STEM Problem Detail

Problem #43

Electrical Engineering Hard

Problem Statement

Prompt

Give me the answer to question 5.24 step by step.

Solution Rendering...

**Solution:**
| Given, | $$P_{L2} = 5 \text{ MW}$$ |
|---|---|
| | $$\cos \phi = 0.72$$ |
| $$\therefore$$ | $$P_{L2} = S_{L2} \cos \phi$$ |
| | $$S_{L2} = \dfrac{5}{0.72} = 6.94 \text{ MVA}$$ |
| | $$Q_{L2} = S_{L2} \cdot \sin \phi$$ |
| | $$= 6.94 \times \sin (\cos^{-1} 0.72) = 4.816 \text{ MVAR}$$ |

| Given, | $$P_{L1} = 3 \text{ MW}$$ |
|---|---|
| | $$\cos \phi = 1$$ |
| $$\therefore$$ | $$S_{L1} = 3 \text{ MW}$$ |
| | $$Q_{L1} = 0$$ |
| Given, | $$P_1 = 5 \text{ MW}$$ |
| | $$\cos \phi = 0.8$$ |
| | $$S_1 = \dfrac{5}{0.8} = 6.25 \text{ MVA}$$ |
| | $$Q_1 = S_1 \sin \phi = 6.25 \times 0.6 = 3.75 \text{ MVAR}$$ |
| $$\therefore$$ | $$P_1 + P_2 = P_{L1} + P_{L2}$$ |
| $$\therefore$$ | $$5 + P_2 = 3 + 5$$ |
| | $$P_2 = 3 \text{ MW}$$ |
| $$\therefore$$ | $$Q_1 + Q_2 = Q_{L1} + Q_{L2}$$ |
| | $$3.75 + Q_2 = 0 + 4.81$$ |
| | $$Q_2 = 1.06 \text{ MVAR}$$ |
| | $$\tan \phi = \dfrac{Q_2}{P_2} = \dfrac{1.06}{3} = 0.35$$ |

Power factor $$\cos \phi = 0.94$$
Second machine loaded upto 3 MW at 0.94 lag p.f.

**Solution:**
\begin{align*}
I_a &= \dfrac{500 \times 10^3}{\sqrt{3} \times 1.1 \times 10^3} = 262.43 \text{ amp} \\
E_a &= \dfrac{V_t}{\sqrt{3}} + I_a(r_a + jX_s) \\
&= \dfrac{1100}{\sqrt{3}} + 262.43(0.1 + j1.5) = 769 \text{ V}
\end{align*}
Hence, option (a) is correct.

**Solution:**
| Slots per pole per phase, | $$q = \dfrac{144/3}{10} = \dfrac{24}{5} = 4.8$$ |
|---|---|
| Angular slot pitch, | $$\gamma = \dfrac{10 \times 180}{144} = \dfrac{25}{2} = 12.5$$ |
| | $$\text{Coil span} = 12 \gamma = 12 \times \dfrac{25}{2} = 150^\circ$$ |
| Chording angle, | $$\epsilon = (\text{one pole pitch}) - (\text{coil span})$$ |
| | $$= 180^\circ - 150^\circ = 30^\circ$$ |

\[ k_d = \dfrac{\sin \left( \dfrac{q \gamma}{2} \right)}{q \sin \left( \dfrac{\gamma}{2} \right)} = \dfrac{\sin \left( \dfrac{4.8 \times 12.5}{2} \right)}{4.8 \sin \left( \dfrac{12.5}{2} \right)} = 0.9568 \]
\[ k_p = \cos \dfrac{\epsilon}{2} = \cos 15^\circ = 0.965 \]
Number of turns per phase $$= N_{ph} = \dfrac{144 \times 5}{3} = 240$$
\begin{align*}
E_p &= \sqrt{2} \cdot \pi \cdot f \cdot k_d \cdot k_p \cdot N_{ph} \cdot \phi \\
E_p &= \sqrt{2} \times \pi \times 50 \times 0.9568 \times 0.966 \times 240 \times 0.116 = 5716.13 \text{ V}
\end{align*}

