinf*L)/nu\n", + "print\"Re=\",Re\n", + "print\"which is less than critical reynolds number,So the flow is laminar.\"\n", + "#NuL is the average nusselt number\n", + "print\"The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)\"\n", + "NuL=0.664*Re**0.5*Pr**(1/3)\n", + "print\"NuL=\",NuL\n", + "#Heat flux is given by h=(k/L)*NuL\n", + "print\"Heat flux in W/(m**2*K) is\"\n", + "h=(k/L)*NuL\n", + "print\"h=\",h\n", + "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n", + "#Since unit width is considerd so B=1\n", + "#Area(A)=L*B\n", + "B=1;\n", + "A=L*B;\n", + "print\"The rate of heat transfer per unit width in W is\"\n", + "Q=h*A*(Tinf-Tw)\n", + "print\"Q=\",Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.2:pg-298" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\n", + "The location x in m where the transition occurs\n", + "x= 0.275\n", + "The average Nusselt number for the laminar zone is\n", + "Nux= 469.518902708\n", + "Heat flux in W/(m**2*K) is\n", + "h= 44.3908780742\n", + "The reynolds number at L=2m is\n", + "ReL= 3636363.63636\n", + "The average heat transfer coefficient over L=2m in W/(m**2*K)\n", + "hbarL= -11.322519\n", + "The rate of heat transfer per unit width in W is\n", + "Q= -2264.5038\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\"\n", + "#Atmospheric air at temprature,Tinf=300K and with a free stream Velocity Uinf=30m/s flows over a flat plate parallel to a side of length(L)=2m.\n", + "Tinf=300;\n", + "Uinf=30;\n", + "L=2;\n", + "#It is maintained at a uniform temprature of Tw=400K.\n", + "Tw=400;\n", + "#The properties of air at the film temprature of 350K are Prandtl number(Pr=0.705),conductivity(k=0.026W/(m*°C)),kinematic viscosity(nu=16.5*10**-6m**2/s)\n", + "Pr=0.705; \n", + "k=0.026;\n", + "nu=16.5*10**-6;\n", + "#We first find the location x(for reynolds number,Re=5*10**5) where the transition occurs\n", + "#Rex is reynolds number\n", + "print\"The location x in m where the transition occurs\"\n", + "Rex=5*10**5;\n", + "x=(nu*Rex)/Uinf\n", + "print\"x=\",x\n", + "#The average Nusselt number for the laminar zone is given by Nux=0.664*Re**0.5*Pr**(1/3)\n", + "print\"The average Nusselt number for the laminar zone is\"\n", + "Nux=0.664*Rex**0.5*Pr**(1/3)\n", + "print\"Nux=\",Nux\n", + "#Heat flux is given by h=(k/x)*Nux\n", + "print\"Heat flux in W/(m**2*K) is\"\n", + "h=(k/x)*Nux\n", + "print\"h=\",h\n", + "#Reynolds number is given by ReL=(Uinf*L)/nu\n", + "print\"The reynolds number at L=2m is\"\n", + "ReL=(Uinf*L)/nu\n", + "print\"ReL=\",ReL\n", + "#The average heat transfer coefficient over L=2m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n", + "print\"The average heat transfer coefficient over L=2m in W/(m**2*K)\"\n", + "hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n", + "print\"hbarL=\",hbarL\n", + "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n", + "#Since unit width is considerd so B=1\n", + "#Area(A)=L*B\n", + "B=1;\n", + "A=L*B;\n", + "print\"The rate of heat transfer per unit width in W is\"\n", + "Q=hbarL*A*(Tw-Tinf)\n", + "print\"Q=\",Q\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.3:pg-314" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\n", + "(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes\n", + "ReL= 1250000.0\n", + "which is greater than critical Reynolds number.\n", + "The average heat transfer coefficient over L=5m in W/(m**2*K)\n", + "hbarL= -5.225778\n", + "The rate of heat transfer per unit width in W is\n", + "Q= -3135.4668\n", + "(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes \n", + "which is less than critical Reynolds number.