diff options
Diffstat (limited to '3472/CH25')
| -rw-r--r-- | 3472/CH25/EX25.1/Example25_1.sce | 40 | ||||
| -rw-r--r-- | 3472/CH25/EX25.2/Example25_2.sce | 34 | ||||
| -rw-r--r-- | 3472/CH25/EX25.3/Example25_3.sce | 53 | ||||
| -rw-r--r-- | 3472/CH25/EX25.4/Example25_4.sce | 37 | ||||
| -rw-r--r-- | 3472/CH25/EX25.6/Example25_6.sce | 40 | ||||
| -rw-r--r-- | 3472/CH25/EX25.7/Example25_7.sce | 62 | ||||
| -rw-r--r-- | 3472/CH25/EX25.8/Example25_8.sce | 36 | ||||
| -rw-r--r-- | 3472/CH25/EX25.9/Example25_9.sce | 57 |
8 files changed, 359 insertions, 0 deletions
diff --git a/3472/CH25/EX25.1/Example25_1.sce b/3472/CH25/EX25.1/Example25_1.sce new file mode 100644 index 000000000..eb15cb6b9 --- /dev/null +++ b/3472/CH25/EX25.1/Example25_1.sce @@ -0,0 +1,40 @@ +// 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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.1 :
+// Page number 437
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V_A = 225.0 // Potential at point A(V)
+R_A = 5.0 // Resistance of line A(ohm)
+V_B = 210.0 // Potential at point B(V)
+R_B = 1.0 // Resistance of line B(ohm)
+V_C = 230.0 // Potential at point C(V)
+R_C = 1.0 // Resistance of line C(ohm)
+V_D = 230.0 // Potential at point D(V)
+R_D = 2.0 // Resistance of line D(ohm)
+V_E = 240.0 // Potential at point E(V)
+R_E = 2.0 // Resistance of line E(ohm)
+
+// Calculations
+V_0 = ((V_A/R_A)+(V_B/R_B)+(V_C/R_C)+(V_D/R_D)+(V_E/R_E))/((1/R_A)+(1/R_B)+(1/R_C)+(1/R_D)+(1/R_E)) // Potential at point O(V)
+I_A = (V_A-V_0)/R_A // Current leaving supply point A(A)
+I_B = (V_B-V_0)/R_B // Current leaving supply point B(A)
+I_C = (V_C-V_0)/R_C // Current leaving supply point C(A)
+I_D = (V_D-V_0)/R_D // Current leaving supply point D(A)
+I_E = (V_E-V_0)/R_E // Current leaving supply point E(A)
+
+// Results
+disp("PART II - EXAMPLE : 18.1 : SOLUTION :-")
+printf("\nPotential of point O, V_0 = %.f V", V_0)
+printf("\nCurrent leaving supply point A, I_A = %.f A", I_A)
+printf("\nCurrent leaving supply point B, I_B = %.f A", I_B)
+printf("\nCurrent leaving supply point C, I_C = %.f A", I_C)
+printf("\nCurrent leaving supply point D, I_D = %.2f A", I_D)
+printf("\nCurrent leaving supply point E, I_E = %.2f A", I_E)
diff --git a/3472/CH25/EX25.2/Example25_2.sce b/3472/CH25/EX25.2/Example25_2.sce new file mode 100644 index 000000000..b736b34c0 --- /dev/null +++ b/3472/CH25/EX25.2/Example25_2.sce @@ -0,0 +1,34 @@ +// 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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.2 :
+// Page number 437-438
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+I = 600.0 // Constant current drawn(A)
+D = 8.0 // Distance b/w two sub-stations(km)
+V_A = 575.0 // Potential at point A(V)
+V_B = 590.0 // Potential at point B(V)
+R = 0.04 // Track resistance(ohm/km)
+
+// Calculations
+x = poly(0,'x') // x(km)
+I_A = ((-V_B+R*I*D+V_A)-(R*I)*x)/(D*R) // Simplifying
+V_P = V_A-I_A*R*x // Potential at P in terms of x(V)
+dVP_dx = derivat(V_P) // dV_P/dx
+x_sol = roots(dVP_dx) // Value of x(km)
+I_A_1 = ((-V_B+R*I*D+V_A)-(R*I)*x_sol)/(D*R) // Current drawn from end A(A)
+I_B = I-I_A_1 // Current drawn from end B(A)
+
+// Results
+disp("PART II - EXAMPLE : 18.2 : SOLUTION :-")
