Question

Three single-phase two-winding transformers, each rated $\mathbf{25}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{34}\mathbf{.}\mathbf{5}\mathbf{/}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{kV}$, are connected to form a three-phase $\mathbf{∆}\mathbf{-}\mathbf{∆}$bank. Balanced positive-sequence voltages are applied to the high-voltage terminals, and a balanced, resistive Y load connected to the low-voltage terminals absorbs $\mathbf{75}\mathbf{}\mathbf{MW}\mathbf{}\mathbf{at}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{kV}$. If one of the single-phase transformers is removed (resulting in an open- $\mathbf{∆}$connection) and the balanced load is simultaneously reduced to $\mathbf{43}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{MW}$( 57.7% of the original value), determine (a) the load voltages ${\mathit{V}}_{\mathbf{a}\mathbf{n}}\mathbf{,}{\mathit{V}}_{\mathbf{b}\mathbf{n}}\mathbf{,}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{V}}_{\mathbf{c}\mathbf{n}}$ (b) load currents ${\mathit{I}}_{\mathbf{a}}\mathbf{,}\mathbf{}\mathbf{}{\mathit{I}}_{\mathbf{b}}\mathbf{,}\mathbf{}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{}{\mathit{I}}_{\mathbf{c}}$ and (c) the supplied by each of the remaining two transformers. Are balanced voltages still applied to the load? Is the open-$\mathbf{∆}$transformer overloaded?

Short answer

(a) The phase voltages are $7.967\angle 0°\mathrm{kV},7.968\angle -120°\mathrm{kV}\mathrm{and}7.968\angle -240°\mathrm{kV}$

(b) The line currents are $1.812\angle 0°\mathrm{kA},1.812\angle -120°\mathrm{kA}\mathrm{and}1.812\angle -240°\mathrm{kA}$

(c) The power supplied by two transformers are$25\angle 30°\mathrm{MVA},\mathrm{and}25\angle -30°\mathrm{MVA}$

The balance voltages are still applied to the load. The transformer will not be overloaded.

Step-by-step solution

Determine the formulas of phase voltages, line currents and power

Write the formula of phase voltage at star connected load.

${\mathbf{V}}_{\mathbf{an}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{ab}}}{\sqrt{\mathbf{3}}}\mathbf{\angle }\mathbf{0}\mathbf{°}$ ……. (1)

Write the formulas of phase voltages across phases ‘b’ and ‘c’.

${\mathbf{V}}_{\mathbf{bn}}\mathbf{=}{\mathbf{V}}_{\mathbf{an}}\mathbf{\angle }\mathbf{-}\mathbf{120}\mathbf{°}$ ……. (2)

${\mathbf{V}}_{\mathbf{c}\mathbf{}\mathbf{n}}\mathbf{=}{\mathbf{V}}_{\mathbf{an}}\mathbf{\angle }\mathbf{-}\mathbf{240}\mathbf{°}$ ……. (3)

Write the formula of phase current.

${\mathbf{I}}_{\mathbf{a}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{3}\mathbf{-}\mathbf{ph}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{ab}}}\mathbf{\angle }\mathbf{0}\mathbf{°}$……. (4)

Write the formula of phase currents through phases ‘b’ and ‘c’.

${\mathbf{I}}_{\mathbf{b}}\mathbf{=}{\mathbf{I}}_{\mathbf{a}}\mathbf{\angle }\mathbf{-}\mathbf{120}\mathbf{°}$ ……. (5)

${\mathbf{I}}_{\mathbf{c}}\mathbf{=}{\mathbf{I}}_{\mathbf{a}}\mathbf{\angle }\mathbf{-}\mathbf{240}\mathbf{°}$ ……. (6)

Write the formula of power delivered by two transformers.

${\mathbf{S}}_{\mathbf{bc}}\mathbf{=}\left(V_{\mathrm{bc}}\angle 30°\right)\left(I_b^{*}\right)$ ……. (7)

${\mathbf{S}}_{\mathbf{ca}}\mathbf{=}\left(V_{\mathrm{ca}}\angle 30°\right)\left(I_a^{*}\right)$ ……. (8)

Determine the phase voltages

Substitute$13.8/\mathrm{kV}$for${\mathrm{V}}_{\mathrm{ab}}$ in equation (1).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{an}}=\frac{13.8\mathrm{kV}}{\sqrt{3}}\angle 0° \\ =7.967\angle 0°\mathrm{kV} \end{gathered}$$

Even after removing one phase from a three-phase transformer, the phase voltages and currents will be balanced.

Therefore, using equations (2) and (3), calculate phase voltages in ‘b’ and ‘c’.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{bn}}={\mathrm{V}}_{\mathrm{an}}\angle -120° \\ =7.967\angle -120°\mathrm{kV} \end{gathered}$$

Calculate for the ‘c’ phase.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{cn}}={\mathrm{V}}_{\mathrm{an}}\angle -240° \\ =7.967\angle -240°\mathrm{kV} \end{gathered}$$

Determine the line currents

(b)

Substitute $13.8\mathrm{kV}$for role="math" localid="1655188210253" ${\mathrm{V}}_{\mathrm{ab}},43.3\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{3-\mathrm{ph}}$, in equation (4).

$$\begin{gathered} I_a=\frac{43.3\mathrm{MVA}}{\sqrt{3}\times 13.8\mathrm{kV}}\angle 0° \\ =1.812\angle 0°\mathrm{kA} \end{gathered}$$

Using equations (5) and (6), calculate line currents in ‘b’ and ‘c’.

$$\begin{gathered} I_b={\mathrm{I}}_{\mathrm{a}}\angle -120° \\ =1.812\angle -120°\mathrm{kA} \end{gathered}$$

Calculate for the ‘c’ phase.

$$\begin{gathered} I_c={\mathrm{I}}_{\mathrm{a}}\angle -120° \\ =1.812\angle -240°\mathrm{kA} \end{gathered}$$

Determine the MVA power supplied by two transformers.

(c)

After removing one phase, calculate the power supplied by one transformer.

Substitute $13.8\angle -120°\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{bc}}\mathrm{and}1.812\angle -120°\mathrm{kA}\mathrm{for}{\mathrm{I}}_{\mathrm{a}}$in equation (7).

$$\begin{gathered} {\mathrm{S}}_{\mathrm{bc}}=\left(13.8\angle -120°+30°\mathrm{kV}\right)\left(1.812\angle -120°\mathrm{kA}\right) \\ =25\angle 30°\mathrm{MVA} \end{gathered}$$

Substitute $13.8\angle 120°\mathrm{kV}\mathrm{for}{\mathrm{V}}_{ca}\mathrm{and}1.812\angle 0°\mathrm{kA}\mathrm{for}{\mathrm{I}}_{\mathrm{a}}$in equation (7).

$$\begin{gathered} S_{ca}=\left(13.8\angle 120°+30°\mathrm{kV}\right)\left(1.812\angle 0°\mathrm{kA}\right) \\ =25\angle -30°\mathrm{MVA} \end{gathered}$$

When one phase is removed, the transformer with the remaining two phases will not be overloaded.