diff options
Diffstat (limited to '3472/CH18')
| -rw-r--r-- | 3472/CH18/EX18.1/Example18_1.sce | 39 | ||||
| -rw-r--r-- | 3472/CH18/EX18.10/Example18_10.sce | 32 | ||||
| -rw-r--r-- | 3472/CH18/EX18.11/Example18_11.sce | 35 | ||||
| -rw-r--r-- | 3472/CH18/EX18.12/Example18_12.sce | 36 | ||||
| -rw-r--r-- | 3472/CH18/EX18.13/Example18_13.sce | 55 | ||||
| -rw-r--r-- | 3472/CH18/EX18.2/Example18_2.sce | 56 | ||||
| -rw-r--r-- | 3472/CH18/EX18.3/Example18_3.sce | 38 | ||||
| -rw-r--r-- | 3472/CH18/EX18.4/Example18_4.sce | 39 | ||||
| -rw-r--r-- | 3472/CH18/EX18.5/Example18_5.sce | 27 | ||||
| -rw-r--r-- | 3472/CH18/EX18.6/Example18_6.sce | 37 | ||||
| -rw-r--r-- | 3472/CH18/EX18.7/Example18_7.sce | 41 | ||||
| -rw-r--r-- | 3472/CH18/EX18.8/Example18_8.sce | 42 | ||||
| -rw-r--r-- | 3472/CH18/EX18.9/Example18_9.sce | 44 |
13 files changed, 521 insertions, 0 deletions
diff --git a/3472/CH18/EX18.1/Example18_1.sce b/3472/CH18/EX18.1/Example18_1.sce new file mode 100644 index 000000000..12a6b7a59 --- /dev/null +++ b/3472/CH18/EX18.1/Example18_1.sce @@ -0,0 +1,39 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.1 :
+// Page number 330
+clear ; clc ; close ; // Clear the work space and console
+funcprot(0)
+
+// Given data
+rating = 1000.0 // Rating of alternator(kW)
+load = 1600.0 // Total load(kW)
+X_fl = 100.0 // Full load speed regulation of alernator X(%)
+Y_fl = 104.0 // Full load speed regulation of alernator Y(%)
+X_nl = 100.0 // No load speed regulation of alernator X(%)
+Y_nl = 105.0 // No load speed regulation of alernator Y(%)
+
+// Calculations
+h = poly(0,"h")
+PB = (Y_nl-X_nl)-h
+PR = rating/(Y_nl-X_nl)*PB // Load shared by machine X(kW) in terms of h
+QQ = (Y_fl-X_fl)-h
+RQ = rating/(Y_fl-X_fl)*QQ // Load shared by machine Y(kW) in terms of h
+h_1 = roots(PR+RQ-load)
+PB_1 = (Y_nl-X_nl)-h_1
+PR_1 = rating/(Y_nl-X_nl)*PB_1 // Load shared by machine X(kW)
+QQ_1 = (Y_fl-X_fl)-h_1
+RQ_1 = rating/(Y_fl-X_fl)*QQ_1 // Load shared by machine Y(kW)
+load_cease = rating/(Y_nl-X_nl) // Y cease supply load(kW)
+
+// Results
+disp("PART II - EXAMPLE : 11.1 : SOLUTION :-")
+printf("\nLoad shared by machine X, PR = %.f kW", PR_1)
+printf("\nLoad shared by machine Y, RQ = %.f kW", RQ_1)
+printf("\nLoad at which machine Y ceases to supply any portion of load = %.f kW", load_cease)
diff --git a/3472/CH18/EX18.10/Example18_10.sce b/3472/CH18/EX18.10/Example18_10.sce new file mode 100644 index 000000000..b3a9645ec --- /dev/null +++ b/3472/CH18/EX18.10/Example18_10.sce @@ -0,0 +1,32 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.10 :
+// Page number 336
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+load_1 = 10000.0 // Total balanced load(kW)
+V = 33000.0 // Voltage(V)
+PF_1 = 0.8 // Lagging power factor
+R = 1.6 // Resistance of feeder(ohm/phase)
+X = 2.5 // Reactance of feeder(ohm/phase)
+load_2 = 4460.0 // Load delivered by feeder(kW)
+PF_2 = 0.72 // Lagging power factor
+
+// Calculations
+I = load_1*1000/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1)) // Total line current(A)
+I_1 = load_2*1000/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Line current of first feeder(A)
