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diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb new file mode 100644 index 00000000..ba64d857 --- /dev/null +++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb @@ -0,0 +1,1051 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 02:One dimensional steady-state heat conduction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.1:pg-33" + ] + }, + { + "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 2, Example 1\n", + "The convective resistance Ro= 1/(ho*A) at the outer surface in KW**-1 is\n", + "Ro= 0.0666666666667\n", + "The conduction resistance Rs= Ls/(ks*A) of steel sheet in KW**-1 is\n", + "Rs= 8e-05\n", + "The conduction resistance Rg= Lg/(kg*A) of glass wool in KW**-1 is\n", + "Rg= 2.66666666667\n", + "The conduction resistance Rp= Lp/(kp*A) of plywood in KW**-1 is\n", + "Rp= 0.266666666667\n", + "The convective resistance Ri= 1/(hi*A) at the outer surface in KW**-1 is\n", + "Ri= 0.111111111111\n", + "The rate of heat flow Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri) in W is\n", + "Q= 12.5353919471\n", + "The temprature at the outer surface of wall is T1=To-(Q*Ro) in °C\n", + "T1= 23.1643072035\n", + "The temprature at the interface b/w steel sheet and glass wool is T2=T1-(Q*Rs) in°C\n", + "T2= 23.1633043722\n", + "The temprature at the interface b/w glass wool and plywood is T3=T2-(Q*Rg) in°C\n", + "T3= -10.2644074867\n", + "The temprature at the inner surface of wall is T4=T3-(Q*Rp)in °C\n", + "T4= -13.6071786725\n", + "Check for Ti(in °C)\n", + "The value is same as given in the problem\n", + "Ti= -15.0\n" + ] + } + ], + "source": [ + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 1\"\n", + "#The length of steel sheet (Ls)=1.5 mm and thermal conductivity (ks)=25 W/(mK) at the outer surface.\n", + "Ls=1.5;\n", + "ks=25;\n", + "#The length of plywood (Lp)=10 mm and thermal conductivity (kp)= .05 W/(mK) at the inner surface.\n", + "Lp=10;\n", + "kp=.05;\n", + "#The length of glass wool (Lg)=20 mm and thermal conductivity (kg)= .01W/(mK) in between steel sheet and plywood.\n", + "Lg=20;\n", + "kg=.01;\n", + "#The temprature of van inside cold Enviroment is (Ti)= -15°C while the outside surface is exposed to a surrounding ambient temprature (To)=24°C \n", + "To=24;\n", + "Ti=-15;\n", + "#The average value of heat transfer coefficients at the inner and outside surfaces of the wall are hi=12 W/(m**2*K) and ho= 20 W/(m**2*K) \n", + "hi=12;\n", + "ho=20;\n", + "#The surface area of wall (A)= .75 m**2 \n", + "A=.75;\n", + "#The convective resistance is Ro= 1/(ho*A) at the outer surface\n", + "print\"The convective resistance Ro= 1/(ho*A) at the outer surface in KW**-1 is\"\n", + "Ro=1/(ho*A)\n", + "print\"Ro=\",Ro\n", + "#The conduction resistance is Rs= Ls/(ks*A) of steel sheet\n", + "print\"The conduction resistance Rs= Ls/(ks*A) of steel sheet in KW**-1 is\"\n", + "Rs=Ls*10**-3/(ks*A)\n", + "print\"Rs=\",Rs\n", + "#The conduction resistance is Rg= Lg/(kg*A) of glass wool\n", + "print\"The conduction resistance Rg= Lg/(kg*A) of glass wool in KW**-1 is\"\n", + "Rg= Lg*10**-3/(kg*A)\n", + "print\"Rg=\",Rg\n", + "#The conduction resistance is Rp= Lp/(kp*A) of plywood\n", + "print\"The conduction resistance Rp= Lp/(kp*A) of plywood in KW**-1 is\"\n", + "Rp= Lp*10**-3/(kp*A)\n", + "print\"Rp=\",Rp\n", + "#The convective resistance is Ri= 1/(hi*A) at the outer surface\n", + "print\"The convective resistance Ri= 1/(hi*A) at the outer surface in KW**-1 is\"\n", + "Ri= 1/(hi*A)\n", + "print\"Ri=\",Ri\n", + "#The rate of heat flow is Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)\n", + "print\"The rate of heat flow Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri) in W is\"\n", + "Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)\n", + "print\"Q=\",Q\n", + "#The tempraure at the outer surface of wall is T1.