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June 19, 2013, 12:30 
Heat Exchanger Resistance

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cfdonline
Join Date: Jun 2013
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Method 1: Direct HT Resistance = r1=(T2/(K2 *A2) +T1/(K1*A1)+ 1/(hA)) Note that here, A2=A1=A=W2*length The remaining portion of oval tube has (behave line fins) a length of S= (W2/2) (top of oval tube) + H2 +(W2W1)/2 on left side and the same on the right side. Assume that entire plate and tube is adiabatic. Parallel resistance is r2 =1/(Fineff*S*T2). Where Fineff =tanh (m*S)/(m*S) Now we can find equivalent R using 1/R = 1/R1 + 1/R2 Method 2: Area of oval tube At= 2*(W1+H1)*L Total resistance =(T2/(K2 *A2) +T1/(K1*A1)+ 1/(hAt)) Which one is more accurate? Any suggestion or refrerences? Thanks 

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June 28, 2013, 07:26 

#2  
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i.e. k1/k2 ? Here's a quick answer: make two dimensionless parameters epsilon=k1/k2 and eta characteristic length tube/characteristic length plate, make sure they are << 1. Denote them as epsilon and eta. Now expand both expression you've written in powers of epsilon and eta. The one with the higher order O(epsilon^n), O(eta^n) is the more accurate. 

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