ECE 476 Fall 2010: Homework 1 Solutions
Problem 1 With |V| = 100V, the instantaneous power, p(t), into a network N has a maximum value 1707 W and a minimum value of -293 W. 1. Find a possible series RL circuit equivalent to N. 2. Find S = P + jQ into N. 3. Find the maximum instantaneous power into L and compare with Q. Solution 1. The first step in finding a possible RL is defining v(t), i(t) and p(t): v(t) =
√ 2Vcos(ωt + θV )
i(t) =
√ 2Icos(ωt + θI )
p(t) = 2V I · cos(ωt + θV ) · cos(ωt + θI ) = V I[cos(θV − θI ) + cos(2ωt + θV + θI )] Letting φ = θV + θI the above equation becomes: p(t) = V I · cos(φ) + V I · cos(2ωt + φ) To find the maximum and minimum values of p(t), take the derivative with respect to time and set equal to zero:
d p(t) = −2ωV I · sin(2ωt + φ) = 0 dt
Using the properties of a sine, it is known that the power will be maximized when 2ωt + φ = 0 and minimized when 2ωt + φ = π. Thus two equations with two unknowns can be created: p(t)max = 1707W = V I · cos(φ) + V I · cos(0) = V I · cos(φ) + V I p(t)min = −293W = V I · cos(φ) + V I · cos(π) = V I · cos(φ) − V I Subtracting the equations and solving for I yields: 2V I = 2000 = V I = 1000 → I = 10 A
1
Adding the equations and solving for φ yields: φ = cos−1
! 1414 = ±45◦ 2V I
Since it is known that an inductive load will yield a current that lags the voltage, φ = −45◦ . Now the impedance of the load can be found:
Thus, R = 7.07Ω and L =
V¯ 100∠0◦ = 7.07 + j7.07Ω Z¯ = = I¯ 10∠ − 45◦ 7.07 ω H.
2. Using the current and voltage found above, the complex power can be found: S¯ = V¯ I¯∗ = (100∠0◦ ) · (10∠45◦ ) = 707 + j707V A → P = 707 W, Q = 707 var 3. The maximum instantaneous power into the inductor is equal to the reactive power absorbed by the circuit: p(t)L,max =
1707 − 293 = 707 = Q 2
Problem 2 A certain 1φ load draws 5 MW at 0.7 power factor lagging. Determine the reactive power required from a parallel capacitor to bring the power factor of the parallel combination up to 0.9. Solution Find the magnitude of the original complex power absorbed by the load: |S¯ original | =
5 = 7.143MV A 0.7
And the reactive power consumed by the load: |Qoriginal | =
q p |S¯ original |2 − P2 = 7.1432 − 52 = 5.10Mvar
Now find the magnitude of the desired complex power, noting that real power will remain unchanged: |S¯ desired | =
5 = 5.556MV A 0.9
And the reactive power consumed by the load and parallel capacitor: |Qdesired | =
q p |S¯ desired |2 − P2 = 5.5562 − 52 = 2.423Mvar
Finally, the reactive power required from the parallel capacitor: Qcap = Qoriginal − Qdesired = 2.678Mvar
2
Problem 3 A 3φ load draws 200 kW at a PF of 0.707 lagging from a 440-V line. In parallel is a 3φ capacitor bank that supplies 50 kvar. Find the resultant power factor and current (magnitude) into the parallel combination. Solution The first step is to find the magnitude of the complex power per phase: |S¯ load,1φ | =
1 200 = 94.295kV A 3 0.707
Next, the reactive power per phase can be found: s Qload,1φ
q 2 = |S¯ load,1φ | − P2 =
94.2952 −
200 2 = 66.69kvar 3
Now find the complex power of the parallel combination: ! 50 ¯ ¯ ¯ S parallel = S load + S capacitor = (66.67 + j66.69) + − j = 66.67 + j50.02kV A 3 The power factor angle and power factor can be found using the complex power found above: φ = tan
−1
Q P
= tan
−1
! 50.02 = 36.88◦ → p f = cos(φ) = 0.799 ≈ 0.8 lagging 66.67
Finally the magnitude of the current into the parallel combination can be found: |S¯ parallel | =
p
P2 − Q 2 =
p
66.672 − 50.022 = 83.348kV A
|S¯ parallel | = 109.366A |I parallel | = √ 3 · 440
Problem 4 A 1φ load draws 10 kW from a 416-V line at a power factor of 0.9 lagging. 1. Find S = P + jQ. 2. Find |I|. 3. Assume that ∠I = 0 and find the instantaneous power p(t). Solution 1. First, find the magnitude of the complex power and the reactive power: P 10 |S¯ | = = = 11.11 kV A → Q = pf 0.9
q 2 |S¯ | − P2 = 4.84 kvar
The complex power is then: S¯ = 10 + j4.48 kVA
3
2. The current can be found using the complex power: 10000 + j4480 I¯ = 416
!∗ ¯ = 26.71A = 24.04 − j11.63 → |I|
3. Assuming ∠I = 0, p(t) is: √ √ p(t) = v(t) · i(t) = ( 2 · 416)( 2 · 26.71) = 22.2kW
Problem 5 A small manufacturing plant is located 2 km down a transmission line, which has a series reactance of 0.5Ω/km. The line resistance is negligible. The line voltage at the plant is 480∠0◦ V (rms), and the plant consumes 120 kW at 0.85 power factor lagging. Determine the voltage and power factor at the sending end of the transmission line by using: 1. A complex power approach. 2. A circuit analysis approach. Solution 1. The first step is to find the power factor angle and the reactive and complex power absorbed by the load: θload = cos−1 (0.85) = 31.77◦ Qload = Pload · tan(θload ) = 74.374 kvar S¯ load = Pload + jQload = 120 + j74.374kV A = 141.18 ∠ 31.79◦ kV A The current into the load can now be found: !∗ ! ◦ ∗ ¯Iload = S load = 141, 180 ∠ 31.79 = 294.3 ∠ − 31.79◦ A Vload 480 ∠ 0◦ The losses in the line will be strictly reactive: Qline = |I|2 Xline = 294.32 · 1 = 86.512 kvar The complex power supplied by the source will be: S¯ source = S¯ load + S¯ line = (120 + j74.34) + ( j86.512) = 200.7 ∠ 53.28◦ kV A The power factor at the source can be found using S¯ source : pf = cos(53.28◦ ) = 0.6 lagging
4
Finally, the voltage at the sending end: S¯ source 200700 ∠ 53.28◦ = 682.4 ∠ 21.5◦ V(rms) = V¯ source = ∗ 294.3 ∠ 31.79◦ I¯load 2. To solve this problem using a circuit analysis approach, apply KVL to the per-phase equivalent of the circuit: V¯ source = ( j1.0) · (294.13 ∠ − 31.79◦ ) + 480 ∠ 0◦ = 682.4 ∠ 21.5◦ V(rms) p f source = cos(θV − θI ) = cos(21.5◦ − (−31.79◦ )) = 0.6 lagging
5
