Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx

(

)cos⎝

2

sin⎝ 2 (

)cos⎝

sin⎝ 2

A

E = ⎞

V+ RAI A ￾ XS I A + XS I A + RAI S

V⎞ = A

E

2 (

)⎝X￾ S I A cos

+ RAI S sin⎝ 2 ￾ RAI A cos⎝ + XS I A sin⎝

If we examine these three cases, we can see that the only difference among them is the sign of the term

sin⎝ . If ⎝ is taken as positive for lagging power factors and negative for leading power factors (in other

words, if ⎝ is the impedance angle), then all three cases can be represented by the single equation:

V⎞ = A

E

2 (

)⎝X￾ S I A cos

￾ RAI S sin⎝ 2 ￾ RAI A cos⎝ ￾ XS I A sin⎝

A MATLAB program that calculates terminal voltage as function of impedance angle is shown below:

% M-file: prob5_20.m

% M-file to calculate and plot the terminal voltage

% of a synchronous generator as a function of impedance

% angle as PF changes from 0.85 lagging to 0.85

% leading.

% Define values for this generator

EA = 509; % Internal gen voltage

I = 361; % Current (A)

R = 0.04; % R (ohms)

X = 0.695; % XS (ohms)

% Calculate impedance angle theta in degrees

theta = -31.8:0.318:31.8;

th = theta * pi/180; % In radians

% Calculate the phase voltage and terminal voltage

VP = sqrt( EA^2 - (X.*I.*cos(th) - R.*I.*sin(th)).^2 ) ...

- R.*I.*cos(th) - X.*I.*sin(th);

VT = VP .* sqrt(3);

% Plot the terminal voltage versus power factor

figure(1);

plot(theta,abs(VT),'b-','LineWidth',2.0);

title ('\bfTerminal Voltage Versus Impedance Angle');

xlabel ('\bfImpedance Angle (deg)');

ylabel ('\bfTerminal Voltage (V)');

%axis([0 260 300 540]);

grid on;

hold off;

The resulting plot of terminal voltage versus impedance angle (with field and armature currents held

constant) is shown