\n", + "ReL= 250000.0\n", + "Heat flux in W/(m**2*K) is\n", + "h= 9.96\n", + "The rate of heat transfer per unit width in W is\n", + "Q= 5976.0\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\"\n", + "#Air at a pressure of 101kPa and temprature,Tinf=20°C flows with a velocity(Uinf) of 5m/s over a flat plate whose temprature is kept constant at Tw=140°C.\n", + "Tw=140;\n", + "Tinf=20;\n", + "Uinf=5;\n", + "#The properties at the film temprature of 80°C are Prandtl number(Pr=0.706),Conductivity(k=0.03W/(m*°C)),kinematic viscosity(nu=2*10**-5m**2/s)\n", + "Pr=0.706;\n", + "k=0.03;\n", + "nu=2*10**-5;\n", + "#ReL is reynolds number and L is length of flat plate\n", + "print\"(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes\"\n", + "L=5;\n", + "ReL=(Uinf*L)/nu\n", + "print\"ReL=\",ReL\n", + "print\"which is greater than critical Reynolds number.\"\n", + "#Thus we have combined laminar and tubulent flow.\n", + "# So The average heat transfer coefficient over L=5m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n", + "print\"The average heat transfer coefficient over L=5m in W/(m**2*K)\"\n", + "hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n", + "print\"hbarL=\",hbarL\n", + "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n", + "#Since width is 1m so B=1\n", + "#Area(A)=L*B\n", + "B=1;\n", + "A=L*B;\n", + "#Q is the rate of heat transfer\n", + "print\"The rate of heat transfer per unit width in W is\"\n", + "Q=hbarL*A*(Tw-Tinf)\n", + "print\"Q=\",Q\n", + "#When the air flow is parallel to the 1m side we have L=1\n", + "print\"(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes \"\n", + "L=1;\n", + "ReL=(Uinf*L)/nu\n", + "print\"which is less than critical Reynolds number.\"\n", + "print\"ReL=\",ReL\n", + "#Thus we have laminar flow\n", + "#Heat flux is given by h=(k/L)*0.664*ReL**0.5*Pr**(1/3)\n", + "print\"Heat flux in W/(m**2*K) is\"\n", + "h=(k/L)*0.664*ReL**0.5*Pr**(1/3)\n", + "print\"h=\",h\n", + "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n", + "#Now width is 5m so B=5\n", + "#Area(A)=L*B\n", + "B=5;\n", + "A=L*B;\n", + "#Q is the rate of heat transfer\n", + "print\"The rate of heat transfer per unit width in W is\"\n", + "Q=h*A*(Tw-Tinf)\n", + "print\"Q=\",Q\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.4:pg-322" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\n", + "(a)Reynolds number is\n", + "ReL= 6000.0\n", + "The boundary layer thickness in m is\n", + "delta= 0.387298334621\n", + "Prandtl no. is\n", + "Pr= 831.024930748\n", + "The thermal boundary layer thickness in m is\n", + "deltaT= 0.387298334621\n", + "(b)Since the prandtl number is high So Nusselt no. is\n", + "NuL= 26.2588270873\n", + "Heat flux in W/(m**2*K) is\n", + "hL= 0.919058948055\n", + "hbarL in W/(m**2*K) is\n", + "hbarL= 1.83811789611\n", + "(c)The rate of heat transfer in W is\n", + "Q= 661.7224426\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\"\n", + "#Castor oil at temprature,Tinf=36°C flows over a heated plate of length,L=6m and breadth,B=1m at velocity,Uinf=0.06m/s\n", + "Tinf=36;\n", + "L=6;\n", + "B=1;\n", + "Uinf=0.06;\n", + "#For a surface temprature at Tw=96°C\n", + "Tw=96;\n", + "#The properties at film temprature 66°C conductivity(k=0.21W/(m*K)),kinematic viscosity(nu=6*10**-5m**2/s),Thermal diffusivity(alpha=7.22*10**-8 m**2/s)\n", + "nu=6*10**-5;\n", + "k=0.21;\n", + "alpha=7.22*10**-8;\n", + "#ReL is reynolds number\n", + "print\"(a)Reynolds number is\"\n", + "ReL=(Uinf*L)/nu\n", + "print\"ReL=\",ReL\n", + "#Therefore the boundary layer is laminar over the entire plate.