- #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.

Task gives the following data:

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

I believe that it's best for the solution to come in sequential steps instead of a long essay.

Any sort of help would me more than appreciated.

**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

## 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.

- 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}##
- Winding losses for secondary coil are therefore ##P_{Cu} = P_{Cun}\cdot (\frac{I_2}{I_{2n}})^2 = 7200\text{W}##
- 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}##
- As a result, ##E_2' = 1.1 \cdot U_2 = 1.1 \cdot E_2n = 440\text{V}##
- 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{%}##.
- 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}##.

**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##.- 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}##

Any sort of help would me more than appreciated.