Question

Estimate the top heat loss coefficient of a collector that has the following specifications:
Collector area= 2m2
Collector slope =35⁰
Number of glass covers = 3
Thickness of each glass cover =4 mm.
Thickness of absorbing plate =0.5 mm.
Space between glass covers = 20 mm.
Space between inner glass cover and absorber =40 mm.
Mean absorber temperature, Tp =80 ⁰C
Ambient air temperature =15 ⁰C
Absorber plate emissivity, εp =0.10.
Glass emissivity, εg =0.88
Wind velocity = 2.5m/s.
Assuming that temperatures of first, second and third covers are 288.53K, 293.4K and 313.93K, respectively.

Answer 1

The heat loss coefficient of the building determines the rate of heat flow through the buildings' envelope when a temperature difference exists between the indoor air and the outdoor air under steady state conditions. Top heat loss coefficient is required for evaluating thermal performance of solar collectors. A correct value of Ut is also important for design, simulation of heat losses or thermal performance evaluation of flat plate collectors with vertical configuration. These are used at high latitudes and are integrated with building walls. Top heat loss coefficient, Ut, has to be computed for various values of different variables like emittance of absorber coating (ep), absorber plate temperature (Tp), ambient temperature (Ta), wind heat transfer coefficient (hw), air gap spacing between absorber plate and glass cover (L) and angle of inclination of collector (b). Top heat loss coefficient of a flat plate collector an be computed by numerical solution of heat balance equations or approximately by empirical equations.

The most popular approximate method for calculation of Ut, using Klein’s equation quoted by Duffie and Beckman is:

Based on the above equation the the top heat loss efficiency of a collector can be determined.