\n", + "#delta is the boundary layer thickness\n", + "print\"The boundary layer thickness in m is\"\n", + "delta=(5*L)/(ReL)**0.5\n", + "print\"delta=\",delta\n", + "#Pr is prandtl number.\n", + "print\"Prandtl no. is\"\n", + "Pr=nu/alpha\n", + "print\"Pr=\",Pr\n", + "#deltaT is thermal boundary layer thickness\n", + "print\"The thermal boundary layer thickness in m is\"\n", + "deltaT=delta/(Pr**(1/3))#NOTE:Answer in the book is incorrect(calculation mistake)\n", + "print\"deltaT=\",deltaT\n", + "#NuL is the nusselt number\n", + "print\"(b)Since the prandtl number is high So Nusselt no. is\"\n", + "NuL=0.339*(ReL)**0.5*Pr**(1/3)\n", + "print\"NuL=\",NuL\n", + "#Heat flux is given by hL=(k/L)*NuL\n", + "print\"Heat flux in W/(m**2*K) is\"\n", + "hL=(k/L)*NuL\n", + "print\"hL=\",hL\n", + "#hbarL is the average heat flux over length L\n", + "print\"hbarL in W/(m**2*K) is\"\n", + "hbarL=2*hL\n", + "print\"hbarL=\",hbarL\n", + "#The rate of heat transfer is Q=h*A*(Tinf-Tw)\n", + "#Area(A)=L*B\n", + "A=L*B;\n", + "print\"(c)The rate of heat transfer in W is\"\n", + "Q=hbarL*A*(Tw-Tinf)\n", + "print\"Q=\",Q\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.5:pg-322" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\n", + "Reynolds number is\n", + "ReL= 11181.8181818\n", + "Therefore the flow is turbulent over the module \n", + "The local heat transfer coefficient at L in W/(m**2*K)is\n", + "hL= 8.32911955901e+30\n", + "The required power generation in W/m**3 is\n", + "qm= 1.6658239118e+31\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\"\n", + "#A flat plate of width B=1m is maintained at a uniform surface temprtaure(Tw)=225°C\n", + "Tw=225;\n", + "B=1;\n", + "#Heating is done by rectangular modules of thickness t=10mm and length l=40mm.\n", + "t=10;\n", + "l=40;\n", + "#atmospheric air at temprature,Tinf=25°C flows over the plate at velocity(Uinf)=30m/s.\n", + "Tinf=25;\n", + "Uinf=30;\n", + "#The thermophysical properties of module are conductivity(km=5.2W/(m*K)),specific heat(cp=320J/(kg/K)),density(rho=2300kg/m**3).\n", + "km=5.2;\n", + "cp=320;\n", + "rho=2300;\n", + "#Assume the air properties at the film temprature of 125°C conductivity(ka=0.031W/(m*K)),kinematic viscosity(nu=22*10**-6m**2/s),Prandtl number(Pr=0.7)\n", + "ka=0.031;\n", + "nu=22*10**-6;\n", + "Pr=0.7;\n", + "#Module is placed at a distance of 800mm from the leading edge\n", + "#The distance from leading edge to the centre-line of the module,L=800+20=820mm.\n", + "L=0.0082;#in metre\n", + "#ReL is the reynolds number \n", + "print\"Reynolds number is\"\n", + "ReL=(Uinf*L)/nu\n", + "print\"ReL=\",ReL\n", + "print\"Therefore the flow is turbulent over the module \"\n", + "#The local heat transfer coefficient at L is calculated using hL=(k/L)*0.0296*(ReL)**(4/5)*(Pr)**(1/3)\n", + "print\"The local heat transfer coefficient at L in W/(m**2*K)is\"\n", + "hL=(ka/L)*0.0296*(ReL)**(4/0.5)*(Pr)**(1/0.3)\n", + "print\"hL=\",hL\n", + "#We consider that the local heat transfer coefficient at L=0.82m remains the same over the module which extends from L=0.80m to 0.84m \n", + "#If qm be the power generation in W/m**2 within the module ,we can write from energy balance qm*(t/0.1000)*(l/0.1000)*(B)=hbarL*(t/0.1000)*(B)*(Tw-Tinf)\n", + "print\"The required power generation in W/m**3 is\"\n", + "qm=(hL*(l/0.1000)*(B)*(Tw-Tinf))/((t/0.1000)*(l/0.1000)*(B))\n", + "print\"qm=\",qm\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.6:pg-327" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\n", + "Reynolds number is\n", + "ReL= 15000000.0\n", + "Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing \n", + "Nux= 0.0308\n", + "Nusselt number is \n", + "NubarL= 0.0308\n", + "Average heat transfer coefficient in W/(m**2*K) is\n", + "hbarL= 0.0003696\n", + "Surface temprature of wing in kelvin is\n", + "Tw= 1217800.46753\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\"\n", + "#An aircraft is moving at a velocity of Uinf=150m/s in air at an altitude where the pressure is 0.7bar and the temprature is Tinf=-5°C.