EN 5001 Electric Machine
Test 1 Transformer
2017-04-29: 45 Minutes.
1. The reference polarity and direction of voltage, electromotive force, current and magnetic
flux for a transformer are specified as shown in Figure 1. Please try to write:
(1) The equations of electromotive force 𝐸1̇ and 𝐸2̇ for the primary and secondary
winding of the transformer;
(2) The transformer basic equations (6 equations).
(8 Remarks)
Figure 1.
Answer:
(1) Since the reference direction of electromotive force and magnetic flux satisfies right hand
srew law. Therefore the electromotive force for the primary and secondary winding can b
expressed as :
̇
̇
𝐸1 = −𝑗4.44𝑓𝑁
1 Φ𝑚
̇
𝐸2 = −𝑗4.44𝑓𝑁2 Φ𝑚
(2)
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EN 5001 Electric Machine
2. A Yd three-phase transformer has the ratings of SN=750 kV·A，U1N /U2N=10000/400 V. Noload test is done at the low voltage side. And the test results are 𝑈2 = 𝑈2𝑁 = 400𝑉, 𝐼20 =
65𝐴 , 𝑃0 = 3.7𝑘𝑊. Short-circuit test is done at high voltage side. The test results
are 𝑈1𝑘 = 450𝑉, 𝐼1𝑘 = 35𝐴 , 𝑃𝑘 = 7.5𝑘𝑊 . It is supposed that 𝑍1 = 𝑍2 ′ , try to calculate
(1) The transformer parameters 𝑅1, 𝑅1𝜎, 𝑅2 ′ , 𝑅2𝜎 ′ , 𝑅𝑚 , 𝑋𝑚 ;
(2) And then calculate the above-mentioned parameters per-unit quantities.
(12 Remarks)
Answer:
(1)
The ratio is calculated as:
𝑈1𝑁𝑝ℎ𝑎𝑠𝑒 𝑈2𝑁 /√3 10000/√3
=
=
= 14.43
𝑈2𝑁𝑝ℎ𝑎𝑠𝑒
𝑈2𝑁
400
when the no load test is done at low voltage side, then the exciting impedance can be
calculated as:
𝑈20𝑝ℎ𝑎𝑠𝑒
𝑈2𝑁
400
|𝑍𝑚 ′ | = |𝑍𝑚0 ′ | =
=
=
= 10.66Ω
𝐼20𝑝ℎ𝑎𝑠𝑒
𝐼20/√3 65/√3
k=
𝑅𝑚 ′ = 𝑅0 ′ =
𝑃0𝑝ℎ𝑎𝑠𝑒
𝐼20𝑝ℎ𝑎𝑠𝑒 2
=
𝑃0/3
𝐼20 2
=
3.7 × 103
= 0.8757Ω
652
𝑋𝑚 ′ = √|𝑍𝑚 |2 − 𝑅𝑚 ′2 = √10.662 − 0.87572 = 10.62Ω
The exciting impedance to the primary side is
|𝑍𝑚 | = 𝑘 2 |𝑍𝑚 ′ | = 14.432 × 10.66 = 2220Ω
|𝑅𝑚 | = 𝑘 2 |𝑅𝑚 ′ | = 14.432 × 0.8757 = 182.3Ω
|𝑋𝑚 | = 𝑘 2 |𝑋𝑚 ′ | = 14.432 × 10.62 = 2211Ω
when the short circuit test is done at high voltage side, then the exciting impedance
can be calculated as:
𝑈2𝑘𝑝ℎ𝑎𝑠𝑒 𝑈1𝑘/√3 450/√3
|𝑍𝐾 | =
=
=
= 7.423Ω
𝐼2𝑘𝑝ℎ𝑎𝑠𝑒
𝐼1𝑘
35
𝑃𝑘/3 7.5 × 103 /3
𝑃𝑘𝑝ℎ𝑎𝑠𝑒
|𝑅𝐾 | =
=
=
= 2.041Ω
352
𝐼2𝑘𝑝ℎ𝑎𝑠𝑒 2 𝐼1𝑘 2
𝑋𝑘 = √|𝑍𝑘 |2 − 𝑅𝑘 2 = √7.4322 − 2.0412 = 7.137Ω
Since 𝑍1 = 𝑍2 ′
1
𝑅1 = 𝑅2 ′ = 𝑅𝑘 = 1.021Ω
2
1
𝑋1𝜎 = 𝑋2𝜎 ′ = 𝑋𝑘 = 3.569Ω
2
(3) For the high voltage winding
𝑈1𝑁 10000
𝑈1𝑁Φ =
=
= 5774𝑉
1.732
√3
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EN 5001 Electric Machine
𝐼1𝑁Φ = 𝐼1𝑁 =
𝑆𝑁
=
750 × 103
= 43.3𝐴
√3𝑈1𝑁 √3 × 10000
The high voltage winding base is calculated as
𝑈1𝑁Φ 5774
𝑍1𝑁 =
=
= 133.3 Ω
𝐼1𝑁Φ
43.3
Therefore the per unit values of the parameters of the transformer are calculated as
𝑅𝑚 ∗ =
𝑅𝑚 182.3
=
= 1.368
𝑍1𝑁 133.3
𝑋𝑚 ∗ =
𝑋𝑚
2211
=
= 16.59
𝑍1𝑁 133.3
𝑅1 ∗ = 𝑅2 ′∗ =
𝑅1
1.021
=
= 0.007659
𝑍1𝑁 133.3
𝑋1𝜎 ∗ = 𝑋2𝜎 ′∗ =
𝑋1𝜎 3.569
=
= 0.02677
𝑍1𝑁 133.3
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