\n", + "#The temprature at the interface b/w steel sheet and glass wool is T2.\n", + "#The temprature at the interface b/w glass wool and plywood is T3.\n", + "#The tempraure at the inner surface of wall is T4.\n", + "print\"The temprature at the outer surface of wall is T1=To-(Q*Ro) in °C\"\n", + "T1=To-(Q*Ro)\n", + "print\"T1=\",T1\n", + "print\"The temprature at the interface b/w steel sheet and glass wool is T2=T1-(Q*Rs) in°C\"\n", + "T2=T1-(Q*Rs)\n", + "print\"T2=\",T2\n", + "print\"The temprature at the interface b/w glass wool and plywood is T3=T2-(Q*Rg) in°C\"\n", + "T3=T2-(Q*Rg)\n", + "print\"T3=\",T3\n", + "print\"The temprature at the inner surface of wall is T4=T3-(Q*Rp)in °C\"\n", + "T4=T3-(Q*Rp)\n", + "print\"T4=\",T4\n", + "#Check for Ti(Temprature inside the van)\n", + "print\"Check for Ti(in °C)\"\n", + "Ti=T4-(Q*Ri)\n", + "print\"The value is same as given in the problem\"\n", + "print\"Ti=\",Ti\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": [ + "## Ex2.2:pg-36" + ] + }, + { + "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 2, Example 2\n", + "The thickness of insulating material L=ki*(((Ti-To)/Q)-(Lb/kb)) in m\n", + "L= 0.056\n" + ] + } + ], + "source": [ + " \n", + "\n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 2\"\n", + "#The Thickness of fire clay bricks (Lb)=.2 m\n", + "Lb=0.2;\n", + "#The thermal conductivity of fire clay bricks(kb)=1.0 W/(m*K)\n", + "kb=1;\n", + "#the Thicknes of insulating material is L\n", + "#The thermal conductivity of insulating material(ki)=.07 W/(m*K)\n", + "ki=.07;\n", + "#The furnace inner brick surface is at temprature Ti=1250 K\n", + "Ti=1250;\n", + "#The furnace outer brick surface is at temprature To=310 K\n", + "To=310;\n", + "#The maximum allowable heat transfer rate(Q) from wall = 900 W/m**2\n", + "Q=900;\n", + "#Q=(Ti-To)/((Lb/kb)+(L/ki)) so L=ki*(((Ti-To)/Q)-(Lb/kb))\n", + "print\"The thickness of insulating material L=ki*(((Ti-To)/Q)-(Lb/kb)) in m\"\n", + "L=ki*(((Ti-To)/Q)-(Lb/kb))\n", + "print\"L=\",L\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.4:pg-36" + ] + }, + { + "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 2, Example 4\n", + "Heat transfer per unit surface area through film qf=(To-Tinf)/((1/h)+(Lf/kf))in W/(m**2)\n", + "qf= 4000.0\n", + "Heat transfer per unit surface area through substrate qs=(To-T1/(Ls/ks)in W/(m**2)\n", + "qs= 1000.0\n", + "Heat flux qo=qs+qf in W/(m**2)\n", + "qo= 5000.0\n", + "If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then\n", + "Rate of heat conduction through the film and substrate q1=(To-T1)/(Ls/ks) in W/m**2\n", + "q1= 1000.0\n", + "The temprature of the top surface of the film T2=(q1*(Lf*10**-3/kf))+To in °C\n", + "T2= 70.0\n", + "Rate of convective heat transfer from the upper surface of film to air q2=h*(T2-Tinf)in W/m**2\n", + "q2= 1250.0\n", + "Heat flux qo=q1+q2 in W/m**2 \n", + "qo= 2250.0\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 4\"\n", + "#To cure the bond at a temprature(To),a radiant source used to provide a heat flux qo W/m**2 \n", + "#The back of the substrate is maintained at a temprature T1\n", + "#The film is exposed to air at a temprature (Tinf)\n", + "#the convective heat transfer coefficient is h.\n", + "#Given To=60°C,Tinf=20°C,h=25 W/(m**K)and T1=30°C\n", + "To=60;\n", + "Tinf=20;\n", + "h=25;\n", + "T1=30;\n", + "#Ls is the thickness of substrate and Lf is thickness of film in mm.\n", + "Ls=1.5;\n", + "Lf=.25;\n", + "#kf and ks are thermal conductivity of film and substrate respectively in W/(m*K)\n", + "kf=.025;\n", + "ks=.05;\n", + "#qo is Heat flux.