+printf("\nPoint of minimum potential along the track, x = %.2f km", x_sol)
+printf("\nCurrent supplied by station A, I_A = %.f A", I_A_1)
+printf("\nCurrent supplied by station B, I_B = %.f A \n", I_B)
+printf("\nNOTE: ERROR: Calculation mistake in the textbook solution")
diff --git a/3472/CH25/EX25.3/Example25_3.sce b/3472/CH25/EX25.3/Example25_3.sce new file mode 100644 index 000000000..b1ccbba6c --- /dev/null +++ b/3472/CH25/EX25.3/Example25_3.sce @@ -0,0 +1,53 @@ +// 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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.3 :
+// Page number 438-439
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l = 400.0 // Length of cable(m)
+i = 1.0 // Load(A/m)
+I_1 = 120.0 // Current at 40m from end A(A)
+l_1 = 40.0 // Distance from end A(A)
+I_2 = 72.0 // Current at 72m from end A(A)
+l_2 = 120.0 // Distance from end A(A)
+I_3 = 48.0 // Current at 200m from end A(A)
+l_3 = 200.0 // Distance from end A(A)
+I_4 = 120.0 // Current at 320m from end A(A)
+l_4 = 320.0 // Distance from end A(A)
+r = 0.15 // Cable resistance(ohm/km)
+V_A = 250.0 // Voltage at end A(A)
+V_B = 250.0 // Voltage at end A(A)
+
+// Calculations
+I = poly(0,"I") // Current from end A(A)
+A_A1 = l_1*r*(I-(1.0/2)*i*l_1) // Drop over length(V)
+I_d_1 = 40.0 // Distributed tapped off current(A)
+I_A1_A2 = I-l_1-l_2 // Current fed in over length(A)
+A1_A2 = (l_2-l_1)*r*(I_A1_A2-(1.0/2)*i*(l_2-l_1)) // Drop over length(V)
+I_d_2 = 80.0 // Distributed tapped off current(A)
+I_A2_A3 = I_A1_A2-(I_2+I_d_2) // Current fed in over length(A)
+A2_A3 = (l_3-l_2)*r*(I_A2_A3-(1.0/2)*i*(l_3-l_2)) // Drop over length(V)
+I_d_3 = 80.0 // Distributed tapped off current(A)
+I_A3_A4 = I_A2_A3-(I_3+I_d_3) // Current fed in over length(A)
+A3_A4 = (l_4-l_3)*r*(I_A3_A4-(1.0/2)*i*(l_4-l_3)) // Drop over length(V)
+I_d_4 = 120.0 // Distributed tapped off current(A)
+I_A4_B = I_A3_A4-(I_4+I_d_4) // Current fed in over length(A)
+A4_B = (l-l_4)*r*(I_A4_B-(1.0/2)*i*(l-l_4)) // Drop over length(V)
+V_drop = A_A1+A1_A2+A2_A3+A3_A4+A4_B // Total voltage drop in terms of I
+I = roots(V_drop) // Current(A)
+I_total = 760.0 // Total load current(A)
+I_B = I_total-I // Current from B(A)
+A_A3 = 2.0*r/1000*(l_1*(I-20)+(l_2-l_1)*(I-200)+(l_3-l_2)*(I-352)) // Potential drop over length A_A3(V)
+V_A3 = V_A-A_A3 // Voltage at the lowest run lamp(V)
+
+// Results
+disp("PART II - EXAMPLE : 18.3 : SOLUTION :-")
+printf("\nPosition of lowest-run lamp, A_3 = %.f m", l_3)
+printf("\nVoltage at the lowest-run lamp = %.1f V", V_A3)
diff --git a/3472/CH25/EX25.4/Example25_4.sce b/3472/CH25/EX25.4/Example25_4.sce new file mode 100644 index 000000000..59b45f361 --- /dev/null +++ b/3472/CH25/EX25.4/Example25_4.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 II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.4 :
+// Page number 439
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l = 450.0 // Length of wire(m)
+V_A = 250.0 // Voltage at end A(V)
+V_B = 250.0 // Voltage at end A(V)
+r = 0.05 // Conductor resistance(ohm/km)
+i = 1.5 // Load(A/m)
+I_C = 20.0 // Current at C(A)
+l_C = 60.0 // Distance to C from A(m)
+I_D = 40.0 // Current at D(A)
