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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

User Demyanov
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1 Answer

6 votes

Solution :

It is given that :

Thermal coefficient = 0.0045/K

Ambient temperature,
$T_a = 25 - 40^\circ$

air velocity, v = 0-5 m/s

Solar radiation,
$G= 400-100 \ W/m^2$


$P=50 \ W$

Model calculations :

Cell temperature (
$T_c$)


$T_c = T_a + \left((0.25)/(5.7+3.8 \ v_w)\right) G$

where
$ v_w - v_a = $ wind speed / air speed


$T_c = 2 \pi + \left((0.25)/(5.7+3.8 * 1)\right) * 400$


$T_c = 35.526 ^\circ$


$\Delta T = T_c -25$

= 35.526 - 25

= 10.526 K

Thermal coefficient = 0.0045 x 10.526

= 0.04737

Pv power =
$(1 -C_T) * P * (G)/(1000)$


$=(1 -0.04737) * 50 * (400)/(1000)$

= 17.0526 W

User Jacek Kaniuk
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