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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/CH10/EX10.14/Example10_14.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3472/CH10/EX10.14/Example10_14.sce')
| -rw-r--r-- | 3472/CH10/EX10.14/Example10_14.sce | 48 |
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diff --git a/3472/CH10/EX10.14/Example10_14.sce b/3472/CH10/EX10.14/Example10_14.sce new file mode 100644 index 000000000..9dafa23dc --- /dev/null +++ b/3472/CH10/EX10.14/Example10_14.sce @@ -0,0 +1,48 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES
+
+// EXAMPLE : 3.14 :
+// Page number 144
+clear ; clc ; close ; // Clear the work space and console
+funcprot(0)
+
+// Given data
+f = 50.0 // Frequency(Hz)
+r = 0.1 // Resistance(ohm/km)
+l = 1.4*10**-3 // Inductance(H/km)
+c = 8.0*10**-9 // Capacitance(F/km)
+g = 4.0*10**-8 // conductance(mho/km)
+V_r = 400.0 // Receiving end voltage(kV)
+x = 200.0 // Length of line(km)
+
+// Calculations
+V_2 = V_r/3**0.5 // Receiving end phase voltage(kV)
+z = r+%i*2*%pi*f*l // Total impedance(ohm/km)
+y = g+%i*2*%pi*f*c // Total susceptance(mho/km)
+Z_c = (z/y)**0.5 // Surge impedance(ohm)
+gamma = (z*y)**0.5 // γ
+// Case(i)
+V_0_plus = V_2/2 // Incident voltage to neutral at receiving end(kV)
+// Case(ii)
+V_0_minus = V_2/2 // Reflected voltage to neutral at receiving end(kV)
+// Case(iii)
+gamma_l = gamma*x // γl
+V_1_plus = (V_2/2)*exp(gamma_l) // Incident voltage to neutral at 200 km from receiving end(kV)
+V_1_minus = (V_2/2)*exp(-gamma_l) // Reflected voltage to neutral at 200 km from receiving end(kV)
+// Case(iv)
+V_1 = V_1_plus+V_1_minus // Resultant voltage to neutral(kV)
+V_L = abs(V_1) // Resultant voltage to neutral(kV)
+V_L_ll = 3**0.5*V_L // Line to line voltage at 200 km from receiving end(kV)
+
+// Results
+disp("PART II - EXAMPLE : 3.14 : SOLUTION :-")
+printf("\nCase(i) : Incident voltage to neutral at receiving end, V_0_plus = %.1f∠%.f° kV", abs(V_0_plus),phasemag(V_0_plus))
+printf("\nCase(ii) : Reflected voltage to neutral at receiving end, V_0_minus = %.1f∠%.f° kV", abs(V_0_minus),phasemag(V_0_minus))
+printf("\nCase(iii): Incident voltage to neutral at 200 km from receiving end, V_1_plus = (%.3f+%.2fj) kV", real(V_1_plus),imag(V_1_plus))
+printf("\nCase(iv) : Resultant voltage to neutral at 200 km from receiving end, V_L = %.2f kV", V_L)
+printf("\n Line to line voltage at 200 km from receiving end = %.2f kV", V_L_ll)
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