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nm P ( 3600 r/min )( 2 poles ) = = 60 Hz 120 120
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There are six slots on this stator, with 80 turns of wire per slot The voltage in the coils in one phase
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E A = 2 N C f = 2 ( 80 t )( 0060 Wb )( 60 Hz ) = 1280 V
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Since the machine is -connected, VL = V = 1280 V 4-3 A three-phase Y-connected 50-Hz two-pole synchronous machine has a stator with 2000 turns of wire per phase What rotor flux would be required to produce a terminal (line-to-line) voltage of 6 kV SOLUTION The phase voltage of this machine should be V = VL / 3 = 3464 V The induced voltage per phase in this machine (which is equal to V at no-load conditions) is given by the equation
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EA = 2 N C f
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3464 V = 00078 Wb 2 (2000 t )(50 Hz )
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Modify the MATLAB in Example 4-1 by swapping the currents flowing in any two phases What happens to the resulting net magnetic field SOLUTION This modification is very simple just swap the currents supplied to two of the three phases
% M-file: mag_field2m % M-file to calculate the net magetic field produced % by a three-phase stator % Set up the basic conditions bmax = 1; % Normalize bmax to 1 freq = 60; % 60 Hz w = 2*pi*freq; % angluar velocity (rad/s) % First, generate the three component magnetic fields t = 0:1/6000:1/60; Baa = sin(w*t) * (cos(0) + j*sin(0)); Bbb = sin(w*t+2*pi/3) * (cos(2*pi/3) + j*sin(2*pi/3)); Bcc = sin(w*t-2*pi/3) * (cos(-2*pi/3) + j*sin(-2*pi/3)); % Calculate Bnet Bnet = Baa + Bbb + Bcc; % Calculate a circle representing the expected maximum % value of Bnet circle = 15 * (cos(w*t) + j*sin(w*t)); % Plot the magnitude and direction of the resulting magnetic % fields Note that Baa is black, Bbb is blue, Bcc is % magneta, and Bnet is red for ii = 1:length(t) % Plot the reference circle plot(circle,'k'); hold on; % Plot the four magnetic fields plot([0 real(Baa(ii))],[0 imag(Baa(ii))],'k','LineWidth',2); plot([0 real(Bbb(ii))],[0 imag(Bbb(ii))],'b','LineWidth',2); plot([0 real(Bcc(ii))],[0 imag(Bcc(ii))],'m','LineWidth',2); plot([0 real(Bnet(ii))],[0 imag(Bnet(ii))],'r','LineWidth',3); axis square; axis([-2 2 -2 2]); drawnow; hold off; end
When this program executes, the net magnetic field rotates clockwise, instead of counterclockwise
4-5
If an ac machine has the rotor and stator magnetic fields shown in Figure P4-1, what is the direction of the induced torque in the machine Is the machine acting as a motor or generator
SOLUTION Since ind = kB R B net , the induced torque is clockwise, opposite the direction of motion The machine is acting as a generator
5: Synchronous Machines
5-1 At a location in Europe, it is necessary to supply 300 kW of 60-Hz power The only power sources available operate at 50 Hz It is decided to generate the power by means of a motor-generator set consisting of a synchronous motor driving a synchronous generator How many poles should each of the two machines have in order to convert 50-Hz power to 60-Hz power SOLUTION The speed of a synchronous machine is related to its frequency by the equation
nm =
120 f e P
To make a 50 Hz and a 60 Hz machine have the same mechanical speed so that they can be coupled together, we see that
nsync =
120(50 Hz ) 120(60 Hz ) = P1 P2
P2 6 12 = = P1 5 10
Therefore, a 10-pole synchronous motor must be coupled to a 12-pole synchronous generator to accomplish this frequency conversion 5-2 A 480-V 200-kVA 08-power-factor-lagging 60-Hz two-pole Y-connected synchronous generator has a synchronous reactance of 025 and an armature resistance of 003 At 60 Hz, its friction and windage losses are 6 kW, and its core losses are 4 kW The field circuit has a dc voltage of 200 V, and the maximum I F is 10 A The resistance of the field circuit is adjustable over the range from 20 to 200 The OCC of this generator is shown in Figure P5-1 (a) How much field current is required to make VT equal to 480 V when the generator is running at no load (b) What is the internal generated voltage of this machine at rated conditions (c) How much field current is required to make VT equal to 480 V when the generator is running at rated conditions (d) How much power and torque must the generator s prime mover be capable of supplying
SOLUTION (a) If the no-load terminal voltage is 480 V, the required field current can be read directly from the open-circuit characteristic It is 455 A (b) This generator is Y-connected, so I L = I A At rated conditions, the line and phase current in this generator is
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