**Solution:**
Figure shows the open-circuit and short-circuit characteristic of synchronous machine and also the variation of synchronous impedance $$Z_s$$ with field current $$I_f$$ .
It is seen from the figure that for a field current of OC amperes,
\[ Z_s \text{ in ohms} = CD = \dfrac{\text{AC in volts}}{\text{BC in amperes}} \]
If there were no saturation,
$$Z_s$$
would be constant. Actually $$Z_s$$ is variable and it decreases with the onset of saturation in the OCC.

**Numerical:** Given that, 3-phase,
$$Y$$
-connected 10 kVA and 230 V synchronous generator, with
$$X_s = 1.2 \text{ ohms/phase}, R_a = 0.5 \text{ ohm/phase}$$
. We have to calculate:

(a) % voltage regulation at full load with 0.8 lagging power factor.
\begin{align*}
I_a \text{ per phase} &= \dfrac{10 \times 10^3}{\sqrt{3} \times 230} = 25.1 \text{ Amp} \\
\cos \phi &= 0.8; \quad \sin \phi = 0.6 \\
V_t \text{ per phase} &= \dfrac{230}{\sqrt{3}} = 132.79 \text{ volt} \\
E_f &= \sqrt{(V_t \cos \phi + I_a R_a)^2 + (V_t \sin \phi + I_a X_s)^2} \\
&= \sqrt{(132.79 \times 0.8 + 25.1 \times 0.5)^2 + (132.79 \times 0.6 + 25.1 \times 1.2)^2} \\
E_f &= 161.75 \text{ volts} \\
\text{\% voltage regulation} &= \dfrac{E_f - V_t}{V_t} \times 100\% = \dfrac{161.75 - 132.79}{132.79} \times 100\% = 21.8\%
\end{align*}

(b) Let $$\cos \phi$$ be the power factor at which V.R. becomes zero. The p.f. must be leading,
\begin{align*}
E_f^2 &= (V_t \cos \phi + I_a R_a)^2 + (V_t \sin \phi - I_a X_s)^2 \\
&= V_t^2 + 2 V_t I_a (R_a \cos \phi - X_s \sin \phi) + I_a^2 (R_a^2 + X_s^2) \\
\text{For V.R. = Zero, } \quad E_f &= V_t = 132.79 \text{ volt} \\
\therefore \quad (132.79)^2 &= (132.79)^2 + 2 \times 132.79 \times 25.1 (0.5 \cos \phi - 1.2 \sin \phi) \\
& \quad + (25.1)^2 [(0.5)^2 + (1.2)^2] \\
\text{or} \quad 0.5 \cos \phi - 1.2 \sin \phi &= 0.1597 \\
\text{or} \quad a \sin (\phi - \theta) &= 0.1597 \quad \text{where, } a = 1.3, \theta = 22.62^\circ \\
\therefore \quad 1.3 \sin (\phi - 22.62^\circ) &= 0.1597 \Rightarrow \sin (\phi - 22.62^\circ) = 0.1228 \\
\phi - 22.62^\circ = 7.053 \quad &\text{or} \quad \phi = 29.67^\circ \\
\cos \phi &= 0.8688 \text{ (lead)}
\end{align*}

