Electrical Machines: Solved Assignments and MATLAB Code

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')