4
ECE 476 – Power System Analysis Fall 2013 Homework 2 Due Date: Thursday September 12, 2013 Problem 1. In PowerWorld Simulator Problem 2.32 (see Figure 1 below) a 8 MW/4 Mvar load is supplied at 13.8 kV through a feeder with an impedance of (1+ j2) Ω. The load is compensated with a capacitor whose output, Q cap , can be varied in 0.5 Mvar steps between 0 and 10.0 Mvars. What value of Q cap minimizes the real power line losses? What value of Q cap minimizes the MVA power flow into the feeder? The complex power drawn by the load is 8 + j 4 MVA. The capacitor can provide [0,10] MVar of compensation. So the resulting complex power drawn by the load is S load =8+ j (4 - Q cap ) MVA. Then, the current into the feeder is ¯ I = ¯ S load ¯ V = 8+ j (4 - Q cap ) 13.8 = 8 - j (4 - Q cap ) 13.8 kA. The complex power loss due to the line impedance is ¯ S loss = | ¯ I | 2 ¯ Z = (8 + j (4 - Q cap ))(8 - j (4 - Q cap )) 13.8 2 (1 + j 2) = 64 + (4 - Q cap ) 2 13.8 2 (1 + j 2). The real power lost is P loss = 64 + (4 - Q cap ) 2 13.8 2 MW, which is minimized when Q cap = 4 MVar. For the second part of the question, the complex power into the feeder is the sum of the complex power lost and absorbed by the combined load as follows: ¯ S f eeder = ¯ S loss + ¯ S load = | ¯ I | 2 ¯ Z + ¯ S load = 64 + (4 - Q cap ) 2 13.8 2 (1 + j 2) + 8 + j (4 - Q cap ). We would like to minimize | ¯ S f eeder | with respect to Q cap . Since this is a very complicated expression, we resort to a graphical solution via MATLAB. The plot below shows that | ¯ S f eeder | is minimized when Q cap =4.5 MVar. The code is also provided. > Qcap = 0:0.5:10; > Sload = 8+1i*(4-Qcap); > > I = conj(Sload/13.8); > > Sloss = conj(I).*I.*(1+1i*2); > > Sfeeder = Sload + Sloss; 2 3 4 5 6 7 8.35 8.4 8.45 8.5 8.55 8.6 Q cap | S f eeder | 1

Homework Solutions

Embed Size (px)

DESCRIPTION

Power system analysis homework solutions

Citation preview

Page 1: Homework Solutions

ECE 476 – Power System Analysis Fall 2013Homework 2

Due Date: Thursday September 12, 2013

Problem 1. In PowerWorld Simulator Problem 2.32 (see Figure 1 below) a 8 MW/4 Mvar load is supplied at 13.8kV through a feeder with an impedance of (1+ j2) Ω. The load is compensated with a capacitor whose output, Qcap,can be varied in 0.5 Mvar steps between 0 and 10.0 Mvars. What value of Qcap minimizes the real power line losses?What value of Qcap minimizes the MVA power flow into the feeder?

The complex power drawn by the load is 8 + j4 MVA. The capacitor can provide [0,10] MVar of compensation.So the resulting complex power drawn by the load is Sload = 8+ j(4−Qcap) MVA. Then, the current into the feederis

I =

(

Sload

V

)∗

=

(

8 + j(4−Qcap)

13.8

)∗

=

(

8− j(4 −Qcap)

13.8

)

kA.

The complex power loss due to the line impedance is

Sloss = |I|2Z =(8 + j(4−Qcap))(8 − j(4−Qcap))

13.82(1 + j2) =

64 + (4−Qcap)2

13.82(1 + j2).

The real power lost is

Ploss =64 + (4 −Qcap)

2

13.82MW,

which is minimized when Qcap = 4 MVar.For the second part of the question, the complex power into the feeder is the sum of the complex power lost and

absorbed by the combined load as follows:

Sfeeder = Sloss + Sload = |I|2Z + Sload =64 + (4−Qcap)

2

13.82(1 + j2) + 8 + j(4−Qcap).

We would like to minimize |Sfeeder| with respect to Qcap. Since this is a very complicated expression, we resort toa graphical solution via MATLAB. The plot below shows that |Sfeeder | is minimized when Qcap = 4.5 MVar. Thecode is also provided.

> Qcap = 0:0.5:10;

> Sload = 8+1i*(4-Qcap);

>

> I = conj(Sload/13.8);

>

> Sloss = conj(I).*I.*(1+1i*2);

>

> Sfeeder = Sload + Sloss;

2 3 4 5 6 7

8.35

8.4

8.45

8.5

8.55

8.6

Qcap

|Sf

eeder|

1

Page 2: Homework Solutions

Assume balanced operation.

Figure 1: Screen for Problem 1.