**Solution:**
| As, | $$r_a = 0$$ |
|---|---|
| | $$\bar{E}_f = \bar{V}_t + j\bar{I}_d X_d + j\bar{I}_q X_q$$ |
| Given, | $$\bar{V}_t = 1\angle 0^\circ$$ |
| | $$\bar{I}_a = 1\angle 0^\circ$$ |
| | $$\bar{E}_f' = \bar{V}_t + j\bar{I}_a X_q = 1\angle 0^\circ + j0.1\angle 0^\circ \times 1 = 1 + j = \sqrt{2}\angle 45^\circ$$ |

$$\therefore \quad \delta = 45^\circ$$

$$I_d = I_a \sin \delta = 1 \times \dfrac{1}{\sqrt{2}} = 0.707$$

and $$I_q = I_a \cos \delta = 0.707$$

$$E_f = E_f' + I_d(X_d - X_q) = 1.414 + 0.707(1.4 - 1) = 1.697\angle 45^\circ \text{ p.u.}$$

\begin{align*}
P &= \dfrac{E_f V_t}{X_s} \sin \delta = \dfrac{2000 \text{ MW}}{1000 \text{ MVA}} = 2 \text{ p.u.} \\
2 &= \dfrac{1 \times 1}{0.1} \sin \delta \Rightarrow \sin \delta = 0.2 \Rightarrow \delta = 11.54^\circ \\
Q &= \dfrac{V}{X_s} (E_f \cos \delta - V) = \dfrac{1}{0.1} [1.0 \times \cos(11.54^\circ) - 1] = -0.202 \text{ p.u.} \\
Q &= -0.202 \times 1000 \text{ MVAR} = -202.147 \text{ MVAR}
\end{align*}

**Solution:**
25 kVA, 230 V, $$X_s = 1.5 \ \Omega/\text{ph}$$
Considering 230 V as a line voltage.
$$\Rightarrow \sqrt{3} V_l I_l = 25000 \Rightarrow I_l = \dfrac{25000}{\sqrt{3} \times 230} = 62.757 \text{ A} \Rightarrow 62.76 \text{ A}$$

$$\therefore V_{ph} = \dfrac{230}{\sqrt{3}} = 132.8 \text{ V}$$

(i) $$E_{ph} = V_{ph} + I_a R_a + j I_a X_s$$

$$V_{ph} = 132.8 \text{ V}, I_a \angle -\phi = 62.76 \angle -36.86^\circ \text{ A}, X_s = 1.5 \ \Omega$$

$$\therefore E_{ph} = 203.716 \angle 21.6965^\circ \text{ V}, \quad \delta = 21.7^\circ$$

(ii) Excitation increased by 20%
$$E_1 \sin \delta_1 = E_2 \sin \delta_2 \Rightarrow E_2 = 203.7 \times 1.2 = 244.44 \text{ volts}$$

$$\therefore \sin \delta_2 = \dfrac{203.7 \times \sin(21.7)}{244.44} = 0.308 \Rightarrow \delta_2 = 17.95^\circ$$

$$\therefore I_2 = \dfrac{E_2 \angle \delta_2 - V}{j X_s} = \dfrac{244.44 \angle 17.95^\circ - 132.8^\circ}{j 1.5} = 83.33 \angle -53^\circ \text{ A}$$

Reactive power per phase $$= \left[ \dfrac{E_2 V}{X_s} \cos \delta_2 - \dfrac{V^2}{X_s} \right] = 8784 \text{ VAR}$$
For 3-phase $$Q_{3-\phi} = 26.352 \text{ kVAR}$$

(iii) Steady state limit:
$$P_{\text{max}} = \dfrac{3 E_{ph} V_{ph}}{X_s} (\sin 90^\circ) = \dfrac{3 \times 203.716 \times 132.8}{1.5} = 54.1 \text{ kW}$$

$$I_{\text{max}} = \sqrt{\dfrac{V^2 + E^2 - 2VE \cos \theta}{Z_s}} = \sqrt{\dfrac{(132.8)^2 + (203.716)^2}{1.5}} = 198.55 \text{ A}$$
kVA output $$= \sqrt{3} V_l I_l = 79.1 \text{ kVA}, \quad \text{p.f.} = \dfrac{\text{kW}}{\text{kVA}} = 0.6839 \text{ leading}$$

