Question

A three-phase 138kVΔ13.8kVY distribution substationtransformer rated 400MVAOA50MVAFA65MVAFOA has an 9%impedance. (a) Determine the rated current on the primary distribution side ofthe transformer at its OA, FA, and FOA ratings. (b) Determine the per-unittransformer impedance using a system base of 100MVA and 13.8kV on theprimary distribution side of the transformer. (c) Calculate the short-circuitcurrent on the primary distribution side of the transformer for a three-phasebolted fault on the primary distribution side. Assume that the prefault voltageis 13.8kV .

Short answer

Answer

(a) The rated current of primary distribution transformer at its OA, FA and FOA are

167.34Aphase , 209.18Aphase and 271.94Aphase respectively.

(b) The transformer impedance at the new base values is 0.225pu

(c) The three-phase short circuit current is 1.859kAphase .

Step-by-step solution

Write the following data from the question:

The MVA rating of OA,$S_{OA}=40MVA$

The MVA rating of FA,$S_{FA}=50MVA$

The MVA rating of FOA, $S_{FOA}=65MVA$

The impedance of transformer, $Z=9\%$.

The high voltage,$V_H=138kV$

The low voltage,$V_L=13.8kV$

Determine the equations to calculate the rated current, new per unit impedance of transformer and short circuit current on the primary distribution side of transformer.

The equation to calculate the rated current for OA is given by,

${\mathit{I}}_{\mathbf{O}\mathbf{A}\mathbf{,}\mathbf{L}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{O}\mathbf{A}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{H}}}$ …… (1)

The equation to calculate the rated current for FA is given by,

${\mathit{I}}_{\mathbf{FA}\mathbf{,}\mathbf{L}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{FA}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{H}}}$ …… (2)

The equation to calculate the rated current for FOA is given by,

${\mathit{I}}_{\mathbf{FOA}\mathbf{,}\mathbf{L}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{FOA}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{H}}}$ …… (3)

The equation to calculate the per unit value of impedance on new bases is given as follows.

${\mathit{Z}}_{\mathbf{p}\mathbf{u}\mathbf{,}\mathbf{n}\mathbf{e}\mathbf{w}}\mathbf{=}{\mathit{Z}}_{\mathbf{p}\mathbf{u}\mathbf{,}\mathbf{o}\mathbf{l}\mathbf{d}}{\left(\frac{V_{base,old}}{V_{base,new}}\right)}^{\mathbf{2}}\left(\frac{S_{base,new}}{S_{base,old}}\right)$ …… (4)

The equation to calculate the three-phase short circuit current in per unit is given as follows,

${\left(I_{SC,3\varphi }\right)}_{\mathbf{p}\mathbf{u}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{Z}}_{\mathbf{p}\mathbf{u}}}$ …… (5)

The equation to calculate the three-phase short circuit current in kA is given as follows,

${\left(I_{SC,3\varphi }\right)}_{\mathbf{k}\mathbf{A}}\mathbf{=}{\left(I_{SC,3\varphi }\right)}_{\mathbf{p}\mathbf{u}}\mathbf{}{\mathit{I}}_{\mathbf{O}\mathbf{A}\mathbf{,}\mathbf{L}}$ …… (6)

Step 3: Determine the equations to calculate the rated current.

(a)

Calculate the rated current for OA.

Substitute 40 MVA for $S_{OA},138kV$for $V_H$into equation (1).

$$\begin{gathered} I_{OA,L}=\frac{40\times {10}^6}{\sqrt{3}\times 138\times {10}^3} \\ {I}_{OA,L}=\frac{40\times {10}^6}{239.023} \\ {I}_{OA,L}=167.34A/phase \end{gathered}$$

Calculate the rated current for FA.

Substitute $50MVA$for $S_{FA},138kV$for $V_H$into equation (2).

$$\begin{gathered} I_{FA,L}=\frac{50\times {10}^6}{\sqrt{3}\times 138\times {10}^3} \\ {I}_{FA,L}=\frac{50\times {10}^3}{239.023} \\ {I}_{FA,L}=209.18A/phase \end{gathered}$$

Calculate the rated current for FOA.

Substitute $65MVA$for $S_{FOA},138kV$for $V_H$into equation (3).

$$\begin{gathered} I_{FOA,L}=\frac{65\times {10}^6}{\sqrt{3}\times 138\times {10}^3} \\ {I}_{FOA,L}=\frac{65\times {10}^3}{239.023} \\ {I}_{FOA,L}=271.94A/phase \end{gathered}$$

Hence the rated current of primary distribution transformer at its OA, FA and FOA are $167.34A/phase,209.18A/phase$and $271.94A/phase$respectively.

calculate the impedance of transformer at the new base values.

(b)

The new MVA base,$S_{base,new}=100MVA$

The new voltage base,$V_{base,old}=13.8kV$

Calculate the value of the impedance.

Substitute $0.09pu$for $Z_{pu,old},13.8kV$for $V_{base,new}$and $V_{base,old}$,$40MVA$for $S_{base,old}$and $100MVA$for $S_{base,new}$into equation (4).

$$\begin{gathered} Z_{pu,new}=0.09{\left(\frac{13.8}{13.8}\right)}^2\left(\frac{100}{40}\right) \\ {Z}_{pu,new}=0.09\times \frac{5}{2} \\ {Z}_{pu,new}=0.225pu \end{gathered}$$

Hence the transformer impedance at the new base values is $0.225pu$

Calculate the short circuit current on the primary distribution side of transformer.

(c)

Let’s assume the per unit pre fault voltage,$V_F=1pu$

Calculate the three-phase short circuit current in per unit.

Substitute $1pu$for $V_F$, and $0.09pu$for $Z_{pu}$into equation (5).

$$\begin{gathered} {\left(I_{SC,3\varphi }\right)}_{pu}=\frac{1}{0.09} \\ {\left(I_{SC,3\varphi }\right)}_{pu}=11.11pu \end{gathered}$$

Calculate the three-phase short circuit current in kA.

Substitute $167.34$for $I_{OA,L}$, and $11.11pu$for ${\left(I_{SC,3\varphi }\right)}_{pu}$into equation (6).

$$\begin{gathered} {\left(I_{SC,3\varphi }\right)}_{kA}=11.11\times 167.34 \\ {\left(I_{SC,3\varphi }\right)}_{kA}=1859.14A/phase \\ {\left(I_{SC,3\varphi }\right)}_{kA}=1.859kA/phase \end{gathered}$$

Hence, the three-phase short circuit current is $1.859kA/phase$.