+I_2 = I-I_1 // Line current of first feeder(A)
+Z_1 = complex(R,X) // Impedance of first feeder(ohm)
+Z_2 = I_1*Z_1/I_2 // Impedance of second feeder(ohm)
+
+// Results
+disp("PART II - EXAMPLE : 11.10 : SOLUTION :-")
+printf("\nImpedance of second feeder, Z_2 = %.2f∠%.1f° ohm \n", abs(Z_2),phasemag(Z_2))
+printf("\nNOTE: ERROR: Changes in the obtained answer from that of textbook is due to wrong values of substitution")
diff --git a/3472/CH18/EX18.11/Example18_11.sce b/3472/CH18/EX18.11/Example18_11.sce new file mode 100644 index 000000000..ba9b941c5 --- /dev/null +++ b/3472/CH18/EX18.11/Example18_11.sce @@ -0,0 +1,35 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.11 :
+// Page number 337
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+P = 9.0 // Load supplied from substation(MW)
+V = 33000.0 // Voltage(V)
+PF_1 = 1.0 // Unity power factor
+Z_A = complex(2.0,8.0) // Impedance of circuit A(ohm)
+Z_B = complex(4.0,4.0) // Impedance of circuit B(ohm)
+
+// Calculations
+V_ph = V/3**0.5 // Voltage at receiving end per phase(V)
+P_A = 1.0/3*P // Power supplied by line A(MW)
+P_B = 2.0/3*P // Power supplied by line B(MW)
+I_A = P_A*10**6/(3**0.5*V) // Current through line A(A)
+I_B = P_B*10**6/(3**0.5*V) // Current through line B(A)
+IA_ZA_drop = I_A*Z_A // I_A Z_A drop(V/phase)
+IB_ZB_drop = I_B*Z_B // I_B Z_B drop(V/phase)
+phase_boost = real(IB_ZB_drop)-real(IA_ZA_drop) // Voltage in phase boost(V/phase)
+quad_boost = imag(IB_ZB_drop)-imag(IA_ZA_drop) // Voltage in quadrature boost(V/phase)
+constant_P = V_ph+IA_ZA_drop // Assumed that sending end voltage at P is kept constant(V/phase)
+
+// Results
+disp("PART II - EXAMPLE : 11.11 : SOLUTION :-")
+printf("\nVoltage in-phase boost = %.2f V/phase", phase_boost)
+printf("\nVoltage in quadrature boost = %.f V/phase", quad_boost)
diff --git a/3472/CH18/EX18.12/Example18_12.sce b/3472/CH18/EX18.12/Example18_12.sce new file mode 100644 index 000000000..6e1d77e2c --- /dev/null +++ b/3472/CH18/EX18.12/Example18_12.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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.12 :
+// Page number 337
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+cap_A = 15000.0 // Capacity of station A(kW)
+cap_B = 10000.0 // Capacity of station B(kW)
+cap_C = 2000.0 // Capacity of station C(kW)
+speed_reg_A = 2.4/100 // Speed regulation of A
+speed_reg_B = 3.2/100 // Speed regulation of B
+slip_C = 4.5/100 // Full load slip
+local_load_B_a = 10000.0 // Local load on station B(kW)
+local_load_A_a = 0 // Local load on station A(kW)
+local_load_both = 10000.0 // Local load on both station(kW)
+
+// Calculations
+// Case(a)
+speed_A = speed_reg_A/cap_A // % of speed drop for A
+speed_C = slip_C/cap_C // % of speed drop for C
+speed_B = speed_reg_B/cap_B // % of speed drop for B
+X = local_load_B_a*speed_B/(speed_A+speed_B+speed_C) // Load on C when local load of B is 10000 kW and A has no load(kW)
+// Case(b)
+Y = local_load_both*(speed_B-speed_A)/(speed_A+speed_B+speed_C) // Load on C when both station have local loads of 10000 kW(kW)
+
+// Results
+disp("PART II - EXAMPLE : 11.12 : SOLUTION :-")
