Electrical Machines: Solved Assignments and MATLAB Code

This document provides solved assignments and MATLAB code for Electrical Machines. It covers topics such as voltage regulation, maximum power angle, and complex power of network. Suitable for students of Electrical Engineering.

Running head: ELECTRICAL MACHINES
ELECTRICAL MACHINES
Name of the Student
Name of the University
Author Note

1ELECTRICAL MACHINES
8.9:
a) Given that,
Generator poles P = 4, frequency f = 50 Hz.
Generator p.u. reactance Xpu = 0.85.
Generator rating = 40 MVA and generator rated voltage = 26 kV.
Given, that the field current of the generator is adjusted from 0.75 to 1.5 times of the rated
voltage.
Hence, the induced voltage is varied from 26*0.75 = 19.5 volts to 26*1.5 = 39 volts.
Now, the generator power delivered is
P = MVA rating * power factor = 40*(10^6)*cosθ
Now, X = Xbase*Xpu.
Xbase = ((rated voltage in kV)^2)/MVA = ((26)^2)/40 = 16.9 Ω.
Hence, X = 16.9*0.85 = 14.365 Ω.
Now, line reactance Xl = 0.5*8 = 4 Ω.
Hence, total reactance = 18.365 Ω.
Line resistance R = 0.07*8 = 0.56 Ω.
Hence, Z = sqrt(R^2 + X^2) = sqrt(0.56^2 + 18.365^2) = 18.3735 Ω.
Now, 40*(10^6)*cosθ = ( induced voltage ) 2
Z = (line voltage)^2/18.3735
Voltage regulation = (induced voltage – network voltage)/network voltage

2ELECTRICAL MACHINES
MATLAB code:
rated_volt = 26e3;
ind_volt = (0.75:0.1:1.5).*rated_volt; % induced voltage
xpu = 0.85;
mva = 40e6; xbase = rated_volt^2/mva;
power = ((ind_volt).^2)./(18.3735);
pf = power./mva;
figure(1)
subplot(2,1,1)
plot(pf(1:3),ind_volt(1:3),'r-')
xlabel('power factor')
ylabel('induced voltage in volts')
net_volt = 24e3; % network voltage
volt_regu = (ind_volt - net_volt)./net_volt;
leading_pf = linspace(0.5,1,length(volt_regu));
subplot(2,1,2)
plot(leading_pf,volt_regu)
xlabel('leading power factor')

3ELECTRICAL MACHINES
ylabel('voltage regulation')
0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85
power factor
2
2.2
2.4
2.6
induced voltage in volts 104
0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
leading power factor
-0.2
0
0.2
0.4
0.6
voltage regulation
Hence, 10% voltage regulation corresponds to 0.8 power factor (exactly 0.814).
b) Now, the maximum power angle is calculated by the following method.
The maximum steady state power P = EV/X
Here, E = voltage behind the direct axis synchronous reactance or excitation voltage
V = voltage of the network = 24 kV.
X = Xd + Xl
Where, Xd = direct-axis synchronous generator reactance and Xl = line reactance.
Now, Xd = Xbase*Xpu.
Xbase = ((rated voltage in kV)^2)/MVA = ((26)^2)/40 = 16.9 Ω.
Hence, Xd = 16.9*0.85 = 14.365 Ω.

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Now, Xl = 0.5*8 = 4 Ω.
Hence, X = 14.365 + 4 = 18.365 Ω.
MATLAB code:
rated_volt = 26e3;
ind_volt = (0.75:0.1:1.5).*rated_volt;
vnet = 24e3; X = 18.365;
max_power = (24e3.*ind_volt)/18.365;
plot(ind_volt,max_power)
xlabel('induced voltage in volts')
ylabel('max power in watts')
Plot:
1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8
induced voltage in volts 104
2.5
3
3.5
4
4.5
5
max power in watts
107

5ELECTRICAL MACHINES
Now, if the excited voltage goes beyond 19 kV then the rating of generator will exceed more
than 40 MVA. Hence, the excitation voltage that can deliver maximum power is 19 KV.
8.12:
a)
The one line diagram of the problem is shown below.
Generator rating = 380 MVA.
Xgpu = 1.2 pu.
Transformer rating = 480 MVA. Primary side/secondary side voltage = 22 kV/340 kV.
% Xtrfr= 15%
Hence, the transformers are step up transformers.
Line length = 45 miles.
Line impedance = 0.07 + j0.5 Ω/mi.
Hence, the total line impedance = (0.07 + j0.5)*45 = 3.15 + j*22.5 Ω.
Now, the base voltage is considered to be 22 kV.
Hence, Xgbase = (kV)^2 /MVA = 22^2/(380) = 1.2737 Ω.
Hence, Xg = Xgbase* Xgpu = 1.2737*1.2 = 1.5284 Ω.

6ELECTRICAL MACHINES
Now, the reactance of the transformer is connected to the secondary side. Hence, the voltage
across reactance is 340 kV.
Now, % X = ((kVA)*X)/(10*(kV)^2) = ((480*10^3)*X)/(10*(340^2))
 15 = ((480*10^3)*X)/(10*(340^2))
 X = ((10*(340^2))*15)/ (480*10^3)
 X= 36.13 Ω.
Hence, transformer reactance Xtrfr = 36.13 Ω.
b) Now, given that pf = 0.88 lagging.
Hence, power P = MVA*pf = 380*0.88 = 334.4 MW.
Now, Generator current Ig = 334.4 MW / 22kV
Now, excitation voltage E = V + jIg*X = 22 kV + Ig*j1.5284 = 41.6 kV ∠ 63.9°
c) Absolute value of excitation voltage = 41.6 kV.
Now, transmitted power to the network is given by
P = ((E*V)sinδ)/X.
V = network voltage = 21 kV.
X = 1.5284 Ω.
Hence, P = (41.6*21*10^6)/ 24.0284 = 571.58sinδ MW
Now, in MATLAB the transmitted power vs power angle is plotted.
MATLAB code:
delta = 0:1:90;

7ELECTRICAL MACHINES
P = 571.58*sind(delta);
plot(delta,P)
xlabel('power angle in degrees')
ylabel('transmitted power in MW')
Plot:
0 10 20 30 40 50 60 70 80 90
power angle in degrees
0
100
200
300
400
500
600
transmitted power in MW
Hence, from the above plot it can be seen that when power transmitted is 350 MW then the
power angle is approximately 50 degrees.

8ELECTRICAL MACHINES
8.5:
Generator rating = 250 MVA.
Generator line voltage = 24 kV.
% reactance = 125%.
Now, % reactance = ((kVA)*X)/(10*(kV)^2) = ((250*10^3)*X)/(10*24^2)
 X = (125*(10*24^2))/(250*10^3) = 2.88 Ω.
Network voltage V = 27 kV.
The line to neutral voltage or phase voltage = 24/sqrt(3) = 13.8564 kV.
Excitation voltage or induced voltage E = 2* line to neutral voltage = 2*13.8564 = 27.7128
kV.
Phase angle δ between network and induced voltage = 55 degrees.
Now, the complex power of network is given by,
Snetwork = Pe + jQe
Where, Pe= ( E∗V ) sinδ
X (1) is the real or active power
And, Qe= ( E∗V ) cosδ
X – V 2
X (2) is the reactive power
Now, putting the values calculated above in (1) and (2) we get,
Pe=373 MVA and Q e = -1.49 MVA.

9ELECTRICAL MACHINES