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"source": [
"# Chapter 5:Convection"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.3:pg-206"
]
},
{
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"Introduction to heat transfer by S.K.Som, Chapter 5, Example 3\n",
"Reynolds number is\n",
"ReL= 55263.1578947\n",
"The average heat transfer coefficient over a length(L)= 1m ,in W/m**2 is\n",
"hbarL= 4.16\n",
"The rate of heat transfer in W/m of width is\n",
"Q= 332.8\n"
]
}
],
"source": [
"import math \n",
"from scipy.integrate import quad\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 3\"\n",
"#Air at temprature (T1=20Â°C) and 1 atmospheric pressure flows over a flat plate with a free stream velocity(Uinf) of 1m/s.\n",
"Uinf=1;\n",
"T1=20;\n",
"#The length of plate is 1m and is heated over its entire length to a constant temprature of T2=100Â°C.\n",
"T2=100;\n",
"#For air at 20Â°C(The mean temprature of 100Â°C and 20Â°C),viscosity(mu=1.9*10**-5kg/(m*s)),density(rho=1.05kg/m**3),conductivity(k=0.03W/(m*K)),specific heat(cp=1.007kJ/(kg*K))\n",
"#Prandtl number is Pr=0.7\n",
"mu=1.9*10**-5;\n",
"rho=1.05;\n",
"k=0.03;\n",
"cp=1.007;\n",
"Pr=0.7;\n",
"#For laminar flow over a plate Nusselt number is Nux=0.332*Rex**0.5*Pr**(1/3)\n",
"#The boundary layer flow over a flat plate will be laminar if Reynolds number is Rex=(rho*Uinf*x)/mu<5*10**5\n",
"#First of all we have to check whether the flow is laminar or not.\n",
"#Let us check at x=1m\n",
"x=1.0;\n",
"print\"Reynolds number is\"\n",
"ReL=(rho*Uinf*x)/mu\n",
"print\"ReL=\",ReL\n",
"#There fore the flow is laminar and we can use the relationships of Nux,\n",
"#Thus Rex=(1.05*1*x)/(1.9*10**-5)=0.5526*10**5*x\n",
"#Therefore we can write Nux=(hx*x/k)=0.332*(0.5526*10**5*x)**0.5*Pr**(1/3)....or hx=2.08*x**(-1/2) W/(m**2*Â°C)\n",
"#hbarL is the average heat transfer coefficient over a length(L)\n",
"print\"The average heat transfer coefficient over a length(L)= 1m ,in W/m**2 is\"\n",
"L=1;\n",
"hbarL=(1.0/L)*quad(lambda x:2.08*x**(-1/2.0),0,L)[0]\n",
"print\"hbarL=\",hbarL\n",
"#Q is the rate of heat transfer\n",
"print\"The rate of heat transfer in W/m of width is\"\n",
"Q=hbarL*L*(T2-T1)\n",
"print\"Q=\",Q"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.4:pg-207"
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {
"collapsed": false
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"outputs": [
{
"name": "stdout",
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"text": [
"Introduction to heat transfer by S.K.Som, Chapter 5, Example 4\n",
"The local heat transfer coefficient hx is hx=27.063*U**0.85\n",
"The minimum flow velocity in m/s is\n",
"U= 15.4806813943\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 4\"\n",
"#Air at atmospheric pressure is required to flow over a circuit board to cool the electronics element mounted on it.\n",
"#Chip has length (L)=3mm and width(B)=3mm located x=0.1m from the leading edge\n",
"L=0.003;#in metre\n",
"B=0.003;#in metre\n",
"x=0.1;\n",
"#The Nusselt no. is given by Nux=0.06*Rex**0.85*Pr**0.33\n",
"#The chip has to dissipate E=50mW of energy while its surface temprature has to be kept below temprature,Ts=45Â°C and free strem Temptrature of air is Tinf=25Â°C\n",
"Ts=45;\n",
"Tinf=25;\n",
"E=50*10**-3;#in watt\n",
