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Solutions Manual

Hadi Saadat

Professor of Electrical Engineering

Milwaukee School of Engineering

Milwaukee, Wisconsin

McGraw-Hill, Inc.

CHAPTER 1 PROBLEMS

1.1 The demand estimation is the starting point for planning the future electric

power supply. The consistency of demand growth over the years has led to numer-

ous attempts to fit mathematical curves to this trend. One of the simplest curves

is

P = P

e

a(t−t

where a is the average per unit growth rate, P is the demand in year t, and P

is

the given demand at year t

Assume the peak power demand in the United States in 1984 is 480 GW with

an average growth rate of 3.4 percent. Using MATLAB , plot the predicated peak

demand in GW from 1984 to 1999. Estimate the peak power demand for the year

We use the following commands to plot the demand growth

t0 = 84; P0 = 480;

a =.034;

t =(84:1:99)’;

P =P0exp(a(t-t0));

disp(’Predicted Peak Demand - GW’)

disp([t, P])

plot(t, P), grid

xlabel(’Year’), ylabel(’Peak power demand GW’)

P99 =P0exp(a(99 - t0))

The result is

1

Predicted Peak Demand - GW

P99 =

The plot of the predicated demand is shown n Figure 1.

Peak

Power

Demand

GW

Year

FIGURE 1

Peak Power Demand for Problem 1.

1.2 In a certain country, the energy consumption is expected to double in 10 years.

11 12 8];

P = data(:,3); % Column array of load

Dt = data(:, 2) - data(:,1); % Column array of demand interval

W = P’*Dt; % Total energy, area under the curve

Pavg = W/sum(Dt) % Average load

Peak = max(P) % Peak load

LF = Pavg/Peak*100 % Percent load factor

barcycle(data) % Plots the load cycle

xlabel(’time, month’), ylabel(’P, MW’), grid

result in

Pavg =

Peak =

LF =

P

MW

time, month

FIGURE 2

Monthly load cycle for Problem 1.

CHAPTER 2 PROBLEMS

2.1. Modify the program in Example 2.1 such that the following quantities can be

entered by the user:

The peak amplitude V

m

, and the phase angle θ

v

of the sinusoidal supply v(t) =

V

m

cos(ωt + θ

v

). The impedance magnitude Z, and its phase angle γ of the load.

The program should produce plots for i(t), v(t), p(t), p

r

(t) and p

x

(t), similar to

Example 2.1. Run the program for V

m

= 100 V, θ

v

= 0 and the following loads:

An inductive load, Z = 1. 25

A capacitive load, Z = 2. 0

A resistive load, Z = 2. 5

(a) From p

r

(t) and p

x

(t) plots, estimate the real and reactive power for each load.

Draw a conclusion regarding the sign of reactive power for inductive and capaci-

tive loads.

(b) Using phasor values of current and voltage, calculate the real and reactive power

for each load and compare with the results obtained from the curves.

(c) If the above loads are all connected across the same power supply, determine

the total real and reactive power taken from the supply.

The following statements are used to plot the instantaneous voltage, current, and

the instantaneous terms given by(2-6) and (2-8).

Vm = input(’Enter voltage peak amplitude Vm = ’);

thetav =input(’Enter voltage phase angle in degree thetav = ’);

Vm = 100; thetav = 0; % Voltage amplitude and phase angle

Z = input(’Enter magnitude of the load impedance Z = ’);

gama = input(’Enter load phase angle in degree gama = ’);

thetai = thetav - gama; % Current phase angle in degree

5

ωt, degrees

v(t) = V

m

cos ωt, i(t) = I

m

cos(ωt − 60)

ωt, degrees

p(t) = v(t)i(t)

ωt - degrees

p

r

(t), Eq. 2.

ωt, degrees

p

x

(t), Eq. 2.

FIGURE 3

Instantaneous current, voltage, power, Eqs. 2.6 and 2.8.

I =

A

Using (2.7) and (2.9), we have

P = (70.71)(56.57) cos(60) = 2000 W

Q = (70.71)(56.57) sin(60) = 3464 Var

Running the above program for the capacitive load Z = 2. 0

Ω will result in

(a) Estimate from the plots

P =

Q =

Similarly, for Z = 2. 5

Ω, we get

P =

Q =

(c) With the above three loads connected in parallel across the supply, the total real

and reactive powers are

P = 2000 + 2165 + 2000 = 6165 W

Q = 3464 − 1250 + 0 = 2214 Var

2.2. A single-phase load is supplied with a sinusoidal voltage

v(t) = 200 cos(377t)

The resulting instantaneous power is

p(t) = 800 + 1000 cos(754t − 36. 87

(a) Find the complex power supplied to the load.