+printf("\nCase(a): Load on C when local load of B is 10000 kW and A has no load, X = %.f kW", X)
+printf("\nCase(b): Load on C when both station have local loads of 10000 kW, Y = %.f kW", Y)
diff --git a/3472/CH18/EX18.13/Example18_13.sce b/3472/CH18/EX18.13/Example18_13.sce new file mode 100644 index 000000000..b53c5fc0d --- /dev/null +++ b/3472/CH18/EX18.13/Example18_13.sce @@ -0,0 +1,55 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.13 :
+// Page number 337-338
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l = 20.0 // Length of cable(km)
+r = 0.248 // Resistance(ohm/km)
+x = 0.50*10**-3 // Inductance(H/m)
+V_gen = 6600.0 // Generation voltage(V)
+f = 50.0 // Frequency(Hz)
+V = 33000.0 // Transmission voltage(V)
+rating = 10.0 // Transformer rating(MVA)
+loss_cu = 100.0 // Copper loss at full load(kW)
+x_tr = 2.5/100 // Transformer reactance
+load = 7.5 // Load to be transmitted(MW)
+PF = 0.71 // Lagging power factor
+
+// Calculations
+R = l*r // Resistance of the cable(ohm)
+I_fl = rating*10**6/(3**0.5*V) // Transformer current at full load(A)
+R_eq = loss_cu*1000/(3*I_fl**2) // Equivalent resistance per phase of transformer(ohm)
+R_total_hv = R+2.0*R_eq // Total resistance per conductor in terms of hv side(ohm)
+X = 2.0*%pi*f*l*x // Reactance of cable per conductor(ohm)
+per_X_tr = V/3**0.5*x_tr/I_fl // % reactance of transformer(ohm)
+X_total_hv = X+2.0*per_X_tr // Total reactance per conductor in terms of hv side(ohm)
+I = load*10**6/(3**0.5*V*PF) // Line current at receiving end(A)
+IR = I*R_total_hv // IR drop(V)
+IX = I*X_total_hv // IX drop(V)
+E_r = V/3**0.5 // Phase voltage at station B(V)
+cos_phi_r = PF
+sin_phi_r = (1-PF**2)**0.5
+E_s = ((E_r*cos_phi_r+IR)**2+(E_r*sin_phi_r+IX)**2)**0.5/1000 // Sending end voltage(kV)
+E_s_ll = 3**0.5*E_s // Sending end line voltage(kV)
+V_booster = 3**0.5*(E_s-E_r/1000) // Booster voltage between lines(kV)
+tan_phi_s = (E_r*sin_phi_r+IX)/(E_r*cos_phi_r+IR) // tanΦ_s
+phi_s = atand(tan_phi_s) // Φ_s(°)
+cos_phi_s = cosd(phi_s) // cosΦ_s
+P_s = 3.0*E_s*I*cos_phi_s // Power at sending end(kW)
+loss = P_s-load*1000 // Loss(kW)
+loss_per = loss/(load*1000)*100 // loss percentage
+
+// Results
+disp("PART II - EXAMPLE : 11.13 : SOLUTION :-")
+printf("\nLoss in the interconnector as a percentage of power received = %.3f percent", loss_per)
+printf("\nRequired voltage of the booster = %.3f kV (in terms of H.V) \n", V_booster)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
+printf("\n kVA rating of booster is not calculated in textbook and here")
diff --git a/3472/CH18/EX18.2/Example18_2.sce b/3472/CH18/EX18.2/Example18_2.sce new file mode 100644 index 000000000..3dc2da888 --- /dev/null +++ b/3472/CH18/EX18.2/Example18_2.sce @@ -0,0 +1,56 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.2 :
+// Page number 330-331
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA = 5000.0 // Rating of alternator(kVA)
+N = 1500.0 // Speed(rpm)
+V = 6600.0 // Voltage(V)
+f = 50.0 // Frequency(Hz)
+PF = 0.8 // Lagging power factor
+x = 0.15 // Short circuit reactance
+
+// Calculations
+E = V/3**0.5 // Phase voltage(V)