"#For air ,density(rho=1.2kg/m**3),viscosity(mu=1.8*10**5kg/(m*s)),conductivity(k=0.03W/(m*K)) and specific heat(cp=1000J/(kg*K))\n",
"rho=1.2;\n",
"mu=1.8*10**5;\n",
"k=0.03;\n",
"cp=1000;\n",
"#Let the minimum flow velocity be U.\n",
"#The local heat transfer coefficient hx where the chip is mounted is determined as hx=(k/x)*0.06*(rho*U*x/mu)**0.85*(mu*cp/k)**0.33\n",
"print\"The local heat transfer coefficient hx is hx=27.063*U**0.85\"\n",
"#from an enrgy balance we can write as E=27.063*U**0.85*L*B*(Ts-Tinf)\n",
"print\"The minimum flow velocity in m/s is\"\n",
"U=(E/(27.063*L*B*(Ts-Tinf)))**(1/0.85)\n",
"print\"U=\",U"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.6:pg-208"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
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"outputs": [
{
"name": "stdout",
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"text": [
"Introduction to heat transfer by S.K.Som, Chapter 5, Example 6\n",
"mdot= 0.0235619449019\n",
"(dTb/dx)in Â°C/m is\n",
"Y= 26.6666666667\n",
"Therefore Exit bulk mean temprature Tb2 in Â°C is\n",
"Tb2= 83.3333333333\n",
"Heat flux(hx) in W/(m**2*Â°C) is \n",
"hx= 100\n",
"Overall Nusselt number is \n",
"NuL= 87.7192982456\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 6\"\n",
"#Air at 1atm pressure and temprature(Tin)=30Â°C enters a tube of 25mm diameter(D) with a velocity(U) of 10m/s\n",
"D=0.025;#in metre\n",
"U=10;\n",
"Tin=30;\n",
"#Tube is heated so that a constant heat flux(q) of 2kW/m**2 is maintained at the wall whose temprature is deltaT=20Â°C above the bulk mean air temprature through the length of tube \n",
"#Let Tw-Tb=T\n",
"q=2000;\n",
"deltaT=20;\n",
"#The length(L)= 2m\n",
"L=2;\n",
"#For air density(rho=1.2kg/m**3),specific heat(cp=1000J/(kg*K))\n",
"rho=1.2;\n",
"cp=1000;\n",
"#From an energy balance of a control volume of air we get mdot*cp*(Tb+(dTb/dx)*deltax-Tb)=q*pi*D*deltax or (dTb/dx)=(q*pi*D)/(mdot*cp)\n",
"#mass flow rate=mdot\n",
"mdot=rho*math.pi*D**2*U;\n",
"print\"mdot=\",mdot\n",
"#let (dTb/dx)=Y\n",
"print\"(dTb/dx)in Â°C/m is\"\n",
"Y=(4*q*math.pi*D)/(mdot*cp)\n",
"print\"Y=\",Y\n",
"#Tb2 is Exit bulk mean temprature\n",
"print\"Therefore Exit bulk mean temprature Tb2 in Â°C is\"\n",
"Tb2=Tin+Y*2\n",
"print\"Tb2=\",Tb2\n",
"#Again we can write at any section of the tube hx*(Tw-Tb)=q or hx=q/(Tw-Tb)\n",
"#hx is heat flux\n",
"print\"Heat flux(hx) in W/(m**2*Â°C) is \"\n",
"hx=q/(deltaT)\n",
"print\"hx=\",hx\n",
"#Since Tw-Tb remains the same,The heat transfer coefficient at all sections are the same\n",
"#Now Overall Nusselt number,NuL=hx*D/k\n",
"#The thermal conductivity of air at mean temprature of (30+83.4)/2=56.7Â°C is k=0.0285 W/(m*K)\n",
"k=0.0285;\n",
"print\"Overall Nusselt number is \"\n",
"NuL=hx*D/k\n",
"print\"NuL=\",NuL"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.7:pg-210"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
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"outputs": [
{
"name": "stdout",
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"text": [
"Introduction to heat transfer by S.K.Som, Chapter 5, Example 7\n",
"Beta(The volumetric coefficient of expansion in K**-1 is\n",
"Beta= 0\n",
"Grashoff number is\n",
"Gr= 0.0\n",
"The average nusselt number is\n",
"NuHbar= 0.13\n",
