Specific energy for lifting liquid
Electricity to lift 1 m³ and 1 t of liquid by a pumping system — ESP, rod pump, injection or transfer pump — against a given head with the overall efficiency, and the daily consumption.
e = ρ · g · H / (3.6 · 10⁶ · η)e_t = 1000 · e / ρW = e · QThe specific energy is the key efficiency indicator of artificial lift and injection: at the same head, a rise means a lower efficiency (pump wear, an off-design duty, cable losses). The theoretical minimum: 1 m³ of water lifted 100 m takes 0.272 kWh.
Source: Pump power output P = ρ · g · Q · H (ISO 9906); 1 kWh = 3.6 MJ
Inputs
You can change a field's unit: the value is converted to the formula's units automatically.
For a well: the dynamic fluid level plus the wellhead pressure in metres of liquid and friction losses; for injection and transfer pumps: the pump head.
For watered-out production, the weighted average of the oil and water densities.
From the supply to the lifted liquid: usually 25–60 % for well pumping systems, 50–75 % for injection pumps.
Unit converter for this formulaLength · Density and °API · Liquid rate · Power · Fraction and percent
- m1
- cm100
- mm1,000
- km0.001
- ft3.28084
- in39.3701
- 1/32 in1,259.84
- 1/64 in2,519.69
- mile0.000621371
- e_t — Energy per tonne–kWh/t
- e_0 — Theoretical minimum (η = 100 %)–kWh/m³
- W — Energy per day–kWh/d
- P — Average power drawn–
More in Energy use
Results are engineering estimates from standard formulas; for design decisions check them against the codes, project documents and specialists' calculations. The formulas carried over from the original set are unchanged, and their errors are described in the notes.