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Problem 1: Generate and plot a vee curve diagram for the turbogenerator you analyzed in Prob lem Set 3. This is a plot of
armature current magnitude |Ia| vs. field current If for real values of output power of 0, 200, 400, 600, 800 and 1000
MW.
Problem 2: This problem concerns a salient pole synchronous generator with the following pa rameters:
VA Rating 150 MVA
Voltage Rating 13.8 kV (line-line, RMS)
7967 V (line-neutral, RMS)
D-Axis Synchronous Reactance 2.5 Ω
Q-Axis Synchronous Inductance 1.5 Ω
Stator Winding Phase Resistance .009 Ω
Field Winding Resistance 1.0 Ω
Magnetic Core Loss at Rated Voltage 1.0 MW
Field Current for Rated Voltage, Open Circuited (Ifnl) 600 A
Number of pole pairs (p) 7
Stator Connection Wye
1. To start, note that this machine will have a stability limit for operation at low field excitation
(corresponding to high absorbed reactive power). For a round rotor machine this limit is
reached at a torque angle of 90◦, but this machine has saliency so you must determine
the value of angle for which stability is reached. Compute and plot the angle and
corresponding value of field current at the stability threshold for this machine, against real
power. Hint: The stability limit is reached when the derivative of torque with respect to
angle is zero. Since torque is proportional to real power, you can use the derivative of power
with angle. For this part of the problem, ignore resistances and core loss.
2. Find the value of reactive power at the underexcited stability limit and plot Q(P ).
3. Find required field current for operation at rated power at unity power factor, and plot a
torque/angle curve for operation at rated terminal voltage and that field current.
4. Now, considering the loss elements: armature resistance loss, core loss and field winding loss,
calculate and plot machine efficiency over the range of 50MW < P < 150MW at unity
power factor.
Problems

Problem 3: A particular salient-pole generator has inductances of:
Ld = 6mH
Lq = 4mH
L0 = 1mH
Find the elements of the phase inductance matrix:
⎡ ⎤
La Lab Lac ⎥
⎣
⎢
L ab b
L L bc ⎦
Lac Lbc Lc
as functions of rotor position φ.You may ﬁnd it convenient to use the periodicity of the machine
to reduce the number of calculations you need to do.

Problem 1: No secrets here. The script to do the vee curve is appended. The script, first, finds the field
winding capability, armature winding capability and then finds the stability limit (using a slightly
different method from Problem Set 3). The torque angle is used as a parameter. Note the limits for
field winding and armature currents are checked in a straightforward fashion and noted on the plot.
The plot is shown in Figure 1. Note the problem statement is flawed: there is no vee curve for the 1,000
MW curve, which is right on the armature capacity and therefore is a single point.
x 10
4 Round Rotor, 1000 MW Vee Curve
2.5
2
1.5
1
0.5
0
Field Current
Figure 1: Vee Curve for Example Generator
Problem 2: Salient Pole Machine
1. To start, note that this machine will have a stability limit for operation at low field
excitation (corresponding to high absorbed reactive power). For a round rotor
machine this limit is reached at a torque angle of 90◦, but this machine has
saliency so you must determine the value of angle for which stability is reached.
Compute and plot the angle and corresponding value of field current at the stability
threshold for this machine, against real power. Hint: The stability limit is reached
when the derivative of torque with respect to angle is zero. Since torque is
proportional to real power, you can use the derivative of power with angle. For this
part of the problem, ignore resistances and core loss.
Armature
Current
(RMS)
0 1000 2000 3000 4000 5000 6000 7000
Solutions

