Solar Pump on the School Farm
At a school farm in Gharyan, a 200 m² plot is to be irrigated by a small solar pump. The panel is rated 18 V and 5.5 A at standard test conditions, about 99 W. The pump is rated 12 V, 60 W and draws 5.0 A when running at 12 V. The two-core cable from the controller to the pump has a total loop resistance of 0.50 Ω. A controller is available: its label says input up to 25 V, output regulated at 12 V, maximum output current 6 A. A student proposes connecting the panel directly to the pump to avoid controller losses. At midday the panel actually delivers 4.5 A at 16 V, below its rated output. There is no battery, and the pump must not see more than 12 V. The school can afford either one controller or a thicker cable, not both this term.
1. At the pump's rated current of 5.0 A, calculate the voltage drop along the cable and the power dissipated in it. If the controller output is 12 V, what voltage actually reaches the pump?
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V = IR = 5.0 × 0.50 = 2.5 V drop around the loop. P = I²R = 25 × 0.50 = 12.5 W lost as heat in the cable. With a 12 V controller output, the pump terminals see 12 − 2.5 = 9.5 V, not 12 V. That is about 21% below rated voltage. A small centrifugal pump driven at 9.5 V will run slower and deliver much less flow; it may not start against head. The controller regulates its own output, not the voltage at the far end of the cable, so the 0.50 Ω loop is a real design fault. Note that 12.5 W lost out of roughly 60 W delivered to the pump is about 17% of the useful power, which is unacceptable for a stand-alone system. The first check the class should make is whether the cable can be shortened: halving its length would halve the resistance and the loss.
2. A student proposes removing the controller and connecting the panel directly to the pump. Evaluate this proposal using the midday figures of 16 V and 4.5 A.
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Direct connection is unsafe. The panel's midday operating voltage is 16 V, above the pump's 12 V rating, and at low load, before the pump starts, a PV panel can sit near its open-circuit voltage, which is higher still. Even under load, the pump would see roughly 16 V minus cable drop, well over 12 V, risking overheating and damage to brushes or insulation. The controller is not just a lossy accessory; it is the voltage regulator that protects the pump. The student is right that a controller wastes some power, but the correct response is to reduce cable resistance, not to remove protection. A direct connection also gives no controlled start-up and no low-voltage protection, though with no battery that matters less; the decisive issue is overvoltage. Keep the controller. The proposal should be rejected in writing, with the 16 V measurement attached, because otherwise someone may try it when the pump fails to start on a hazy day.
3. Using the midday panel output of 72 W, decide whether the pump can receive its rated 60 W through the existing cable. Recommend one purchase: the controller or thicker cable.
Read the worked answer
Available panel power is 16 × 4.5 = 72 W. At 12 V and 5.0 A the pump needs 60 W, and the cable would dissipate 12.5 W at that current, so the requirement is 72.5 W - already 0.5 W above the panel's midday output, before controller losses. The system is therefore marginal or short: the pump will run below rated speed at midday and will not cope on a hazy day. The controller is mandatory for voltage regulation, so the purchase order should be: controller first, then thicker cable as soon as funds allow. If only one purchase is possible this term, buy the controller and shorten or double up the existing cable; doubling the conductor area halves loop resistance to 0.25 Ω, cutting loss to 6.25 W. That brings the requirement to 66.25 W, which 72 W can meet with a small margin. The engineering answer is therefore controller plus a cable fix, not either/or, and the flow rate should be measured after the fix rather than assumed.