+I = kVA*1000/(3**0.5*V) // Full load current of alternator(A)
+V_drop = E*x // Synchronous reactance drop(V)
+X = V_drop/I // Synchronous reactance per phase(ohm)
+P = 120*f/N // Number of poles
+n = N/60 // Speed(rps)
+phi = acosd(PF) // Φ(°)
+// Case(a)
+theta_a = 2.0 // For a 4 pole m/c. 1 mech degree = 2 elect degree
+E_s_a = E*sind(theta_a) // Synchronizing voltage(V)
+I_s_a = E_s_a/X // Synchronizing current(A)
+P_s_a = E*I_s_a // Synchronizing power per phase(W)
+P_s_a_total = 3.0*P_s_a // Total synchronizing power(W)
+P_s_a_total_kw = P_s_a_total/1000.0 // Total synchronizing power(kW)
+T_s_a = P_s_a_total/(2*%pi*n) // Synchronizing torque(N-m)
+// Case(b)
+sin_phi = sind(phi)
+OB = ((E*PF)**2+(E*sin_phi+V_drop)**2)**0.5 // Voltage(V)
+E_b = OB // Voltage(V)
+alpha_phi = atand((E*sin_phi+V_drop)/(E*PF)) // α+Φ(°)
+alpha = alpha_phi-phi // α(°)
+E_s_b = 2.0*E_b*sind(2.0/2) // Synchronizing voltage(V)
+I_s_b = E_s_b/X // Synchronizing current(A)
+P_s_b = E*I_s_b*cosd((alpha+1.0)) // Synchronizing power per phase(W)
+P_s_b_total = 3.0*P_s_b // Total synchronizing power(W)
+P_s_b_total_kw = P_s_b_total/1000.0 // Total synchronizing power(kW)
+T_s_b = P_s_b_total/(2*%pi*n) // Synchronizing torque(N-m)
+
+// Results
+disp("PART II - EXAMPLE : 11.2 : SOLUTION :-")
+printf("\nCase(a): Synchronizing power for no-load, P_s = %.1f kW", P_s_a_total_kw)
+printf("\n Synchronizing torque for no-load, T_s = %.f N-m", T_s_a)
+printf("\nCase(b): Synchronizing power at full-load, P_s = %.1f kW", P_s_b_total_kw)
+printf("\n Synchronizing torque at full-load, T_s = %.f N-m \n", T_s_b)
+printf("\nNOTE: ERROR: Calculation mistakes in textbook")
diff --git a/3472/CH18/EX18.3/Example18_3.sce b/3472/CH18/EX18.3/Example18_3.sce new file mode 100644 index 000000000..31876c981 --- /dev/null +++ b/3472/CH18/EX18.3/Example18_3.sce @@ -0,0 +1,38 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.3 :
+// Page number 331-332
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 6600.0 // Voltage(V)
+R = 0.045 // Resistance(ohm)
+X = 0.45 // Reactance(ohm)
+Load = 10000.0*10**3 // Total load(W)
+PF = 0.8 // Lagging power factor
+I_a = 437.5 // Armature current(A)
+
+// Calculations
+I = Load/(3**0.5*V*PF) // Load current(A)
+I_working = PF*I // Working component of current(A)
+I_watless = (1-PF**2)**0.5*I // Watless component of current(A)
+I_second = (I_a**2+I_watless**2)**0.5 // Load current supplied by second alternator(A)
+PF_second = I_a/I_second // Lagging power factor of second alternator
+V_ph = V/3**0.5 // Terminal voltage per phase(V)
+I_R = I_second*R // Voltage drop due to resistance(V)
+I_X = I_second*X // Voltage drop due to reactance(V)
+sin_phi_second = (1-PF_second**2)**0.5
+E = ((V_ph+I_R*PF_second+I_X*sin_phi_second)**2+(I_X*PF_second-I_R*sin_phi_second)**2)**0.5 // EMF of the alternator(V/phase)
+E_ll = 3**0.5*E // Line-to-line EMF of the alternator(V)
+
+// Results
+disp("PART II - EXAMPLE : 11.3 : SOLUTION :-")
+printf("\nArmature current of other alternator = %.1f A", I_second)
+printf("\ne.m.f of other alternator = %.f V (line-to-line)", E_ll)
+printf("\nPower factor of other alternator = %.3f (lagging)", PF_second)