Ia
E 1
=
=
3V
V + jXqIa
δ = angle(E1)
Id = Ia sin δ
E a f = abs(E1) + (Xd − Xq)Id
Real power output for a generator is:
3 2
V Ea f
P = 3 sin δ + V
1 1
−
Xq X d
sin 2δ
X d 2
The derivative of power with angle is then simply:
dP V Ea f 1
= 3 cos δ + 3V 2 − 1
dδ X d Xq X d
cos 2δ
At the stability limit, dP = 0, and this may be solved for internal voltage:
dδ
cos 2δ
af
E = −V
X d
−1
Xq cos δ
Using this shorthand:
P0 = 3V 2 1 1
−
Xq X d
we have this nonlinear expression to solve:
P 1
cos δ − sin 2δ + cos 2δsin δ = 0
P0 2
Now, this looks awful but in fact is quite easily solved by most mathematical assistants. MATLAB, for
example, has a routine called ’fzero’ which makes quick work of it. Once δ is found, Ea f may be
determined and the operating point is easily determined.
2. Reactive power at the underexcited stability limit is plotted in Figure 2 Note also the values of field current
and torque angle in Figure 3.
3. Calculation of required field current and calculation of the torque-angle curve is carried out in the third
script appended.
For this, see Figure 4
4. Calculation of efficiency is fairly straightforward. Take the notion that efficiency is:
P
η = P + Pa + Pf + Pc
The calculation is carried out in the normal fashion, assuming that, for unity power factor:
P
Ea f
f
I = I fn l
V
a
= R f I 2
Pa = 3RaI2
Pf
f

Generator
Reactive
Power
x 10
7 Generator Underexcited Reactive Capability (Problem 4.2)
0
−5
−10
Stability thermal
0 5 10
−15
15
x 10
7
Generator Real Power
Figure 2: Reactive and Thermal Reactive Power Limit

Stability Limit for Underexcited Operation
1500
1000
500
0
−500
5
80
60
40
20
0
I
,
A
f
N−
m
Angle,
degrees
5
0 10 15
x 107
0 10
UnderexcitedPower, W
15
x 107
Figure 3: Underexcited Stability Limit
x 10
6 Torque vs. Angle for Example Machine
3.5
3
2.5
2
1.5
1
0.5
0
0 0.5 1 1.5 2 2.5 3 3.5
Radians
Figure 4: Torque-Angle Curve for Example Machine (Generator Coordinates)

Efficiency for Example Machine
0.964
0.966
0.968
0.97
0.972
0.974
0.976
0.978
Efficienc
y
0.5 0.6 0.7 0.8 0.9 1.1 1.2 1.3
1
Output Power
1.4 1.5
x 10
8
Figure 5: Example Machine Eﬃciency

Carrying out this last operation, one ﬁnds a few elements of the answer:
2
�
L 0
�
�
�
Problem 3: This is more scutwork than one might like. Note that calculation of the phase induc tance matrix is:
L = T −1L T
ph dq
So the second part of that is:
cos φ cos(φ − 2π ) cos(φ + 2π )
d 3 3
2
L T = −sin φ sin(φ − 2π ) —sin(φ + 2π )
dq 3 3
3 1
1
2
1
2 2
L 0 0
0 Lq 0
0 0 L0
d
L cos φ d
L cos(φ − 2π ) d
L cos(φ + 2π )
3 3
2 −L q sin(φ − 2π
= −L q sin φ 3 3
−Lq sin(φ + 2π )
3 L L L 0
2
0
2
0
2
T hen
ph
L =
La
L ab b
Lac Lbc
Lab Lac
L L
Lc
bc
cos φ
= cos(φ −2π
3
—sin φ 1 d d 3
sin(φ −2π 1 −L q sin φ −L q sin(φ − 2π
3 3
d 3
L cos φ L cos(φ − 2π ) L cos(φ + 2π )
3
−Lq sin(φ + 2π )
cos(φ + 2π —sin(φ + 2π 1 L0
2
L0
2
L0
2
3 3
2 2
3
L a = L d cos φ + L q sin φ +
2
2
�
2π 2π
Lab = Ld cos φcos(φ − ) + Lq sin φsin(φ − ) +
3 3 3
L 0
�
2
Doing the trig, one ﬁnds:
a d q d q 0
L =
1
((L + L ) + (L −L ) cos(2φ) + L )
3
1
�
2π
Lb =
Lc =
(Ld + Lq ) + (Ld −Lq ) cos 2(φ − ) + L0
3 � 3
2π
(Ld + Lq ) + (Ld −Lq ) cos 2(φ + ) + L0
1
3 3
1 1 2π 1
Lab
= − (Ld + Lq ) + (Ld −Lq ) cos(2φ − ) + L0
6 6 3 3
1
L ac d q d q
6 6
= −
1
(L + L ) +
1
(L −L ) cos(2φ +
2π
) + L0
3 3
L bc d q d q
= −
1
(L + L ) +
1
(L −L ) cos2φ +
1
L 0
6 6 3