\n", + "Tinf=-5;\n", + "Uinf=150;\n", + "#The top surface of the wing absorbs solar radiation at a rate of Qr=900W/m**2.\n", + "Qr=900;\n", + "#Considering the wing as a flat plate of length(L)=2m and to be of solid construction with a single uniform surface temprature .\n", + "L=2;\n", + "#The properties of air at 268K and 0.7 bar are conductivity(k=0.024W/(m*K)),kinematic viscosity(nu=2*10**-5m**2/s),Prandtl number(Pr=0.72)\n", + "k=0.024;\n", + "nu=2*10**-5;\n", + "Pr=0.72;\n", + "#ReL is reynolds number\n", + "print\"Reynolds number is\"\n", + "ReL=Uinf*L/nu\n", + "print\"ReL=\",ReL\n", + "#Rec is critical reynolds number\n", + "print\"Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing \"\n", + "#Nusselt number is given by Nux=0.0308*ReL**(4/5)*Pr**(1/3)\n", + "Nux=0.0308*ReL**(4/5)*Pr**(1/3);\n", + "print\"Nux=\",Nux\n", + "#NubarL is average nusselt number over length L\n", + "print\"Nusselt number is \"\n", + "NubarL=(5/4)*Nux\n", + "print\"NubarL=\",NubarL\n", + "#Average heat transfer coefficient is given by hbarL=(k/L)*NubarL\n", + "print\"Average heat transfer coefficient in W/(m**2*K) is\"\n", + "hbarL=(k/L)*NubarL\n", + "print\"hbarL=\",hbarL\n", + "#From an energy balance the airfoil at steady state,Qr*As=2*hbarL*As*(Tw-Tinf) where Qr=radiation flux,As=upper or lower surface area.\n", + "#Therefore we can write Surface temprature of wing, Tw=Tinf+(Qr/(2*hbarL))\n", + "print\"Surface temprature of wing in kelvin is\"\n", + "Tw=(273+Tinf)+(Qr/(2*hbarL))\n", + "print\"Tw=\",Tw\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.7:pg-331" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\n", + "Reynolds number is\n", + "Re= 141.176470588\n", + "Nusselt number is\n", + "NuD= 6.85819682626\n", + "The average Heat transfer coefficient in W/(m**2*K) is\n", + "hbar= 4629.28285773\n", + "Heat transfer per unit length in W/m is\n", + "qL= 14.5433210172\n", + "If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\n", + "NuD= 7.66669771975\n", + "The average Heat transfer coefficient in W/(m**2*K) is\n", + "hbar= 5175.02096083\n", + "Heat transfer per unit length in W/m is\n", + "qL= 16.2578078327\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\"\n", + "#A fine wire having a diameter(D)=0.04mm is placed in an air stream at temprature,Tinf=25°C having a flow velocity of Uinf=60m/s perpendicular to wire.\n", + "D=0.04;\n", + "Tinf=25;\n", + "Uinf=60;\n", + "#An electric current is passed through the wire ,raising its surface temprature to Tw=50°C\n", + "Tw=50;\n", + "#For air at the film temprature of 37.5°C,conductivity(k=0.027 W/(m*K)),kinematic viscosity(nu=17*10**-6m**2/s) and Prandtl number(Pr=0.71)\n", + "k=0.027;\n", + "nu=17*10**-6;\n", + "Pr=0.71;\n", + "#Re is reynolds number\n", + "print\"Reynolds number is\"\n", + "Re=Uinf*(D*10**-3)/nu\n", + "print\"Re=\",Re\n", + "#C and n are constants\n", + "#The values of C and n are found for Re=141 are C=0.683 and n=0.466\n", + "#NuD is nusselt number\n", + "print\"Nusselt number is\"\n", + "NuD=(0.683)*Re**0.466*Pr**(1/3)\n", + "print\"NuD=\",NuD\n", + "#hbar