Power losses

[gazdkw82]

Going through some mock questions ready for 202 exam tomorrow. Our regular teacher didn't turn up today so we had a stand in. We had a question that no one (including teacher) could work out.

Question is

A load draws 9KW but works at only 85%. How much power will be lost in 4 hours.

My workings were

9000/100x85 = power actually being used = 7650

9000-7650 = actual power loss = 1350.

1350 x 4 = power loss over 4 hours = 5400kw

Unfortunately this is not the answer

I can't remember what the answer was now.

Can anyone help with how to approach this?

 

[Richard Burns]

The wording is somewhat odd as a draw would be a current not kW.
Possibly they intend to mean that the load is 9kW but only 85% efficient so therefore the power consumed is 10,588kW and so 6353kWh is lost without gain over four hours.

 

[gazdkw82]

Sorry. It was my bad explanation and memory

I've tried to attach a picture of the exact question

 

[davesparks]

Working backwards from the four possible answers the closest I can get is 5.28 which would equate to a load of 8.8kW rather than 9kW

Every once in a while the question might actually be wrong, in an exam this will normally be dealt with by not counting that question when the final result is calculated.

 

[gazdkw82]

Are you saying this question could be wrong?

 

[davesparks]

I'm saying it is a possibility, but I could easily have missed something so I'd wait until a few more people have had a look and a go at this before reaching that conclusion.

 

[Richard Burns]

I have realised my decimal points were ridiculous in my post above, sorry.
I am in agreement with davesparks I have done a spreadsheet of efficiency from 80-90% and loads from 8kW to 10.5kW and taken the assumption that the value given is either the energy supplied to heat to water or the energy taken by the supply and the only matching answer I can find is the 5.28kWh at 8.8kW supplied energy.

 

[hightower]

Remember, kW is after power factor is applied, or True Power for another term. So if something is running at 9000kW that to me means that's with the power factor taken in to account. To get Apparent Power is just V*I, or True Power / Power Factor.

So in this case your Apparent Power is 10588VA.

Remember, True Power is given in Watts (W), Apparent Power in Volt Amps (VA), and Reactive Power in Volt Amps Reactive (VAr).

So, going forward from that, 10588 - 9000 is 1588 power lost by having a poor power factor, and over 4 hours this would be 6352.

That's my take on things off the top of my head, but I've barely woken up yet so....