\n", + "#qo=qf+qs where qf and qs are rate of heat transfer per unit surface area through the film and the substrate respectively.\n", + "print\"Heat transfer per unit surface area through film qf=(To-Tinf)/((1/h)+(Lf/kf))in W/(m**2)\"\n", + "qf=(To-Tinf)/((1/h)+(Lf*10**-3/kf))\n", + "print\"qf=\",qf\n", + "print\"Heat transfer per unit surface area through substrate qs=(To-T1/(Ls/ks)in W/(m**2)\"\n", + "qs=(To-T1)/(Ls*10**-3/ks)\n", + "print\"qs=\",qs\n", + "print\"Heat flux qo=qs+qf in W/(m**2)\"\n", + "qo=qs+qf\n", + "print\"qo=\",qo\n", + "#If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then qo=q1+q2\n", + "#q1 is rate of heat conduction through the film and substrate\n", + "#q2 is rate of convective heat transfer from the upper surface of film to air\n", + "print\"If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then\"\n", + "print\"Rate of heat conduction through the film and substrate q1=(To-T1)/(Ls/ks) in W/m**2\"\n", + "q1=(To-T1)/(Ls*10**-3/ks)\n", + "print\"q1=\",q1\n", + "#To determine q2 we need to find the temprature(T2) of the top surface \n", + "print\"The temprature of the top surface of the film T2=(q1*(Lf*10**-3/kf))+To in °C\"\n", + "T2=(q1*(Lf*10**-3/kf))+To\n", + "print\"T2=\",T2\n", + "print\"Rate of convective heat transfer from the upper surface of film to air q2=h*(T2-Tinf)in W/m**2\"\n", + "q2=h*(T2-Tinf)\n", + "print\"q2=\",q2\n", + "print\"Heat flux qo=q1+q2 in W/m**2 \"\n", + "qo=q1+q2\n", + "print\"qo=\",qo\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.6:pg-38" + ] + }, + { + "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 2, Example 6\n", + "(a)The expression for the temprature distribution in the plate T=160+2*10**3*(x-100*x**2)\n", + "(b)The maximum temprature occurs at x=5mm or 0.005m away from the left surface towards the right\n", + "The maximum temprature(Tmax) in °C is\n", + "Tmax= 165.0\n", + "(c(i))The rate of heat transfer at the left face in MW/m**2 is\n", + "Q= -1\n", + "The minus sign indicates that the heat flow in the negative direction\n", + "(c(ii))The rate of heat transfer at the right face in MW/m**2 is\n", + "Q= 1.2\n", + "The minus sign for (q/A)@x=0 and the plus sign for (q/A)@x=0.02 indicates that heat is lost from both the surfaces to surroundings\n", + "(c(iii))The rate of heat transfer at the centre in MW/m**2 is\n", + "Q= 0.4\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 6\"\n", + "#The thickness of plate 20mm or .02m with uniform heat generation qg=80MW/m**3\n", + "#The thermal conductivity of wall(k)is 200W/m*K.\n", + "k=200;\n", + "#The left and right faces are kept at tempratures T=160°C and T=120°C respectively\n", + "#We start with the equation (d**2T/dx**2)+(qg/k)=0 or T=-(qg/2k)*x**2 +(c1*x)+c2 \n", + "#With qg=80*10**6W/m**3 and k=200W/(m*K)\n", + "#Applying boundary condition at x=0,T=160°C and x=0.02,T=120°C we get constant c2=160°C and c2=2000m**-1\n", + "#Hence T=160+2*10**3*(x-100*x**2)----->eq.1\n", + "print\"(a)The expression for the temprature distribution in the plate T=160+2*10**3*(x-100*x**2)\" \n", + "#For maximum temprature differentiating eq.1 with respect to x and equating it to zero...we get dT/dx=(2*10**3)-(4*10**5*x)=0,which gives x=0.005m=5mm\n", + "print\"(b)The maximum temprature occurs at x=5mm or 0.005m away from the left surface towards the right\"\n", + "x=0.005;\n", + "#The maximum temprature is Tmax\n", + "print\"The maximum temprature(Tmax) in °C is\"\n", + "Tmax=160+2*10**3*(x-100*x**2)\n", + "print\"Tmax=\",Tmax\n", + "#The rate of heat transfer(q/A) is given by -k*(dT/dx)\n", + "#Let dT/dx=X and (q/A)=Q\n", + "print\"(c(i))The rate of heat transfer at the left face in MW/m**2 is\"\n", + "#For left face x=0\n", + "x=0;\n", + "X=(2*10**3)-(4*10**5*x);\n", + "Q=-k*X/10**6\n", + "print\"Q=\",Q\n", + "print\"The