Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust
Question Statement:
A solar photovoltaic pumping system has a solar panel array with total area of $10\text{ m}^2$ operating at an electrical conversion efficiency of $15\%$. The average global solar irradiance incident on the array is $800\text{ W/m}^2$. If the motor-pump overall efficiency is $60\%$, calculate the hydraulic power output in $\text{W}$.
Answer Key:
720
Step-by-Step Solution & Explanation
Total solar power incident on array:
$P_{\text{in}} = I \times A = 800\text{ W/m}^2 \times 10\text{ m}^2 = 8000\text{ W}$
Electrical power generated by PV array:
$P_{\text{elec}} = P_{\text{in}} \times \eta_{\text{pv}} = 8000 \times 0.15 = 1200\text{ W}$
Hydraulic power delivered by motor-pump set:
$P_{\text{hyd}} = P_{\text{elec}} \times \eta_{\text{mp}} = 1200 \times 0.60 = 720\text{ W}$