Transformers and Active/Reactive power question from HW

In summary, the transformer is connected for step up operation with a load of 144Ω. The input voltage is 220 V and the secondary voltage is 6600 V ∠0°. The secondary current is 45.8333 A ∠-46° and the primary current is 1375 A ∠-46°.
  • #1
JoeMarsh2017
24
0

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
 
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  • #2
One-cycle-averaged power is conveniently calculated by using complex representations of voltage and current such as [tex]\tilde S = \frac{{\tilde V{{\tilde I}^*}}}{2} = P + iQ[/tex], 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, [tex]\tilde V = 6600{e^{i0}} = 6600,{\rm{ }}\tilde I = 45.8333{e^{i\left( { - {{46}^ \circ }} \right)}}[/tex]

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.
 
  • #3
210.1 kW
217.6 kVAR

These are the answers from the book
 
  • #4
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.
 
  • #5
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
 
  • #6
JoeMarsh2017 said:
Reactive=217.34 =210.13tan **sin** 46
 
  • #7
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
 
  • #8
JoeMarsh2017 said:
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..
 

1. What are Transformers and how do they work?

Transformers are electrical devices that are used to transfer energy from one circuit to another through electromagnetic induction. They consist of two or more coils of wire wrapped around a core material, usually iron. When an alternating current flows through one of the coils, it creates a changing magnetic field which induces a current in the other coil. This allows for the transfer of power between circuits without direct electrical connection.

2. What is Active power and how is it different from Reactive power?

Active power, also known as real power, is the power that is actually consumed by a load and is responsible for performing work. It is measured in watts (W) and is what we typically think of as the "useful" power in an electrical system. Reactive power, on the other hand, is the power that is required to establish and maintain the electromagnetic fields in an electrical system. It is measured in volt-amperes reactive (VAR) and does not perform any useful work.

3. Why is it important to maintain a balance between Active and Reactive power?

In an electrical system, the total power that is supplied must be equal to the total power that is consumed. This means that the balance between Active and Reactive power is crucial for maintaining the stability and efficiency of the system. If there is an imbalance, it can lead to voltage fluctuations, increased losses, and potential equipment damage.

4. How do Transformers affect the balance between Active and Reactive power?

Transformers play a critical role in maintaining the balance between Active and Reactive power. They are designed to have a certain power factor, which is the ratio of Active power to Apparent power (the combination of Active and Reactive power). By properly selecting and using transformers, the power factor of a system can be improved, which helps to reduce losses and improve overall system efficiency.

5. What are some common applications of Transformers and Active/Reactive power?

Transformers and Active/Reactive power play a crucial role in various industries and applications. They are commonly used in power generation, transmission, and distribution systems to transfer electricity over long distances. They are also used in industrial and commercial settings to step up or step down voltage levels for specific equipment or processes. In addition, transformers are used in electronic devices, such as chargers and power supplies, to convert AC power to DC power. Maintaining a balance between Active and Reactive power is important in all of these applications to ensure the efficient and reliable operation of the electrical system.

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