PUMPS
turbine well pump. This would be considered a very efficient pump, but it’s the point where a new vertical turbine should be, if not better. In the field, OPE can vary greatly depending on the age of the pump, changing operating conditions of flow and pressure, water levels, pump wear or irrigation system requirements, but 60% is a realistic number.
Example 3 – Cost to irrigate 1 acre of crop Given:
Original equation: kWh/acre foot = 1.0241 × TDH/ OPE
Assume the following pump parameters: TDH = 300 ft OPE = 60% or 0.60 kWh/acre foot = 1.0241 × 300/0.60 = 512.5 kWh/acre foot kWh cost from the electric company = $0.14
Therefore:
512.5 kWh/acre foot × $0.14 = $71.75 to pump 1 acre foot If your crop requires 3 ft of water per acre each season, then $71.75 × 3 ft = $215.25 to irrigate 1 acre of crop
You may notice that there is no flow rate in this equation. All pumps have different flows, but all things being equal with TDH and OPE, a 20 HP pump will take five times as long as a 100 HP pump to deliver the same volume of water. In this case, it would take a 20 HP pump five times as long to pump an acre foot than a 100 HP pump with identical TDH and OPEs.
Most crops require
anywhere from 1 to 5 feet of water per acre annually.
If you can get the kW meter reading under normal pumping conditions and figure your TDH, OPE (from a pump test) and the cost/kWh, you will be able to determine the cost to move water to your crop.
Bill Green
irrigationtoday.org
Spring 2020 | Irrigation TODAY 21
Page 1 |
Page 2 |
Page 3 |
Page 4 |
Page 5 |
Page 6 |
Page 7 |
Page 8 |
Page 9 |
Page 10 |
Page 11 |
Page 12 |
Page 13 |
Page 14 |
Page 15 |
Page 16 |
Page 17 |
Page 18 |
Page 19 |
Page 20 |
Page 21 |
Page 22 |
Page 23 |
Page 24 |
Page 25 |
Page 26 |
Page 27 |
Page 28 |
Page 29 |
Page 30 |
Page 31 |
Page 32 |
Page 33 |
Page 34 |
Page 35 |
Page 36 |
Page 37 |
Page 38 |
Page 39 |
Page 40
