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clc;
clear;
//Example 5.15
//CASE I:
Cp=4*10^3; //[J/kg.K]
t1=295; //[K]
t2=375; //[K]
sp=1.1; //Specific gravity of liquid
v1=1.75*10^-4; //Flow of liquid in [m^3/s]
rho=sp*1000 //[kg/m^3]
m_dot=v1*rho //[kg/s]
Q=m_dot*Cp*(t2-t1) //[W]
T=395; //[K]
dT1=T-t1 //[K]
dT2=T-t2 //[K]
dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
U1A=Q/dTlm //[W/K]
//CASE-II
v2=3.25*10^-4 //Flow in [m^3/s]
T2=370 //[K]
m_dot=v2*rho //[kg/s]
Q=m_dot*Cp*(T2-t1) //[W]
dT1=T-t1 //[K]
dT2=T-T2 //[K]
dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
U2A=Q/dTlm //[W/K]
//since u is propn to v
//hi =C*v^0.8
U2_by_U1=U2A/U1A
ho=3400 //Heat transfer coeff for condensing steam in [W/sq m.K]
C=poly(0,"C")
//Let C=1 and v=v1
//C=1;
v=v1; //=1.75*10^-4 m^3/s
hi=C*v^0.8
U1=1/(1/ho+1/hi) //
//When v=v2
v=v2;
hi=C*v^0.8
U2=1/(1/ho+1/hi) //
//Since U2=1.6U1
//On solving we get:
C=142497
v=v1
hi=C*v^0.8
U1=1/(1/ho+1/hi) //
A=U1A/U1 //Heat transfer area in [sq m]
printf("\n Overall heat transfer coefficient is %f W/sq m.K and\n\nHeat transfer area is %f sq m",U1,A);
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