is the average Heat transfer coefficient\n", + "print\"The average Heat transfer coefficient in W/(m**2*K) is\"\n", + "hbar=(k/(D*10**-3))*NuD\n", + "print\"hbar=\",hbar\n", + "#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n", + "print\"Heat transfer per unit length in W/m is\"\n", + "qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n", + "print\"qL=\",qL\n", + "#NuD is nusselt number\n", + "print\"If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\"\n", + "NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr)**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\n", + "print\"NuD=\",NuD\n", + "#hbar is the average Heat transfer coefficient\n", + "print\"The average Heat transfer coefficient in W/(m**2*K) is\"\n", + "hbar=(k/(D*10**-3))*NuD\n", + "print\"hbar=\",hbar\n", + "#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n", + "print\"Heat transfer per unit length in W/m is\"\n", + "qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n", + "print\"qL=\",qL\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.8:pg-334" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\n", + "Reynolds number for mercury is\n", + "ReHg= 1000.0\n", + "Reynolds number for oil is\n", + "Reoil= 15.3846153846\n", + "The hydrodynamic entry length for mercury in m is\n", + "LeHg= 1.25\n", + "The hydrodynamic entry length for oil in m is\n", + "Leoil= 0.0192307692308\n", + "The thermal entry length for mercury in m is \n", + "LtHg= 0.02375\n", + "The thermal entry length for oil in m is\n", + "Ltoil= 1.63461538462\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\"\n", + "#Mercury and a light oil flowing at Uinf=4mm/s in a smooth tube having diameter(D)=25mm at a bulk temprature of 80°C.\n", + "Uinf=4*10**-3;#in metre\n", + "D=25*10**-3;#in metre\n", + "#The pertinent properties of the fluid at that temprature are kinematic viscosity of mercury(nuHg=1*10**-7m**2/s),kinematic viscosity of oil(nuoil=6.5*10**-6m**2/s)\n", + "#Prandtl number of mercury(PrHg=0.019),Prandtl number of oil(Proil=85).\n", + "nuHg=1*10**-7;\n", + "nuoil=6.5*10**-6;\n", + "PrHg=0.019;\n", + "Proil=85;\n", + "#ReHg is Reynolds number for mercury\n", + "print\"Reynolds number for mercury is\"\n", + "ReHg=Uinf*D/nuHg\n", + "print\"ReHg=\",ReHg\n", + "#Reoil is Reynolds number for oil\n", + "print\"Reynolds number for oil is\"\n", + "Reoil=Uinf*D/nuoil\n", + "print\"Reoil=\",Reoil\n", + "#The hydrodynamic length are given by L=0.05*Re*D\n", + "#LeHg is the hydrodynamic entry length for mercury\n", + "print\"The hydrodynamic entry length for mercury in m is\"\n", + "LeHg=0.05*ReHg*D\n", + "print\"LeHg=\",LeHg\n", + "#Leoil the hydrodynamic entry length for oil\n", + "print\"The hydrodynamic entry length for oil in m is\"\n", + "Leoil=0.05*Reoil*D\n", + "print\"Leoil=\",Leoil\n", + "#The thermal entry length are given by L=0.05*Re*Pr*D\n", + "#LtHg is the thermal entry length for mercury\n", + "print\"The thermal entry length for mercury in m is \"\n", + "LtHg=0.05*ReHg*PrHg*D\n", + "print\"LtHg=\",LtHg\n", + "#Ltoil is the thermal entry length for oil\n", + "print\"The thermal entry length for oil in m is\"\n", + "Ltoil=0.05*Reoil*Proil*D\n", + "print\"Ltoil=\",Ltoil\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.9:pg-336" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\n", + "Reynold number is\n", + "Re= 348.623853211\n", + "Therefore the flow is laminar.The hydrodynamic entrance length in m is\n", + "Leh= 0.0697247706422\n", + "The thermal entrance length in m is\n", + "Let= 0.0488073394495\n", + "The heat transfer