{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Chapter 01:Fundamental concepts"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.1:pg- 4"
]
},
{
"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 1, Example 1\n",
"The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2 \n",
"q= 50.0\n"
]
}
],
"source": [
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 1\"\n",
"#The temprature of two faces of the slabs are T1=40Â°C & T2=20Â°C \n",
"#The thickness of the slab(L) is 80mm or .08m\n",
"#The thermal conductivity(k)of the material is .20 W/(m*K)\n",
"T1=40;\n",
"T2=20;\n",
"L=.08;\n",
"k=.20;\n",
"#The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L\n",
"print\"The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2 \"\n",
"q=k*(T1-T2)/L\n",
"print\"q=\",q"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.2:pg- 4"
]
},
{
"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 1, Example 2\n",
"The thickness of masonry wall is Lm in m\n",
"Lm= 0.5\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\"\n",
"#The thermal conductivity(km)of masonry wall is .8 W/(mK)\n",
"#The thermal conductivity(kc)of composite wall is .2 W/(mK)\n",
"#The thickness of composite wall(Lc) is 100 mm or .1 m\n",
"km=.8;\n",
"kc=.2;\n",
"Lc=.1;\n",
"#The thickness of masonry wall(Lm) is to be found. \n",
"#The steady state heat flow(qm)through masonry wall is km(T1-T2)/L\n",
"# The steady state heat flow(qc)through composite wall is kc(T1-T2)/L\n",
"#As the steady rate of heat flow through masonry wall is 80% that through composite wall and both the wall have same surface area and same temp. difference so qm/qc=0.8=(km/kc)*(Lc/Lm)\n",
"#The thickness of masonry wall is Lm.\n",
"print\"The thickness of masonry wall is Lm in m\"\n",
"Lm=(km/kc)*(Lc/(0.8))\n",
"print\"Lm=\",Lm"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.4:pg-8"
]
},
{
"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 1, Example 4\n",
"The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2\n",
"q= 16000\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 4\"\n",
"#The average forced convective heat transfer coefficient(hbr) is 200 W/( m**2 Â°C)\n",
"#The fluid temprature(Tinf) upstream of the cold surface is 100Â°C\n",
"#The surface temprature(Ts) is 20Â°C\n",
"hbr=200;\n",
"Tinf=100;\n",
"Ts=20;\n",
"#The rate of heat transfer per unit area is q\n",
"print\"The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2\"\n",
"q=hbr*(Tinf-Ts)\n",
"print\"q=\",q"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.5:pg-9"
]
},
{
"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 1, Example 5\n",
"The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is \n",
"A= 0\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 5\"\n",
"#The average heat transfer coefficient(hbr) is 800 W/(m**2Â°C)\n",
"#The surface temprature of heat exchanger is 75Â°C and air temprature is 25Â°C so deltaT=(75-25)\n",
"#The amount of heat exchanged(Q) is 20 MJ/h\n",
"#The heat exchanger surface area(A) is given by A=Q/(hbr*âˆ†T)\n",
"hbr=800;\n",
"deltaT=(75-25);\n",
"Q=20;\n",
"print\"The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is \"\n",
"A = (Q*10**6)/(3600*hbr*deltaT)\n",
"print\"A=\",A"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.6:pg-9"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\n",
"The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\n",
"Q= 224.0\n",
"The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\n",
"Q= 224.0\n",
"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2Â°C)\n",
"hbr= 11.2\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\"\n",
"#The temprature of the plate(Ts) is 225Â°C\n",
"#The ambient temprature (Tinf) is 25Â°C\n",