Problem 2. A three-phase line, which has an impedance of (2 + j4) Ω per phase, feeds two balanced three-phaseloads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + j40) Ω per phase,and the other is ∆-connected with an impedance of (60 – j45) Ω per phase. The line is energized at the sending endfrom a 60-Hz, three-phase, balanced voltage source of 120

√3 V (rms, line-to-line). Determine:

First, convert the ∆-connected load to its Y -connected equivalent:

Z2 =60− j45

3= 20− j15 Ω

1. The current, real power, and reactive power delivered by the sending-end source.

The per-phase total impedances of the line and loads is

Z = Zline + Z1||Z2 = 2 + j4 +

(

1

(30 + j40)+

1

(20− j15)

)−1

= 24∠0 Ω.

Then, the source current is

Is =V

Z=

120∠0

24= 5∠0 A.

The complex power delivered by the source is

Ss = 3V I∗ = 3(120∠0)(5∠0) = 1800∠0 VA,

with Ps = 1800 W and Qs = 0 MVar.

2. The line-to-line voltage at the load.

The phase voltage at the load is

VL = Vs − ZlineI∗s = 120∠0 − (2 + j4)(5∠0) = 110− j20 V = 111.80∠− 10.3 V.

Therefore the line-to-line voltage at the load is

VL,L−L =√3VL∠30

=√3111.80∠− 10.3 + 30 = 193.65∠19.7 V

3. The current per phase in each load.

The per-phase current through the Y -connected load is

I1 =VL

Z1

=111.80∠− 10.3

30 + j40= 2.236∠− 63.4 A.

Page 3: Homework Solutions

The per-phase current through the Y -connected equivalent of the ∆-connected load is

I2,φ =VL

Z2

=111.80∠− 10.3

20− j15= 4.472∠26.57 A.

So the per-phase current of the ∆-connected load is

I2,∆ =I2,φ√3∠30 =

4.472√3

∠26.57 + 30 = 2.582∠56.57 A.

4. The total three-phase real and reactive powers absorbed by each load and by the line.

The 3φ complex power absorbed by the Y -connected load is

S1 = 3VLI∗1 = 3(111.80∠− 10.3)(2.236∠− 63.4)∗ = 450.3 + j599.7 VA.

The 3φ complex power absorbed by the ∆-connected load is

S2 = 3VL,L−LI∗2,∆ = 3(193.65∠19.7)(2.582∠56.57)∗ = 1200− j900 VA.

The complex power absorbed by the line impedance is

Sline = 3ZlineI2 = 3(2 + j4)52 = 150 + j300 VA.

The sum of the three quantities above is 1800 + j0 VA, which matches the value obtained in Part 1.

Check that the total three-phase complex power delivered by the source equals the total three-phase

power absorbed by the line and loads.

Problem 3. Two three-phase generators supply a three-phase load through separate three-phase lines. The loadabsorbs 30 kW at 0.8 power factor lagging. The line impedance is (1.4 + j1.6) Ω per phase between generator G1and the load, and (0.8 + j1) Ω per phase between generator G2 and the load. If generator G1 supplies 15 kW at 0.8power factor lagging, with a terminal voltage of 460 V line-to-line, determine:

1. The voltage at the load terminals.

The complex power supplied by G1 is

S1,3φ =15

0.8∠ cos−1 0.8 = 18.75∠36.87 kVA.

The current supplied by G1 is

IG1=

[

S1,3φ

3VG1

]∗

=

18.75∠36.87

3(

460√3

)

∠0

= 23.53∠− 36.87 A.

So the phase voltage at the load is

VL = VG1− Zl1 IG1

=

(

460√3

)

∠0 − (1.4 + j1.6)(23.53∠− 36.87) = 216.9∠− 2.74 V,

and the line-to-line voltage at the load is

VL,L−L =√3(216.9)∠27.26 V,

2. The voltage at the terminals of generator G2.

The complex power absorbed by the load is

SL,3φ =30

0.8∠ cos−1 0.8 = 37.5∠36.87 kVA.

Page 4: Homework Solutions

The current into the load is

IL =

[

SL,3φ

3VL

]∗

=

[

37.5∠36.87

3(216.9∠− 2.74)

]∗

= 57.63∠− 39.61 A.

The current into the load is the sum of the currents supplied by the two generators, so

IG2= IL − IG1

= 57.63∠− 39.61 − 23.53∠− 36.87 = 34.15∠− 41.5 A.

And finally the voltage at the terminal of G2 is the sum of the voltage drop across the load and Zl2 :

VG2= VL + Zl2 IG2

= 216.9∠− 2.74 + (0.8 + j1)(34.15∠− 41.5) = 259.8∠− 0.638 V.

And the line-to-line voltage at the terminal of G2 is

VG2,L−L =√3(259.8)∠29.36 V.

3. The real and reactive power supplied by generator G2. The complex power supplied by G2 is

SG2,3φ = 3VG2I∗G2

= 3(259.8∠− 0.638)(34.15∠41.5) = 20.1 + j17.4 kVA.

Hence, the real power supplied is PG2,3φ = 20.1 kW and the reactive power supplied is QG2,3φ = 17.4 kVar.