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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3472/CH30/EX30.5 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3472/CH30/EX30.5')
| -rw-r--r-- | 3472/CH30/EX30.5/Example30_5.sce | 46 |
1 files changed, 46 insertions, 0 deletions
diff --git a/3472/CH30/EX30.5/Example30_5.sce b/3472/CH30/EX30.5/Example30_5.sce new file mode 100644 index 000000000..cb7d7aaf1 --- /dev/null +++ b/3472/CH30/EX30.5/Example30_5.sce @@ -0,0 +1,46 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.5 :
+// Page number 514-515
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 6600.0 // Alternator voltage(V)
+kVA = 10000.0 // Alternator rating(kVA)
+x_1 = 15.0 // Reactance to positive sequence current(%)
+x_2 = 75.0 // Reactance to negative sequence current(%)
+x_0 = 30.0 // Reactance to zero sequence current(%)
+R_earth = 0.3 // Earth resistance(ohm)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+E_g = V/3**0.5 // Phase voltage(V)
+// Case(a)
+I = kVA*1000/(3**0.5*V) // Full load current of each alternator(A)
+X = x_1*V/(100*3**0.5*I) // Positive sequence reactance(ohm)
+Z_g1 = %i*X // Equivalent positive sequence impedance(ohm)
+Z_g2 = Z_g1*x_2/100 // Equivalent negative sequence impedance(ohm)
+Z_g0 = Z_g1*x_0/100 // Equivalent zero sequence impedance(ohm)
+Z_1 = Z_g1/3 // Positive sequence impedance(ohm)
+Z_2 = Z_g2/3 // Negative sequence impedance(ohm)
+Z_0 = Z_g0/3 // Zero sequence impedance(ohm)
+I_a_a = 3*E_g/(Z_1+Z_2+Z_0) // Fault current(A)
+// Case(b)
+Z_0_b = Z_g0 // Impedance(ohm)
+I_a_b = 3*E_g/(Z_1+Z_2+Z_0_b) // Fault current(A)
+// Case(c)
+Z_0_c = R_earth*3+Z_g0 // Impedance(ohm)
+I_a_c = 3*E_g/(Z_1+Z_2+Z_0_c) // Fault current(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.5 : SOLUTION :-")
+printf("\nCase(a): Fault current if all the alternator neutrals are solidly earthed, I_a = %.fj A", imag(I_a_a))
+printf("\nCase(b): Fault current if only one of the alternator neutrals is solidly earthed & others isolated = %.fj A", imag(I_a_b))
+printf("\nCase(c): Fault current if one of alternator neutrals is earthed through resistance & others isolated = %.f A\n", abs(I_a_c))
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
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