% Round-Rotor Machine Vee Curve Generator
%Assumes generator operation
%Machine Data F i r s t
om
=2*pi *60; VA =
1e9; Xd = 1.353;
V= sqrt(2/3)*26000; pfr =
. 8 ;
M= 22. 5e- 3;
%t his i s a 60 Hz machine
%armature rating
% synchronous reactance (ohms)
%voltage: peak phase voltage
%rating point
% field-phase mutual reactance
Pv = [200e6 400e6 600e6 800e6];
%find armature current capability
I a r = VA/(1.5 * V); %remember we are working in peak
%find maximum f i e l d capability:
psir = acos(pfr); %power factor angle
%Ia i s the complex current at rating point I a = Iar*cos(psir) +
j * Iar* s in( ps ir) ;
Eafr = V - j*Xd*Ia;
Eafm = abs(Eafr); %t his i s max magnitude of f i e l d voltage Ifm = Eafm/(om*M);
fprintf(’Rated I a = %g MaxI f = %gn’,Iar, Ifm) figure(1)
c l f hold on
for i = 1:length(Pv) P = P v ( i ) ;
Q= sqrt(VA^2-P^2); Ia_min =
(P+j*Q)/(1.5*V);
Eaf_min = V+ j*Xd*Ia_min; i f
angle(Eaf_min) > pi/2,
dm
i n = pi /2;
else
% under-excited: check for s t a b i l i t y
dmin = angle(Eaf_min);
end
Ia_max = (P-j*Q)/(1.5*V); Eaf_max =
V+j*Xd*Ia_max; i f abs(Eaf_max) <
Eafm,
dmax = angle(Eaf_max);
else
% over-excited: check for Ifmax
dmax = asin(P*Xd/(1.5*V*Eafm));
end
fprintf(’P= %g Ia_min = %g Ia_max = %g dmin = %g
dmax = %gn’, P, abs(Ia_min), abs(Ia_ma

% now that we have limits we go parameterize by delta
delt = dmin:(dmax-dmin)/100:dmax; Eaf = P*Xd . / (1.5*V .*
sin(delt));
Q= 1.5*(V^2/Xd - (V/Xd) .* Eaf .* cos(delt)); Ia = sqrt(P^2 + Q.^2)/(1.5*V);
I f = Eaf . / (om*M);
plot(If, Ia . / sqrt(2))
end
%now for the zero power curve Iaz = [V/Xd 0 (Eafm-V)/Xd];
Ifz = [0 V/(om*M) Eafm/(om*M)]; plot(Ifz, Iaz . / sqrt(2))
%and nowwe plot the limit lines Ia l = [Iar Iar 0];
I f l = [0 Ifm Ifm] plot(Ifl, Ia l . / sqrt(2), ’ - - ’ )
ti tl e( ’ Round Rotor, 1000 M
W Vee Curve’ ) ylabel(’Armature Current
(RMS)’) xlabel(’Field Current’)
grid on hold off

