You can't. Measure the amperes simply tells you what the current is.
Measuring the current in each phase (or do you mean 'line'?) will not give you sufficient information to work out what you are asking for.
To calculate the kWh consumed by a 3-phase motor, you'll need to know both the power factor and operating hours. The formula is: kWh = (√3 x Volts x Amps x Power Factor x Hours) / 1,000. Without the power factor and hours of operation, a precise kWh calculation cannot be provided using just voltage and current.
To convert amps to kilowatt-hours (kWh) for a three-phase system, you need to know the voltage. The formula for three-phase power is: Power (kW) = sqrt(3) x Voltage (V) x Current (A) x Power Factor. Once you have the kW, you can convert to kWh by multiplying by the number of hours the equipment operates.
To calculate three-phase meter reading in kWh with CT coils, you need to first determine the current ratio provided by the CT coils. Multiply the current readings obtained from the CT coils with the current ratio to get the actual current values. Then, multiply the actual current values with the voltage and power factor to get the power in each phase. Finally, sum the power in each phase and multiply by the time in hours to get the kWh reading.
It depends on the voltage. Please restate the question and provide the voltage. In general, however, simply divide total power by KV to get KA. Remember the KWH is an integral, so you need to back calculate KW. If the month is a 30 day month, then KW is KWH / 30 / 1440. Then, if the load is star, simply divide by three; if the load is delta, divide by three and multiply by 1.732, the square root of 3. Example: 480 three phase running star. 8000 / 30 / 1440 is 185 amperes. 185 divided by 3 is 62 amperes per phase. For delta, that becomes 107 amperes.
For a balanced load, you don't have to worry about phase values when you want to determine the power (or, in this case, the energy), whether delta or wye. Rather, you always use line values:P = 1.732 VL IL cos (phase angle)For an unbalanced load, however, you need to measure the phase voltage and phase current and power factor for each of the three phases, and add them together:P = [VpIp cos (phase angle)]phase A +[VpIpcos (phase angle)]phase B+[VpIp cos (phase angle)]phase CTo then calculate the energy expended in kilowatt hours, you need to multiply the total power (as calculated above), expressed in kilowatts, by the time for which the load is operating, expressed in hours.
There are zero amps in 32 kWh. Watts are the product of amps times volts. Without stating what the voltage is, this calculation can not be made. I = W/E is the formula to find amperage but there has to be a time constant if kWh is used. Usually the question is asked as to how many amps are in 32 kW.
Single phase power from a L-L voltage of 208 volts is calc'd: P = V(phase to phase) * I (individual phase current) / sqrt(3) Total power from all three legs is the above P * 3. P is in watts; continuous sampling will result in watt hour measurements; 1000 x this is kWh's.
Ohm's law is P=VI for single phase and 1000 watts per KW where P,V,I denote power,voltage,current so KW = amps *110/1000. So 9.1 amps gives ~1 KW If three phase, it is adjusted by the square root of 3.
10 amps at 120 volts is 1200 watts or 1.2 Kw, so in 1 hour it will use 1.2 Kwh
In exactly the same way as you do so for a single-phase load. It's the product of the power of the load, expressed in kilowatts, and the during of operation, expressed in hours. (And the symbol of kilowatt hours is 'kW.h', not 'kwh')
When you get your electric bill, look to see how many kilo watt hours (kWh) you used for the month. Then how much the bill cost. take the price divided by the (kWh). This is truly what your paying per (kWh). One (kWh) is 1000 watts being used for a hour, so one 100 watt light being used for 10 hours = 1 (kWh) or a 1500 watt hair dryer being used for an hour = 1.5 (kWh). so if your appliance power use is listed in amps, take amps x volts to figure out the watts. Plus determine how often the appliance runs. this will give You an idea of cost.