Queendom Tools

Irrigation Pumping Plant Efficiency Calculator

Pumping plant calculator

One chain: flow and head → water hp → brake hp (pump and drive efficiency) → input hp (motor efficiency) → input kW → kWh per 1,000 gallons → overall pumping plant efficiency. Electric motor drive only.

Water horsepower, whp— hp
Brake horsepower, bhp— hp
Input horsepower, ihp— hp
Input power— kW
Energy per 1,000 gallons pumped— kWh
Overall pumping plant efficiency, OPE— %

Formulas used

Constants as printed in the source: 3,960, 0.746 kW per hp and 0.00314. Values are computed at full precision and rounded only for display. Source: USDA NRCS, National Engineering Handbook Part 623 Irrigation, Chapter 8 Irrigation Pumping Plants (210-VI-NEH, Amend. 78, January 2016), Example 8-9, pages 8-79 and 8-80.

For preliminary estimation only, not a substitute for review by a qualified/licensed engineer.

How much of the power you pay for reaches the water

A meter reads input kilowatts at the motor, but the water only receives water horsepower: flow times head divided by 3,960. Between the two sit three separate losses, in the pump, in the belt drive and in the electric motor. Each one is a percentage, and they multiply, so a plant with respectable-looking individual efficiencies can still deliver a much smaller share to the water.

This calculator is written for irrigation engineers, farm managers and pumping plant designers. It takes pump flow, head and the efficiencies of the pump, the drive and the electric motor and turns them into water, brake and input horsepower, input kilowatts, kWh per 1,000 gallons and overall pumping plant efficiency in one chain, following the USDA NRCS method.

The preloaded numbers are USDA NRCS Example 8-9: 650 gpm, 145 ft of head, 75% pump, 90% drive and 90.1% motor efficiency, which gives 23.80 water hp, 35.26 brake hp, 39.13 input hp, 29.19 kW, 0.749 kWh per 1,000 gallons and an overall efficiency of 60.8%. You supply the three efficiencies yourself; the page does not pick them from a table.

How to use it

  1. Enter the pump flow Q in US gallons per minute.
  2. Enter the total dynamic head TDH in feet.
  3. Enter pump efficiency Ep as a percent.
  4. Enter drive efficiency Ed as a percent. Enter 100 if the pump is direct-coupled with no belt.
  5. Enter electric motor efficiency Em as a percent. The six results and the formulas appear immediately below.

FAQ

What is overall pumping plant efficiency?

It is water horsepower divided by input horsepower, expressed as a percent — 100 × whp ÷ ihp — so it accounts for pump, drive and motor losses together; in USDA NRCS Example 8-9 (650 gpm, 145 ft head, 75% pump, 90% drive, 90.1% motor) it comes out at 60.8%.

How is water horsepower calculated from flow and head?

whp = Q × TDH ÷ 3,960, with Q in US gallons per minute and TDH in feet — the 3,960 constant is the printed value used in USDA NRCS National Engineering Handbook Part 623, Chapter 8; at 650 gpm and 145 ft the water horsepower is 23.80 hp.

How many kWh does it take to pump 1,000 gallons?

kWh per 1,000 gallons = TDH × 0.00314 ÷ (pump efficiency × drive efficiency × motor efficiency, as fractions); for the USDA NRCS Example 8-9 values of 145 ft head and 75%/90%/90.1% the result is 0.749 kWh per 1,000 gallons.

What drive efficiency should I enter for a direct drive?

Enter 100, since a direct-coupled pump has no belt loss; a direct-drive case of 1,000 gpm, 200 ft head, 80% pump and 92% motor gives an overall pumping plant efficiency of 73.6%.