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Understanding Transformer Phasor Diagrams: A Step-by-Step Guide
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[QUOTE="Peter Alexander, post: 5997989, member: 635102"] 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. [B]1. The problem statement, all variables, and given/known data[/B] Task gives the following data: [LIST] [*]##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 [/LIST] 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##. [h2]Homework Equations[/h2] All relevant equations will be present in the attempt at a solution. [h2]The Attempt at a Solution[/h2] I believe that it's best for the solution to come in sequential steps instead of a long essay. [LIST=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}## [*]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}##. [*][B]From here on, I'm starting to get lost[/B]. 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}## [/LIST] 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. [/QUOTE]
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Understanding Transformer Phasor Diagrams: A Step-by-Step Guide
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