minus sign indicates that the heat flow in the negative direction\"\n", + "print\"(c(ii))The rate of heat transfer at the right face in MW/m**2 is\"\n", + "#For right face x=0.02\n", + "x=0.02;\n", + "X=(2*10**3)-(4*10**5*x);\n", + "Q=-k*X/10**6\n", + "print\"Q=\",Q\n", + "#(q/A)@x=0 implies rate of heat transfer at the position where x=0. \n", + "print\"The minus sign for (q/A)@x=0 and the plus sign for (q/A)@x=0.02 indicates that heat is lost from both the surfaces to surroundings\"\n", + "print\"(c(iii))The rate of heat transfer at the centre in MW/m**2 is\"\n", + "#For centre x=0.01\n", + "x=0.01;\n", + "X=(2*10**3)-(4*10**5*x);\n", + "Q=-k*X/10**6\n", + "#A check for the above results can be made from an energy balance of the plate as |(q/A)|@x=0+|(q/A)|@x=0.02=qG*0.02\n", + "print\"Q=\",Q\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.9:pg- 48" + ] + }, + { + "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 2, Example 9\n", + "The critical thickness of insulation in metre is\n", + "X= 0.182340462632\n", + "The heat transfer rate Q per metre of tube length in W/m is \n", + "Q= 26.807459995\n", + "X= 0.559673817307\n", + "The heat transfer rate per metre of tube length Q in W/m is \n", + "Q= 28.1996765363\n", + "X= 1.79194526577\n", + "The heat transfer rate per metre of tube length Q in W/m is \n", + "Q= 21.1086798197\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 9\"\n", + "#A thin walled copper tube of outside metal radius r=0.01m carries steam at temprature, T1=400K.It is inside a room where the surrounding air temprature is Tinf=300K.\n", + "T1=400;\n", + "Tinf=300;\n", + "r=0.01;\n", + "#The tube is insulated with magnesia insulation of an approximate thermal conductivity of k=0.07W/(m*K)\n", + "k=0.07;\n", + "#External convective Coefficient h=4W/(m**2*K)\n", + "h=4;\n", + "#Critical thickness(rc) is given by k/h\n", + "print\"The critical thickness of insulation in metre is\"\n", + "rc=k/h\n", + "#We use the rate of heat transfer per metre of tube length as Q=(Ti-Tinf)/((ln(r2/r1)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L))) where length,L=1m\n", + "L=1;\n", + "#When 0.002m thick layer of insulation r1=0.01m,r2=0.01+0.002=0.012m\n", + "r1=0.01;#inner radius\n", + "r2=0.012;#outer radius\n", + "#Let ln(r2/r1)=X\n", + "X=math.log(r2/r1)/math.log(2.718);\n", + "print\"X=\",X\n", + "#The heat transfer rate per metre of tube length is Q\n", + "print\"The heat transfer rate Q per metre of tube length in W/m is \"\n", + "Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n", + "print\"Q=\",Q\n", + "#When critical thickness of insulation r1=0.01m,r2=0.0175m\n", + "r2=0.0175;#outer radius\n", + "r1=0.01;#inner radius\n", + "#Let ln(r2/r1)=X\n", + "X=math.log(r2/r1)/math.log(2.718);\n", + "print\"X=\",X\n", + "#The heat transfer rate per metre of tube length is Q \n", + "print\"The heat transfer rate per metre of tube length Q in W/m is \"\n", + "Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n", + "print\"Q=\",Q\n", + "#When there is a 0.05 m thick layer of insulation r1=0.01m,r2=.01+0.05=0.06m\n", + "r1=0.01;#inner radius\n", + "r2=0.06;#outer radius\n", + "#Let ln(r2/r1)=X\n", + "X=math.log(r2/r1)/math.log(2.718);\n", + "print\"X=\",X\n", + "#The heat transfer rate per metre of tube length is Q \n", + "print\"The heat transfer rate per metre of tube length Q in W/m is \"\n", + "Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n", + "#It is important to note that Q increases by 5.2% when the insulation thickness increases from 0.002m to critical thickness. \n", + "#Addition of insulation beyond the critical thickness decreases the value of Q (The heat loss).