+l_D = 100.0 // Distance to D from A(m)
+l_E = 200.0 // Distance to E from A(m)
+
+// Calculations
+x = poly(0,"x") // Current to point D from end A(A)
+AD = (I_C+x)*r*l_C+x*r*(l_D-l_C) // Drop in length AD
+BD = (i*r*V_A**2/2)+(I_D-x)*r*(450-l_D) // Drop in length BD
+x_sol = roots(AD-BD) // Current(A)
+I_F = x_sol-I_D // Current supplied to load from end A(A)
+l_F = l_E+(I_F/i) // Point of minimum potential at F from A(m)
+V_F = V_B-(375.0-I_F)*(250-(l_F-200))*r/1000 // Potential at F from end B(V)
+
+// Results
+disp("PART II - EXAMPLE : 18.4 : SOLUTION :-")
+printf("\nPoint of minimum potential occurs at F from A = %.2f metres", l_F)
+printf("\nPotential at point F = %.2f V", V_F)
diff --git a/3472/CH25/EX25.6/Example25_6.sce b/3472/CH25/EX25.6/Example25_6.sce new file mode 100644 index 000000000..e8d6e7dc7 --- /dev/null +++ b/3472/CH25/EX25.6/Example25_6.sce @@ -0,0 +1,40 @@ +// 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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.6 :
+// Page number 440-441
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l_AB = 100.0 // Length between A & B(m)
+l_BC = 150.0 // Length between B & C(m)
+l_CD = 200.0 // Length between C & D(m)
+l_AD = 350.0 // Length between A & D(m)
+l_AE = 200.0 // Length between A & E(m)
+l_ED = 250.0 // Length between E & D(m)
+I_B = 10.0 // Current at B(A)
+I_C = 20.0 // Current at C(A)
+I_D = 50.0 // Current at D(A)
+I_E = 39.0 // Current at E(A)
+
+// Calculations
+x = poly(0,"x") // Current in section AB(A)
+ABCDEA = x*l_AB+(x-I_B)*l_BC+(x-I_B-I_C)*l_CD+(x-I_B-I_C-I_D)*l_ED+(x-I_B-I_C-I_D-I_E)*l_AE // KVL around loop ABCDEA
+x_sol = roots(ABCDEA) // Current in section AB(A)
+V_AD = x_sol*l_AB+(x_sol-I_B)*l_BC+(x_sol-I_B-I_C)*l_CD // Voltage drop from A to D in terms of ρ/a_1(V)
+R_AD = (l_AB+l_BC+l_CD)*(l_AE+l_ED)/(l_AB+l_BC+l_CD+l_AE+l_ED) // Resistance of n/w across terminals AD in terms of ρ/a
+I_AD = V_AD/(R_AD+l_AD) // Current in interconnector AD(A)
+V_A_D = I_AD*l_AD // Voltage drop between A & D in terms of ρ/a_2
+a2_a1 = V_A_D/V_AD
+length_with = (l_AB+l_BC+l_CD+l_AE+l_ED+l_AD) // Length of conductor with interconnector(m)
+length_without = (l_AB+l_BC+l_CD+l_AE+l_ED) // Length of conductor without interconnector(m)
+volume_with = a2_a1*length_with/length_without // Weight of copper with interconnector
+
+// Results
+disp("PART II - EXAMPLE : 18.6 : SOLUTION :-")
+printf("\nRatio of weight of copper with & without interconnector = %.3f : 1 (or) 1 : %.2f", volume_with,1/volume_with)
diff --git a/3472/CH25/EX25.7/Example25_7.sce b/3472/CH25/EX25.7/Example25_7.sce new file mode 100644 index 000000000..4924ab19e --- /dev/null +++ b/3472/CH25/EX25.7/Example25_7.sce @@ -0,0 +1,62 @@ +// 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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.7 :
+// Page number 441-442
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+r_out = 0.05 // Resistance of each outer per 100 metre length(ohm)
+r_neutral = 0.10 // Resistance of each neutral per 100 metre length(ohm)
+V_A = 200.0 // Potential at point A(V)
+V_B = 200.0 // Potential at point B(V)
+l_AC = 100.0 // Length between A & C(m)
+l_CD = 150.0 // Length between C & D(m)
