Simple Replacement of Water Pump Systems with Constant Speed up to 75 kW
Pumps
Measure Goal
The goal is to estimate the electricity savings triggered by the simple replacement of water pump systems with constant speed up to 75 kW without demand analysis. The pump is simply replaced by a new pump designed for the same nominal operating point. The efficiency gain results from switching to a new pump generation (better MEI) and possibly a new motor.
Symbols, Terms and Units
| Symbol | Description | Unit |
|---|---|---|
E | annual electricity consumption | kWh/a |
ΔEeco | cumulative electricity savings | kWh |
f | Factor | — |
Ns | Standard effective duration | a |
H | Height | m |
Δp | Differential pressure at pump connections | kPa |
Ph | hydraulic power | kW |
Pm | mechanical power | kW |
t | annual operating hours | h/a |
Q | Flow rate | m³/h |
τ | Motor load | % |
η | Efficiency | — |
Calculation Formulas
Creditable Electricity Savings
The creditable electricity savings result from the difference between the current and the new annual electricity consumption, multiplied by the reduction factor f_eco of 0.75 and the standard effective duration N_s.
Hydraulische Nennleistung
The nominal hydraulic power is calculated from the differential pressure and flow rate, or from height, water density and flow rate.
Annual Electricity Consumption
The annual electricity consumption is expressed as a function of the nominal hydraulic power, annual operating hours and the nominal efficiency of the pump and motor.
Examples and Scenarios
Replacement of two pump systems (ESOB, single-stage) with a constant flow rate of 400 m³/h at a height of 15 m, operating 18 hours per day. The 4-pole drive motors have a rated power of 22.5 kW and are from the year 2002.
Inputs
2x Pumpen, 22.5 kW Motor (4-polig, Baujahr 2002), 6'570 h/a
Downloads
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