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 cfdonline10 June 19, 2013 12:30

Heat Exchanger Resistance

Assume that there a flat oval tube (thickness T1) of 50 mm width (W1)and 8 mm height (H1)with water flowing at a rate of 2.5 liters/min. Directly below this tube, there a plate (thickness T2) of width of 35 mm (W2). I want to come up with thermal resistance for this assuming there is heat source directly below plate.
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 +(W2-W1)/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

 ryaron June 28, 2013 07:26

Quote:
 Originally Posted by cfdonline10 (Post 434868) Assume that there a flat oval tube (thickness T1) of 50 mm width (W1)and 8 mm height (H1)with water flowing at a rate of 2.5 liters/min. Directly below this tube, there a plate (thickness T2) of width of 35 mm (W2). I want to come up with thermal resistance for this assuming there is heat source directly below plate. 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 +(W2-W1)/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
What are the ratios of the conductivities ?
i.e. k1/k2 ?