Really simple power question from a lowly chem eng ...
Really simple power question from a lowly chem eng ...
(OP)
Hi Electrical people! I was hoping someone could explain how to calculate the power usage of (in this case) a large air-compressor.
I've been told the voltage across the unit is 453V and the current draw is 596A . Now I am assuming this is a 3 phase unit, and the current stated is per phase- but I honestly do not know.
The power factor I've been given is 0.91, but no matter how I try to calculate the power (whether it be using RMS voltage and RMS current ....) I cannot come up with the answers they get which are (and there are two different ones) 407.4 kW and 409.4kW
The way I calculated it was...
P=Pf * (V*I*3)/2 =368.5kW (where the 3 comes from 3 phase power, the two is from RMS voltage and Current (root 2 squared), and Pf is the power factor. Perhaps I'm leaving something out?
Any insight would be appriciated!
Cheers
James
I've been told the voltage across the unit is 453V and the current draw is 596A . Now I am assuming this is a 3 phase unit, and the current stated is per phase- but I honestly do not know.
The power factor I've been given is 0.91, but no matter how I try to calculate the power (whether it be using RMS voltage and RMS current ....) I cannot come up with the answers they get which are (and there are two different ones) 407.4 kW and 409.4kW
The way I calculated it was...
P=Pf * (V*I*3)/2 =368.5kW (where the 3 comes from 3 phase power, the two is from RMS voltage and Current (root 2 squared), and Pf is the power factor. Perhaps I'm leaving something out?
Any insight would be appriciated!
Cheers
James
Read the FAQ ... ht





RE: Really simple power question from a lowly chem eng ...
Second, you can either use three times phase current times phase-neutral voltage, or (since phase-phase voltage is sqrt(3) times phase-neutral) the formula P = sqrt(3)xUxIxPF, which results in 426 kW. At least on my calculator.
This is absorbed (electric) power. Shaft power will then be 426 kW times efficiency. You need an efficiecy around 95 percent to get the 407 kW shaft power you mention in your post. That is quite high - but not impossible with the new motors.
Gunnar Englund
www.gke.org
RE: Really simple power question from a lowly chem eng ...
Read the FAQ ... ht tp://www.e ng-tips.co m/faqs.cfm ?pid=731&a mp;fid=376
RE: Really simple power question from a lowly chem eng ...
RE: Really simple power question from a lowly chem eng ...
You know I luv ya. But not always. I was given three significant figures. Where did you get the fourth?
Gunnar Englund
www.gke.org
RE: Really simple power question from a lowly chem eng ...
lolol
Actually I was allowed to assume ".0" on both the given voltage and current because the two stated possible answers had the first decimel point;
"407.4 kW and 409.4kW"
Now why would there be to possible answers..
RE: Really simple power question from a lowly chem eng ...
RE: Really simple power question from a lowly chem eng ...
Apologies if it was too basic a question for these forums, but it has been a fair few years since I've had to deal with electrical calcs in 3phase AC circuits. Thanks for all your help though!
RE: Really simple power question from a lowly chem eng ...
Gunnar Englund
www.gke.org