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clear;
clc;
printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.1 Page 680 \n'); //Example 11.1
// Tube Length to achieve a desired hot fluid temperature
//Operating Conditions
Tho = 60+273 ;//[K] Hot Fluid outlet Temperature
Thi = 100+273 ; //[K] Hot Fluid intlet Temperature
Tci = 30+273 ;//[K] Cold Fluid intlet Temperature
mh = .1 ;//[kg/s] Hot Fluid flow rate
mc = .2 ;//[kg/s] Cold Fluid flow rate
Do = .045 ;//[m] Outer annulus
Di = .025 ;//[m] Inner tube
//Table A.5 Engine Oil Properties T = 353 K
cph = 2131 ;//[J/kg.K] Specific Heat
kh = .138 ; //[W/m.K] Conductivity
uh = 3.25*10^-2 ; //[N.s/m^2] Viscosity
//Table A.6 Saturated water Liquid Properties Tc = 308 K
cpc = 4178 ;//[J/kg.K] Specific Heat
kc = 0.625 ; //[W/m.K] Conductivity
uc = 725*10^-6 ; //[N.s/m^2] Viscosity
Pr = 4.85 ;//Prandtl Number
q = mh*cph*(Thi-Tho);
Tco = q/(mc*cpc)+Tci;
T1 = Thi-Tco;
T2 = Tho-Tci;
Tlm = (T1-T2)/(2.30*log10(T1/T2));
//Through Tube
Ret = 4*mc/(%pi*Di*uc);
printf("\n Flow through Tube has Reynolds Number as %i. Thus the flow is Turbulent", Ret);
//Equation 8.60
Nut = .023*Ret^.8*Pr^.4;
hi = Nut*kc/Di;
//Through Shell
Reo = 4*mh*(Do-Di)/(%pi*uh*(Do^2-Di^2));
printf("\n Flow through Tube has Reynolds Number as %i. Thus the flow is Laminar", Reo);
//Table 8.2
Nuo = 5.63;
ho = Nuo*kh/(Do-Di);
U = 1/[1/hi+1/ho];
L = q/(U*%pi*Di*Tlm);
printf("\n Tube Length to achieve a desired hot fluid temperature is %.1f m",L);
//END
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