coefficient in W/(m**2*K) is \n", + "h= 32.7\n", + "The mass flow rate of air in kg/s is\n", + "mdot= 2.38761041673e-05\n", + "Therefore the constant surface heat flux qw in W/m**2 is\n", + "qw= 95.95\n", + "The tube surface temprature at the exit plane in °C is \n", + "Twe= 127.934250765\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\"\n", + "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n", + "Tbi=75;\n", + "D=4*10**-3;#in metre\n", + "U=2;\n", + "#The tube length is L=1.0m and a constant heat flux is imposed by the tube surface on the air over the entire length.\n", + "L=1;\n", + "#An exit bulk mean temprature(Tbo)=125°C is required.\n", + "Tbo=125;\n", + "#The properties of air 100°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))\n", + "rho=0.95;\n", + "Pr=0.70;\n", + "k=0.03;\n", + "mu=2.18*10**-5;\n", + "cp=1.01*10**3;\n", + "#Re is reynolds number\n", + "print\"Reynold number is\"\n", + "Re=rho*U*D/mu\n", + "print\"Re=\",Re\n", + "#Leh is the hydrodynamic entrance length\n", + "print\"Therefore the flow is laminar.The hydrodynamic entrance length in m is\"\n", + "Leh=0.05*Re*D\n", + "print\"Leh=\",Leh\n", + "#Let is the thermal entrance length\n", + "print\"The thermal entrance length in m is\"\n", + "Let=0.05*Re*Pr*D\n", + "print\"Let=\",Let\n", + "#The length of tube is given as 1m.A reasonable approach is to consider the flow to be fully developed for both velocity and tempratures over the entire profile lengths.\n", + "#For a fully developed flow with constant surface heat flux,Nusselt number is Nu=4.36\n", + "Nu=4.36;\n", + "#h is the heat transfer coefficient\n", + "print\"The heat transfer coefficient in W/(m**2*K) is \"\n", + "h=Nu*(k/D)\n", + "print\"h=\",h\n", + "#Here h=hL Since the heat transfer coefficient is constant over the entire length of tube.\n", + "#hL is the local heat transfer coefficient\n", + "hL=h;\n", + "#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)\n", + "#mdot is mass flow rate\n", + "print\"The mass flow rate of air in kg/s is\"\n", + "mdot=rho*(math.pi/4)*D**2*U\n", + "print\"mdot=\",mdot\n", + "#qw is the constant surface heat flux\n", + "print\"Therefore the constant surface heat flux qw in W/m**2 is\"\n", + "qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)\n", + "print\"qw=\",qw\n", + "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n", + "print\"The tube surface temprature at the exit plane in °C is \"\n", + "Twe=Tbo+(qw/hL)\n", + "print\"Twe=\",Twe\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.10:pg-338" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\n", + "Reynold number is\n", + "Re= 348.623853211\n", + "Therefore the flow is laminar.The hydrodynamic entrance length in m is\n", + "Leh= 0.0697247706422\n", + "The thermal entrance length in m is\n", + "Let= 0.0488073394495\n", + "The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed\n", + "The inverse of graetz number Gr_1 is\n", + "Gr_1= 0.040977443609\n", + "Therefore the local heat transfer coefficient in W/(m**2*K) is\n", + "hL= 35.25\n", + "The mass flow rate of air in kg/s is\n", + "mdot= 2.38761041673e-05\n", + "Therefore surafce heat flux qw in W/m**2 is\n", + "qw= 2398.75\n", + "The tube surface temprature at the exit plane in °C is \n", + "Twe= 193.04964539\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\"\n", + "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n", + "Tbi=75;\n", + "D=4*10**-3;\n", + "U=2;\n", + "#The heated tube length is L=0.04m and a constant heat flux is imposed by the tube surface on the air over the entire length.