Understanding Transformer Phasor Diagrams

In summary, the problem is that the phasor diagram provided as a solution does not match the computed data. The phasor diagram describes the voltage drop on the secondary side of the transformer, but the data given does not match this. The problem is then solved by solving for the voltage drop on the primary side, using known values for the resistance and reactance.
  • #1
Peter Alexander
26
3
Hello everyone! I'm currently studying transformers and a task related to phasor diagrams shows up. I'm having lots of problems with comprehension of the subject, so I'd like to ask for some help. I don't understand how the phasor diagram given as a solution could possibly be drawn out of the computed data.

1. The problem statement, all variables, and given/known data
Task gives the following data:
  • ##S_n = 1\text{MVA}, \quad 10\text{kV} / 0.4\text{kV}, \quad f=50\text{Hz}##
  • ##P_{Cun} = 15\text{kW}, \quad P_0 = 5\text{kW}##
  • ##u_K = 6\text{%}##
  • Current through the secondary coil ##I_2 = 1000A## at ##\cos{\phi_2} = 0.5##, inductive
And requires computation of losses in copper on secondary side (this represents windings), iron (this represents the core) as well as actual value of ##U'_2##.

Homework Equations


All relevant equations will be present in the attempt at a solution.

The Attempt at a Solution


I believe that it's best for the solution to come in sequential steps instead of a long essay.
  1. Compute the rated currents ##I_{1n} = \frac{S_n}{U_1\sqrt{3}} = 57.34\text{A}## and ##I_{2n} = \frac{S_n}{U_2\sqrt{3}} = 1443.4\text{A}##
  2. Winding losses for secondary coil are therefore ##P_{Cu} = P_{Cun}\cdot (\frac{I_2}{I_{2n}})^2 = 7200\text{W}##
  3. Core losses are computed as ##P_{Fe} = P_{Fen}\cdot\frac{f'}{f_n}\cdot(\frac{B'}{B_n})^2## but since ##U_1' = 1.1 U_1## we can deduce that ##f'B' = 1.1 f_n B_n## consequently leading to ##P_{Fe}=P_{Fen}\cdot\frac{f'}{f_n}\cdot (1.1\cdot\frac{f_n}{f'})^2 = 5042\text{W}##
  4. As a result, ##E_2' = 1.1 \cdot U_2 = 1.1 \cdot E_2n = 440\text{V}##
  5. For ##U'_2##, we need the following values: ##u_K = 6\text{%}##, ##u_R = \frac{P_Cun}{S_n} = 1.5\text{%}## and ##u_X = \sqrt{u_K^2 - u_R^2} = 5.81\text{%}##.
  6. We require a ratio ##\frac{I_2}{I_{2n}} = 0.6929## which is preserved even on the primary side, making ##I_1 = 0.6929 \cdot I_{1n} = 39.73\text{A}##.
  7. From here on, I'm starting to get lost. From known ratio, we can compute ##u'_R = 1.039\text{%}##, ##u'_X = 4.025\text{%}## and ##u_K = 4.157\text{%}##, where all we did was to multiply previous values by the factor ##0.6929##.
  8. This equation is given on the datasheet and it describes the voltage drop on secondary side ##\Delta U##. It can also be derived from the Kapp triangle using points on the diagram (will attach it below). Computational procedure is $$\Delta U = U_2 \cdot (
    u'_{R}\cos\phi_{2}+u'_{X}\sin\phi_{2}+U_{1}-\sqrt{U_{1}^{2}-(u'_{X}\cos\phi_{2}-u'_{R}\sin\phi_{2})^{2}}\cong4\text{\%}) = 17.6\text{V}$$ meaning that ##U'_2 = E'_2 - \Delta U = 422.4\text{V}##
Now if we take a look at the phasor diagram (should be attached below), proportions don't add up, because if ##U'_1 = 1100\text{V}##, then how can ##U'_2##, which should be ##422.4\text{V}##, be drawn as if it measures ##\approx 660\text{V}##?

Any sort of help would me more than appreciated.
 

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  • #2
upload_2018-5-21_14-12-2.png

From the attached Single-Phase Transformer Diagram where:
Vp=primary rated voltage
Rp=primary windings resistance
Xp=primary windings magnetic flux leakage reactance
Xm= main magnetic flux reactance
Rfe= equivalent resistance of magnetic losses
Ip=primary load current
Io=primary windings main magnetic flux current
I's=secondary current referred to primary
V's=secondary terminals voltage referred to primary
R's=secondary windings resistance referred to primary
X's=secondary windings leakage reactance referred to primary
E1=e.m.f of primary windings
E'2=e.m.f. of seconadry windings referred to primary [E'2=E1]
Zp=Rp+jXp ; Z's=R's+jX's
Po=Rp*Io^2+E1^2/Rfe where E1^2/Rfe=Pcore
Pcore=Core loss = Hysteresis loss + Eddy current loss
Pcore=Kh.f.Bmax^n+Ke.f2.Bmax^2
If we could neglect Rp*Io^2 then Po=Pcore
If we shall neglect hysteresis loss then only eddy current loss will be:
Po=Ke.f^2.Bmax^2
Bmax=k.E1/f/Areacore
Po/Pon=Ke.f^2.Bmax^2/Ke.f^2.Bmax^2=(f/fn*Bmax/Bmaxn)^2 or:
Po/Pon=[f/fn.E1/E1n.fn/f]^2=(E1/E1n)^2
E=Vp-Zp*Ip=V's+Z's*I's
In order to calculate E we need to know Zp or Z's.
In a short-circuit case E1=E2≈0 then Zsc=Zp+Zs'
Zsc=Vp^2/S.uk%/100
So what we can know it is only Zsc and we don't know what is Zp and Zs' separate.
Let's say Zp=Z's=Zsc/2
Now we could calculate E1/E1n and recalculate Po.
The resistance and reactance it could be considered constant.
 

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  • #3
upload_2018-5-22_19-32-42.png
 

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  • #4
This was actually incredibly helpful! Thank you for your time and patience! I finally understand how it works, thank you!
 

Related to Understanding Transformer Phasor Diagrams

1. What is a transformer phasor diagram?

A transformer phasor diagram is a graphical representation of the voltage and current relationships in a transformer. It shows the magnitude and phase angles of the voltage and current on the primary and secondary sides of the transformer.

2. What information can be obtained from a transformer phasor diagram?

A transformer phasor diagram can provide information such as the voltage and current ratios between the primary and secondary sides, the power flow direction, and the phase relationship between the voltage and current.

3. How is a transformer phasor diagram constructed?

A transformer phasor diagram is constructed by representing the voltages and currents on the primary and secondary sides of the transformer as vectors. The length of the vector represents the magnitude and the angle represents the phase angle.

4. What does a lagging and leading power factor look like on a transformer phasor diagram?

A lagging power factor is represented by a phasor that is behind the reference phasor, while a leading power factor is represented by a phasor that is ahead of the reference phasor. This indicates the phase relationship between the voltage and current.

5. How can a transformer phasor diagram be used in troubleshooting?

A transformer phasor diagram can be used to troubleshoot transformer issues by comparing the actual phasor diagram with the expected phasor diagram. Any discrepancies can indicate problems such as incorrect voltage or current values, or issues with the transformer itself.

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