"Heat flux hbar in W/(m**2*Â°C)\n",
"hbar= 0.00169\n",
"The heat loss from the plate in W is\n",
"Q= 0.4394\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 7\"\n",
"#A wall is exposed to nitrogen at one atmospheric pressure and temprature,Tinf=4Â°C.\n",
"Tinf=4;\n",
"#The wall is H=2.0m high and B=2.5m wide and is maintained at temprature,Ts=56Â°C\n",
"Ts=56;\n",
"H=2;\n",
"B=2.5;\n",
"A=H*B;#area is(A)\n",
"#The average nusselt number NuHbar over the height of the plate is given by NuHbar=0.13*(Gr*Pr)**(1/3)\n",
"#The properties of nitrogen at mean film temprature(Tf) is (56+4)/2=30Â°C are given as density(rho=1.142kg/m**3) ,conductivity(k=0.026W/(m*K)),\n",
"#kinematic viscosity(nu=15.630*10**-6 m**2/s) ,Prandtl number(Pr=0.713)\n",
"rho=1.142;\n",
"k=0.026;\n",
"nu=15.630*10**-6;\n",
"Pr=0.713;\n",
"Tf=30;\n",
"#We first have to detrmine the value of Grashoff number,Gr.In consideration of nitrogen as an ideal gas,we can write\n",
"#Beta(The volumetric coefficient of expansion)=1/T\n",
"print\"Beta(The volumetric coefficient of expansion in K**-1 is\"\n",
"Beta=1/(273+Tf)\n",
"print\"Beta=\",Beta\n",
"#Now Gr=(g*Beta*(Ts-Tinf)*H**3)/nu**2\n",
"g=9.81;#acceleration due to gravity\n",
"print\"Grashoff number is\"\n",
"Gr=(g*Beta*(Ts-Tinf)*H**3)/nu**2\n",
"print\"Gr=\",Gr\n",
"print\"The average nusselt number is\"\n",
"NuHbar=0.13*(Gr*Pr)**(1/3)\n",
"print\"NuHbar=\",NuHbar\n",
"#hbar is the heat flux\n",
"print\"Heat flux hbar in W/(m**2*Â°C)\"\n",
"hbar=NuHbar*k/H\n",
"print\"hbar=\",hbar\n",
"#Q is the heat loss from the plate\n",
"print\"The heat loss from the plate in W is\"\n",
"Q=hbar*A*(Ts-Tinf)\n",
"print\"Q=\",Q"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.8:pg-211"
]
},
{
"cell_type": "code",
"execution_count": 1,
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"outputs": [
{
"name": "stdout",
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"text": [
"Introduction to heat transfer by S.K.Som, Chapter 5, Example 8\n",
"The heat flux in W/(m**2*K) is\n",
"hbar 120.0\n",
"The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt\n",
"Q= 1.88495559215\n",
"The current in Ampere is\n",
"I= 17.7245385091\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 8\"\n",
"#Eletric current passes through a L=0.5m long horizontal wire of D=0.1mm diameter.\n",
"L=0.5;\n",
"D=0.1*10**-3;\n",
"#The wire is to be maintained at temprature,Twire=400K and the air is at temprature,Tair=300K.\n",
"Twire=400;\n",
"Tair=300;\n",
"#The resistance of the wire(R) is 0.012ohm per meter.Nusselt number(NuL) over the length of wire to be 0.4.\n",
"NuL=0.4;\n",
"R=0.012;\n",
"#At mean temprature of Tf=350K, The thermal conductivity of air is k=0.03W/(m*K)\n",
"k=0.03;\n",
"#Nusselt number is NuL=hbar*D/k\n",
"#hbar is the heat flux\n",
"print\"The heat flux in W/(m**2*K) is\"\n",
"hbar=NuL*k/D\n",
"print\"hbar\",hbar\n",
"#Q is the heat loss from the wire\n",
"print\"The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt\"\n",
"Q=hbar*math.pi*D*L*(Twire-Tair)\n",
"print\"Q=\",Q\n",
"#At steady state the ohmic loss in the wire equals the heat loss from its surface Therfore I**2*R=Q\n",
"#I is the current flow.\n",
"print\"The current in Ampere is\"\n",
"I=(Q/(R*L))**0.5\n",
"print\"I=\",I"
]
}
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