had a simular ? on NE test

[sparky=t]

Q- an irrigation pump runs continuous for 24 hrs., how many KWH would be billed for that period?

Given- 3 phase 30 HP, 480 V pump motor

T-430-250 30 HP @ 460 V = 40 amps
P=E X I
480 X 40 = 19200 Watts
19200 x 24 Hrs = 460800 Watts
460800 / 1000 = 460.8 KWH

did I miss something?

 

[HEYDOG]

480 * 1.732 x 40 *24 = 798,106/1,000 = 798 KWH

 

[sparky=t]

$@%** duh, what was i thinking, or not.......... thanks

 

[Cold Fusion]

... did I miss something?

It's a poor question. They did not give you either the motor efficiency nor the power factor. Without one or the other, all you can calculate is the KVA. Can't get the KW or KWH.

If you figure 90% efficiency:
30hp x 746 /.9 /1000 = 24.9kw
24.9kw x 24 = 597KWH


If you figure pf = .85
480 x 40 x 1.732 x .85 = 28.3kw
28.3kw x 24 = 678kwh

I don't know what the test writers are trying to get you to do to figure it out.

cf