Question

A balanced three-phase network supplies a balanced load through a three-phase wye-wye connected transformer as shown in the following figure. Transformer Supply Network B8 Load Y-Y The wye-wye connected transformer has the following parameter: S 25 MVA Xir 15% 240/132 kV The three-phase load parameters are: Pload 20 MW pfload 0.8 The three-phase supply network line voltage and short circuit current are: Vnet 225 kV Inet_short= 15 kA. A. Calculate primary and secondary current rating. B. Calculate trans former actual reactance at the primary of the transformer. C. Calculate trans former actual reactance at the secondary of the transformer.

Answer 1

Ans)

A)

Given $S=25\: \: MVA,\: \: V_{p}/V_{s}=240/132\: \: kV,\: \: X_{tr}=15\: \: o/o$

the primary rated current is (as it is star connected line current is same as phase current)

$I_{p}=\frac{S}{\sqrt{3}V_{p}}=\frac{25\: \: M}{\sqrt{3}*240k}=60.141\: \: A$

$\mathbf{I_{p}=60.141\: \: A}$

secondary current is

$I_{s}=\frac{S}{\sqrt{3}V_{s}}=\frac{25\: \: M}{\sqrt{3}*132k}=109.35\: \: A$

$\mathbf{I_{s}=109.35\: \: A}$

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B)

Reactance referred to primary is

$X_{tr}=X_{tr,pu}*\frac{V_{p}^{2}}{S}=\frac{15}{100}*\frac{(240k)^{2}}{25M}=345.6\: \: \Omega$

$\mathbf{X_{tr}=345.6\: \: \Omega }$

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C)Reactance referred to secondary is

$X_{tr}=X_{tr,pu}*\frac{V_{s}^{2}}{S}=\frac{15}{100}*\frac{(132k)^{2}}{25M}=104.54\: \: \Omega$

$\mathbf{X_{tr}=104.54\: \: \Omega}$