Transformer Phasor Diagrams

  • #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.
 

Attachments

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Answers and Replies

  • #2
Babadag
537
147
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
Babadag
537
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upload_2018-5-22_19-32-42.png
 

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  • #4
Peter Alexander
26
3
This was actually incredibly helpful! Thank you for your time and patience! I finally understand how it works, thank you!
 

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