diff options
Diffstat (limited to '3472/CH47')
-rw-r--r-- | 3472/CH47/EX47.1/Example47_1.sce | 31 | ||||
-rw-r--r-- | 3472/CH47/EX47.2/Example47_2.sce | 26 |
2 files changed, 57 insertions, 0 deletions
diff --git a/3472/CH47/EX47.1/Example47_1.sce b/3472/CH47/EX47.1/Example47_1.sce new file mode 100644 index 000000000..93557b0b0 --- /dev/null +++ b/3472/CH47/EX47.1/Example47_1.sce @@ -0,0 +1,31 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART IV : UTILIZATION AND TRACTION
+// CHAPTER 9: ELECTRIC TRACTION SYSTEMS AND POWER SUPPLY
+
+// EXAMPLE : 9.1 :
+// Page number 817-818
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+L = 3.0 // Length of section ACB of rail(km)
+L_B_A = 2.0 // Distance of B from A(km)
+I_load = 350.0 // Loading(A/km)
+r_rail = 0.035 // Resistance of rail(ohm/km)
+r_feed = 0.03 // Resistance of negative feeder(ohm/km)
+
+// Calculations
+x_val = integrate('I_load*(L-x)','x',0,L_B_A)
+I = x_val/(L_B_A-0) // Current in negative feeder(A)
+x = L-(I/I_load) // Distance from feeding point(km)
+C = integrate('r_rail*I_load*x','x',0,x)
+V = r_feed*L_B_A*I // Voltage produced by negative booster(V)
+rating = V*I/1000 // Rating of the booster(kW)
+
+// Results
+disp("PART IV - EXAMPLE : 9.1 : SOLUTION :-")
+printf("\nMaximum potential difference between any two points of the rails, C = %.2f V", C)
+printf("\nRating of the booster = %.1f kW", rating)
diff --git a/3472/CH47/EX47.2/Example47_2.sce b/3472/CH47/EX47.2/Example47_2.sce new file mode 100644 index 000000000..4b81630ae --- /dev/null +++ b/3472/CH47/EX47.2/Example47_2.sce @@ -0,0 +1,26 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART IV : UTILIZATION AND TRACTION
+// CHAPTER 9: ELECTRIC TRACTION SYSTEMS AND POWER SUPPLY
+
+// EXAMPLE : 9.2 :
+// Page number 820
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+D = 50.0 // Distance between poles(m)
+w = 0.5 // Weight of trolley wire per metre(kg)
+T = 520.0 // Maximum tension(kg)
+
+// Calculations
+l = D/2 // Half distance b/w poles(m)
+d = w*l**2/(2*T) // Sag(m)
+wire_length = 2*(l+(2*d**2/(3*l))) // Length of wire required(m)
+
+// Results
+disp("PART IV - EXAMPLE : 9.2 : SOLUTION :-")
+printf("\nMaximum sag, d = %.4f metres", d)
+printf("\nLength of wire required = %.f metres", wire_length)
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