Calculations for motors above 250HP,@ 480vac

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sjwister

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I have been seeing alot of motors on single line diagrams for motor sizes that are larger than that listed in the NEC 2005 handbook (which goes upto 200hp). How do you calculate the size of CB's and conductors for them?

On the sinlge line diagram I am looking at, it is showing a 250hp, a 450hp, and a 800hp motor going back to a 480vac motor control center. How can I calculate the correct sizes for CB's, power and grounding conductors?

Any help would be appreciated.

Thanks
SJWister
 
For motors above 500Hp, or medium voltage motors, use this as a rule of thumb -

Hp x 1000 = KVA

Example:

850Hp x 1000 = 850KVA

850KVA/(1.732 x 460) = 1067 A


BEWARE:

Not sure why your NEC 2005 Handbook only goes to 200Hp, mine goes to 500hp. Keep in mind that motors over 500Hp at 480V do not fall under NEMA MG1and therefore are going to be a custom built motor. Starting characteristics and motor capabilities are not guaranteed to match the standards.

As with any motor, make sure you look at VD, and starting capability of the system. Large motors on 480V systems can cause an extensive VD at the bus when started, and it is unlikly the motor will start at all without using a VFD or soft starter.
 
Actually you can only use S'mise's formula. HP x 1000 only equals HP x 1000 and will never equal kVA.

HP x 746 = watts
watts / (supply voltage x sqrt(supply phase) x PF x Eff) = amperes

which is the same as S'mise's formula.
 
Snorks said:
Actually you can only use S'mise's formula. HP x 1000 only equals HP x 1000 and will never equal kVA.
Click to expand...

Really! No wonder I said as a rule of thumb........

And suppose I tell you that I have a 600Hp motor, don't know the power factor, don't know the efficiency. So then, what is the current?

I'll bet I get closer using my rule of thumb, then by using some imaginary power factor and efficiency.

BTW: The use of Hp x 1000 = KVA is something that comes from a accepted quick and dirty method of doing a fault current calculation called the MVA method, developed by Moon H. Yuen in the 60's while working at Bechtel; published and accepted by IEEE.
 
250 hp @ 480volts, with a SF of 1.15 = FLA of 302amps, 3/0 2 phase
450 hp @ 480 volts, with a SF of 1.15 = FLA of 515amps, 400 2/ phase
800hp @ 480volts, I never would use 480volts, 2140 & above.
Just my $.02
 
davidr43229 said:
250 hp @ 480volts, with a SF of 1.15 = FLA of 302amps, 3/0 2 phase
450 hp @ 480 volts, with a SF of 1.15 = FLA of 515amps, 400 2/ phase
800hp @ 480volts, I never would use 480volts, 2140 & above.
Just my $.02
Click to expand...
can you verify where is the data from? Do you have the corresponding PF and EFF?
thanks..
 
kingpb said:
BTW: The use of Hp x 1000 = KVA is something that comes from a accepted quick and dirty method of doing a fault current calculation called the MVA method, developed by Moon H. Yuen in the 60's while working at Bechtel; published and accepted by IEEE.
Click to expand...
I agree. My instructor taught us this as he said if we just need some estimate finding the fault. It's so easy to remember.
 
When Moon Yuen worked for me after he left Bechtel (he was the smart one), he had me calculate KW and amps for some motors with typical power factors of 0.80-0.85 and efficiencies of 0.88-0.92.

After doing about 10 motors he let me "discover" that 0.85 x 0.88 is very close to the 0.746 kW/HP factor and that 1 HP is very close to 1 kVA; usually within 3%. (Moon was a great engineer, a better teacher and great mentor.)

(1 HP x 0.746 HP/KW) / (0.83pf x 0.90 eff) = 0.997 kVA

If the motor is a modern energy efficient design, the actual FLA will be less than the value calculated using this estimation. But IMHO it is good enough to size cables and MCC's when nameplate data is not available.
 
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