\n", + "L=0.04;\n", + "#An exit bulk mean temprature(Tbo)=125°C is required.\n", + "Tbo=125;\n", + "#The properties of air 100°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))\n", + "rho=0.95;\n", + "Pr=0.70;\n", + "k=0.03;\n", + "mu=2.18*10**-5;\n", + "cp=1.01*10**3;\n", + "#Re is the reynolds number \n", + "print\"Reynold number is\"\n", + "Re=rho*U*D/mu\n", + "print\"Re=\",Re\n", + "#Leh is the hydrodynamic entrance length\n", + "print\"Therefore the flow is laminar.The hydrodynamic entrance length in m is\"\n", + "Leh=0.05*Re*D\n", + "print\"Leh=\",Leh\n", + "#Let is thermal entrance length\n", + "print\"The thermal entrance length in m is\"\n", + "Let=0.05*Re*Pr*D\n", + "print\"Let=\",Let\n", + "print\"The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed\" \n", + "#We calculate the inverse graetz number at x=L=0.04m\n", + "x=0.04;\n", + "#Gr_1 is inverse of graetz number\n", + "print\"The inverse of graetz number Gr_1 is\"\n", + "Gr_1=(x/D)*(1/(Re*Pr))\n", + "print\"Gr_1=\",Gr_1\n", + "#For constant surface heat flux nusselt number is Nu=4.7 and Graetz number is Gr=4.1*10**-2\n", + "Nu=4.7;\n", + "Gr=4.1*10**-2;\n", + "#hL is the local heat transfer coefficient\n", + "print\"Therefore the local heat transfer coefficient in W/(m**2*K) is\"\n", + "hL=Nu*(k/D)\n", + "print\"hL=\",hL\n", + "#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)\n", + "#mdot is the mass flow rate\n", + "print\"The mass flow rate of air in kg/s is\"\n", + "mdot=rho*(math.pi/4)*D**2*U\n", + "print\"mdot=\",mdot\n", + "#qw is the surface heat flux\n", + "print\"Therefore surafce heat flux qw in W/m**2 is\"\n", + "qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)\n", + "print\"qw=\",qw\n", + "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n", + "print\"The tube surface temprature at the exit plane in °C is \"\n", + "Twe=Tbo+(qw/hL)\n", + "print\"Twe=\",Twe\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex7.11:pg-339" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\n", + "Reynold number is\n", + "Re1= 13541.3214942\n", + "Nusselt number is\n", + "Nubar1= 56.808608087\n", + "The heat transfer transfer coefficient in W/(m**2*°C) \n", + "hbar1= 522.6391944\n", + "Outlet fluid temprature in first iteration is Tbo2 in °C is\n", + "Tb2 in °C is\n", + "Tb2= -30.4912413164\n", + "Reynold number is\n", + "Re2= 13938.8493187\n", + "Nusselt number is\n", + "The heat transfer transfer coefficient in W/(m**2*°C) \n", + "hbar2= 784.03829067\n", + "Outlet fluid temprature in second iteration is Tbo3 in °C is\n", + "Tbo3= -16.646852652\n", + "Tb3 in °C is\n", + "The Exit fluid temprature after second iteration is obtained as Tbo=-16.67°C\n", + "Tb3= -28.323426326\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + " import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\"\n", + "#Liquid sulphur di oxide in a saturated state flows inside a L=5m long tube and D=25mm internal diameter with a mass flow rate(mdot) of 0.15 kg/s.\n", + "#The tube is heated at a constant surface temprature(Tw) of -10°C and the inlet fluid temprature is Tbi=-40°C\n", + "Tw=-10;\n", + "Tbi=-40;\n", + "mdot=0.15;\n", + "D=0.025;#in metre\n", + "L=5;\n", + "#The properties to be used shoud be estimated at a temprature which is arithmetic mean of Tbi and Tbo.\n", + "#Since (outlet fluid temprature Tbo) is not known a priori,the solution has to be based on an iterative method starting with a guess value of Tb1=(Tbi+Tbo)/2\n", + "#Here we denote bulk mean temprature as Tb.The superscript refers to the no. of trials\n", + "#For first trial,guess Tbo1=-20°C;so Tb1=-30°C\n", + "#We have the property values as follows at a temprature of -30°C.