EE 351
Homework 6
Due at 11:59pm on Thursday, May 21th. Submit to https://canvas.uw.edu/courses/1883403/assignments/11435558. It is fine to submit scan or a picture of handwritten answers, but we must be able to read it. Indicate your final answer clearly, show your work. Each problem is 10 points (all subproblems weighted equally).
Note: all voltages are line-to-line unless specified otherwise.
Problem 1
A Y-connected balanced three phase source has a line-to-neutral voltage on phase \(b\) of
\[V_{bn} = 100\angle 45° \text{ V}.\]Find
a) Line-to-neutral voltage \(V_{an}\)
b) The line-to-line voltage \(V_{bc}\)
Solutions.
a) Since \(V_{bn}\) lags \(V_{an}\) by \(120°\):
\[V_{bn} = V_{an}\angle{-120°}\]and
\[V_{an} = V_{bn}\angle{+120°} = 100\angle 165° \text{ V}\]b) We have
\[V_{bc} = \sqrt{3}\, V_{bn}\angle{+30°} = 173.2\angle 75° \text{ V}\]Problem 2
The current and voltage of a Y-connected load are (read the subscripts carefully)
\[V_{ab} = 200\angle -40° \text{ V}, \quad I_{c} = 10\angle 80° \text{ A}.\]Find the active power consumed by the load. Note, one way to do this is to shift everything to the same phase. For example, one can compute the line-to-neural voltage \(V_{an}\) and the line current \(I_{a}\), then compute the power consumed.
Solutions.
\[V_{an} = \frac{200}{\sqrt{3}}\angle(-40^\circ - 30^\circ) = \frac{200}{\sqrt{3}}\angle{-70°} \text{ V}\] \[V_{cn} = \frac{200}{\sqrt{3}}\angle(-70° + 120°) = \frac{200}{\sqrt{3}}\angle{50°} \text{ V}\]The angle \(\phi\) is the phase difference between \(V_{cn}\) and \(I_c\):
\[\phi = \theta_V - \theta_I = 50° - 80° = -30°\] \[P = 3\, V_{cn} I_c \cos\phi = 3 \cdot \frac{200}{\sqrt{3}} \cdot 10 \cdot \cos(-30^\circ)= 3000 \text{ W}\]Problem 3
A three-phase, \(12 \text{ kV}\) system is connected to a balanced three-phase load. The transmission line current \(I_a\) has a magnitude of \(2 \text{ A}\) and has the same angle as the line-to-line voltage \(V_{bc}\). Compute the impedance of the load for the following cases (Take \(V_{an}\) as the reference, that is, it’s angle is \(0^{\circ}\)):
a) For a Y-connected load
b) For a delta-connected load
Solutions.
With \(V_{an}\) as reference:
\[V_{bn} = \frac{12000}{\sqrt{3}}\angle -120^\circ \text{ V}=4000\sqrt{3}\angle -120^\circ \text{ V}, \quad V_{cn} = 4000\sqrt{3}\angle 120^\circ \text{ V}\] \[V_{bc} = \sqrt{3}\,V_{bn}\angle 30^\circ = 12000\angle -90^\circ \text{ V}\]Since \(I_a\) has the same angle as \(V_{bc}\):
\[I_a = 2\angle -90^\circ \text{ A}\]a) Each phase impedance sees \(V_{an}\) and carries \(I_a\):
\[Z_Y = \frac{V_{an}}{I_a} = \frac{4000\sqrt{3}\angle 0^\circ}{2\angle -90^\circ} = j3464 \text{ }\Omega\]b) The current $I_{ab}$ relates to line current $I_a$ by:
\[I_{ab} = \frac{2}{\sqrt{3}}\angle(-90^\circ + 30^\circ) = \frac{2}{\sqrt{3}}\angle -60^\circ \text{ A}\]The voltage across the delta phase is the line voltage $V_{ab}$:
\[V_{ab} = 12000\angle 30^\circ \text{ V}\]So
\[Z_\Delta = \frac{V_{ab}}{I_{ab}} = j10392 \text{ }\Omega\]Problem 4
a) A balanced Y-connected load of \(Z = 2 - j0.4 \ \Omega\) is connected across a three-phase source of \(208 \text{ V}\). Compute the active power delivered to the load.
b) Now suppose the load is delta-connected. Compute the reactive power delivered to the load.
Solutions.
a) The phase (line-to-neutral) voltage is:
\[V_\phi = \frac{V_L}{\sqrt{3}} = \frac{208}{\sqrt{3}} \text{ V}\]The phase (line) current magnitude:
\[\vert I_a \vert = \frac{V_\phi}{ \vert Z \vert} = 58.88 \text{ A}\]Total active power delivered to the load:
\[P = 3\vert I_a\vert ^2 R = 3 \times \frac{(208)^2}{3 \times 4.16} \times 2 = 20.8 \text{ kW}\]b) Now each impedance sees the full line-to-line voltage \(V = 208 \text{ V}\), so the current is:
\[\vert I \vert = \frac{V }{\vert Z\vert } = \frac{208}{2.040} = 102.0 \text{ A}\]Total reactive power delivered to the load:
\[Q = 3\vert I \vert ^2 X = -12.48 \text{ kVAR}\]Problem 5
A balanced \(480 \text{ V}\) three-phase source supplies a Y-connected load in parallel with a delta-connected load. The impedance in each phase of the Y-connected load is \(Z_Y = 20 + j7 \ \Omega\) and the impedance in each phase of the delta-connected load is \(Z_\Delta = 24 + j9 \ \Omega\). Calculate the active and reactive powers supplied by this source.
Solutions.
Convert \(\Delta\)$ to \(Y\):
\[Z_{Y,eq} = \frac{Z_\Delta}{3} = \frac{24 + j9}{3} = 8 + j3 \ \Omega\]And the total equivalent impedance is
\[Z_{eq} = \frac{Z_{Y1} \cdot Z_{Y2}}{Z_{Y1} + Z_{Y2}}=5.715 + j2.102 \ \Omega\]Taking \(V_{an}\) as reference:
\[V_{an} = \frac{480}{\sqrt{3}}\angle 0^\circ = 277.1\angle 0^\circ \text{ V}\] \[I_a = \frac{V_{an}}{Z_{eq}} = 45.51\angle -20.17^\circ \text{ A}\] \[S = 3\,V_{an}\,I_a^* = 37{,}847\angle 20.17^\circ \text{ VA}=35.5 \text{ kW} + j 13.1 \text{kVAr}\]Problem 6
Continuing with the previous question. Now suppose (each phase of) the transmission line has an impedance of \(Z_l=2+j3 \ \Omega\). The load impedances are the same. Calculate the active power lost in the line impedance.
Solutions.
From the previous solution, the equivalent Y impedance of the combined load is:
\[Z_{load} = 5.715 + j2.102 \ \Omega\]Adding the line impedance in series:
\[Z_{total} = Z_l + Z_{load} = (2 + j3) + (5.715 + j2.102) = 7.715 + j5.102 \ \Omega\]The active power lost in the impedance is
\(P_{loss}=3 \vert I_a \vert^2 R_l=3 \vert \frac{V_{an}}{Z_{total}} \vert^2 2 =5.39 \text{ kW}\).
Problem 7
A synchronous generator connected directly to an infinite bus is generating at its maximum active power. The line-to-line voltage of the infinite bus is \(600 \text{ V}\) and the synchronous reactance of the generator is \(2 \ \Omega\). Compute the reactive power supplied by the generator.
Solutions.
Maximum active power occurs when:
\[\delta = 90°\]The three-phase reactive power supplied by the generator is:
\[Q = \frac{3\left(E_f V_t \cos\delta - V_{t}^2\right)}{X_s}\]Substituting \(\delta = 90°\) gives \(\cos 90° = 0\) and
\[Q = \frac{3\left(E_f V_{t} \cdot 0 - V_{t}^2\right)}{X_s} = -\frac{3\,V_{t}^2}{X_s}=-\frac{3 \frac{600}{\sqrt{3}}^2}{2} = -\frac{360{,}000}{2} = -180 \text{ kVAR}\]Problem 8
A synchronous generator is connected to an infinite bus through a transmission line. The synchronous reactance of the generator is \(6 \, \Omega\) and the inductive reactance of the transmission line is \(4 \, \Omega\). The infinite bus voltage is \(138 \, \text{kV}\) (line-to-line). Suppose the generator is delivering \(1 \, \text{GW}\) of active power and no reactive power to the infinite bus. Compute the following:
a) The terminal voltage of the generator
b) The equivalent field voltage (i.e., \(E_f\))
Solutions.
Since the generator delivers zero reactive power to the infinite bus, the current at the bus is in phase with the bus voltage. Taking \(V_\infty\) as the reference phasor:
\[V_\infty = \frac{138{,}000}{\sqrt{3}} = 79{,}674 \ \text{V} \quad \text{(per phase)}\] \[I = \frac{P}{3 V_\infty} = \frac{10^9}{3 \times 79{,}674} = 4{,}184 \angle 0^\circ \ \text{A}\]a)
\[V_t = V_\infty + jX_L I = 79{,}674 + j\,16{,}736 \ \text{V}\]The magnitude is:
\[\vert V_t \vert = 81{,}414 \ \text{V}\]Converting to line-to-line:
\[V_{t,LL} = 141.0 \ \text{kV}\]b)
\[{E}_f = V_t + jX_s I = 79{,}674 + j\,41{,}840 \ \text{V}\]The magnitude is:
\[\vert E_f \vert = 90{,}000 \ \text{V}\]Converting to line-to-line:
\[E_{f,LL} = 155.9 \ \text{kV}\]Problem 9
A synchronous generator is connected directly to an infinite bus. The voltage of the infinite bus is \(16 \, \text{kV}\). The equivalent field voltage of the generator is \(15 \, \text{kV}\). The synchronous reactance of the machine is \(4 \, \Omega\). Compute the maximum power the generator can deliver to the infinite bus. If we want to increase this power by \(25\%\), find the new equivalent field voltage.
Solutions.
\[P_{max} = 3 \frac{E_f V}{X_s} = 3\frac{\frac{15 \times 10^3}{\sqrt{3}}\frac{16 \times 10^3}{\sqrt{3}}}{4}= 60 \ \text{MW}\]The new target power is:
\[P_{new} = 1.25 \times P_{max} = 75 \ \text{MW}\]To achieve this, $E_f$ should increase by 1.25 times as well,
\[E_{f,new} = 1.25 \cdot 15 = 18.75 \ \text{kV}\]