A flat plate collector is inclined at a slope of 50 degrees

A flat plate collector is inclined at a slope of 50 degrees from the horizontal. The cover plate temperature is 31.5 degrees C and the ambient temperature is 10 degrees C. The air wind speed is 5.0 m/s. Estimate the collector heat transfer coefficient off the top of the plate to the wind if:

a. The collector is free standing and has dimensions of 2.5x10 m.

b. The collector is flush mounted on a building with a volume of 564 m^3

Solution

GIVEN:- Tc = 31.5C = 304.65K, Ta = 10C = 283.15K, V = 5m/s, Air density(d) = 1kg/m3, Cp = 1.005 KJ/kg.K

SOLUTION:-

Typical flat plate collector efficiency E = 75%

(a) As = 2.5x10 = 25m2

By formula Heat loss rate to air Qloss = m x Cp x (Tc-Ta) = (As x V x d) x Cp x (Tc-Ta) = (25x5x1) x 1.005 x 21.5

i.e. Qloss = 2700.93 KJ/s which is the (100-E=25%) of the total solar energy rate Qs incident on the collector, So the heat rate absorbed by the collector = Qc = Qs - Qloss = 2700.93/0.25 - 2700.93 = 8102.79 KJ/s

By Formula Qc = U x As x (Tc-Ta) i.e. U = 8102.79 / (25 x 21.5) = 15.07 KJ/m2.K

So Collector heat transfer coefficient U = 15.07 KJ/m2.K (ANSWER)

(b) Building volume = 564m3 (Assume building height h=50m, side s = (546/h)^0.5 = 3.36m) So surface area of flat plate collector = As = 3.36 x 50 = 168 m2

Now Qloss to air = 168 x 5 x 1 x 1.005 x 21.5 = 18150.3 KJ/s

So Qc = Qs - Qloss = 18150.3/0.25 - 18150.3 = 54450.9 KJ/s

Also Qc = U x As x (Tc-Ta) i.e. U = 54450.9 / (168x21.5) = 15.075 KJ/m2.K (ANSWER)

IN BOTH CASES COLLECTOR HEAT TRANSFER COEFFICIENT REMAINS THE SAME.