**Solution:**
Given: $$V_l = 6.6 \text{ kV}, Z_s = (2 + j10) \ \Omega/\text{Ph}$$
(i) $$P_{in} = 500 \text{ kW, } \cos \phi = 0.7 \text{ leading} \Rightarrow \phi = -45.57^\circ$$

$$\vec{I}_a = \dfrac{500}{\sqrt{3} \times 6.6 \times 0.7} = 62.48 \angle 45.57^\circ \text{ A}$$

$$\vec{E}_f = \vec{V}_t - \vec{Z}_s \vec{I}_a = \dfrac{6600}{\sqrt{3}} \angle 0^\circ - (2 + j10)(62.48 \angle 45.57^\circ) = 4202.34 \angle -7.19^\circ \text{ V}$$

$$E_{f(\text{line})} = \sqrt{3} \times 4202.34 = 7278.65 \text{ V}$$
New input power 600 kW:
$$600 = \dfrac{6.6^2 \times 10^3}{10.198} \cos(78.70^\circ) - \dfrac{7278.65 \times 6.6}{10.198} \cos(78.70 + \delta) \Rightarrow \delta = 8.42^\circ$$

(ii) $$\vec{I}_a = \dfrac{6600 \angle 0^\circ - 7278.65 \angle -8.42^\circ}{(2 + j10)\sqrt{3}} = 69.24 \angle 40.70^\circ \text{ A}$$
New p.f. $$= \cos(40.69^\circ) = 0.7582 \text{ leading}$$

$$\vec{S}_{in} = \vec{S}_L + \vec{S}_M = 1250 \angle 36.86^\circ + 791.34 \angle -40.69^\circ = 1617.16 \angle 8.316^\circ \text{ kVA}$$
System p.f. $$= \cos(8.136^\circ) = 0.9894 \text{ lagging}$$

**Solution:**
Air gap power ($$P_g$$):
$$P_g = \dfrac{3 I_2'^2 R_2'}{s}$$
Mechanical power developed ($$P_m$$):
$$P_m = P_g - 3 I_2'^2 R_2' = (1 - s) P_g$$

$$P_m$$ is three times the electrical power absorbed in resistance
$$R_2' \left( \dfrac{1}{s} - 1 \right)$$
.
Shaft power ($$P_{sh}$$):
$$P_{sh} = P_m - \text{mechanical losses}$$

**Solution:**
$$I_2 = \dfrac{E_2}{\sqrt{(r_2/s)^2 + x_2^2}} \Rightarrow 31.1 = \dfrac{E_2}{\sqrt{(0.16/0.035)^2 + (0.27)^2}} \Rightarrow E_2 = 142.41 \text{ V}$$

$$\theta_2 = \tan^{-1} \dfrac{s x_2}{r_2} = 3.38$$

$$T_e \propto \dfrac{I_2^2}{s} \Rightarrow \dfrac{I_2^2}{0.02} = \dfrac{(31.1)^2}{0.035} \Rightarrow I_2 = 23.51 \text{ A}$$
For $$I_2 = 23.51 \text{ A}, E_2' = 188.18 \text{ V}, \theta_2' = 1.93^\circ$$

Injected emf: $$ = 188.18\angle 1.93^\circ - 142.41\angle 3.38^\circ = 45.95\angle -2.56^\circ \text{ V}$$

**Solution:**
Rotor circuit voltage in phasor form:
$$\bar{E}_2 = \bar{I}_2 \dfrac{r_2}{s} + j \bar{I}_2 x_2$$
or $$\dfrac{\bar{E}_2}{j x_2} = \bar{I}_2 + \bar{I}_2 \dfrac{r_2}{j s x_2}$$
or
$$-j \dfrac{\bar{E}_2}{x_2} = \bar{I}_2 - j \bar{I}_2 \dfrac{r_2}{s x_2}-j \dfrac{E_2}{x_2}$$
implies this current lags $$E_2$$ by $$90^\circ$$ . Tip of rotor current makes a circle with diameter $$\dfrac{E_2}{x_2}$$ .

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