Electric Machinery Fundamentals Power & Energy 2 potx
Electric Machinery Fundamentals Power & Energy_2 potx
...
plot(amps,abs(VS(1,:)/1000),'b-','LineWidth' ,2. 0);
hold on;
plot(amps,abs(VS (2, :)/1000),'k ','LineWidth' ,2. 0);
plot(amps,abs(VS(3,:)/1000),'r ','LineWidth' ,2. 0); ... ','LineWidth' ,2. 0);
title ('\bfSecondary Voltage Versus Load');
xlabel ('\bfLoad (A)');
ylabel ('\bfSecondary Vo...
Electric Machinery Fundamentals Power & Energy_3 docx
...
plot(amps,VR(1,:),'b-','LineWidth' ,2. 0);
hold on;
plot(amps,VR (2, :),'k ','LineWidth' ,2. 0);
plot(amps,VR(3,:),'r ','LineWidth' ,2. 0);
title ('\bfVoltage ...
plot(amps,abs(VSP(1,:)),'b-','LineWidth' ,2. 0);
hold on;
plot(amps,abs(VSP (2, :)),'k ','LineWidth' ,2. 0);
plot(amps...
Electric Machinery Fundamentals Power & Energy_4 pptx
... part of the autotransformer
72
1
22 2
1
M
V
B
RC
ω
=
+
and
2
222
1
M
RC V
B
RC
ω
ω
−
=
+
Therefore, the forced solution to the equation is
()
,
22 2 22 2
sin cos
11
MM
Cf
VRCV
vt ... the beginning of the half-
cycle:
()
0
22 2 22 2
0 sin 0 cos 00
11
MM
RC
C
VRCV
vAe
RC RC
ω
ωω
−
=+ − =
++
22 2
0
1
M
RC V
A
RC
ω
ω
−=
+
22 2
1
M
RC V
A
RC
ω
ω
=
+...
Electric Machinery Fundamentals Power & Energy_5 docx
...
plot(t,vb,'r:','Linewidth' ,2. 0);
plot(t,vc,'k ','Linewidth' ,2. 0);
title('\bfPhase Voltages');
xlabel('\bfTime (s)');
ylabel('\bfVoltage (V)');
grid ...
ylabel('\bfVoltage (V)');
grid on;
legend('Phase a','Phase b','Phase c');
hold off;
86
1
22 2
1
M
V
B
RC
ω
=
+...
Electric Machinery Fundamentals Power & Energy_6 doc
...
plot(Pload,PB,'k ','LineWidth' ,2. 0);
plot(Pload,PC,'r ','LineWidth' ,2. 0);
plot([0 10],[3 3],'k','LineWidth',1.0);
plot([0 10],[0 0],'k:'); ...
plot(I,abs(VT_unity),'k ','LineWidth' ,2. 0);
plot(I,abs(VT_lead),'r ','LineWidth' ,2. 0);
title ('\bfTerminal Voltage Versus...
Electric Machinery Fundamentals Power & Energy_8 ppt
...
plot(If,abs(Ia_nl),'k-','Linewidth' ,2. 0);
hold on;
plot(If,abs(Ia_half),'b ','Linewidth' ,2. 0);
plot(If,abs(Ia_full),'r:','Linewidth' ,2. 0);
xlabel('\bfField ... A
jX X=+ +EE I
1 52
The resulting plot is shown below
350 400 450 500 550 600 650 700
20 0
21 0
22 0
23 0
24 0
25 0
26 0
E
A
(V)
I
A
(A)
Synchr...
Electric Machinery Fundamentals Power & Energy_9 doc
... .
()( ) ()
22
22
20
SL S L L SL
RR X X VVR RR
+++ − +=
()( ) ()
22
2
SL S L LSL
RR X X RRR
+++ = +
()
2
22 2
22 2
SSLLSL SLL
RRRRXX RRR++++=+
()
2
22 2
2
SL SL L
RR ...
MMM
M
RX X X XX
X
RXX
++
=
++
.
If
1M
XR>> and
1M
XX>> then
22 2
111MMM
XX R X X X>> + and
()
2
22
11
MM
XX X R+≈>>
, so
2
1
TH 1
2
M
M
XX...
Electric Machinery Fundamentals Power & Energy_11 ppt
... (8-56).
23 0
figure(1);
plot(t_ind,n,'b-','LineWidth' ,2. 0);
hold on;
xlabel('\bf\tau_{ind} (N-m)');
ylabel('\bf\itn_{m} \rm\bf(r/min)');
title ('\bfSeries ... these conditions is
()
22 4.8 V
900 r/min 803 r/min
25 2 V
A
o
Ao
E
nn
E
== =
The power converted from electrical to mechanical form is
()()
conv
22 4.8 V 101.3 A 2...
Electric Machinery Fundamentals Power & Energy_12 pptx
... (['Vt = ' num2str(Vt(ii)) ' V']);
disp (['If = ' num2str(i_f(ii)) ' A']);
% Plot the curves
figure(1);
plot(i_f,Ea,'b-','LineWidth' ,2. 0); ...
current');
axis ([0 5 0 150]);
set(gca,'YTick',[0 10 20 30 40 50 60 70 80 90 100 110 120 130 140
150]')
set(gca,'XTick',[0 0.5 1.0 1.5 2. 0 2. 5 3....
Electric Machinery Fundamentals Power & Energy_13 pptx
...
()()
22
22
/
/
M
F
M
Rs jX jX
Z
R s jX jX
+
=
++
()()
100 2. 40 60
2. 96 2. 46
10 02. 4060
F
jj
Zj
jj
+
==+Ω
++
()
()
()
22
22
/2
/2
M
B
M
R s jX jX
Z
R s jX jX
=
−+ +
()()
1 .28 2 2. 40 ...
disp (['Ea = ' num2str(Ea(ii)) ' V']);
disp (['Vt = ' num2str(Vt(ii)) ' V']);
disp (['If = ' num2str(i_f(ii)) ' A&a...