\n", + "rhob1=1520.64;#density in kg/m**3\n", + "nub1=0.371*10**-6;#kinematic viscosity in m**2/s\n", + "kb1=0.23;#conductivity in W/(m*°C)\n", + "Prb1=3.31;#Prandtl number\n", + "mub1=nub1*rhob1;#viscosity in kg/(m*s)\n", + "cpb1=1361.6;#specific heat in J/(kg*K)\n", + "#muw=nuw*rhow at Tw=10°C\n", + "nuw=0.288*10**-6;#kinematic viscosity at Tw in m**2/s\n", + "rhow=1463.61;#density at Tw in kg/m**3\n", + "muw=nuw*rhow;#viscosity at Tw in kg/(m*s)\n", + "#The reynolds number is found as Re1=(4*mdot)/(math.pi*D*mub1)\n", + "print\"Reynold number is\"\n", + "Re1=(4*mdot)/(math.pi*D*mub1)\n", + "print\"Re1=\",Re1\n", + "#Hence the flow is turbulent\n", + "#Now using equation, nusselt number is,Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14\n", + "print\"Nusselt number is\"\n", + "Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14\n", + "print\"Nubar1=\",Nubar1\n", + "#The heat transfer transfer coefficient hbar1=(kb1/D)*Nubar1\n", + "print\"The heat transfer transfer coefficient in W/(m**2*°C) \"\n", + "hbar1=(kb1/D)*Nubar1\n", + "print\"hbar1=\",hbar1\n", + "#The outlet fluid temprature can be found by making use of eqn Tbo2=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar1)/(mdot*cpb1))\n", + "print\"Outlet fluid temprature in first iteration is Tbo2 in °C is\"\n", + "Tbo2=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar1)/(mdot*cpb1))\n", + "#Tb2 is the bulk mean temprature.\n", + "print\"Tb2 in °C is\"\n", + "Tb2=(Tbi+Tbo2)/2\n", + "print\"Tb2=\",Tb2\n", + "#Since the value differs from the assumed value of Tb1=-30°C,WE require furtheriteration,Therfore we start second trial with Tb2=-28.36°C\n", + "#We have the property value at a temprature of -28.36°C as follows\n", + "rhob2=1514;#density in kg/m**3\n", + "nub2=0.362*10**-6;#kinematic viscosity in m**2/s\n", + "kb2=0.229;#conductivity in W/(m*°C)\n", + "Prb2=3.23;#Prandtl number\n", + "mub2=nub2*rhob2;#viscosity in kg/(m*s)\n", + "cpb2=1362;#specific heat in J/(kg*K)\n", + "#muw=nuw*rhow at Tw=10°C\n", + "nuw=0.288*10**-6;#viscosity at Tw in m**2/s\n", + "rhow=1463.61;#density at Tw in kg/m**3\n", + "muw=nuw*rhow;#kinematic viscosity at Tw in kg/(m*s)\n", + "#The reynolds number is found as Re2=(4*mdot)/(math.pi*D*mub2)\n", + "print\"Reynold number is\"\n", + "Re2=(4*mdot)/(math.pi*D*mub2)\n", + "print\"Re2=\",Re2\n", + "#Now using equation, nusselt number is,Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14\n", + "print\"Nusselt number is\"\n", + "Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14\n", + "#The heat transfer transfer coefficient hbar2=(kb2/D)*Nubar2\n", + "print\"The heat transfer transfer coefficient in W/(m**2*°C) \"\n", + "hbar2=(kb2/D)*Nubar2\n", + "print\"hbar2=\",hbar2\n", + "#The outlet fluid temprature can be found by making use of eqn Tbo3=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar2)/(mdot*cpb2))\n", + "print\"Outlet fluid temprature in second iteration is Tbo3 in °C is\"\n", + "Tbo3=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar2)/(mdot*cpb2))\n", + "print\"Tbo3=\",Tbo3\n", + "#Tb3 is the bulk mean temprature.\n", + "print\"Tb3 in °C is\"\n", + "Tb3=(Tbi+Tbo3)/2\n", + "#We see that difference between Tbo2 and Tbo3 and that between Tb2 and Tb3 is marginal.Therfore we can stop iteration and present the result as Tbo=-16.67°C\n", + "print\"The Exit fluid temprature after second iteration is obtained as Tbo=-16.67°C\"\n", + "print\"Tb3=\",Tb3" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |