Question

Consider a three-phase generator rated $\mathbf{300}\mathbf{}\mathbf{MW}\mathbf{}\mathbf{,}\mathbf{}\mathbf{23}\mathbf{kV}$supplying a system load of $\mathbf{240}\mathbf{}\mathbf{MW}$and 0.9 power factor lagging at $\mathbf{230}\mathbf{}\mathbf{kV}$through a $\mathbf{330}\mathbf{}\mathbf{MW}\mathbf{,}\mathbf{}\mathbf{23}\mathbf{∆}\mathbf{/}\mathbf{230}\mathbf{}\mathbf{Y}\mathbf{}\mathbf{kV}$ step-up transformer with a leakage reactanceof 0.11 per unit. (a) Neglecting the exciting current and choosing base values at the load of 100MW and 230kV , find the phasor currents role="math" localid="1655190109478" ${\mathit{I}}_{\mathbf{A}}\mathbf{,}\mathbf{}{\mathit{I}}_{\mathbf{B}}\mathbf{}\mathbf{and}\mathbf{}{\mathit{I}}_{\mathbf{C}}$supplied to the load in per unit. (b) By choosing the load terminalvoltage ${\mathit{V}}_{\mathbf{A}}$as reference, specify the proper base for the generator circuitand determine the generator voltageas well as the phasor currentsrole="math" localid="1655190101018" ${\mathit{I}}_{\mathbf{A}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{I}}_{\mathbf{B}}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}{\mathit{I}}_{\mathbf{C}}$from the generator. (Note:Take into account the phase shift of the transformer.) (c) Find the generator terminal voltage in kV and the real power supplied by the generator in MW . (d) By omitting the transformer phase shift altogether, check to see whether you get the same magnitude ofgenerator terminal voltage and real power delivered by the generator.

Short answer

(a) The phase currents are$I_{A\left(pu\right)}=2.4\angle -25.84°\mathrm{pu},I_{B\left(pu\right)}=2.4\angle -145.84°\mathrm{pu}\mathrm{and}{I}_{C\left(pu\right)}=2.4\angle 94.16°\mathrm{pu}$

(b) The generator voltage is $V_A=1\angle 30°\mathrm{pu}$ and the phase currents are$I_{a\left(pu\right)}=2.4\angle -55.84°\mathrm{pu}\mathrm{and}{I}_{b\left(pu\right)}=2.4\angle -175.84°\mathrm{pu}\mathrm{and}{I}_{c\left(pu\right)}=2.4\angle 64.16°\mathrm{pu}$

(c) The generator terminal voltage is $23.86\angle -26.02°\mathrm{kV}$and the real power supplied by the generator is $216\mathrm{M}\mathrm{W}$.

(d) The terminal voltage and power will remain same, if phase shift is omitted.

Step-by-step solution

Determine the formulas ofactual current, base current, per-unit current, power angle, per-unit reactance, generator terminal voltage and real power supplied by generator

Write the formula of per-unit current.

${\mathbf{I}}_{\mathbf{A}\left(\mathrm{pu}\right)}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{actual}}}{{\mathbf{I}}_{\mathbf{actual}}}$……. (1)

Write the formulas of base current.

role="math" localid="1655190935371" ${\mathbf{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}\mathbf{\times }{\mathbf{V}}_{\mathbf{base}}}$……. (2)

Write the formulas of actual current.

role="math" localid="1655191088960" ${\mathbf{I}}_{\mathbf{actual}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{actual}}}{\sqrt{\mathbf{3}}\mathbf{\times }{\mathbf{V}}_{\mathbf{actual}}}$ ……. (3)

Write the formula of power angle.

$\mathbf{\theta }\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(p.f\right)$……. (4)

Write the formula of phase current in phase ‘B’.

${\mathbf{I}}_{\mathbf{B}\left(\mathrm{pu}\right)}\mathbf{=}{\mathbf{I}}_{\mathbf{A}\left(\mathrm{pu}\right)}\mathbf{\angle }\mathbf{-}\mathbf{120}\mathbf{°}$ …… (5)

Write the formula of phase current in phase ‘C’.

${\mathbf{I}}_{\mathbf{C}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}\mathbf{=}{\mathbf{I}}_{\mathbf{A}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}\mathbf{\angle }\mathbf{120}\mathbf{°}$ …… (6)

Write the formula of new reactance of transformer in per-unit.

${\mathbf{X}}_{\mathbf{pu}\left(\mathrm{new}\right)}\mathbf{=}{\mathbf{X}}_{\mathbf{pu}\left(\mathrm{new}\right)}\mathbf{\times }\frac{{\mathbf{S}}_{\mathbf{base}\left(\mathrm{new}\right)}}{{\mathbf{S}}_{\mathbf{base}\left(\mathrm{old}\right)}}$ …… (7)

Write the formula of generator terminal voltage.

${\mathbf{V}}_{\mathbf{t}}\mathbf{=}{\mathbf{V}}_{\mathbf{A}}\mathbf{\angle }\mathbf{-}\mathbf{30}\mathbf{°}\mathbf{+}{\mathbf{X}}_{\mathbf{pu}}{\mathbf{I}}_{\mathbf{A}}$ …… (8)

Write the formula of real power supplied by generator.

${\mathbf{P}}_{\mathbf{p}\mathbf{u}}\mathbf{=}{\mathbf{V}}_{\mathbf{A}}{\mathbf{I}}_{\mathbf{a}}^{\mathbf{*}}$ …… (9)

Determine the per-unit phasor currents.

