How do I calculate the power used for a three phase motor running at 575 volts and drawing an average of 575 amps? This is for the arguement of running an electric feedpump motor vs running a turbine feedpump and sending the power out the door. We know how much steam the turbine uses, we need to know how much electricity the motor uses.
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Motor Power Consumption
- Thread starter bethw
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I too am mechanical delving into the world of electrical engineering, and the first question I would expect you would be asked is if that is 575 A/phase or 575 A total.
If it is per phase, you are looking at 575V x 575A x root3 = 572.7 kVA. Assuming a Power Factor of .8, that is 458.1 kW.
If the 575 is total, then it is simple 575V x 575A = 330.6 kVA. A PF of .8 puts it a 264.5 kV.
Hope that I didn't muddle it up too bad. True EEs, correct me if I'm wrong.
If it is per phase, you are looking at 575V x 575A x root3 = 572.7 kVA. Assuming a Power Factor of .8, that is 458.1 kW.
If the 575 is total, then it is simple 575V x 575A = 330.6 kVA. A PF of .8 puts it a 264.5 kV.
Hope that I didn't muddle it up too bad. True EEs, correct me if I'm wrong.
I'm not sure how you can get 'total' amps, but the calculation for kVA is correct. Your assumption of 80%PF is probably on the conservative side for a motor; most motors run in the 90%PF range when they are run near full load.
Which brings me to my question...no nameplate info? The best way to determine kW of a motor is by looking at the nameplate for HP, kW, and/or FLA, some of which should be there.
Which brings me to my question...no nameplate info? The best way to determine kW of a motor is by looking at the nameplate for HP, kW, and/or FLA, some of which should be there.
- Thread starter
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bethw,
HP in the nameplate is the power. If you want power in KW, then KW = 0.746 x HP.
Current mentioned is the line current in each line. If your nameplated current is 450 A and your actual working current is 575 A, then you are overloading your motor which will lead to its premature failure.
HP in the nameplate is the power. If you want power in KW, then KW = 0.746 x HP.
Current mentioned is the line current in each line. If your nameplated current is 450 A and your actual working current is 575 A, then you are overloading your motor which will lead to its premature failure.
If you want to know the power actually used by the motor, you should measure the power with a wattmeter or measure the power factor and then use the V·I·sqrt(3)·pf formula. The nameplate HP gives full load output power (you need to divide by efficiency to get input power), but if the motor is not fully loaded mechanically, this will not be the power used.
Read your pump curve, and see what HP is required at the flows you are running, and then compare your electrical data, as recommended above, to what the pump should be using at that operating point.
If for nothing else, it should serve as a check and balance on your electrical work.
Part of your evaluation regarding whether or not you want to sell the power vs run plant equipment has to do with whether or not the same steam that the BFPT's would use to do this work is best used in the BFPT, or the main turbine.
In some cases the steam not extracted messes up the volumetric efficiency of the main steam turbine stages down stream of the extraction point, and contributes to exit losses at the hoods into the condenser. In other cases, it is actually beneficial. Depends upon your particular turbine.
Bottom line, if your BFPT efficiency is at all near the efficiency of the lower stages of the main turbine, ignoring things like exit losses, etc., it is more efficient to do the work with extracted steam, since to make this into electricity and run the motor adds all the electrical losses to the system before the work is applied to the pump. On the other hand, BFPT's are seldom as efficient as MST's. So, get out the Mollier Dia.
rmw
If for nothing else, it should serve as a check and balance on your electrical work.
Part of your evaluation regarding whether or not you want to sell the power vs run plant equipment has to do with whether or not the same steam that the BFPT's would use to do this work is best used in the BFPT, or the main turbine.
In some cases the steam not extracted messes up the volumetric efficiency of the main steam turbine stages down stream of the extraction point, and contributes to exit losses at the hoods into the condenser. In other cases, it is actually beneficial. Depends upon your particular turbine.
Bottom line, if your BFPT efficiency is at all near the efficiency of the lower stages of the main turbine, ignoring things like exit losses, etc., it is more efficient to do the work with extracted steam, since to make this into electricity and run the motor adds all the electrical losses to the system before the work is applied to the pump. On the other hand, BFPT's are seldom as efficient as MST's. So, get out the Mollier Dia.
rmw
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