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here is the question i need an answer to.

There is well that is costing $109,587.60 per year for annual operating costs. The Power cost $.15/KWH. The motor tag indicates the motor is 125 hp and 90% Efficient. Referring to the pump curve information, in indicates that the pump is designed to pump 1200 gallons per minute at 290 feet of head. This well runs 24 hours per day, 7 days per week. The pressure gauge at the well head is indicating 125 psi when the pump is running. What is the % of efficiency between the calculated flow and the flow indicated on the pump curve?

Answer 1

$109,587.60 / $.15/KWH = 730580 kW-hr

730580 kW-hr / 365*24 hr/yr = 83.4 kW

(this is not needed, as the number can be calculated other ways, see below)

125 hp (93.2 kw) and 90% Efficient means it uses 103.6 kw of power.

The two numbers don't track. $$ number gives us 83.4 kW, efficiency and HP gives us 103.6 kW.

I'll use the 103.6 number.

1200 gallons is 5.4552 m³ x 998 kg/m³ = 5444 kg

290 ft is 88.4m

To calculate mechanical power output, se don't need the pressure, just the amount of water lifted to what height.

So to lift that weight energy of 5444 * 9.8 * 88.4 = 4.716 MJ

This is every minute, so power is 4.716 MJ / 60 = 78.6 kW

So electric power is 103.6 kw, power output is 78.6 kW, and efficiency is 75.9%

Answer 2

A year has 8760 hours, but leap year has 8784, so an average year has 8766 hours.

If you divide the annual operating cost by the $0.15 cost per kWh, the power usage is 730,586 kWh per year. If you divide that by 8766, the power supplied to the motor is 83.34 kW. If the motor is 90% efficient, the motor delivers 83.34/0.9 = 92.6 kW to the pump. Since 1 Hp = 0.7457 kW, the pump is using 92.6/.7457 = 124.2 Hp.

Assuming water is what is being pumped, the power transferred to the moving water is 1200 X 290 / 3960 = 87.88 Hp at the design point. If the actual discharge pressure is 125 PSI, that is 125 X 2.3067 = 288.3 feet of head at 32F (actual temperature unknown). The actual operating point is practically at the design point, so assume that the pump is delivering 87.88 Hp. The pump efficiency is then 100 X 87.88 / 124.2 = 70.8%.

Oops

I should have multiplied motor input power by 0.9 rather than divide. Motor output is 75 kW or 100.6 Hp. That makes the efficiency 100 X 87.88 / 100.6 = 87.4%. That is just the pump.

If you want the efficiency of electrical input vs water output, the output in kW is 87.88 X .7457 = 65.5 kW. The overall efficiency is 100 X 65.5 / 83.34 = 78.6%.

You can not assume that the motor is running at its rated power. You must calculate the motor power from metered power usage.

Re other answer:

In addition to problem with motor load power, 1200 USA gallons is 4.542 cubic meters. Are the units USA or British?