A 5kVA motor operates from a 240V 50Hz supply at a lagging p.f. of 0.6 and is in parallel with a heating and lighting load of 3kW. What is the overall p.f. and what capacitance is required to improve this to 0.95 lagging? What is the apparent power before and after correction? What is the total current flow before and after correction?
I have no idea but wouldn't someone have to know the impedance of all the stuff to be able to answer that? Or is it that you can calculate that from the power factor (size and direction)?
Like, if the heating load was purely resistive then I don't believe that would play a part in the calculations. I believe motors and lighting ballasts are typically inductive and hence why you would use capacitors to help improve the pf.
All this sounds like one of those 'If a train is traveling at 60 mph in a Westerly direction and a car at 30mph in an Easterly, what time did they set off?' (But I'm probably wrong). ;-)
Just from a practical pov. Would one normally try to correct the pf of each load so if they are switched in and out the pf would still stay the best it can be?
Cheers, T i m
You don't need the impedances; everything you need to know to solve this is provided in the question.
If they were irrelevant to the problem, they wouldn't have been included!
So, this is a puzzle really, rather than a technical question where you are actually looking for technical help. Do other NG users think as I do, that this should really be signalled more clearly in the header?
It's an exam question, really so I take your point. But it'll be the last one I'll post, anyway. (no doubt that'll be a relief for a lot of people here) ;-)
Hmm, would that be a constant pf independent of motor load? :-)
It's unusual to specify a motor in terms of kVA and assume that this is the load it takes from the supply.
Real part of current = 0.6 x 5,000 / 240 = 12.5A Reactive part = sqrt( (5,000/240)^2 - 12.5^2) ) = 16.67A
3kW load current = 3,000 / 240 = 12.5ATherefore total real load is 25A. Complex current we would get if pf is 0.95 = 25 / 0.95 = 26.3A, Reactive current allowed at pf 0.95 = sqrt( 26.3^2 - 25^2 ) = 8.22A
A correction capacitor must therefore supply leading current of 16.67 -
8.22 = 8.45AXc = 1/wC = 240 / 8.45 ; C = 8.45 / (2 x pi x 50 * 240) = 112uF
Am I far out?
Probably silly question.
A friend of mine who is not that great at his current job is thinking of retraining as an electrician. Would this question be typical in basic installation qualifications or something more advanced.
If basic, then (gulp) I fear for him ...
Shame this will be the last one, perhaps the header could mention puzzle or exam question. If anyone is really bothered they can easily create a message filter that traps the words "puzzle" of "exam".
It'll only be a relief for those few who think they know so much, but in reality have forgotten it all.
Most electricians don't have to worry about motors and power factor. However, the 17th edition regulations and the corresponding exam covers a very wide subject range. It's not a trivial subject and he'll have to spend a lot of time learning.
I'm just about to hop into bed for the night, but I'll set your mind at rest - you got the cap value near enough bang on; the rest I'll have to check tomorrow. Good point about the phrasing of the question, btw., hadn't struck me until you mentioned it. Nighty Night!
If he is doing industrial design it might be.
If he is doing houses there isn't much chance he will need to know.
You might be if it were mine, I like to use heat pumps for heating so they are reactive loads and not constant either. Of course you would correct the PF for each one.
The mind boggles as to what you are building.... Brian
Scuse me I'm just off to dig half a hole. Brian
Does the heater change resistance as it heats up? Brian
I can well remember a visit at the place I worked back in the 60s from a techy from the Leccy board as then was complaining that whatever we were doing was not liked by the local sub station. What we were doing was running a hundred or so tvs with autotransformers and half wave rectifiers on a soak test rack. I think we had to have it split b between mains phases to get it sorted out. Now being the simple person I am, I cannot see why this would be any different to 100 houses watching Coronation street, but there we go. Brian
I certainly recall TV sets of the type you describe, though often the heaters were supplied by one half wave rectifier, and the electronics by the other "half" to try and equalise the current taken on +ve and -ve voltage excursions.
There are two reasons, both have an effect on the substation, one is the harmonic nature of the waveform with even harmonics, and the other was the net DC current supplied back to the substation.
A DC current would bias the magnetising flux to become non-symetrical and the magnetising current would increase depending how close the flux level got to saturation.
I din't think anyone bothered to rectify heater supplies. They were either
6.3v ac or fed in a long series chain.
They were in a long chain and typically of the "U" variety denoting
0.1A. Earlier chassis would have the prefix "P" denoting 0.3A. BICBWThe half wave feed was an alternative to using a humungous dropper resistor.
or "line cord"!
HomeOwnersHub website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.