\n", + "print\"Q=\",Q\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.10:pg-49" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 2, Example 10\n", + "the thickness of insulation in metre is\n", + "t= 0.019\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 10\"\n", + "#A copper pipe having 35mm outer diameter(Do) and 30mm inner diameter(Di) carries liquid oxygen to the storage site of a space shuttle at temprature,T1=-182°C\n", + "#mass flow rate is ,mdot=0.06m**3/min. \n", + "Di=0.03;#in metre\n", + "Do=0.035;# in metre\n", + "T1=-182;\n", + "mdot=0.06;\n", + "#The ambient air is at temprature(Ta)=20°C and has a dew point(T3)=10°C.\n", + "Ta=20;\n", + "T3=10;\n", + "#The thermal conductivity(k) of insulating material is 0.02W/(m*k)\n", + "k=0.02;\n", + "#The convective heat transfer coefficient on the outside is h=17W/(m**2*K)\n", + "h=17;\n", + "#The thermal conductivity of copper kcu=400W/(m*K)\n", + "kcu=400;\n", + "#We can write Q=((Ta-T1)/(R1+R2+R3))=((Ta-T3)/(R3)),Rearranging we get ((R1+R2+R3)/(R3))=((Ta-T1)/(Ta-T3))--------eq.1\n", + "#The conduction Resistance of copper pipe(R1)=ln(0.035/0.03)/(2*math.pi*L*kcu)=3.85*10**-4/(2*math.pi*L)K/W\n", + "#The conduction resistance of insulating material (R2)=ln(r3/0.035)/(2*math.pi*L*k)=(1/(2*math.pi*L))((50*ln(r3/0.035)))K/W where r3 is the outer radius of insulation in metres.\n", + "#The convective resistance at the outer surface(R3)=1/(2*math.pi*L*h*r3)=(1/2*math.pi*L)*(mdot/r3)K/W\n", + "#Substituting the values in eq.1 we have 1+((50*ln(r3/0.035)+(3.85*10**-4))/(mdot/r3))=20-(-182)/(20-10)\n", + "#A rearrangement of the above equation gives r3*ln(r3)+3.35*r3=0.023\n", + "#The equation is solved by trial and error method which finally gives r3=0.054m\n", + "r3=0.054;#outer radius of insulation\n", + "#Therefore the thickness of insulation is given by t=r3-Do\n", + "print\"the thickness of insulation in metre is\"\n", + "t=r3-Do\n", + "print\"t=\",t\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.11:pg-52" + ] + }, + { + "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 2, Example 11\n", + "The rate of heat generation of thermal energy in W/m**3 is\n", + "qG= 407436654.315\n", + "The temprature of wire at the centre in K is \n", + "To= 656.631455962\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 11\"\n", + "#An electrical resistance wire 2.5mm or 2.5*10**-3m in diameter(D) and L=0.5m long has a measured voltage drop of E=25V for a current flow of I=40A.\n", + "D=2.5*10**-3;\n", + "I=40;\n", + "E=25;\n", + "L=0.5;\n", + "ro=D/2;#ro is radius of wire\n", + "#The thermal conductivity(k) of wire material is 24W/(m*K) \n", + "k=24;\n", + "#The rate of generation of thermal energy per unit volume is given by qG=(E*I)/(L*math.pi*D**2/4)\n", + "print\"The rate of heat generation of thermal energy in W/m**3 is\"\n", + "qG=(E*I)/(L*math.pi*D**2/4)\n", + "print\"qG=\",qG\n", + "#The temprature at the centre is given by To=Tw+((qG*ro**2)/(4*k)) where Tw=650K is surface temprature \n", + "Tw=650;\n", + "print\"The temprature of wire at the centre in K is \"\n", + "To=Tw+((qG*ro**2)/(4*k))\n", + "#Note:The answer in the book is incorrect(value of D has been put instead of ro)\n", + "print\"To=\",To\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.12:pg-56" + ] + }, + { + "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 2, Example 12\n", + "The inner(A1) and outer surfaces(A2) areas of the tank in m**2 are\n", + "The convective resistance(Ri) at the inner surface in K/W is \n", + "Ri= 0.000159154943092\n", + "The conduction resistance(Rs)of the tank in K/W is\n", + "Rs= 0.0\n", + "The convective resistance(Roc) at the outer surface in K/W is\n", + "Roc= 0.00127323954474\n", + "The radiative heat transfer coefficient hr in