no accurate result can be found but it may be nearly 938 kWh per month
CFM is a rate of flow. KW is a rate of energy. You really cannot convert them.
Try this site: http://www.eia.doe.gov/cneaf/electricity/epm/table5_6_a.html
If the town has natural gas for heating, then the power needed per household is between 5 kW and 10 kW. If only electric heat is available, then the combination of electric resistance (auxiliary - from 7 to 10 kW for small residential units) heating, combined with electric clothes dryer (from 2 to 5 kW), electric water heater (from 2.5 to 7.5 kW), and stove (5 kW to 7.5 kW) totaling a min of 15 kW to a max of 20 kW. That would put the max power requirements between 5,000 kW and 10,000 kW. As a matter of reality, a diversity factor of .6 to .7 would reduce the max power to 3,500 kW and 7,000 kW.
As the motor is drawing 9.7×110 = 1,067 watts (or 1.067 kW), and delivering 1.25×746 watts (or .9325 kW) of mechanical energy, it should release 1,067-932.5 = 134.5 watts (or .1345 kW) of heat.
600 KW
very much :-)
kW (kilowatt) is a unit of power. It refers to how quickly (at what rate) energy is transferred or converted - in this case, probably it refers to how quickly electrical energy is converted to heat. 1 watt = 1 joule / second. 1 kW = 1 kJ (or 1000 joule) / second.
To convert bhp to kW simply multiply the bhp figure by .7457
There are 14 000 W in 14 kW. A watt is the unit for power, which is the rate of doing work.
40956
Electrical consumption is measured in kW per hr. 312000/1000 = kW. To answer this question your rate per kW hour from the utility company must be stated. Once known multiply that rate times the kW used in one hour.
16 kW at 240 volts runs at 16000/240 amps, or 67 amps.