Heat is to be transferred from water to air through an aluminum wall. It is prod
ID: 1717260 • Letter: H
Question
Heat is to be transferred from water to air through an aluminum wall. It is produce that rectangular fins 0.05 in. Thick and 1/4 in long and spaced 0.08 in a part be added to the aluminum surface to aid in transferring heat. The heat transfer coefficients on the air and water sides are 3 Btu/hr-ft2-T and 25C Btu/hr-ft2-T, respectively. Evaluate the percentage of increase in heat transfer if these files are added to (a) the air side, (b) the water side, or (c) both sides. What conclusions may be reached regarding this result?Explanation / Answer
If the fin is insulated at the end then from the Data book and fin is fixed at the air side
Heat Transfer Q = Sq root of h x P x k x Acx(Tb -Ta )tanh(mL)
Where P = Perimeter h= heat transfer coefficient k = themal condutivity ,Ac Area of cross section
Tb = Temperature of the base of the fin
Talpha = surrounding temperature
P = 2 ( 0.05 +0.08)
= 0.26 in
Ac = Area of cross section= 0.05 x 0.08 = 0.004 in^2
m =sq root of h x P /k x Ac
= 3 x 0.26 / 124 x 0.004
= 1.572
Q = Sq root of (3 x 0.26 x 124 x 0.004 x (Tb-Ta ) tanh(mL)
mL = 1.572 x 0.25
= 0.393
Q = sq root of 0.3868)(Tb - Ta)tanh(0.393)
Assuming Tb = 80 Ta = 28 oC
Q = Sq root of (0.3868 ) (80 -28 ) tanh(0.393 )
= 0.621 x 52 x tanh (0.393)
= 7 .76 watts
When Fin is fixed in the Water side
Heat transfer Q = sq root of hxPxxkxAc (Tb -Ta )tanh(mL)
P = 2 ( 0.05 +0.08)
= 0.26 in
Ac = Area of cross section= 0.05 x 0.08 = 0.004 in^2
m =sq root of h x P /k x Ac
= 250 x 0.26 / 124 x 0.004
= 131.2
Q = Sq root of (250 x 0.26 x 124 x 0.004 x (Tb-Ta ) tanh(mL)
mL = 131.2 x 0.25
= 32.76
Q = sq root of 32.24)(Tb - Ta)tanh(32.76)
Assuming Tb = 80 Ta = 28 oC
Q = Sq root of (32.24 ) (80 -28 ) tanh(32.76 )
5.67 (52 ) (0.9809)
= 294 x 0.9809
Q = 289 W
when fin is fixed water side heat trasfer increased from 7.76 W to 289 W
If the both sides fins fixed then heat transfer still more = 296 W
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