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

To determine the overall heat loss coefficient of an FPC during nighttime, calculate the rate of heat transfer using water temperatures, flow rate, ambient temperature, and area, and then adjust for unit consistency.

Step-by-step explanation:

The question relates to an FPC (Flat Plate Collector) and the calculation of the overall heat loss coefficient of the system during a nighttime test. To find the coefficient, we must evaluate the heat loss based on the inlet and outlet temperature of the water, the flow rate, ambient temperature, and the area of the collector.

The rate of heat transfer (Q) can be given by:

• Q = Flow rate x Density x Specific Heat Capacity x Temperature Change

With an initial temperature of 60°C and an exit temperature of 49°C, and an ambient temperature of 8°C, the temperature difference between the inlet water and ambient conditions is crucial for heat loss determination.

The heat loss per second (q) can be calculated by:

• q = Q / Area

Finally, the overall heat loss coefficient (U-value) is found by:

• U = q / (Inlet Temperature - Ambient Temperature)

Here, flow rate conversion from liters per second to cubic meters per second and unit consistency are important to ensure the U-value is correctly calculated.

Answer 2

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

Step-by-step 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 (
`(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 )