+l_DB = 200.0 // Length between D & B(m)
+l_AF = 200.0 // Length between A & F(m)
+l_FE = 100.0 // Length between F & E(m)
+l_EB = 150.0 // Length between E & B(m)
+I_C = 20.0 // Current at point C(A)
+I_D = 30.0 // Current at point D(A)
+I_F = 60.0 // Current at point F(A)
+I_E = 40.0 // Current at point E(A)
+
+// Calculations
+x = poly(0,"x") // Current in positive outer alone(A)
+equ_1 = r_out*(l_DB*(I_D-x))-r_out*(l_AC*(I_C+x)+l_CD*x)
+x_sol = roots(equ_1) // Current in positive outer alone(A)
+y = poly(0,"y") // Current in negative outer alone(A)
+equ_2 = r_out*((I_E-y)*l_FE+(I_E+I_F-y)*l_AF)-r_out*(l_EB*y)
+y_sol = roots(equ_2) // Current in negative outer alone(A)
+I_pos_out = I_C+x_sol // Current entering positive outer(A)
+I_neg_out = I_E+I_F-y_sol // Current returning via negative outer(A)
+I_middle = I_neg_out-I_pos_out // Current in the middle wire towards G(A)
+r_CD = r_out*l_CD/100.0 // Resistance between C & D(ohm)
+r_D = r_out*l_DB/100.0 // Resistance between D & B(ohm)
+r_IH = r_neutral*l_FE*0.5/100.0 // Resistance between I & H(ohm)
+r_IJ = r_neutral*l_FE*0.5/100.0 // Resistance between I & J(ohm)
+r_GH = r_neutral*l_AF*0.5/100.0 // Resistance between G & H(ohm)
+r_AF = r_out*l_AF/100.0 // Resistance between A & F(ohm)
+I_CD = x_sol // Current flowing into D from C(A)
+I_out_D = I_D-x_sol // Current flowing into D from outer side(A)
+I_GH = I_C+I_middle // Current flowing into H from G(A)
+I_IH = I_F-I_GH // Current flowing into H from I(A)
+I_BJ = I_E-(I_D-I_IH) // Current flowing into J from B(A)
+I_FE = y_sol-I_E // Current flowing into E from F(A)
+I_IJ = I_D-I_IH // Current flowing into J from I(A)
+V_C = V_A-(I_pos_out*r_out-I_middle*r_neutral) // Potential at load point C(A)
+V_D = V_C-(I_CD*r_CD+I_IH*r_IH-I_GH*r_GH) // Potential at load point D(A)
+V_F = V_A-(I_middle*r_neutral+I_GH*r_neutral+I_neg_out*r_AF) // Potential at load point F(A)
+V_E = V_F-(-I_IH*r_IH+I_IJ*r_IJ-I_FE*r_out) // Potential at load point E(A)
+
+// Results
+disp("PART II - EXAMPLE : 18.7 : SOLUTION :-")
+printf("\nPotential difference at load point C = %.3f V", V_C)
+printf("\nPotential difference at load point D = %.3f V", V_D)
+printf("\nPotential difference at load point E = %.3f V", V_E)
+printf("\nPotential difference at load point F = %.3f V", V_F)
diff --git a/3472/CH25/EX25.8/Example25_8.sce b/3472/CH25/EX25.8/Example25_8.sce new file mode 100644 index 000000000..5a9061b52 --- /dev/null +++ b/3472/CH25/EX25.8/Example25_8.sce @@ -0,0 +1,36 @@ +// 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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.8 :
+// Page number 442-443
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 440.0 // Voltage between outer(V)
+I_pos = 210.0 // Ligting load current on positive side(A)
+I_neg = 337.0 // Ligting load current on negative side(A)
+I_power = 400.0 // Power load current(A)
+P_loss = 1.5 // Loss in each balancer machine(kW)
+
+// Calculations
+P = I_power*V/1000.0 // Power(kW)
+load_pos = I_pos*V*0.5/1000.0 // Load on positive side(kW)
+load_neg = I_neg*V*0.5/1000.0 // Load on negative side(kW)
+loss_total = 2*P_loss // Total loss on rotary balancer set(kW)