(b) Find the instantaneous current i(t) and the rms value of the current supplied to

the load.

(c) Find the load impedance.

(d) Use MATLAB to plot v(t), p(t), and i(t) = p(t)/v(t) over a range of 0 to 16. 67

ms in steps of 0.1 ms. From the current plot, estimate the peak amplitude, phase

angle and the angular frequency of the current, and verify the results obtained in

part (b). Note in MATLAB the command for array or element-by-element division

is ./.

p(t) = 800 + 1000 cos(754t − 36. 87

= 800 + 1000 cos 36. 87

cos 754t + sin 36. 87

sin 754t

= 800 + 800 cos 754t + 600 sin 754t

= 800[1 + cos 2(377)t] + 600 sin 2(377)t

p(t) is in the same form as (2.5), thus P = 600 W, and Q = 600, Var, or

S = 800 + j600 = 1000

VA

(b) Using S =

V

m

I

m

, we have

I

m

Vm = 200;

t=0:.0001:0.01667; % wt from 0 to 2*pi

v=Vmcos(377t); % Instantaneous voltage

p = 800 + 1000cos(754t - 36.87*pi/180);% Instantaneous power

i=p./v; % Instantaneous current

wt=180/pi377t; % converting radian to degree

xline = zeros(1, length(wt)); % generates a zero vector

subplot(221), plot(wt, v, wt, xline), grid

xlabel(’wt, degrees’), title(’v(t)’)

subplot(222), plot(wt, p, wt, xline), grid

xlabel(’wt, degrees’), title(’p(t)’)

subplot(223), plot(wt, i, wt, xline), grid

xlabel(’wt, degrees’), title(’i(t)’), subplot(111)

The result is shown in Figure 4. The inspection of current plot shows that the peak

amplitude of the current is 10 A, lagging voltage by 36. 87

, with an angular fre-

quency of 377 Rad/sec.

2.3. An inductive load consisting of R and X in series feeding from a 2400-V rms

supply absorbs 288 kW at a lagging power factor of 0.8. Determine R and X.

V

I

R X

FIGURE 5

An inductive load, with R and X in series.

θ = cos

The complex power is

S =

kVA

The current given from S = V I

, is

I =

360 × 10

− 36. 87 A

Therefore, the series impedance is

Z = R + jX =

V

I

= 12.8 + j 9 .6 Ω

Therefore, R = 12.8 Ω and X = 9.6 Ω.

2.4. An inductive load consisting of R and X in parallel feeding from a 2400-V

rms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R and X.

V

I

R X

FIGURE 6

An inductive load, with R and X in parallel.

The complex power is

S =

kVA

= 288 kW + j 216 kvar

R =

|V |

P

288 × 10

X =

|V |

Q

216 × 10

2.5. Two loads connected in parallel are supplied from a single-phase 240-V rms

source. The two loads draw a total real power of 400 kW at a power factor of 0.

lagging. One of the loads draws 120 kW at a power factor of 0.96 leading. Find the

complex power of the other load.

θ = cos

The total complex load is

S =

kVA

= 400 kW + j 300 kvar

The 120 kW load complex power is

S =

kVA

= 120 kW − j 35 kvar

2.7. Two impedances, Z

= 0.8 + j 5 .6 Ω and Z

= 8 − j16 Ω, and a single-

phase motor are connected in parallel across a 200-V rms, 60-Hz supply as shown

in Figure 8. The motor draws 5 kVA at 0.8 power factor lagging.

¡¿

¬ø

I

I

I

I

V

j 5. 6

−j 16

S

= 5 kVA

at 0.8 PF lag

M

a

a

FIGURE 8

Circuit for Problem 2.7.

(a) Find the complex powers S

, S

for the two impedances, and S

for the motor.

(b) Determine the total power taken from the supply, the supply current, and the

overall power factor.

(c) A capacitor is connected in parallel with the loads. Find the kvar and the ca-

pacitance in μF to improve the overall power factor to unity. What is the new line

current?

(a) The load complex power are

S

|V |

Z

0. 8 − j 5. 6

= 1000 + j 7000 VA

S

|V |

Z

8 + j 16

= 1000 − j 2000 VA

S

= 4000 + j 3000 VA

Therefore, the total complex power is

S

t

= 6 + j8 = 10

kVA

(b) From S = V I

, the current is

I =

− 53. 13 A

and the power factor is cos 53. 13

= 0. 6 lagging.

