Question

Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers,each specified by ${\mathbf{X}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{}\mathbf{\Omega }$(on the low-voltage side), negligible resistance and magnetizing current, and turns ratio $\mathbf{\eta }\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}\mathbf{10}$. The transformer bank is delivering $\mathbf{100}\mathbf{}\mathbf{MW}$at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV.

(a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus,role="math" localid="1655206659086" ${\mathbf{V}}_{\mathit{a}\mathbf{\text{'}}\mathit{n}\mathbf{\text{'}}}$as the reference. Account for the phase shift, and assume positive-sequence operation.

(b) Find the phase shift between the primary and secondary voltages.

Short answer

(a) Therefore, the primary current is $5435\angle -66.86°\mathrm{A}$, the primary voltage is $7253.3\angle -13.93°\mathrm{kV}$and the complex power is $118.27\angle 52.94°\mathrm{MVA}$

(b) The phase shift is $-13.93°$.

Step-by-step solution

Given data.

The reactance of three-phase transformer is$0.24\mathrm{\Omega }$.

The turns ratio is 10.

The transformer delivering 100 MW at 0.8 power factor and 230kV .

Determine the formulas of phase voltage, complex power delivered by transformer, current at secondary of transformer, turns ratio, current at primary of transformer,

Write the formula of phase voltage.

${\mathbf{V}}_{\mathbf{ph}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{L}\mathbf{-}\mathbf{L}}}{\sqrt{\mathbf{3}}}$……. (1)

Write the formula ofsingle-phase complex power.

${\mathbf{S}}_{\mathbf{1}\mathbf{-}\mathbf{ph}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{out}}}{\mathbf{3}\mathbf{\times }\mathbf{p}\mathbf{.}\mathbf{f}}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{p}\mathbf{.}\mathbf{f}$……. (2)

Write the formula of secondary current.

role="math" localid="1655207281512" ${\mathbf{I}}_{\mathbf{s}}\mathbf{=}{\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{1}\mathbf{-}\mathbf{ph}}}{{\mathbf{V}}_{\mathbf{ph}}}\mathbf{\right)}}^{\mathbf{*}}$ ……. (3)

Write the formula of turns ratio for star-delta.

${\mathbf{\eta }}_{\mathbf{1}}\mathbf{=}\sqrt{\mathbf{3}}\mathbf{\eta }$……. (4)

Write the formula of primary current.

${\mathbf{I}}_{\mathbf{P}}\mathbf{=}{\mathbf{\eta }}_{\mathbf{1}}\mathbf{\angle }\mathbf{-}\mathbf{30}\mathbf{°}{\mathbf{I}}_{\mathbf{s}}$ ……. (5)

Write the formula of primary voltage.

${\mathit{V}}_{\mathbf{p}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{ph}}}{{\mathbf{\eta }}_{\mathbf{1}}}\mathbf{+}{\mathbf{X}}_{\mathbf{T}}{\mathbf{I}}_{\mathbf{p}}$ ……. (6)

Write the formula of complex power.

${\mathbf{S}}_{\mathbf{p}}\mathbf{=}\mathbf{3}{\mathbf{V}}_{\mathbf{p}}{\left(I_p\right)}^{\mathbf{*}}$ ……. (7)

Determine the primary current magnitude, primary voltage magnitudeand complex power supplied by the generator.

(a)

Determine the phase voltage.

Substitute 230 kV for $V_{L-L}$in equation (1).

$$\begin{gathered} V_{ph}=\frac{230\mathrm{kV}}{\sqrt{3}} \\ =132.8\mathrm{kV} \end{gathered}$$

Determine the single-phase complex power delivered by transformer.

Substitute 100 MW for $P_{out}$and 0.8 for $p.f$in equation (2).

$$\begin{gathered} S_{1-ph}=\frac{100\mathrm{MW}}{3\times 0.8}{\cos }^{-1}0.8 \\ =41.67\angle 36.86°\mathrm{MVA} \end{gathered}$$

Substitute 10 for $\eta$in equation (4).

${\eta }_1=10\sqrt{3}$

Substitute $41.67\angle 36.86°\mathrm{MVA}\mathrm{for}{S}_{1-ph}\mathrm{and}132.8\mathrm{kV}\mathrm{for}{V}_{ph}$in equation (3).

$$\begin{gathered} I_s={\left(\frac{41.67\angle 36.86°\mathrm{MVA}}{132.8\mathrm{kV}}\right)}^{*} \\ =313.8\angle -36.86°\mathrm{kA} \end{gathered}$$

Substitute localid="1655260368325" $10\sqrt{3}\mathrm{for}{\eta }_1,313.8\angle -36.86°\mathrm{kA}\mathrm{for}{I}_s$and in equation (5).

localid="1655261366864" $$\begin{gathered} I_p=10\sqrt{3}\times 313.8\angle -30°\mathrm{kA} \\ =5435\angle -66.86°\mathrm{A} \end{gathered}$$

Substitute $10\sqrt{3}\mathrm{for}{\eta }_1,132.8kV\mathrm{for}{V}_{ph},j0.08\Omega \mathrm{for}{X}_T$andlocalid="1655260577308" role="math" $5435\angle -66.86°A\mathrm{for}{I}_p$in equation (6).

$$\begin{gathered} V_p=\frac{132.8}{10\sqrt{3}}+\mathrm{j}0.08\left(5435\angle -66.86°\right) \\ =7253.3\angle -13.93°\mathrm{kV} \end{gathered}$$

Substitute$7253.3\angle -13.93°kV\mathrm{for}{V}_p\mathrm{and}5435\angle -66.86°A$for $I_p$in equation (7).

$$\begin{gathered} S_p=3\left(7253.3\angle -13.93°\right)\left(5435\angle -66.86°\right) \\ =118.27\angle 52.94°\mathrm{MVA} \end{gathered}$$

Therefore, the primary current is $5435\angle -66.86°\mathrm{A}$, the primary voltage is $7253.3\angle -13.93°\mathrm{kV}$and the complex power is $118.27\angle 52.94°\mathrm{MVA}$.

Determine the phase shift between primary and secondary voltages.

(b)

The primary voltage is $7253.3\angle -13.93°\mathrm{kV}$and the secondary voltage is $230\angle 0°\mathrm{kV}$.

Therefore, the phase shift is $-13.93°$.