diff --git a/3472/CH18/EX18.4/Example18_4.sce b/3472/CH18/EX18.4/Example18_4.sce new file mode 100644 index 000000000..2b03931f3 --- /dev/null +++ b/3472/CH18/EX18.4/Example18_4.sce @@ -0,0 +1,39 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.4 :
+// Page number 332-333
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+X = 10.0 // Reactance(ohm)
+I_a = 220.0 // Armature current(A)
+PF = 1.0 // Unity power factor
+V = 11000.0 // Phase voltage(V)
+emf_raised = 0.2 // EMF rasied by 20%
+
+// Calculations
+I_X = I_a*X // Reactance drop(V)
+E_0 = (V**2+I_X**2)**0.5 // EMF(V)
+E_00 = (1+emf_raised)*E_0 // New value of induced emf(V)
+U = ((E_00**2-I_X**2)**0.5-V)/X // Current(A)
+I_1 = (I_a**2+U**2)**0.5 // Current(A)
+PF_1 = I_a/I_1 // Lagging power factor
+I_X_2 = (E_00**2+V**2)**0.5 // Reactance drop(V)
+I_2 = I_X_2/X // Current corresponding to this drop(A)
+PF_2 = E_00/I_X_2 // Leading power factor
+P_max = V*I_2*PF_2/1000 // Maximum power output(kW)
+
+// Results
+disp("PART II - EXAMPLE : 11.4 : SOLUTION :-")
+printf("\nNew value of machine current = %.1f A", I_1)
+printf("\nNew vaue of power factor, p.f = %.4f (lagging)", PF_1)
+printf("\nPower output at which alternator break from synchronism = %.f kW", P_max)
+printf("\nCurrent corresponding to maximum load = %.f A", I_2)
+printf("\nPower factor corresponding to maximum load = %.4f (leading) \n", PF_2)
+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
diff --git a/3472/CH18/EX18.5/Example18_5.sce b/3472/CH18/EX18.5/Example18_5.sce new file mode 100644 index 000000000..c99c94415 --- /dev/null +++ b/3472/CH18/EX18.5/Example18_5.sce @@ -0,0 +1,27 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.5 :
+// Page number 333
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 10000.0 // Voltage(V)
+rating = 10000.0 // Full load rating(kW)
+V_drop_per = 0.2 // Voltage drop of 20% for 10000 kW
+
+// Calculations
+V_drop = V_drop_per*rating // Voltage drop(V)
+sin_theta_2 = (V_drop/2)/V // Sin(θ/2)
+theta_2 = asind(sin_theta_2) // θ/2(°)
+theta = 2.0*theta_2 // Phase angle between busbar sections, θ(°)
+
+// Results
+disp("PART II - EXAMPLE : 11.5 : SOLUTION :-")
+printf("\nPhase angle between busbar sections, θ = %.2f° \n", theta)
+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
diff --git a/3472/CH18/EX18.6/Example18_6.sce b/3472/CH18/EX18.6/Example18_6.sce new file mode 100644 index 000000000..fab1f3b8d --- /dev/null +++ b/3472/CH18/EX18.6/Example18_6.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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.6 :
+// Page number 334
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+load_1 = 20000.0 // Total load(kW)
+V = 11000.0 // Voltage(V)
+PF_1 = 1.0 // Unity power factor
+load_2 = 8000.0 // Load supplied(kW)
+PF_2 = 0.8 // Lagging power factor
+R = 0.5 // Resistance(ohm/phase)
+X = 0.8 // Reactance(ohm/phase)
+
+// Calculations
+I_1 = load_1*1000/(3**0.5*V*PF_1) // Load current(A)
+I_2 = load_2*1000/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Current supplied by local generators(A)
+I_3 = I_1-I_2 // Current through interconnector(A)
