Watt usage (kWh) of Heat Pump?

How can I determine the watt usage (kWh) of our heat pump? We have a Pentair UltraTemp 140 (P/N 460934). I am big on data collection and wanted to get an idea of exactly how much it's costing us to heat our pool (our utility bills are high, wife likes it warm). Is there a spec sheet that says how much kWh this thing uses? does that vary? is there some way I can measure it reliably using something like CT clamps somehow?

Thanks!
 
The energy use is somewhat variable. In the manual or brochure they will usually give you a BTUh and COP for a certain operating condition (temp & humidity). You can then calculate kw by the following formula:

Input Power (kw) = BTUh / (COP * 3412.142)

To get to kwh, just multiply the above number by the number of hours the heater is running.

The table is near the end: https://www.pentair.com/content/dam...emp_High_Performance_Heat_Pump_BR_English.pdf
 
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80° F Air Temp, 80% Relative Humidity, 80° F Water Inlet Temp.
Input = 7.2 kW (24,600 btu/hr), 33 amps, C.O.P = 5.8, output = 143,000 btu/hr.

80° F Air Temp, 63% RH, 80° F Water Inlet Temp.
Input 7.0 kW, 32 amp, C.O.P = 5.6, Output 134,000 btu/hr.
The lower relative humidity results in a lower heat output because part of the heat comes from the air and part comes from condensing water from the air. That's why heat pumps drip water continuously while heating. It's the opposite of cooling by evaporating water.
Condensing 5 gallons of water provides 37,534 btu.

Drainage and Condensation
Condensation occurs from the evaporator coil while the unit is running, and drains at a steady rate (usually three to five gallons per hour), depending upon ambient air temperature and humidity. The more humid the ambient conditions, the more condensation will be produced.

50° F Air Temp, 63% RH, 80° F Water Inlet Temp.
Input 6.4 kW, 29 amp, C.O.P = 4.0, Output 88,000 btu/hr.

You can measure the volts and amps to get the power usage if you want the exact number.

The numbers are in the manual if you need them.
 
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