Question

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

Answer 1

Answer:

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

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

Calculate the overall heat loss coefficient

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 )