Heat Transfer problem. A flat rectangular plate 120cm long and 200cm wide at 10

Question

Heat Transfer problem.

A flat rectangular plate 120cm long and 200cm wide at 10 degree C is immersed in an air stream of 30 degree C and 1 atm with a free stream velocity of 6 m/s according to Fig. A. Calculate total heat rate (q) in Watts on one side of the plate. As a thermal design engineer, you hypothesis that by changing the orientation of the plate according to Fig B, you can enhance the heat rate. Calculate the total heat rate for Fig. B. What is the percentage of improvement in heat transfer rate? Take the average fluid properties to be p= 1.19 kg/m^3, Cp-1010 J/kg*K, y=15*10^-6 m^2/s, k=0.0251 W/m*K, Pr=0.71, Re_c= 5*10^-5.

Explanation / Answer

Reynolds number Re = V*L / neu

In first case, characteristic length L = 120 cm = 1.2 m

Re = 6*1.2 / (15*10-6) = 4.8*105

Since Re < Recr (5*105), the flow is laminar.

Avergaed Nusselt number Nu = 0.664 * Re1/2 * Pr1/3

Nu = 0.664 * (4.8*105)1/2 * 0.711/3

Nu = 410.4

Heat transfer coeff h = Nu*k/L

h = 410.4*0.0251 / 1.2

h = 8.58 W/m2-K

Total heat rate q = hA(Tamb - Tsurf)

= 8.58*(1.2*2)*(30 - 10)

q = 412 Watts

For figure B, we get characteristic length L = 200 cm = 2 m

Re = 6*2 / (15*10-6) = 8*105

Since Re > Recr (5*105), the flow is turbulent.

Avergaed Nusselt number Nu = 0.037 * Re4/5 * Pr1/3

Nu = 0.037 * (8*105)4/5 * 0.711/3

Nu = 1742.2

Heat transfer coeff h = Nu*k/L

h = 1742.2*0.0251 / 2

h = 21.86 W/m2-K

Total heat rate q = hA(Tamb - Tsurf)

= 21.86*(1.2*2)*(30 - 10)

q = 1049.5 Watts

% improvement in heat transfer rate = (1049.5 - 412) / 412 *100 = 154.7%

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