To convert 'kwh' to 'kvah' you first need to measure the length of time. You will then convert this amount to hours by dividing by 3,600. You will then divide this amount by the length of time.
we can calculate kwh,kva,kvah by following formulas kva=[(kw)2 + (kvah)2]1/2 kw=[(kvah)2- (kva)2]1/2 kvah=[(kva)2-(kw)2]1/2
1000
A 1000 watt device operated continusouly for 1 hour would equal 1 Kwh.
KWH = KW times hours If you run a 750 KW load (lights, motors, so forth) for 1 hour, you have 750 KWH. If you run it for 1/2 hour, 750 KW X .5 hours = 375 KWH. If you run it for 5 hours, 750 KW X 5 = you do the math.
The question is incomplete, because there are no mention about CT & PT ratios. 600VA 5 can not be CT ratio.
To reduce kWh by capacitor is when a motor is put in. The terminal voltage is reducing and current is increasing it is connected parallel with the motor.
Kwh will equal to Kvah.
trivector meters are used to measure kVAh and also kVA of maximum demand.it has a kwh meter and reactive kvah meter in a case with special summator mounted between them.
On a KWh billing systemOn a kVAh billing systemtariff 1 euro 20.50/KW Each kWh@14c7tariff 2 euro 19.20/KVA Each kVAh@13c5
To calculate kWh a time component is needed for the hours.
Call your supplier, find out how much for one kWh, then 60 x that price.
P.F = Kwh/Kvarh
A kV.A.h (not 'kvah') is a kilovolt ampere hour. You can think of it as being the vector sum of 'active' (kW.h) and 'reactive' (kvar.h) energies.
A kV.A.h (not 'kvah') is a kilovolt ampere hour. You can think of it as being the vector sum of 'active' (kW.h) and 'reactive' (kvar.h) energies.
MCF * BTU = MMBTU MMBTU * KWH = Heat rate
You multiply the power by time, ie kW*h
You can't. Measure the amperes simply tells you what the current is.
Efficiency = (860.4*100)/(Heat rate in kCal/kWh) or Efficiency = (860.4*4.18*100)/(Heat rate in kJ/kWh) Ex 1: if heat rate is 2500 kCal/kWh, then efficiency is 34.416% Ex 2: if heat rate is 9000 kJ/kWh, then efficiency is 39.96%