Answer:
a
Explanation:
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A horizontal poly crystalline solar panel module has to be investigated by natural cooling. For crystal silicon, the thermal coefficient approximately 0.0045/K is used. Investigate the effect of air velocity on the cooling performance of PV panels at 0-5 m/s air velocities, 25-40 ºC ambient temperatures, and 400-1000 W/ m2 solar radiation
Solution :
It is given that :
Thermal coefficient = 0.0045/K
Ambient temperature, [tex]$T_a = 25 - 40^\circ$[/tex]
air velocity, v = 0-5 m/s
Solar radiation, [tex]$G= 400-100 \ W/m^2$[/tex]
[tex]$P=50 \ W$[/tex]
Model calculations :
Cell temperature ([tex]$T_c$[/tex])
[tex]$T_c = T_a + \left(\frac{0.25}{5.7+3.8 \ v_w}\right) G$[/tex]
where [tex]$ v_w - v_a = $[/tex] wind speed / air speed
∴ [tex]$T_c = 2 \pi + \left(\frac{0.25}{5.7+3.8 \times 1}\right) \times 400$[/tex]
[tex]$T_c = 35.526 ^\circ$[/tex]
[tex]$\Delta T = T_c -25$[/tex]
= 35.526 - 25
= 10.526 K
Thermal coefficient = 0.0045 x 10.526
= 0.04737
Pv power = [tex]$(1 -C_T) \times P \times \frac{G}{1000}$[/tex]
[tex]$=(1 -0.04737) \times 50 \times \frac{400}{1000}$[/tex]
= 17.0526 W