initial commit / add all books

1 files changed, 37 insertions, 0 deletions

diff --git a/3472/CH30/EX30.13/Example30_13.sce b/3472/CH30/EX30.13/Example30_13.sce
new file mode 100644
index 000000000..f9de20838
--- /dev/null
+++ b/3472/CH30/EX30.13/Example30_13.sce
@@ -0,0 +1,37 @@
+// 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 : 4.13 :

+// Page number 522

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+kV = 6.9          // Alternator rating(kV)

+MVA = 10.0        // Alternator rating(MVA)

+X_st = 0.15       // Sub-transient reactance(p.u)

+X_2 = 0.15        // Negative sequence reactance(p.u)

+X_0 = 0.05        // Zero sequence reactance(p.u)

+X = 0.397         // Grounding reactor(ohm)

+

+// Calculations

+MVA_base = 10.0                         // Base MVA

+kV_base = 6.9                           // Base kV

+Z_base = kV_base**2/MVA_base            // Base impedance(ohm)

+Z_n = X/Z_base                          // Grounding reactor(p.u)

+Z_1 = %i*X_st                           // Positive sequence impedance(p.u)

+Z_2 = %i*X_2                            // Negative sequence impedance(p.u)

+Z_0 = %i*(X_0+3*Z_n)                    // Zero sequence impedance(p.u)

+E_a = 1.0                               // Phase voltage(p.u)

+I_a_pu = 3*E_a/(Z_1+Z_2+Z_0)            // Sub-transient current in the faulty phase(p.u)

+I_base = kV_base*1000/(3**0.5*Z_base)   // Base current(A)

+I_a = abs(I_a_pu)*I_base                // Sub-transient current in the faulty phase(A)

+

+// Results

+disp("PART III - EXAMPLE : 4.13 : SOLUTION :-")

+printf("\nSub-transient current in the faulty phase, I_a = %.f A\n", I_a)

+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")