% 6.685 2013 Problem Set 4, Problem 2
%F i r s t , line up parameters VA =
150e6;
V= 13800/sqrt(3); om=
2*pi*60;
%Machine Rating
% Phase voltage, RMS
% frequency
%direct axis reactance
%quadrature axis reactance
% rated voltage open c i r c u it f i e l d current
Xd = 2.5;
Xq = 1.5;
AFNL = 600;
p = 7;
%Find St abilit y Limit for underexcited operation
%This w i l l be angle delta as a function of real power P
% warning off MATLAB:fzero:UndeterminedSyntax % to suppress a whole lo t of wierd warnings
%convenient shorthand
% establish a range of real power
%space for Q
P_0 = 3*V^2*(1/Xq-1/Xd); P =
(.01:.01:1) .* VA;
Qs = zeros(size(P)); Qc =
zeros(size(P)); E_af =
zeros(size(P)); ds =
zeros(size(P));
for i = 1:length(P)
Pr = P(i)/P_0; %here i s how we use the notation
d = f z e r o ( ’ e f ’ , [0 pi/ 2 ], [ ] , P r ) ; %t his gives angle at s t a b i l i t y lim it Eaf = -V*(Xd/Xq-
1)*cos(2*d)/cos(d); %and corresponding internal voltage E_af(i) = Eaf;
d s ( i ) = d;
Qs(i) = (3*V*Eaf/Xd) * cos(d) + 1.5*V^2*(1/Xq-1/Xd) * cos(2*d) - 1.5*V^2*(1/Xq+1/Xd); Qc(i) = -sqrt(VA^2-P(i)^2);
end
dpdd = (3*V/Xd) .* E_af .* cos(ds) +3*V^2*(1/Xq-1/Xd) .* cos(2 .* ds);
figure(1)
plot(P, Qs, P, Qc) legend(’Stability’, ’thermal’)
title(’Generator Underexcited Reactive Capability (Problem 4 . 2 ) ’ ) ylabel(’Generator Reactive
Power’)
xlabel(’Generator Real Power’)
%axis([0 2e8 -1e8 0])
%axis square grid on
I _ f = (AFNL/V) .* E_af;

figure(2) subplot 211 plot(P, I _ f )
t i t l e ( ’ S t a b i l i t y Limit for Underexcited Operation’) y l a b e l ( ’ I _ f , A’)
grid on subplot 212
plot(P, (180/pi) .* ds) ylabel(’Angle, degrees’) xlabel(’Underexcited Power, W’) grid on

% 6.685 2013 Problem Set 4, Problem 2, parts 3 and 4
%F i r s t , line up parameters VA = 150e6;
V= 13800/sqrt(3); om= 2*pi*60; %Machine Rating
% Phase voltage, RMS
% frequency
Xd = 2.5;
Xq = 1.5;
AFNL = 600;
p = 7;
Ra = .009;
Rf = 1.0; Pc =
1e6;
%direct axis reactance
%quadrature axis reactance
% rated voltage open c i r c u it f i e l d current
%number of pole pairs
%armature resistance
%f i e l d resistance
%core loss
% F i r s t , find f i e l d current for rated operation, unity power factor
I a = VA/(3*V);
E_1 = V + j*Xq*Ia; delta =
angle(E_1); Id = I a * sin(delta);
% t his would be current at rated operation
% spot on the q axis
% t his i s torque angle
%d- axis current
Eaf = abs(E_1) + (Xd-Xq)*Id; %internal voltage
I _ f = Eaf*AFNL/V;
fprintf(’Problem 4, part 3: Field Current = %gn’, I _ f ) ; f print f ( ’ D etails of that: E_1 = %g, delta = %gn’,
abs(E_1), delta); fprintf(’More de t ails: I_d = %g, E_af = %gn’, I d , E af ) ;
delt = 0:pi/100:pi;
P = (3*V*Eaf/Xd) .* sin(delt) + 1.5*V^2*(1/Xq - 1/Xd) .* sin(2 .* d e l t ) ;
T = (p/om) .* P;
figure(1) plot(delt, T)
title(’Torque v s . Angle for Example Machine’) ylabel(’N-m’)
xl abel ( ’ Radi ans’ ) grid on
%now for machine efficiency
P = 50e6:5e5:150e6; Ia = P . /
(3*V);
E_1 = V + Xq .* I a ;
% run over t his range
%armature current
%voltage on q axis

delta = angle(E_1);
Id = Ia .* sin(delta);
%torque angle
% d- axis current
Eaf = E_1 + (Xd-Xq) .* Id;
I_f = (AFNL/V) .* Eaf; % required field current
P_a = 3*Ra .* Ia .^2; P_f = Rf .* I_f .^2;
% armature winding loss
%field winding loss
eff = P . / (P + P_a + P_f + Pc);
figure(2) plot(P, eff)
title(’Efficiency for Example Machine’) ylabel(’Efficiency ’)
xlabel(’Output Power’)
grid on

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