W/(m**2*K) is\n", + "hr= 5.26265474661\n", + "Therefore the radiative resistance(Ror) at the outer surface in K/W is\n", + "Ror= 0.00241938642385\n", + "The equivalent resistance in K/W is\n", + "Ro= 0.000834218925785\n", + "The total resistance in K/W is\n", + "Rtotal= 0.000993373868877\n", + "The rate of heat transfer,Q in W is\n", + "Q= 20133.406592\n", + "The outer surface temprature in °C is\n", + "T2= 3.2043311804\n", + "which is sufficiently close to the assumption.So there is no need of further iteration\n", + "The total heat transfer(Qt) during a 24-hour period in KJ is\n", + "Qt= 1739526.32955\n", + "Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is\n", + "mice= 5208.16266333\n" + ] + } + ], + "source": [ + "\n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 12\"\n", + "#A spherical tank of internal diameter, Di=5m ,made of t=25mm thick stainless steel(thermal conductivity,k=15W/(m*K)),is used to store water at temprature(Ti)=0°C.Do is outer diameter\n", + "Di=5;\n", + "t=25;\n", + "Do=5+2*(t/1000);#in metre\n", + "k=15;\n", + "Ti=0;\n", + "#The tank is located in a room whose temprature is (To)=20°C.\n", + "To=20;\n", + "#Emmisivity is 1.\n", + "#The convection heat transfer coefficients at the inner and outer surfaces of the tank are hi=80W/(m**2*K) and ho=10W/(m**2*K)\n", + "hi=80;\n", + "ho=10;\n", + "#The heat of fusion of ice at atmospheric pressure is deltahf=334kJ/kg.The stefan-boltzman constant(sigma) is 5.67*10**-8W/m**2.\n", + "sigma=5.67*10**-8;\n", + "deltahf=334;\n", + "#The inner surface area is (A1) and outer surface area is (A2)of the tank\n", + "print\"The inner(A1) and outer surfaces(A2) areas of the tank in m**2 are\"\n", + "A1=math.pi*Di**2\n", + "A2=math.pi*Do**2\n", + "#The individual thermal resistances can be determined as\n", + "#The convective resistance is (Ri)\n", + "print\"The convective resistance(Ri) at the inner surface in K/W is \"\n", + "Ri=1/(hi*A1)\n", + "print\"Ri=\",Ri\n", + "#The conduction resistance is(Rs)\n", + "print\"The conduction resistance(Rs)of the tank in K/W is\"\n", + "Rs=(Do-Di)/(2*k*math.pi*Di*Do)\n", + "print\"Rs=\",Rs\n", + "#The convective resistance is(Roc)\n", + "print\"The convective resistance(Roc) at the outer surface in K/W is\"\n", + "Roc=1/(ho*A2)\n", + "print\"Roc=\",Roc\n", + "#The radiative resistance(Ror) at the outer surface is Ror=1/(A2*hr) \n", + "#The radiative heat transfer coefficient hr is determined by hr=sigma*(T2**2+293.15**2)*(T2+293.15)\n", + "#But we do not know the outer surface temprature T2 of the tank and hence we cannot determine the value of hr.\n", + "#Therfore we adopt an iterative procedure.We assume T2=4°C=277.15K.Putting the value in hr=sigma*(T2**2+293.15**2)*(T2+293.15) we get\n", + "T2=277.15;\n", + "print\"The radiative heat transfer coefficient hr in W/(m**2*K) is\"\n", + "hr=sigma*(T2**2+293.15**2)*(T2+293.15)\n", + "print\"hr=\",hr\n", + "print\"Therefore the radiative resistance(Ror) at the outer surface in K/W is\"\n", + "Ror=1/(A2*hr)\n", + "print\"Ror=\",Ror\n", + "#The two parallel resistances Roc and Ror can be replaced by an equivalent resistance Ro as X=(1/Ro)=(1/Roc)+(1/Ror)\n", + "print\"The equivalent resistance in K/W is\"\n", + "X=(1/Roc)+(1/Ror);\n", + "Ro=1/X\n", + "print\"Ro=\",Ro\n", + "#Now the resistance Ri,Rs and Ro are in series an hence the total resistance becomes Rtotal=Ri+Rs+Ro \n", + "print\"The total resistance in K/W is\"\n", + "Rtotal=Ri+Rs+Ro \n", + "print\"Rtotal=\",Rtotal\n", + "#The rate of heat transfer is given by Q=(To-Ti)/Rtotal\n", + "print\"The rate of heat transfer,Q in W is\"\n", + "Q=(To-Ti)/Rtotal\n", + "print\"Q=\",Q\n", + "#The outer surface(T2) is calculated as T2=To-Q*Ro\n", + "print\"The outer surface temprature in °C