(c) For overall unity power factor, Q

C

= 8000 Var, and the capacitive impedance

is

Z

C

|V |

S

C

j 8000

= −j5 Ω

and the capacitance is

C =

= 530. 5 μF

The new current is

I =

0 A

2.8. Two single-phase ideal voltage sources are connected by a line of impedance of

0 .7 + j 2 .4 Ω as shown in Figure 9. V

V and V

V. Find

the complex power for each machine and determine whether they are delivering or

receiving real and reactive power. Also, find the real and the reactive power loss in

the line.

π∏

∫∑

π∏

∫∑

V

I

0 .7 + j 2 .4 Ω

V

FIGURE 9

Circuit for Problem 2.8.

I

0 .7 + j 2. 4

= 42 + j56 = 70

A

S

= V

I

= 28000 − j 21000 VA

S

= V

I

= −24570 + j 32760 VA

E1 Q-1 Q-2 Q-L

Examination of Figure 10 shows that the flow of reactive power along the intercon-

• 1 THE POWER SYSTEM: AN OVERVIEW
• 2 BASIC PRINCIPLES
  • THE PER-UNIT SYSTEM
• 4 TRANSMISSION LINE PARAMETERS
• 5 LINE MODEL AND PERFORMANCE
• 6 POWER FLOW ANALYSIS
• 7 OPTIMAL DISPATCH OF GENERATION
• 8 SYNCHRONOUS MACHINE TRANSIENT ANALYSIS
• 9 BALANCED FAULT
• 10 SYMMETRICAL COMPONENTS AND UNBALANCED FAULT
• 11 STABILITY
• 12 POWER SYSTEM CONTROL
  • Source # 1 Voltage Mag. =
  • Source # 1 Phase Angle = -
  • Source # 2 Voltage Mag. =
  • Source # 2 Phase Angle =
  • Line Resistance =
  • Line Reactance =
  • 90.0000 -105.5173 129.1066 23.
  • 91.0000 -93.9497 114.9856 21.
  • 92.0000 -82.1021 100.8646 18.
  • 93.0000 -69.9745 86.7435 16.
  • 94.0000 -57.5669 72.6225 15.
  • 95.0000 -44.8794 58.5015 13.
  • 96.0000 -31.9118 44.3804 12.
  • 97.0000 -18.6642 30.2594 11.
  • 98.0000 -5.1366 16.1383 11.
  • 99.0000 8.6710 2.0173 10.
• 100.0000 22.7586 -12.1037 10.
• 101.0000 37.1262 -26.2248 10.
• 102.0000 51.7737 -40.3458 11.
• 103.0000 66.7013 -54.4668 12.
• 104.0000 81.9089 -68.5879 13.
• 105.0000 97.3965 -82.7089 14.
• 106.0000 113.1641 -96.8299 16.
• 107.0000 129.2117 -110.9510 18.
• 108.0000 145.5393 -125.0720 20.
• 109.0000 162.1468 -139.1931 22.
• 110.0000 179.0344 -153.3141 25.
• 111.0000 196.2020 -167.4351 28.
• 112.0000 213.6496 -181.5562 32.
• 113.0000 231.3772 -195.6772 35.
• 114.0000 249.3848 -209.7982 39.
• 115.0000 267.6724 -223.9193 43.
• 116.0000 286.2399 -238.0403 48.
• 117.0000 305.0875 -252.1614 52.
• 118.0000 324.2151 -266.2824 57.
• 119.0000 343.6227 -280.4034 63.
• 120.0000 363.3103 -294.5245 68.

Q

var

Source # 1 Voltage magnitude

Q

Q

Q

L

FIGURE 10

Reactive power versus voltage magnitude.

2.10. A balanced three-phase source with the following instantaneous phase volt-

ages

v

an

= 2500 cos(ωt)

v

bn

= 2500 cos(ωt − 120

v

cn

= 2500 cos(ωt − 240

supplies a balanced Y-connected load of impedance Z = 250

Ω per phase.

(a) Using MATLAB , plot the instantaneous powers p

a

, p

b

, p

c

and their sum versus

ωt over a range of 0 : 0.05 : 2π on the same graph. Comment on the nature of the

instantaneous power in each phase and the total three-phase real power.

(b) Use (2.44) to verify the total power obtained in part (a).

We use the following commands

wt=0:.02:2*pi;

pa=25000cos(wt).cos(wt-36.87*pi/180);

pb=25000cos(wt-120pi/180).cos(wt-120pi/180-36.87*pi/180);

pc=25000cos(wt-240pi/180).cos(wt-240pi/180-36.87*pi/180);

p = pa+pb+pc;