## Fan Coil Unit Power Consumption

## Fan Coil Unit Power Consumption

(OP)

I have attached a datasheet of Fan coil unit. Actually what we are looking for is the electrical power consumption (Electrical Input kW or kVA) required for the given BTUH of a HVAC unit where we don't have any other data sheet available. In the datasheet, manufacturer simply divided the BTU/hr by 3413 to get the power consumption. Is that gives the correct electrical input kW? or it is just the cooling capacity.

## RE: Fan Coil Unit 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

--------------------

"Why not the best?"

Jimmy Carter

## RE: Fan Coil Unit Power Consumption

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.

## RE: Fan Coil Unit Power Consumption

For a large delta T, such as subzero cooling, the EER may be less than 1.

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Fan Coil Unit Power Consumption

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.

## RE: Fan Coil Unit Power Consumption

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

## RE: Fan Coil Unit Power Consumption

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.

## RE: Fan Coil Unit Power Consumption