is\"\n", + "T2=To-Q*Ro\n", + "print\"T2=\",T2\n", + "print\"which is sufficiently close to the assumption.So there is no need of further iteration\"\n", + "#The total heat transfer is (Qt),during a 24-hour period\n", + "print\"The total heat transfer(Qt) during a 24-hour period in KJ is\"\n", + "Qt=Q*24*3600/1000\n", + "print\"Qt=\",Qt\n", + "#the amount of ice in kG which melts during a 24 hour period is (mice)\n", + "print\"Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is\"\n", + "mice=Qt/deltahf\n", + "print\"mice=\",mice\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.13:pg-68" + ] + }, + { + "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 2, Example 13\n", + "P/A in m**-1 is\n", + "X= 400.0\n", + "m in m**-1 is\n", + "m= 3.28797974611\n", + "M in W/K is\n", + "M= 0.0955478103936\n", + "thetab in °C is \n", + "thetab= 100\n", + "Heat loss from rod in Watt, for different value of length(in m) is \n", + "Q= 0.705573384326\n", + "Q= 1.32655528327\n", + "Q= 2.53006298798\n", + "Q= 5.56299390901\n", + "Q= 8.29001416681\n", + "Q= 9.45769063154\n", + "Q= 9.52862252977\n", + "Q= 9.55478103936\n", + "For an infintely long rod heat loss in W is\n", + "We see that since k is large there is significant difference between the finite length and the infinte length cases\n", + "However when the length of the rod approaches 1m,the result become almost same.\n", + "Qinf= 9.55478103936\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 13\"\n", + "#A very long,10mm diameter(D) copper rod(thermal conductivity,k=370W/(m*K))is exposed to an enviroment at temprature,Tinf=20°C.\n", + "D=0.01;\n", + "k=370;\n", + "Tinf=20;\n", + "#The base temprature of the radius maintained at Tb=120°C.\n", + "Tb=120;\n", + "#The heat transfer coefficient between the rod and the surrounding air is h=10W/(m*K**2)\n", + "h=10;\n", + "#The rate of heat transfer for all finite lengths will be given by P/A=(4*pi*D)/(pi*D**2)\n", + "#Let P/A=X\n", + "print\"P/A in m**-1 is\"\n", + "X=(4*math.pi*D)/(math.pi*D**2)\n", + "print\"X=\",X\n", + "#m is defined as ((h*p)/(k*A)]**0.5 \n", + "print\"m in m**-1 is\"\n", + "m=(h*X/k)**0.5\n", + "print\"m=\",m\n", + "#Let Y=h/(m*k)\n", + "Y=h/(m*k)\n", + "#Let M=(h*P*k*A)**0.5\n", + "P=(math.pi*D);#perimeter of the rod\n", + "A=(math.pi*D**2)/4;#Area of the rod\n", + "print\"M in W/K is\"\n", + "M=(h*P*k*A)**0.5\n", + "print\"M=\",M\n", + "#thetab is the parameter that defines the base temprature\n", + "print\"thetab in °C is \"\n", + "thetab=Tb-Tinf\n", + "print\"thetab=\",thetab\n", + "#Heat loss from the rod is defined as Q=(h*P*k*A)*thetab*(((h/m*k)+math.tanh(m*L)]/(1+(h/m*k)*math.tanh(m*L)]}\n", + "print\"Heat loss from rod in Watt, for different value of length(in m) is \"\n", + "L=0.02#Length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "L=0.04#length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "L=0.08#length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "L=0.20#length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "L=0.40#length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "L=0.80#length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "L=1.00#length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "L=10.00#length of rod\n", + "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n", + "print\"Q=\",Q\n", + "#For an infinitely long rod we use heat loss as ,Qinf=(h*P*k*A)**0.5*thetab\n", + "print\"For an infintely long rod heat loss in W is\"\n", + "Qinf=(h*P*k*A)**0.5*thetab\n", + "print\"We see that since k is large there is significant difference between the finite length and the infinte length cases\"\n", + "print\"However when the length of the rod approaches 1m,the result become almost same.