Fan Coil Unit Power Consumption

Not enough information given for an evaporator fan coil power consumption.
That figure (20.2 kW) may be the electrical equivalent of the BTUs supplied or absorbed by the fan coil unit under the stated conditions. It is common to be able to pump more BTUs per unit of power input that could be generated with electric heat at moderate temperatures.
The compressor power input is given is given as 10.26 kW.
The fan input power is generally much less than the compressor input power.
Think of the electrical equivalent as the amount of electric heat needed to cancel the cooling effect.

Bill
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"Why not the best?"
Jimmy Carter

NickParker,
In fact 20.20kW is the direct conversion from British Thermal Unit (Btu) to Joules (Wsec). Therefore, one can say that 20.20kW is the equivalent
of Btu/Hr. It has nothing to do with the input electrical power consumption to the unit. In order to calculate the input electrical power, we have to assume a reasonable
value for Cooling EER (Energy Eff Ratio) for the unit. Normally EER is 10-11. The name plate of the unit shall indicate the EER value. Otherwise you can get it from the
manufacturer too. If you take EER=10, then the electrical power requirement is (Btu/Hr/ 10= Watts)=69,000/10= 6900Watts =6.9kW. The unit will consume 6.9kW of electrical power.
Hope this helps.

Te EER is a rating applied to air conditioners. The conversion from electrical energy to BTUs of cooling depends on the delta T between the evaporator and the condenser as well as on the overall efficiency.
For a large delta T, such as subzero cooling, the EER may be less than 1.

Bill
--------------------
"Why not the best?"
Jimmy Carter

Based upon the voltage, phase, and frequency indicated in the data sheet, I am assuming that you're based in the United States.

Nearly all HVAC equipment manufacturers will provide the minimum circuit ampacity (MCA) for a given piece of equipment. Since the NEC requires that the MCA be calculated at 125% of the full-load current, you can easily retrieve the full-load current (And, subsequently, the apparent power) by multiplying the MCA by 80%. Usually, the maximum overcurrent protective device (MOCPD) rating is also provided alongside with the MCA. I would not recommend referencing the kW value indicated within the data sheet. It's shown as "required capacity", which is kind of nonsensical in this context. Besides, even if you did know the input kW, there would be no way to determine the apparent power (kVA) without knowing power factor.

I'm really surprised that neither the MCA nor the MOCPD are given.

That data sheet looks like the mechanical specs for the fan coil unit.
As I understand your problem you are just looking for the capacity required for the fans.
Not enough information is given
If this is for the condenser fans on a skid that includes the compressor, and you need to power the compressor skid, you should have more information.
You can use the 3413 constant to determine the BTUs that a heating element will produce but it is not valid for ectrical input of a refrigeration unit. A refrigeration unit is not creating BTUs as an electric heater does. The refrigeration unit is pumping BTUs from one temperature ambient to another temperature ambient.
The ratio between the evaporation temperature in absolute degrees and the condensing temperature in absolute degrees is an important factor.
That said, the bottom line of the spec sheet shows the compressor power input as 10.26 kW.
That figure is more valid for electrical supply than the value of 20.20 kW given as the BTU per Hr capacity.
We still don't know the power factor nor the fan ratings.

Bill
--------------------
"Why not the best?"
Jimmy Carter

If it is about only of the fan of the indoor unit
if we take the formula from The Engineering Toolbox Fan Power Consumption
the ideal power consumption for a fan (without losses) can be expressed as:
Pi=dp*q
Pi = ideal power consumption (W)
dp = total pressure increase in the fan (Pa, N/m2)
q = air volume flow delivered by the fan (m3/s)
q =1723 CFM=0.813165 m^3/sec
dp = 73 Pa
Pi= 59.36105 w cos(φ)=0.7 eff.=0.7
Pinput=59.36/.7/.7=121 VA.

By the way, a HVAC specialist told me you usually you have to consider the intake air filter. The filter may require another 10 mmH2O but up to 3 times more if the filter is dust filled. So the fan static pressure could be 0.293+3*394= 1.475 inwg [367.5 Pa] and finally the fan power will be about 350-400 W.