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24 kV = (382 + j140 ) = 112 + j 411 14 kV
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The secondary current I S is given by
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90 kW = 4348 A (2300 V )(09 ) I S = 4348 258 A IS =
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(a) The voltage at the power source of this system (referred to the secondary side) is
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Vsource = VS + I S Z line + I S Z EQ Vsource = 2300 0 V + ( 4348 258 A )(112 + j 411 ) + ( 4348 258 A )( 012 + j 05 ) Vsource = 2441 37 V
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Therefore, the voltage at the power source is
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Vsource = (2441 37 V )
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14 kV = 1424 37 kV 24 kV
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(b) To find the voltage regulation of the transformer, we must find the voltage at the primary side of the transformer (referred to the secondary side) under full load conditions:
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VP = VS + I S Z EQ VP = 2300 0 V + (4348 258 A )(012 + j 05 ) = 2314 043 V
There is a voltage drop of 14 V under these load conditions Therefore the voltage regulation of the transformer is
VR =
2314 - 2300 100% = 06% 2300
The power supplied to the load is POUT = 90 kW The power supplied by the source is
PIN = Vsource I S cos = (2441 V )(4348 A ) cos 295 = 9237 kW
Therefore, the efficiency of the power system is
3-5
POUT 90 kW 100% = 100% = 974% PIN 9237 kW
When travelers from the USA and Canada visit Europe, they encounter a different power distribution system Wall voltages in North America are 120 V rms at 60 Hz, while typical wall voltages in Europe are 220-240 V at 50 Hz Many travelers carry small step-up / step-down transformers so that they can use their appliances in the countries that they are visiting A typical transformer might be rated at 1-kVA and 120/240 V It has 5001 turns of wire on the 120-V side and 1000 turns of wire on the 240-V side The magnetization curve for this transformer is shown in Figure P3-2, and can be found in file p32mag at this book s Web site
Note that this turns ratio was backwards in the first printing of the text This error should be corrected in all subsequent printings 33
(a) Suppose that this transformer is connected to a 120-V, 60 Hz power source with no load connected to the 240-V side Sketch the magnetization current that would flow in the transformer (Use MATLAB to plot the current accurately, if it is available) What is the rms amplitude of the magnetization current What percentage of full-load current is the magnetization current (b) Now suppose that this transformer is connected to a 240-V, 50 Hz power source with no load connected to the 120-V side Sketch the magnetization current that would flow in the transformer (Use MATLAB to plot the current accurately, if it is available) What is the rms amplitude of the magnetization current What percentage of full-load current is the magnetization current (c) In which case is the magnetization current a higher percentage of full-load current Why SOLUTION (a) When this transformer is connected to a 120-V 60 Hz source, the flux in the core will be given by the equation
(t ) =
VM cos t N P
(3-104)
The magnetization current required for any given flux level can be found from Figure P3-2, or alternately from the equivalent table in file p32mag The MATLAB program shown below calculates the flux level at each time, the corresponding magnetization current, and the rms value of the magnetization current
% M-file: prob3_5am % M-file to calculate and plot the magnetization % current of a 120/240 transformer operating at
% 120 volts and 60 Hz This program also % calculates the rms value of the mag current % Load the magnetization curve It is in two % columns, with the first column being mmf and % the second column being flux load p32mag; mmf_data = p32(:,1); flux_data = p32(:,2); % Initialize values S = 1000; Vrms = 120; VM = Vrms * sqrt(2); NP = 500;
% % % %
Apparent power (VA) Rms voltage (V) Max voltage (V) Primary turns
% Calculate angular velocity for 60 Hz freq = 60; % Freq (Hz) w = 2 * pi * freq; % Calculate flux versus time time = 0:1/3000:1/30; % 0 to 1/30 sec flux = -VM/(w*NP) * cos(w * time); % Calculate the mmf corresponding to a given flux % using the MATLAB interpolation function mmf = interp1(flux_data,mmf_data,flux); % Calculate the magnetization current im = mmf / NP; % Calculate the rms value of the current irms = sqrt(sum(im^2)/length(im)); disp(['The rms current at 120 V and 60 Hz is ', num2str(irms)]); % Calculate the full-load current i_fl = S / Vrms; % Calculate the percentage of full-load current percnt = irms / i_fl * 100; disp(['The magnetization current is ' num2str(percnt) '% of full-load current']); % Plot the magnetization current figure(1) plot(time,im); title ('\bfMagnetization Current at 120 V and 60 Hz'); xlabel ('\bfTime (s)'); ylabel ('\bf\itI_{m} \rm(A)'); axis([0 004 -05 05]); grid on;
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