Electrical

The idea of homework is you do your own research.

CR4 isn't a site of homework cheats.

If you do not want to help you can keep quiet.

No need to be offensive and condescending.

Wonder if you know the answer or should we infer that your rudeness is a sign of an all knowing "guru". Ho Ho

I don't think in stating a fact that this isn't a homework site in offensive and that one should do ones own homework is quite pertinent.

Read other comments.

Not worth answering?

What is the difference then going to ask your parents a question?

As engineers and as in any other job we are always learning. By telling someone not to ask questions, school related or not. You are not only preventing someone from learning but you are also causing a safety issue (if it was work place related) by holding back information.

Oh and listen to your mother, I assume you have one. "If you can't say anything nice. F*** off".

Blogs are not intended from personal issues, and people saying "I don't know". If you don't know the answer shut up. And yes I know that I am taking up virtual space by saying this.

But :P

:)

Why not do it the old fashioned way, measure it?

Martin

The powerplant capacity is 4.2MW - the MW is the unit of power. However teh mean output power- that has to be measured (the power demand is by load the generator will only provide the maximum it can supply) No one can tell the answer to this question. It is rather the other wasy round

Compute the loads, individual maximum loads, diversity factor,...

calculate the probable maximum load (demand)

Rate the generator according to the probable peak demand.

Compressor motor is 3.2MW -

The output power = Input electrical power x motor efficiency x compressor efficiency.

Again input electrical power is not known (the motor rating would be significantly more than the actual power drawn/ demanded)

Go again for reverse calculation

Compressor output - flow, fluid, pressure values

Convert into the power (Output head)

divide by the two efficiencies and you would know the demand.

_{(The other posts of the OP denotes that this is not a homework question)}

Your first question is not worth answering.

Second question also contain ignorance. Compressor is not producing power so output power is not measured, it produces compressed gas which is measured.

For compressor two other terms are used- 'gas power' and 'shaft power'. Gas power is theoretical power input required. Shaft power is after thermodynamic efficiency. After considering mechanical losses at gears, bearings etc (2 to 3% depending on size), required power at coupling is derived. Motor rating (in your case 3.2 MW) is generally 10 to 15% higher than power at coupling.