Transformers and Active/Reactive power question from HW

[JoeMarsh2017]

Homework Statement

A 7200-240 V, 60 Hz transformer is connected for step up operation, and a 144∠46° Ω Load is connected to the secondary. Assume the transformer is ideal and the input voltage is 220 V at 60 Hz.
Determine
a)secondary voltage
b)secondary current
c)primary current
d)input impedance
e)1)active
2)reactive
3)apparent power input to the transformer

Homework Equations

since its operating as "Step Up" transformer...the turns ratio aka "a' = 30

240-7200
1:30 step up

The Attempt at a Solution

a)secondary voltage =220*30= 6600 V∠0°
b)secondary current = 6000V/144Ω ∠46°= 45.8333 A∠-46°
c)primary current= a x I sec = 45.83333 * 30 = 1375 A∠-46°
d)Z input= 220 V / 1375A = 0.16∠46°Ω

Active / reactive and apparent Power are all listed as answers in the back of the book
I am having trouble keeping these numbers straight in my head! Please help me understand my errors on this problem...

210.1 kW =active power is my true power
217.6 kVAR = Is my reactive power
302.5 kVA 1375*220 = 302500 /1000(for the K) = 302.5 kVA

Can you kindly show me what I am forgetting about active and reactive power guys?

Thanks Joe

 

[goodphy]

One-cycle-averaged power is conveniently calculated by using complex representations of voltage and current such as \tilde S = \frac{{\tilde V{{\tilde I}^*}}}{2} = P + iQ, where P and Q are an active and reactive power respectively, and absolute value of S is an apparent power.

Here, I choose that V and I are for the secondary side of the transformer so, \tilde V = 6600{e^{i0}} = 6600,{\rm{ }}\tilde I = 45.8333{e^{i\left( { - {{46}^ \circ }} \right)}}

Calculation shows that P ≅ 105 kW, Q ≅ 109 kW, and the apparent power is 151 kW.

Your values for powers are twice than what I calculated. I guess you missed a factor of 1/2 somehow.

 

[JoeMarsh2017]

210.1 kW
217.6 kVAR

These are the answers from the book

 

[magoo]

Since you are dealing with effective or rms voltages, the power equation is:

S = V * I' = P + j Q

I' should be I-conjugate

There is no factor of 2 involved.

 

[JoeMarsh2017]

GUYS, I figured it out...My problem was that once I solved for "S"..and I had the angle @ 46°...then I can solve for active and reactive power
Power triangle means
s=302.5 theta=46°
Active=210.13 =302.5cos 46°
Reactive=217.34 =210.13tan 46

Answer

 

[cnh1995]

Reactive=217.34 =210.13tan **sin** 46

 

[JoeMarsh2017]

210.13 tan 46 gets you 217.59 which is what I was looking for..

Answers from the book are
210.1 kW Active Power
217.6 kVAR Reactive Power

This is per the power triangle

 

[cnh1995]

210.13 tan 46 gets you 217.59 which is what I was looking for..

Answers from the book are
210.1 kW Active Power
217.6 kVAR Reactive Power

This is per the power triangle

Oh..Right!
You took Q=PtanΦ and I misread it as Q=StanΦ.
Q=S*sinΦ=P*tanΦ..