Question 5 [15 marks]

A 230V, 50Hz single phase supply feeds an industrial plant with the following loads: (a) incandescent lamps drawing a current of 8A at unity power factor; (b) fluorescent lamps drawing a current of 5A at 0.8 leading power factor; (c) a motor drawing a current of 7A at 0.75 lagging power factor.

1. Find the real and reactive power drawn by loads (a), (b) & (c).
2. Find the total real, reactive and apparent power that the plant draws from the s upply.
3. Draw a power triangle for the plant and show its operating power factor angle.
   
4. Find the value of capacitance required to correct the plant power factor to unity.

No solutions are given, so I have no idea whether my methods are right or wrong, could someone please provide answers or explanations for these questions.

Is this correct for 5.1?
Consider each load to be a parallel branch,
then work out Z for each branch using E = IZ; Z = E/I
Power factor = P/S = (I^2*R)/(I^2*Z)
In each branch the currents are the same, so power factor = R / Z.
Therefore we can now work out R, and subsequently P, Q, S for each branch.

5.2
I thought of doing this two ways, adding up the true powers for each branch (can you do this, eg (Ia ^ 2) * Ra + (Ib ^ 2) * Rb) and then the same for the reactive powers, and then finding apparent power from these.

Also thought of finding overall impedance and finding the apparent power first, but then I don't know what to do from there.

5.3 is easy if I could work out the powers.

5.4 is not too bad applying methods

SOURCE: UCT EEE2038W June 2009 Exam

 

Part 1.

You don't need to use impedances.

The individual real & reactive powers can be found by using

P=V*I*cos(θ) S=V*I*sin(θ)

where
pf=cos(θ) as given for each load,
I is given for each load,
V=230V.
You need to work out sin(θ) - which you can do given cos(θ)=pf

Part 2.

Add the individual load Real and Reactive powers to give P total & S total. The total Apparent power Q total is given by

Q total=√(P total^2+S total^2)

Part 3.

Draw what you calculated in Part 2.

 

Thanks very much, this seems like a much better approach except I think you have some letters mixed up... Isn't it Q = V*I*sin(θ).

Eg Q is reactive power. S is apparent power, I think you have mixed these two!

 

Thanks very much, this seems like a much better approach except I think you have some letters mixed up... Isn't it Q = V*I*sin(θ).

Eg Q is reactive power. S is apparent power, I think you have mixed these two!

You are correct - my mistake. I should watch my "P's & Q's"!

 

You are correct - my mistake. I should watch my "P's & Q's"!

LOL. good for a laugh in the morning.