Electrical Load List (Motor List) Terms
Electrical load lists are documents that list major draws and users of electrical power; mainly motors. You may encounter the following terminology and short forms in these documents:
- Connected load (CL) = motor nameplate rating = electrical load at power input terminals
- Demand Factor (DF) = Actual maximum demand load (design load)/connected load
- Max Demand = maximum power that you imagine could be consumed by the load = connected load * DF
- Utilization Factor (UF) = Actual operating time/possible operating time. (e.g. UF = 1 for always on, 0 always off)
- Peak Demand = CL * DF * UF
Remember that the terms above are the electrical loads demanded at the motor. The process will receive less energy due to inefficiencies inherent in the motor, as well as the device the motor powers (pump/compressor/fan/etc. efficiency).
And if examining the overall plant:
- Plant Diversity (Coincidence) Factor (PDF) = Factor applied to Total Plant Net Run Demand, based on coincidence probability of loads operating concurrently
- Power distribution losses = Peak Run Demand x Plant Diversity Factor x % Power Distribution losses (3-5%)
Utilization Factor Guidelines (not rules):
If an equipment is used only rarely and intermittently, then the UF can be based on the % of the year where it is used.
What about equipment that is used continuously? (i.e. all the time except for shutdowns or maintenance problems). Reliable equipment, such as API 610 pumps, should be online at least 95% of time if using correctly. Maybe even 98-99%.
If there are two pumps one online and one spare, there are two options.The simple method is to set the UF of one pump at 1.00 and other at 0.00 as an approximation.
To be more accurate, set the main pump’s UF high like say at 0.95. The missing 5% accounts for trips and plant outages. Then put the second pump UF = ~0.1, 0.05 since occasionally must start up spare pump and shut down main pump.
For compressors, they tend to be more finicky than pumps, so the main compressor may only have UF = 0.8-0.9ish.
In the design condition, a pump must produce 200 gpm flow over 86 psi difference. (Discharge pressure – suction pressure = 86 psi).
This is a hydaulic horse power requirement of ~ 10 hp. (hhp = flow * dP / 1714 with these units). The process needs to receive an input of 10 hp at the pump design condition.
Suppose that the selected pump itself is only 50% efficient. Determine the brake horsepower: bhp = hhp / pump efficiency = 10 / 0.5 = 20 hp.
The motor is only 90% efficient. The motor hp = bhp / motor efficiency = 20/0.9 = 22 hp.
The nearby common motor sizes are 20 hp, 25 hp, and 30 hp. Therefore, suppose the pump is given a 25 hp motor.
Therefore CL = 25 hp. Max demand = 22 hp = CL * DF, so DF must = 0.89. (Check: Max demand = CL * DF = 25 * 0.89 = 22, good).
Say the pump will be used almost all the time, assign a UF = 0.95. Peak demand = CL * DF * UF – 25 * 0.89 * 0.95 = 21 hp, or ~16 kVa.