"#The change in plate temprature with time is dT/dt=-.02K/s\n",
"#The plate area (A)=.1m**2 , mass(m)= 4Kg and specific heat(cp)=2.8KJ/(Kg*K)\n",
"#The average free convective heat coefficient(hbr) is to be found\n",
"Ts=225;\n",
"Tinf=25;\n",
"#|dT/dt|=0.2,because it is modulus function and it converts negative values to positive value.\n",
"#Let |dT/dt|=X\n",
"X=0.02;\n",
"A=.1;\n",
"m=4;\n",
"cp=2.8;\n",
"print\"The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\"\n",
"Q=hbr*A*(Ts-Tinf)\n",
"print\"Q=\",Q\n",
"print\"The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\"\n",
"print\"Q=\",Q\n",
"print\"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2Â°C)\"\n",
"hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))\n",
"print\"hbr=\",hbr"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.7:pg-10"
]
},
{
"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 1, Example 7\n",
"The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2\n",
"F= 556.4411381\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 7\"\n",
"#The temprature(T) of brick wall after sunset is 50Â°C\n",
"#The emissity value(emi)=0.9\n",
"#The radiant heat flux per square meter =E/A Where E is radiant heat energy and A is area of brick wall.\n",
"#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).\n",
"T=50;\n",
"emi=.9;\n",
"sigma=5.6697*10**-8;\n",
"print\"The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2\"\n",
"#Let E/A=F\n",
"F=emi*sigma*(T+273.15)**4\n",
"print\"F=\",F"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.8:pg-11"
]
},
{
"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 1, Example 8\n",
"The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\n",
"F= 618.267931222\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 8\"\n",
"#The temprature(T) of asphalt pavement = 50Â°C\n",
"#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).\n",
"T=50;\n",
"sigma=5.6697*10**-8;\n",
"#The emitted radiant energy per unit surface area is given by (Eb/A)=sigma*T**4\n",
"print\"The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\"\n",
"#Let Eb/A=F\n",
"F=sigma*(50+273.15)**4\n",
"print\"F=\",F"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.9:pg-12"
]
},
{
"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 1, Example 9\n",
"The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2\n",
"F= 213.333333333\n",
"The surface tempratures of wall on 60Â°C side is T1 =Ta-(Q/(A*hbr1)) in Â°C\n",
"T1= 54.6666666667\n",
"The surface tempratures of wall on 20Â°C side is T2 =Tb+(Q/(A*hbr2)) in Â°C\n",
"T2= 41.3333333333\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 9\"\n",
"#The Thickness(L) of wall= 150 mm or 0.15 m.\n",
"#The wall on one side is exposed to air at temprature(Ta)= 60Â°C and on the other side to air at temprature(Tb) = 20Â°C\n",
"#The average convective heat transfer coefficients are hbr1=40 W/(m**2Â°C) on the 60Â°C and hbr2= 10 W/(m**2Â°C) on 20Â°C side.\n",
"#The thermal conductivity(k)=.8 W/(mÂ°C)\n",
"L=0.15;\n",
"Ta=60;\n",
"Tb=20;\n",
"hbr1=40;\n",
"hbr2=10;\n",
"k=0.8;\n",
"#Area(A=1 m**2 )since unit surface area is required.\n",
"A=1;\n",
"#The rate of heat transfer per unit surface area of wall is given by (Q/A)=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))\n",
"print\"The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2\"\n",
"#Let Q/A=F\n",
"F=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))\n",
"print\"F=\",F\n",
"#The surface tempratures of wall on 60Â°C side is T1 and on 20Â°C side is T2\n",
"print\"The surface tempratures of wall on 60Â°C side is T1 =Ta-(Q/(A*hbr1)) in Â°C\"\n",
"T1 =Ta-(F/hbr1)\n",
"print\"T1=\",T1\n",
"print\"The surface tempratures of wall on 20Â°C side is T2 =Tb+(Q/(A*hbr2)) in Â°C\"\n",
