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clc;
clear;
//Example 5.10
m_dot=7250 //[kg/h] of nitrobenzene
Cp=2.387; //[kJ/kg.K]
mu=7*10^-4; //[kg/m.s]
k=0.151; //[W/m.K]
vis=1;
Ft=0.9; //LMTD correction factor
T1=400 //[K]
T2=317 //[K]
t1=333 //[K]
t2=300 //[K]
dT1=T1-t1 //[K]
dT2=T2-t2 //[K]
dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
//For nitrobenzene
Q=m_dot*Cp*(T1-T2) //[kJ/h]
Q=Q*1000/3600 //[W]
n=170 //No. of tubes
L=5 //[m]
Do=0.019 //[m]
Di=0.015 //[m]
Ao=n*%pi*Do*L //[sq m]
Uo=Q/(Ao*Ft*dTlm) //[W/sq m.K]
Ud=Uo //[W/sq m.K]
B=0.15 //Baffle spacing [m]
Pt=0.025 //Tube pitch in [m]
C_dash=Pt-Do //Clearance in [m]
id=0.45 //[m]
//Shell side cross flow area
as=id*C_dash*B/Pt //[sq m]
//Equivalent diameter of shell
De=4*(Pt^2-(%pi/4)*(Do^2))/(%pi*Do) //[m]
//Mass velocity on shell side
Gs=m_dot/as //[kg/(m.h)]
Gs=Gs/3600 //[kg/m^2.s]
mu=7*10^-4 //[kg/m.s]
Cp=Cp*1000 //[J/kg.K]
Nre=De*Gs/mu //Reynolds number
Npr=Cp*mu/k //Prandtl number
//From empirical eqn:
mu_w=mu //
Nnu=0.36*Nre^0.55*Npr^(1/3)
ho=Nnu*k/De //[W/sq m.K]
hi=1050 //Given [W/sq m.K]
Uo=1/(1/ho+(1/hi)*(Do/Di)) //[W/sq m.K]
Uc=Uo //W/sq m.K
//Suitability of heat exchanger
Rd_given=9*10^-4 //[W/sq m.K]
Rd=(Uc-Ud)/(Uc*Ud) //[W/sq m.K]
printf("\n Rd calculated(%f W/m^2.K) is mazimum allowable scale resistance\n",Rd);
printf("\n\nAs Rd calculated (%f W/sq m.K)(OR 1.1*10^-3) is more than Rd given(%f W/sq m,K),the given heat exchanger is suitable\n",Rd,Rd_given);
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