+angle_I_3 = phasemag(I_3) // Current through interconnector leads reference phasor by angle(°)
+V_drop = (R+%i*X)*I_3 // Voltage drop across interconnector(V)
+V_ph = V/3**0.5 // Phase voltage(V)
+V_S = V_ph+V_drop // Sending end voltage(V/phase)
+V_S_ll = 3**0.5*V_S // Sending end voltage(V)
+angle_V_S_ll = phasemag(V_S_ll) // Angle of sending end voltage(°)
+PF_S = cosd(angle_I_3-angle_V_S_ll) // Power factor at sending station
+
+// Results
+disp("PART II - EXAMPLE : 11.6 : SOLUTION :-")
+printf("\nVoltage at this latter station = %.f∠%.2f° V (line-to-line)", abs(V_S_ll),angle_V_S_ll)
+printf("\nPower factor at this latter station = %.4f (leading)", PF_S)
diff --git a/3472/CH18/EX18.7/Example18_7.sce b/3472/CH18/EX18.7/Example18_7.sce new file mode 100644 index 000000000..43f7fe3dc --- /dev/null +++ b/3472/CH18/EX18.7/Example18_7.sce @@ -0,0 +1,41 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.7 :
+// Page number 334
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 33000.0 // Voltage(V)
+R = 0.7 // Resistance(ohm/phase)
+X = 3.5 // Reactance(ohm/phase)
+load_1 = 60.0 // Load on generator at station X(MW)
+PF_1 = 0.8 // Lagging power factor
+load_2 = 40.0 // Local load taken by consumer(MW)
+PF_2 = 0.707 // Lagging power factor
+
+// Calculations
+V_ph = V/3**0.5 // Phase voltage(V)
+I_1 = load_1*10**6/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1)) // Load current on generator at X(A)
+I_2 = load_2*10**6/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Current due to local load(A)
+I_3 = I_1-I_2 // Current through interconnector(A)
+angle_I_3 = phasemag(I_3) // Current through interconnector leads reference phasor by angle(°)
+V_drop = (R+%i*X)*I_3 // Voltage drop across interconnector(V)
+V_Y = V_ph-V_drop // Voltage at Y(V)
+angle_V_Y = phasemag(V_Y) // Angle of voltage at Y(°)
+phase_diff = angle_I_3-angle_V_Y // Phase difference b/w Y_Y and I_3(°)
+PF_Y = cosd(phase_diff) // Power factor of current received by Y
+P_Y = 3*abs(V_Y*I_3)*PF_Y/1000.0 // Power received by station Y(kW)
+phase_XY = abs(angle_V_Y) // Phase angle b/w voltages of X & Y(°)
+
+// Results
+disp("PART II - EXAMPLE : 11.7 : SOLUTION :-")
+printf("\nLoad received from station X to station Y = %.f kW", P_Y)
+printf("\nPower factor of load received by Y = %.4f (lagging)", PF_Y)
+printf("\nPhase difference between voltage of X & Y = %.2f° (lagging) \n", phase_XY)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH18/EX18.8/Example18_8.sce b/3472/CH18/EX18.8/Example18_8.sce new file mode 100644 index 000000000..275567475 --- /dev/null +++ b/3472/CH18/EX18.8/Example18_8.sce @@ -0,0 +1,42 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.8 :
+// Page number 335
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V_tie = 11000.0 // Tie line Voltage(V)
+Z = (3.5+%i*7.0) // Impedance of tie line(ohm/conductor)
+V = 6600.0 // Bus bar voltage(V)
+Z_per = (2.5+%i*7.5) // Percentage impedance on 1000kVA rating
+kVA = 2500.0 // Load receieved by other(kVA)
+
+// Calculations
+V_ph = V/3**0.5 // Phase voltage(V)
+I_fl_LV = 100.0*V_tie/V_ph // LV side Full load current of each transformer(A)