"T2 =Tb+(F/hbr2)\n",
"print\"T2=\",T2"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.10:pg-13"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\n",
"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\n",
"F1= 332.029390022\n",
"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\n",
"F2= 332.132667923\n",
"The values of temprature that are considered are <298 K\n",
"Satisfactory solutions for Temprature in K is\n",
"T2= 292.5\n",
"Approximate Rate of Heat Transfer in W/m**2 is\n",
"F1= 332.029390022\n",
"F2= 332.132667923\n"
]
}
],
"source": [
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\"\n",
"#The spacecraft panel has thickness(L)=.01 m\n",
"#The spacecraft has inner temprature (Ti)=298 K\n",
"#The spacecraft has outer temprature(T2)\n",
"#The panel is exposed to deep space where temprature(To)= 0K\n",
"#The material has Thermal conductivity(k)= 5.0 W/(m*K)\n",
"#The emissivity(emi)=0.8\n",
"#The inner surface of the panel is exposed to airflow resulting in an average heat transfer coefficient(hbri)=70 W/(m**2*K)\n",
"L=0.01;\n",
"Ti=298.0;\n",
"To=0.0;\n",
"k=5.0;\n",
"emi=0.8;\n",
"hbri=70.0;\n",
"#The stefan Boltzman constant(sigma)= 5.67*10**-8 W/(m**2/K**4)\n",
"sigma=5.67*10**(-8);\n",
"#Heat transfer from the outer surface takes place only by radiation is given by Q/A=emi*sigma*(T2**4-T0**4)in W/m**2=F1\n",
"#heat transfer from the outer surface can also be written as Q/A=(Ti-To)/((1/hbri)+(L/k)+(1/hr))=F2\n",
"#Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)\n",
"#so hr=4.536*10**-8*T2**3\n",
"print\"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\"\n",
"print\"F1=\",F1\n",
"print\"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\"\n",
"print\"F2=\",F2\n",
"print\"The values of temprature that are considered are <298 K\"\n",
"for i in range(285,292):\n",
" T2=i\n",
" hr=4.536*10**(-8)*i**3\n",
" F1=emi*sigma*(T2**4-To**4)\n",
" F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))\n",
"if F1==F2:\n",
" T2=i\n",
"else: \n",
" T2=292.5\n",
" hr=4.536*10**(-8)*T2**3\n",
" F1=emi*sigma*(T2**4-To**4)\n",
" F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))\n",
"print\"Satisfactory solutions for Temprature in K is\"\n",
"print\"T2=\",T2\n",
"print\"Approximate Rate of Heat Transfer in W/m**2 is\"\n",
"print\"F1=\",F1\n",
"print\"F2=\",F2"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex1.11:pg-15"
]
},
{
"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 1, Example 11\n",
"L= 1\n",
"A= 0.251327412287\n",
"The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\n",
"F= 121.586773684\n"
]
}
],
"source": [
"\n",
"import math \n",
"\n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 11\"\n",
"#The horizontal steel pipe has outer diameter(D)=80 mm or.08 m\n",
"#The pipe is maintained at a temprature(T1)=60Â°C where the air and wall temprature(T2)=20 Â°C \n",
"#The average free convective heat transfer coefficient(hbr)=6.5 W/(m**2/K) b/w the outer surface of the pipe and air\n",
"D=.08;\n",
"T1=60;\n",
"T2=20;\n",
"hbr=6.5;\n",
"#Length(L=1) since per unit length is considered\n",
"L=1;\n",
"#The surface area of pipe is given by A=(math.pi*D*L)\n",
"print\"L=\",L\n",
"A=(math.pi*D*L);\n",
"#The surface emissivity(emi) of steel = 0.8\n",
"#The stefan -Boltzman constant(sigma)= 5.7*10**-8 W/(m**2*K**4)\n",
"print\"A=\",A\n",
"sigma=5.67*10**-8;\n",
"emi=.8;\n",
"#The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4)\n",
"print\"The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\"\n",
"#Let Q/L=F\n",
"F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)**4-(T2+273.15)**4)\n",
"print\"F=\",F"
]
}
],
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