The single-line diagram and the per-phase equivalent circuit, with all parameters in per unit, are given below:

(a)

Substitute 100 M Wfor $S_{base}$and 230 kVfor ${\mathrm{V}}_{\mathrm{base}}$in equation (2).

$$\begin{gathered} I_{base}=\frac{100\mathrm{MW}}{\sqrt{3}\times 230\mathrm{kV}} \\ =250.02\mathrm{A} \end{gathered}$$

Substitute 240 M W for $S_{actual}$ and 230 kVfor $V_{actual}$in equation (3).

$$\begin{gathered} I_{actual}=\frac{240\mathrm{MW}}{\sqrt{3}\times 230\mathrm{kV}} \\ =602.45\mathrm{A} \end{gathered}$$

Substitute 602.45 Afor ${\mathrm{I}}_{\mathrm{actual}}$ and 250.02 A for ${\mathrm{I}}_{base}$in equation (1).

$$\begin{gathered} I_{A\left(pu\right)}=\frac{602.45\mathrm{A}}{250.02\mathrm{A}} \\ =2.4\mathrm{pu} \end{gathered}$$

Substitute 0.9for $p.f$in (4).

$$\begin{gathered} \theta ={\cos }^{-1}0.9 \\ =25.84° \end{gathered}$$

Therefore, for lagging load, the current is $I_{A\left(pu\right)}=2.4\angle -25.84°\mathrm{pu}$.

The currents in ‘B’ and ‘C’ phases using equations (5) and (6).

$$\begin{gathered} I_{B\left(pu\right)}=2.4\angle -120°-25.84°\mathrm{pu} \\ =2.4\angle -145.84°\mathrm{pu} \end{gathered}$$

Current in phase ‘C’ is,

$$\begin{gathered} I_{C\left(pu\right)}=2.4\angle 120°-25.84°\mathrm{pu} \\ =2.4\angle 94.16°\mathrm{pu} \end{gathered}$$

Determine the generator voltage and currents.

(b)

Let, the reference voltage at load in per-unit is $V_A=1\angle 0°\mathrm{pu}$.

The generator voltage leads the load voltage by $30°$.

Therefore, the generator voltage will be $V_A=1\angle 30°\mathrm{pu}$.

The generator current lags the load current by $30°$.

Therefore, the generator current in phase ‘a’ is,

$$\begin{gathered} I_{a\left(pu\right)}=2.4\angle -25.84°-30°\mathrm{pu} \\ =2.4\angle -55.84°\mathrm{pu} \end{gathered}$$

The generator current in phase ‘b’ is,

$$\begin{gathered} I_{b\left(pu\right)}=2.4\angle -145.84°-30°\mathrm{pu} \\ =2.4\angle -175.84°\mathrm{pu} \end{gathered}$$

Current in phase ‘c’ is,

$$\begin{gathered} I_{c\left(pu\right)}=2.4\angle 94.16°-30°\mathrm{pu} \\ =2.4\angle 64.16°\mathrm{pu} \end{gathered}$$

Determine the generator terminal voltage in kV and MW power supplied by generator.

(c)

Determine the new reactance of transformer.

Substitute $0.11\mathrm{pu}\mathrm{for}{\mathrm{X}}_{pu\left(old\right)},100\mathrm{M}\mathrm{W}\mathrm{for}{\mathrm{S}}_{base\left(new\right)}\mathrm{and}330\mathrm{M}\mathrm{W}\mathrm{for}{\mathrm{S}}_{base\left(old\right)}$in equation (7).

$$\begin{gathered} {\mathrm{X}}_{pu\left(old\right)}=0.11\mathrm{pu}\times \frac{100}{330} \\ =\frac{1}{30}\mathrm{pu} \end{gathered}$$

Substitute $1\angle -30°\mathrm{pu}\mathrm{for}{\mathrm{V}}_A,\frac{1}{30}\mathrm{pu}\mathrm{for}{\mathrm{X}}_{pu}\mathrm{and}2.4\angle -25.84°\mathrm{for}{\mathrm{I}}_A$in equation (8).

$$\begin{gathered} V_{t\left(pu\right)}=1\angle -30°+\left(\frac{1}{30}\right)\left(2.4\angle -55.84°\right) \\ =1.0374\angle -26.02°\mathrm{pu} \end{gathered}$$

The actual generator voltage will be,

$$\begin{gathered} V_t=\left(1.0374\angle -26.02°\right)\left(23\mathrm{kV}\right) \\ =23.86\angle -26.02°\mathrm{kV} \end{gathered}$$

Substitute $1.0374\angle -26.02°\mathrm{pu}\mathrm{for}{\mathrm{V}}_t\mathrm{and}2.4\angle -55.84°\mathrm{for}{\mathrm{I}}_A$ in equation (9).

$$\begin{gathered} P_{\left(pu\right)}=\mathrm{Re}\left[\left(1.0374\angle -26.02°\mathrm{pu}\right){\left(2.4\angle -55.84°\right)}^{*}\right] \\ =2.16\mathrm{pu} \end{gathered}$$

The real power in MW is,

$$\begin{gathered} P=P_{base}{P}_{pu} \\ =100\mathrm{MW}\times 2.16\mathrm{pu} \\ =216\mathrm{MW} \end{gathered}$$

Determine the generator terminal voltage in kV and MW power supplied by generator.

(d)

Determine the generator terminal voltage when phase shift is omitted.

Substitute$1\angle 0°\mathrm{pu}\mathrm{for}{\mathrm{V}}_A,\frac{1}{30}\mathrm{pu}\mathrm{for}{\mathrm{X}}_{pu}\mathrm{and}2.4\angle -25.84°\mathrm{for}{\mathrm{I}}_A$ in equation (8).

$$\begin{gathered} V_{t\left(pu\right)}=1\angle 0°+\left(\frac{1}{30}\right)\left(2.4\angle -55.84°\right) \\ =1.0374\angle -3.62°\mathrm{pu} \end{gathered}$$

Therefore, the terminal voltage will be same, and hence power will also same.