+load_main = P+load_pos+load_neg+loss_total // Load on main machine(kW)
+I = load_main*1000/V // Current(A)
+I_M = I-610.0 // Current through balancer machine(A)
+I_G = 127.0-I_M // Current through generator(A)
+output_G = I_G*V*0.5/1000.0 // Output of generator(kW)
+input_M = I_M*V*0.5/1000.0 // Input to balancer machine(kW)
+
+// Results
+disp("PART II - EXAMPLE : 18.8 : SOLUTION :-")
+printf("\nLoad on the main machine = %.2f kW", load_main)
+printf("\nOutput of generator = %.2f kW", output_G)
+printf("\nInput to balancer machine = %.2f kW", input_M)
diff --git a/3472/CH25/EX25.9/Example25_9.sce b/3472/CH25/EX25.9/Example25_9.sce new file mode 100644 index 000000000..1ec686ae2 --- /dev/null +++ b/3472/CH25/EX25.9/Example25_9.sce @@ -0,0 +1,57 @@ +// 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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.9 :
+// Page number 444
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V_a = 11.0*10**3 // Line voltage at A(V)
+Z_AB = complex(1.0,0.8) // Impedance between A & B(ohm)
+Z_AC = complex(3.0,2.0) // Impedance between A & C(ohm)
+Z_BD = complex(3.0,4.0) // Impedance between B & D(ohm)
+Z_CD = complex(1.0,0.7) // Impedance between C & D(ohm)
+I_B = 60.0 // Current at B(A)
+I_C = 30.0 // Current at C(A)
+I_D = 50.0 // Current at D(A)
+pf_B = 0.8 // Power factor at B
+pf_C = 0.9 // Power factor at C
+pf_D = 0.707 // Power factor at D
+
+// Calculations
+sin_phi_B = (1-pf_B**2)**0.5
+I_B1 = I_B*(pf_B-%i*sin_phi_B) // Load current(A)
+sin_phi_C = (1-pf_C**2)**0.5
+I_C1 = I_C*(pf_C-%i*sin_phi_C) // Load current(A)
+sin_phi_D = (1-pf_D**2)**0.5
+I_D1 = I_D*(pf_D-%i*sin_phi_D) // Load current(A)
+V_A = V_a/3**0.5 // Phase voltage at A(V)
+I_AC = I_C1 // Current in section AC when C & D is removed(A)
+I_BD = I_D1 // Current in section BD when C & D is removed(A)
+I_AB = I_B1+I_D1 // Current in section AB when C & D is removed(A)
+V_AC_drop = I_AC*Z_AC // Voltage drop at section AC(V)
+V_AB_drop = I_AB*Z_AB // Voltage drop at section AB(V)
+V_BD_drop = I_BD*Z_BD // Voltage drop at section BD(V)
+V_drop_D = V_BD_drop+V_AB_drop // Total drop upto D(V)
+pd_CD = V_drop_D-V_AC_drop // Potential difference between C & D(V)
+Z_C_D = Z_AB+Z_BD+Z_AC // Impedance of network looking from terminal C & D(ohm)
+I_CD = pd_CD/(Z_C_D+Z_CD) // Current flowing in section CD(A)
+I_AC = I_CD+I_C1 // Current flowing in section AC(A)
+I_BD = I_D1-I_CD // Current flowing in section BD(A)
+I_AB = I_BD+I_B1 // Current flowing in section AB(A)
+V_drop_AC = I_AC*Z_AC // Drop caused by current flowing in section AC(V/phase)
+V_drop_AC_line = V_drop_AC*3**0.5 // Drop caused by current flowing in section AC(V)
+V_C = V_a-V_drop_AC_line // Voltage at C(V)
+
+// Results
+disp("PART II - EXAMPLE : 18.9 : SOLUTION :-")
+printf("\nCurrent in section CD, I_CD = (%.2f%.2fj) A", real(I_CD),imag(I_CD))
+printf("\nCurrent in section AC, I_AC = (%.2f%.2fj) A", real(I_AC),imag(I_AC))
+printf("\nCurrent in section BD, I_BD = (%.2f%.2fj) A", real(I_BD),imag(I_BD))
+printf("\nCurrent in section AB, I_AB = (%.2f%.2fj) A", real(I_AB),imag(I_AB))
+printf("\nVoltage at load point C = %.2f∠%.2f° kV", abs(V_C)/1000,phasemag(V_C))
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