Complex power/power factor calculations

  • #1

Homework Statement


(old exam question attached)

Homework Equations


Complex power: S = VI* / 2 = P + jQ where P is the average power and Q is the reactive power
Magnitude of complex power |S| = Vmax*Imax/2
V = ZI

The Attempt at a Solution


We can calculate θv - θi for each load by using arccos(power factor).

arccos 0.8 = 36.86 degrees
arccos 0.75 = 41.4 degrees

Because I'm fairly new to these types of problems I need to make sure I'm understanding this correctly before I can go any further:

- The first load is given in kW and the other is given in kVA. Does this mean that we can assume that the first load is purely resistive?
- Can I go ahead and express the complex power of the second load as 40/_41.4 -> 7.5 + 6.6j?

I'm thinking that the answers to these questions will indicate what I should do next.

Regarding part b), how exactly is system power defined? Is it the sum of the power delivered to all of the elements? If so, could I simply add the complex power delivered to the impedance Zs to the complex power delivered to both loads?

Thanks for any help you guys can offer!

EDIT: I just realized I have enough information to calculate the current flowing through the kVA load using S = VI* / 2. Alone, this doesn't help much, but by following a similar procedure for the first load (after determining what the complex power is, exactly) I can apply KCL and determine the current through the branch with the impedance (i1 + i2 = i3).

EDIT 2: Another thought - could I find the current through the impedance directly by using the complex power delivered to both loads as Stotal = S1 + S2 = 40 + (7.5 + 6.6j) and S = VI* / 2? This would solve part a), after finding the voltage of the impedance Zs and using KVL in the leftmost mesh.
 

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Answers and Replies

  • #2
115
2
First load is not pure resisitive as power factor is not [itex]1[/itex]:
[itex]\begin{cases}P_1=40\,kW\\PF=0.8\end{cases} \Rightarrow Q_1=P_1\frac{\sqrt{1-(PF)^2}}{PF}=30\,kVA\\
\begin{cases}Q_2=40\,kVA\\PF=0.75\end{cases} \Rightarrow P_2=Q_2\frac{PF}{\sqrt{1-(PF)^2}}=\frac{120}{\sqrt{7}}\,kW
[/itex]

I think [itex]V=220\angle 0[/itex] means [itex]V_{rms}=220[/itex] not [itex]V_{max}=220[/itex] so [itex]S=VI^*[/itex]
 

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