+R_eq = V_ph*real(Z_per)/(100*I_fl_LV) // Equivalent resistance of transformer(ohm/phase)
+X_eq = 3.0*R_eq // Equivalent reactance of transformer(ohm/phase)
+R_phase = real(Z)*(V/V_tie)**2 // Resistance of line per phase(ohm)
+X_phase = imag(Z)*(V/V_tie)**2 // Resistance of line per phase(ohm)
+R_total = 2.0*R_eq+R_phase // Total resistance per phase(ohm)
+X_total = 2.0*X_eq+X_phase // Total resistance per phase(ohm)
+Z_total = R_total+%i*X_total // Total impedance(ohm/phase)
+I = kVA*1000/(3**0.5*V) // Load current(A)
+V_drop = I*Z_total // Voltage drop per phase(V)
+V_A = V_ph
+V_AA = V_A+V_drop // Sending end voltage per phase(V)
+V_increase = abs(V_AA)-V_A // Increase in voltage required(V/phase)
+percentage_increase = V_increase/V_A*100 // Percentage increase required(%)
+phase_diff = phasemag(V_AA) // Angle at which V_A & V_B are displaced(°)
+
+// Results
+disp("PART II - EXAMPLE : 11.8 : SOLUTION :-")
+printf("\nCase(a): Percentage increase in voltage = %.2f percent", percentage_increase)
+printf("\nCase(b): Phase angle difference between the two busbar voltages = %.2f° \n", phase_diff)
+printf("\nNOTE: ERROR: Several calculation mistakes in the textbook")
diff --git a/3472/CH18/EX18.9/Example18_9.sce b/3472/CH18/EX18.9/Example18_9.sce new file mode 100644 index 000000000..4f8584bb6 --- /dev/null +++ b/3472/CH18/EX18.9/Example18_9.sce @@ -0,0 +1,44 @@ +// 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 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.9 :
+// Page number 335-336
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+X = 2.80 // Combined reactance(ohm/phase)
+load_1 = 7000.0 // Consumer load at station A(kW)
+PF_1 = 0.9 // Lagging power factor
+V = 11000.0 // Voltage(V)
+load_2 = 10000.0 // Load supplied by station B(kW)
+PF_2 = 0.75 // Lagging power factor
+
+// Calculations
+V_ph = V/3**0.5 // Phase voltage(V)
+I_1 = load_1*10**3/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1)) // Current at A due to local load(A)
+I_2 = load_2*10**3/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Current at B due to local load(A)
+IA_X = 0.5*(load_1+load_2)*1000/(3**0.5*V) // Current(A)
+Y_1 = 220.443/V_ph // Solved manually referring textbook
+X_1 = (1-Y_1**2)**0.5
+angle_1 = atand(Y_1/X_1) // Phasor lags by an angle(°)
+IA_Y = (6849.09119318-V_ph*X_1)/X // Current(A)
+Y_X = IA_Y/IA_X
+angle_2 = atand(Y_X) // Angle by which I_A lags behind V_A(°)
+PF_A = cosd(angle_2) // Power factor of station A
+angle_3 = acosd(PF_2)+angle_1 // Angle by which I_2 lags V_A(°)
+I_22 = load_2*10**3/(3**0.5*V*PF_2)*exp(%i*-angle_3*%pi/180) // Current(A)
+I = 78.7295821622-%i*(IA_Y-177.942225747) // Current(A)
+I_B = I_22-I // Current(A)
+angle_4 = abs(phasemag(I_B))-angle_1 // Angle by which I_B lags behind V_B(°)
+PF_B = cosd(angle_4) // Power factor of station B
+
+// Results
+disp("PART II - EXAMPLE : 11.9 : SOLUTION :-")
+printf("\nPower factor of station A = %.4f (lagging)", PF_A)
+printf("\nPower factor of station B = %.4f (lagging)", PF_B)
+printf("\nPhase angle between two bus bar voltages = %.f° (V_B lagging V_A)", angle_1)
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