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2 years ago

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An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th

e collector at a flow rate of 0.06 l/s. The ambient temperature is 8 C and the exit temperature is 49 C. Determine the overall heat loss coefficient.

1 answer:

sp2606 [1]2 years ago

7 0

Answer:

<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

<u>Calculate the overall heat loss coefficient </u>

Note : heat lost by water = heat loss through convection

m*Cp*dT = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp ( integrate the relation )

In ( $\frac{49-8}{60-8}$ ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

= - 14.943 W/m^2K ( heat loss coefficient )

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