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// ELECTRICAL MACHINES
// R.K.Srivastava
// First Impression 2011
// CENGAGE LEARNING INDIA PVT. LTD
// CHAPTER : 3 : TRANSFORMERS
// EXAMPLE : 3.12
clear ; clc ; close ; // Clear the work space and console
// GIVEN DATA
N1 = 1000; // 1st Test at No-load condition f1 Frequency, Speed in RPM
Vo1 = 250; // 1st Test at No-load condition f1 Frequency, Voltage in Volts
Io1 = 0.5; // 1st Test at No-load condition f1 Frequency, Current in Amphere
Wo1= 230; // 1st Test at No-load condition f1 Frequency, Power in Watts
N2 = 900; // 2nd Test at No-load condition f2 Frequency, Speed in RPM
Vo2 = 225; // 2nd Test at No-load condition f2 Frequency, Voltage in Volts
Io2= 0.5; // 2nd Test at No-load condition f2 Frequency, Current in Amphere
Wo2 = 200; // 2nd Test at No-load condition f2 Frequency, Power in Watts
p = 6; // Number of poles of single phase alternator
N = 220; // Number of the turns of single phase alternator
R = 0.66; // Resistance of the single phase alternator in Ohms
// CALCULATIONS
f1 = (N1*p)/120; // 1st case Supply Frequency in Hertz
Ratio1 = Vo1/f1; // 1st case Ratio of the Volatge and Frequency in Volts/Hertz
f2 = (N2*p)/120; // 2nd case Supply Frequency in Hertz
Ratio2 = Vo2/f2; // 2nd case Ratio of the Volatge and Frequency in Volts/Hertz
c = (Wo1-(Io1^2)*R)/f1; // No-load corrected losses Eq 1 in Watts
d = (Wo2-(Io2^2)*R)/f2; // No-load corrected losses Eq 2 in watts
x = [ 1 f1 ; 1 f2 ]; // No-load corrected losses Eq 1 in watts
y = [ c ; d ]; // No-load corrected losses Eq 2 in watts
E = x\y; // Solution of constants A in Watts/Hertz and B in watts/Hertz-Sqare in matrix form
A = E(1,1); // Solution of constant A in Watts/Hertz
B = E(2,1); // Solution of constant B in watts/Hertz-Sqare
Ph = f1*A; // Hysteresis loss at 50 Hertz in Watts
Pe = (f1^2)*B; // Eddy current loss at 50 Hertz in Watts
// DISPLAY RESULTS
disp("EXAMPLE : 3.12 : SOLUTION :-") ;
printf("\n (a) Hysteresis loss at %.f Hertz , Ph = %.3f W \n ",f1,Ph);
printf("\n (b) Eddy current loss at %.f Hertz , Pe = % .2f W \n",f1,Pe);
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