\" \n", + "print\"Qinf=\",Qinf\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.14:pg-81" + ] + }, + { + "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 2, Example 14\n", + "The thermal conductivity of Rod B kB in W/(m*K) is \n", + "kB= 18.0\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 14\"\n", + "#Considering two very long slender rods of the same diameter but of different materials.\n", + "#The base of each rod is maintained at 100°C while the surfaces of the rod are exposed to 20°C\n", + "#By traversing length of each rod with a thermocouple it was observed that tempratures of rod were equal at the position xA=0.15m and xB=0.075 from base.\n", + "xA=0.15;\n", + "xB=0.075;\n", + "#Thermal conductivity of rod A is known to be kA=72 W/(m*K)\n", + "kA=72;\n", + "#In case of a very long slender rod we use the tip boundary condition thetaL=0 as L--->infinity\n", + "#Therfore we can write for the locations where the tempratures are equal thetab*e**(-mA*xA)=thetab*e**(-mB*xB) or xA/xB=mB/mA,Again mB/mA=(kA/kB)**0.5\n", + "#So kB=kA*(xB/xA)**2\n", + "#The thermal conductivity of Rod B iskB\n", + "print\"The thermal conductivity of Rod B kB in W/(m*K) is \"\n", + "kB=kA*(xB/xA)**2\n", + "print\"kB=\",kB\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", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.15:pg-83" + ] + }, + { + "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 2, Example 15\n", + " perimeter of each fin in m is\n", + "P= 0.202\n", + "Cross sectional area of fin in m**2 is\n", + "A= 0.0001\n", + "M= 0.834805366538\n", + "m= 36.2958855016\n", + "Temprature parameter at fin base in K is\n", + "thetab= 100\n", + "Fixed temprature at fin tip in K is\n", + "thetaL= 50\n", + "Heat loss from the plate at 400K in W is\n", + "Qb= 13028.1662009\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 15\"\n", + "#A stack that is b=300mm wide and l=100mm deep contains N=60 fins each of length L=12mm.\n", + "L=0.012;#in metre\n", + "b=0.3;#in metre\n", + "l=0.1;#in metre\n", + "N=60;\n", + "#The entire stack is made of aluminum which is everywhere t=1.0 mm thick.\n", + "t=0.001;#in metre\n", + "#The temprature limitations associated with electrical components are Tb=400K and TL=350K.\n", + "#Tb is base temprature and TL is end temprature\n", + "Tb=400;\n", + "TL=350; \n", + "#Given convection heat transfer coefficient(h=150W/(m**2*K)),Surrounding Temprature(Tinf=300K),thermal conductivity of aluminium(kaluminium=230W/(m*K))\n", + "h=150;\n", + "Tinf=300;\n", + "kal=230;\n", + "#Here both the ends of the fins are at fixed tempratures .Therefore we use M=(h*P*k*A)**0.5 and m=((h*P)/(k*A))**0.5,thetab=Tb-Tinf,thetaL=TL-Tinf\n", + "#from the given data perimeter of each fin is given by P= 2*(l+t)in m and area of each fin is A=t*l\n", + "print\" perimeter of each fin in m is\"\n", + "P= 2*(l+t)\n", + "print\"P=\",P\n", + "print\"Cross sectional area of fin in m**2 is\"\n", + "A=t*l\n", + "print\"A=\",A\n", + "#M is defined as (h*P*kal*A)**0.5 and m is defined as ((h*P)/(kal*A))**0.5\n", + "M=(h*P*kal*A)**0.5\n", + "m=((h*P)/(kal*A))**0.5\n", + "print\"M=\",M\n", + "print\"m=\",m\n", + "#thetab and thetaL are the parameters that define the fin tempratures at base and tip respectively.\n", + "print\"Temprature parameter at fin base in K is\"\n", + "thetab=Tb-Tinf\n", + "print\"thetab=\",thetab\n", + "print\"Fixed temprature at fin tip in K is\"\n", + "thetaL=TL-Tinf\n", + "print\"thetaL=\",thetaL\n", + "#Heat loss from the plate is Qb\n", + "print\"Heat loss from the plate at 400K in W is\"\n", + "Qb=(N*(h*P*kal*A)**0.5*thetab*((math.cosh(m*L)-(thetaL/thetab))/(math.sinh(m*L))))+(((l*b)-(N*A))*h*thetab)+(l*b*h*thetab)\n", + "print\"Qb=\",Qb\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |