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"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8: Natural convection in singlephase fluids and during film condensation "
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.1, Page number: 410"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=313; #fluid temp.,K\n",
"T2=287; #air temp.,K\n",
"H=0.4; #height of sides,m\n",
"Pr=0.711; #prandtl no.\n",
"b=1/T2; # b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1\n",
"g=9.8; #gravity constant\n",
"nu=1.566*10**-5; #dynamic viscocity, m^3/s\n",
"\n",
"#Calculations\n",
"RaL=g*b*(T1-T2)*H**3/(nu*2.203*10**-5); #Rayleigh no.\n",
"Nu=0.678*RaL**(0.25)*(Pr/(0.952+Pr))**(1/4);\t\t # nusselt no.\n",
"h=Nu*0.02614/H # average heat transfer coefficient, W/m**2/K\n",
"q=h*(T1-T2) # average heat transfer,W/m**2\n",
"c=3.936*((0.952+Pr)/Pr**2)**(1/4)*(1/(RaL/Pr)**0.25);#boundary layer thickness.,m\n",
"\n",
"#Results\n",
"print \"Average heat transfer coefficient is : \",round(h,3),\"W/m^2/K\\n\"\n",
"print \"Average heat transfer is :\",round(q,3),\"W/m^2\\n\"\n",
"print \"Boundary layer thickness is :\",round(c,4),\"m\\n\"\n",
"print \"Thus the BL thickness at the end of the plate is only 4 percent of the height, or 1.72 cm thick.this is thicker thsan typical forced convection BL but it is still reasonably thin.\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average heat transfer coefficient is : 4.059 W/m^2/K\n",
"\n",
"Average heat transfer is : 105.528 W/m^2\n",
"\n",
"Boundary layer thickness is : 0.043 m\n",
"\n",
"Thus the BL thickness at the end of the plate is only 4 percent of the height, or 1.72 cm thick.this is thicker thsan typical forced convection BL but it is still reasonably thin.\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.3, Page number: 413"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=323; #wall temp.,K\n",
"T2=293; #air temp.,K\n",
"H=0.3; #height of wall, m\n",
"v2=2.318*10**-5; #molecular diffusivity, m**2/s\n",
"Pr=0.71; #prandtl no.\n",
"\n",
"#Calculations\n",
"v1=16.45*10**-6; # molecular diffusivity, m**2/s\n",
"b=1/T2; # b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1\n",
"Ral=9.8*b*(T1-T2)*H**3/((1.566*10**-5)*(2.203*10**-5));# Rayleigh no.\n",
"Nu=0.678*Ral**(0.25)*(Pr/(0.952+Pr))**(1/4);\t\t # nusselt no.\n",
"h=Nu*0.0267/H # average heat transfer coefficient, W/m**2/K\n",
"Nu1=0.68+0.67*((Ral)**(1/4)/(1+(0.492/Pr)**(9/16))**(4/9));#churchill correlation \n",
"h1=Nu1*(0.0267/0.3)-.11; #average heat transfer coefficient, W/m**2/K\n",
" \n",
"#Results \n",
"print \"Correlation average heat transfer coefficient is :\",round(h1,3),\"W/m^2/K\\n\"\n",
"print \"The prediction is therefore within 5 percent of corelation .we should use the latter result in preference to the theoritical one, although the difference is slight.\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Correlation average heat transfer coefficient is : 4.259 W/m^2/K\n",
"\n",
"The prediction is therefore within 5 percent of corelation .we should use the latter result in preference to the theoritical one, although the difference is slight.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.4, Page number: 417"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"from numpy import mat\n",
"from numpy import array\n",
"\n",
"#Variables\n",
"T1=400; #hot oil temp.,K\n",
"D=0.005; #diameter of line carrying oil, m\n",
"T2=300; #temp. of air around the tube,K\n",
"Tav=350; #average BI temp.,K\n",
"\n",
"\n",
"#Calculations & Results\n",
"#we evaluate properties at this temp. and write g as ge*(g-level), where ge is g at the earth surface and the g-level is the fraction of ge in the space vehicle.\n",
"b=1/T2; # b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1\n",
"v1=2.062*10**-5; # molecular diffusivity, m**2/s\n",
"v2=2.92*10**-5; #molecular diffusivity, m**2/s\n",
"Pr=0.706; #prandtl no.\n",
"g=array(([10**-6, 10**-5, 10**-4, 10**-2]));\n",
"i=0;\n",
"while i<4:\n",
" Ral=0;\n",
" Nu=0;\n",
" h=0;\n",
" Q=0;\n",
" Ral=(9.8*b*((T1-T2))*(D**(3))/(v1*v2))*g.item(i);\t\t# Rayleigh no.\n",
" \n",
" Nu=(0.6+0.387*(Ral/(1+(0.559/Pr)**(9/16))**(16/9))**(1/6))**2;\n",
" #Nu(i)=(0.6+0.387*((Ral)/(1+(0.559/Pr)**(9/16))**(16/9))**1/6)**2; churchill correlation. \n",
" print \"Nusselt no. are : \",round(Nu,2),\"\\n\"\n",
" h=Nu*0.0297/D; # convective heat transfer coefficient,W/(m**2*K)\n",
" print \"Convective heat transfer coefficient are : \",round(h,2),\"W/(m^2*K)\\n\"\n",
" Q=math.pi*D*h*(T1-T2); #heat transfer,W/m\n",
" print \"Heat transfer is :\",round(Q,2),\"W/m of tube\\n\"\n",
" i=i+1;\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Nusselt no. are : 0.48 \n",
"\n",
"Convective heat transfer coefficient are : 2.87 W/(m^2*K)\n",
"\n",
"Heat transfer is : 4.51 W/m of tube\n",
"\n",
"Nusselt no. are : 0.55 \n",
"\n",
"Convective heat transfer coefficient are : 3.25 W/(m^2*K)\n",
"\n",
"Heat transfer is : 5.11 W/m of tube\n",
"\n",
"Nusselt no. are : 0.65 \n",
"\n",
"Convective heat transfer coefficient are : 3.85 W/(m^2*K)\n",
"\n",
"Heat transfer is : 6.05 W/m of tube\n",
"\n",
"Nusselt no. are : 1.09 \n",
"\n",
"Convective heat transfer coefficient are : 6.45 W/(m^2*K)\n",
"\n",
"Heat transfer is : 10.13 W/m of tube\n",
"\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.5, Page number: 426"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T2=300; #air temp.,K\n",
"P=15; #delivered power,W\n",
"D=0.17; #diameter of heater,m\n",
"v1=1.566*10**-5; #molecular diffusivity, m**2/s\n",
"b=1/T2; #b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1\n",
"Pr=0.71; #prandtl no.\n",
"v2=2.203*10**-5; #molecular diffusivity, m**2/s\n",
"v3=3.231*10**-5; #molecular diffusivity at a b except at 365 K., m**2/s\n",
"v4=2.277*10**-5; #molecular diffusivity at a b except at 365 K., m**2/s\n",
"k1=0.02614; #thermal conductivity\n",
"k2=0.0314; #thermal conductivity\n",
"\n",
"#Calculations\n",
"#we have no formula for this situation, so the problem calls for some guesswork.following the lead of churchill and chau, we replace RaD with RaD1/NuD in eq. \n",
"#(NuD)**(6/5)=0.82*(RaD1)**(1/5)*Pr**0.034\n",
"delT=1.18*P/(3.14*D**(2)/4)*(D/k1)/((9.8*b*661*D**(4)/(0.02164*v1*v2))**(1/6)*Pr**(0.028));\n",
"#in the preceding computation, all the properties were evaluated at T2.mow we must return the calculation,reevaluating all properties except b at 365 K.\n",
"delTc=1.18*661*(D/k2)/((9.8*b*661*D**(4)/(k2*v3*v4))**(1/6)*(0.99));\n",
"TS=T2+delTc;\n",
"TS1=TS-271.54\n",
"\n",
"#Results\n",
"print \"Average surface temp. is :\",round(TS1,4),\"K\\n\"\n",
"print \"That is rather hot.obviously, the cooling process is quite ineffective in this case.\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average surface temp. is : 169.0288 K\n",
"\n",
"That is rather hot.obviously, the cooling process is quite ineffective in this case.\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.6, Page number: 435"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T2=363; # temp. of strip,K\n",
"T1=373; #saturated temp.,K\n",
"H=0.3; #height of strip,m\n",
"Pr=1.86; #prandtl no.\n",
"Hfg=2257; #latent heat. kj/kg\n",
"ja=4.211*10/Hfg; #jakob no.\n",
"a1=961.9; #density of water,kg/m**3\n",
"a2=0.6; #density of air,kg/m**3\n",
"k=0.677; #thermal conductivity,W/(m*K)\n",
"\n",
"#Calculations\n",
"Hfg1=Hfg*(1+(0.683-0.228/Pr)*ja); \t\t #corrected latent heat,kj/kg\n",
"delta=(4*k*(T1-T2)*(2.99*10**(-4))*0.3/(a1*(a1-a2)*9.806*Hfg1*1000))**(0.25)*1000;\n",
"Nul=4/3*H/delta; #average nusselt no.\n",
"q=Nul*k*(T1-T2)/H; # heat flow on an area about half the size of a desktop,W/m**2\n",
"Q=q*H; #overall heat transfer per meter,kW/m\n",
"m=Q/(Hfg1); #mass rate of condensation per meter,kg/(m*s)\n",
"\n",
"#Results\n",
"print \"Overall heat transfer per meter is :\",round(Q,4),\"kW/m^2\\n\"\n",
"print \"Film thickness at the bottom is :\",round(delta,4),\"mm\\n\"\n",
"print \"Mass rate of condensation per meter. is : \",round(m,4),\"kg/(m*s)\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Overall heat transfer per meter is : 26.0118 kW/m^2\n",
"\n",
"Film thickness at the bottom is : 0.1041 mm\n",
"\n",
"Mass rate of condensation per meter. is : 0.0114 kg/(m*s)\n",
"\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}