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3V E A sin XS
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(5-18)
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The torque angle can be found by calculating E A :
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E A = 277 0 + j ( 025 )( 2406 0 A ) E A = 283 12 V Thus the torque angle = 12
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(b) The maximum power that this generator can deliver to a unity power factor load can be found from Equation (5-18):
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E A = V + R A I A + jX S I A
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3V E A sin 3 ( 277 V )( 283 V ) sin 90 = 941 kW 025 3 ( 277 V )( 283 V ) sin12 = 196 kW 025 XS
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Pmax =
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(c) The current torque angle is = 12 , and the maximum angle is 90 , so it is very far from the static stability limit In terms of power, the generator is currently supplying
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3V E A sin XS
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Therefore the general is supplying about 1/5 of the static stability limit (d)
% % % %
A program to plot the torque as a function of is shown below:
M-file: prob5_4dm M-file to calculate and plot the power supplied by a synchronous generator as a function of the torque angle delta
% Define values for this generator EA = 283; % Internal gen voltage VP = 277; % Phase voltage XS = 025; % XS (ohms) % Calculate the power vs torque angle delta = 0:1:90; P = 3 * EA * VP / XS * sin( delta * pi / 180 ); % Plot the output power versus torque angle plot(delta,P/1000,'b-','LineWidth',20); title ('\bfOutput Power Versus Torque Angle'); xlabel ('\bfTorque Angle (deg)'); ylabel ('\bfOutput Power (kW)'); grid on;
The resulting plot is shown below: 75
Problems 5-5 to 5-14 refer to a two-pole Y-connected synchronous generator rated at 300 kVA, 480 V, 60 Hz, and 085 PF lagging Its armature resistance RA is 004 The core losses of this generator at rated conditions are 10 kW, and the friction and windage losses are 13 kW The open-circuit and short-circuit characteristics are shown in Figure P5-2
5-5
(a) What is the saturated synchronous reactance of this generator at the rated conditions (b) What is the unsaturated synchronous reactance of this generator (c) Plot the saturated synchronous reactance of this generator as a function of load 76
SOLUTION (a) The rated armature current for this generator is
IA = IL =
S 300 kVA = = 361 A 3 VT 3 (480 V )
The field current required to produce this much short-circuit current may be read from the SCC It is 258 A The open circuit voltage at 258 A is 435 V, so the open-circuit phase voltage (= E A ) is 435 / 3 = 251 V The approximate saturated synchronous reactance X S is
XS =
251 V = 0695 361 A
(b) The unsaturated synchronous reactance X Su is the ratio of the air-gap line to the SCC At I F = 258 A, the air-gap line voltage is 447 V, and the SCC is 361 A
X Su =
(447 V ) /
361 A
= 0715
(c) This task can best be performed with MATLAB The open-circuit characteristic is available in a file called p52occ, and the short-circuit characteristic is available in a file called p52scc Each of these files are organized in two columns, where the first column is field current and the second column is either open-circuit terminal voltage or short-circuit current A program to read these files and calculate and plot X S is shown below
% % % % M-file: prob5_5cm M-file to calculate and plot the saturated synchronous reactance of a synchronous generator
% Load the open-circuit characteristic It is in % two columns, with the first column being field % current and the second column being terminal % voltage load p52occ; if_occ = p52(:,1); vt_occ = p52(:,2); % Load the short-circuit characteristic It is in % two columns, with the first column being field % current and the second column being line current % (= armature current) load p52scc; if_scc = p52(:,1); ia_scc = p52(:,2); % Calculate Xs If = 0001:01:10; vt = interp1(if_occ,vt_occ,If); ia = interp1(if_scc,ia_scc,If); Xs = (vt / sqrt(3)) / ia; % Plot the synchronous reactance figure(1)
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