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Good answers, but more simply,you can convert kilowatt hours to amperes if you know the voltage at which your loads operate and assume a steady state load (a big assumption). If, for example you consume 100 kwh in a day at 240VAC and assume a steady state load, that's 100 kwh divided by 24 hours, or a 4.17 kw load. 4.17 kw divided by 240V is 17 ...
132 KW/hr divided by 24 hours = 5.5 KW/hr divided by 240V = 22.92 amps ave per hour and therefore if I assume most of the usage is during an eight hour period, I could guess that: 132 KW/hr divided by 8 hours = 16.5 KW/hr divided by 240V = 68.75 amps ave per hour
I am trying to calculate the FLA for a motor given the motor kW rating (this is a German motor and they like to give their ratings in kW rather than hp.)but my numbers are not matching up with those given by the manufacturer. I am trying to calculate the FLA as an exercise for myself to try to figure out how to derive them from a given kW.
Yes, kind of, if you ignore power factor. At power factor of 1.0, kva=kw. If the load is mainly resistive, e.g. lighting and heat, power factor will be 1 or very close to that. When you have motors or other less simple loads, then the waveforms of volts and amps may not line up. That's when you get a power factor below 1. KVA is then more than kW.
Nov 21, 2009. #9. infinity said: You are permitted to calculate the size of a branch circuit for an electric range according to Table 220.55. Look at note #4 to the table. One appliance in column C not over 12 kw is permitted to be calculated at a maximum demand of 8 kw. 8000/240 = 33.33 amps.
If the load factor were very high, you might assume the peak kW is closer to the average kW. Not the case here. However, for amp calculation example purposes, let's assume 90% pf. With a peak kW of 146.56 we get the peak kVA as 146.56 / 0.9 = 162.84 kVA. At 208Y/120 this yields 162.84 / sqrt(3) / 0.208 = 452 amps.
Here it is: 1 ton of cooling = 12,000 BTUs of heat = 3.513725 kilowatt-hours 5 tons of cooling = 5 X 3.513725 = 60,000 BTUs = 17.57 kWH. You're assuming there is a 1:1 conversion between cooling tons and applied energy. If an AC worked like a resistance heater, that would be a good assumption. But they don't work that way.
A 200 amp 120/240 single phase panel in a home is capable of providing 48,000 watts (48 KW) of power. A 200 amp 208Y/120 three phase panel is capable of providing 71,000 watts (71 KW) of power. A 200 amp 480Y/277 three phase panel is capable of providing 166,000 watts (166 KW) of power.
Cost of electricity is for kW, not Amps. Amps is essentially irrelevant. So yes, there are 746 watts/HP, that has nothing to do with amps, volts, phase etc. That is the power that is used. You are metered on kWH, (kilo Watt Hours), so to determine the cost of operating something electrical, simply take the watts (kW) x operating time (Hours).
Please help me make sure this calculation is correct for a 12.9 KW range. 12.9 KW-12.0KW= .9 KW .9KW x 5%=.o45% or 100.045% 100.045% x 8kw= 8.0036kw 8.0036kw rounds down to 8KW or 8000 W/240V = 33.3 amps wire size # 8 thhn awg. breaker 40amp Thank you