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Diffstat (limited to '1820/CH6')
| -rwxr-xr-x | 1820/CH6/EX6.1/Example6_1.sce | 39 | ||||
| -rwxr-xr-x | 1820/CH6/EX6.10/Example6_10.sce | 64 | ||||
| -rwxr-xr-x | 1820/CH6/EX6.2/Example6_2.sce | 36 | ||||
| -rwxr-xr-x | 1820/CH6/EX6.3/Example6_3.sce | 28 | ||||
| -rwxr-xr-x | 1820/CH6/EX6.4/Example6_4.sce | 49 | ||||
| -rwxr-xr-x | 1820/CH6/EX6.5/Example6_5.sce | 45 | ||||
| -rwxr-xr-x | 1820/CH6/EX6.6/Example6_6.sce | 42 | ||||
| -rwxr-xr-x | 1820/CH6/EX6.7/Example6_7.sce | 32 |
8 files changed, 335 insertions, 0 deletions
diff --git a/1820/CH6/EX6.1/Example6_1.sce b/1820/CH6/EX6.1/Example6_1.sce new file mode 100755 index 000000000..5d557a2a8 --- /dev/null +++ b/1820/CH6/EX6.1/Example6_1.sce @@ -0,0 +1,39 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+K_1 = 2.5 ; // Factor
+K_2 = 1.7 ; // Factor
+
+// CALCULATIONS
+// For case (b)
+I_d = poly(0,'I_d') ; // since P_loss(dc) = P_loss (ac)
+I_L = poly(0,'I_L') ; // i.e 2*I_d^2*R_dc = 3*I_L^2*R_ac
+I_d = sqrt(3/2)*I_L ; // Ignoring skin effects R_dc = R_ac
+I_d1 = 1.225*I_L ; // Refer Equ 6.23
+
+// For case (a)
+V_d = poly(0,'V_d') ; // Defining a ploynomial V_d
+E_p = poly(0,'E_p') ; // since P_dc = P_ac (or) V_d*I_d = 3*E_p*I_L
+V_d = 2.45*E_p ; // Refer Equ 6.25
+
+// For case (c)
+ins_lvl = (K_2*(V_d/2))/(K_1*E_p) ; // Ratio of dc insulation level to ac insulation level
+ins_lvl_1 = (K_2*2.45/2)/K_1 ; // simplifying above equ
+dc_i = poly(0,'dc_i') ; // dc_i = dc insulation level
+ac_i = poly(0,'ac_i') ; // ac_i = ac insulation level
+dc_i = ins_lvl_1 * ac_i ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.1 : SOLUTION :-") ;
+printf("\n (a) Line-to-line dc voltage of V_d in terms of line-to-neutral voltage E_p , V_d = \n") ; disp(V_d) ;
+printf("\n (b) The dc line current I_d in terms of ac line current I_L , I_d = \n"); disp(I_d1) ;
+printf("\n (c) Ratio of dc insulation level to ac insulation level = \n") ; disp(dc_i/ac_i) ;
+printf("\n (or) dc insulation level = \n") ; disp(dc_i) ;
diff --git a/1820/CH6/EX6.10/Example6_10.sce b/1820/CH6/EX6.10/Example6_10.sce new file mode 100755 index 000000000..c53593ff8 --- /dev/null +++ b/1820/CH6/EX6.10/Example6_10.sce @@ -0,0 +1,64 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.10 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_C = 6.2292 ; // commutating reactance when all 3 breakers are closed
+I_db = 1600 ; // dc current base in A
+V_db = 125 * 10^3 ; // dc voltage base in V
+I_d = I_db ; // Max continuous current in A
+V_d = 100 * 10^3 ; // dc voltage in V
+alpha = 0 ; // Firing angle α = 0 degree
+
+// CALCULATIONS
+// For case (a)
+R_c = (3/%pi) * X_C ;
+R_cb = V_db/I_db ; // Resistance base in Ω
+V_d_pu = V_d/V_db ; // per unit voltage
+I_d_pu = I_d/I_db ; // per unit current
+R_c_pu = R_c/R_cb ; // per unit Ω
+E_pu = (V_d_pu + R_c_pu * I_d_pu)/cosd(alpha) ; // Open ckt dc voltage in pu
+V_d0 = E_pu * V_db ; // Open ckt dc voltage in V
+
+// For case (b)
+E = V_d0/2.34; // Open ckt ac voltage on wye side of transformer in V
+
+// For case (c)
+E_1LN = 92.95 * 10^3 ; // voltage in V
+E_1B = E_1LN ;
+E_LN = 53.44 * 10^3 ; // voltage in V
+a = E_1LN/E_LN ;
+n = a ; // when LTC on neutral
+X_c_pu = 2 * R_c_pu ;
+E_1_pu = E_1LN / E_1B ; // per unit voltage
+cos_delta = cosd(alpha) - ( (X_c_pu * I_d_pu)/( (a/n) *E_1_pu) ) ;
+delta = acosd(cos_delta) ;
+u = delta - alpha ;
+
+// For case (d)
+cos_theta = V_d/V_d0 ; // pf of rectifier
+theta = acosd(cos_theta) ;
+
+// For case (e)
+Q_r = V_d*I_d*tand(theta) ; // magnetizing var I/P
+
+// For case (f)
+d_V = E_LN - E ; // necessary change in voltage in V
+p_E_LN = 0.00625 * E_LN ; // one buck step can change in V/step
+no_buck = d_V / p_E_LN ; // No. of steps of buck
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.10 : SOLUTION :-") ;
+printf("\n (a) Open circuit dc Voltage , V_d0 = %.2f V \n",V_d0);
+printf("\n (b) Open circuit ac voltage on wye side of transformer , E = %.2f V \n",E);
+printf("\n (c) Overlap angle , u = %.2f degree \n",u)
+printf("\n (d) Power factor , cosθ = %.3f \n",cos_theta);
+printf("\n and θ = %.2f degree \n ",theta);
+printf("\n (e) Magnetizing var input to rectifier , Q_r = %.4e var \n",Q_r);
+printf("\n (f) Number of 0.625 percent steps of buck required , No. of buck = %.f steps \n",no_buck);
diff --git a/1820/CH6/EX6.2/Example6_2.sce b/1820/CH6/EX6.2/Example6_2.sce new file mode 100755 index 000000000..a6f4a7fbe --- /dev/null +++ b/1820/CH6/EX6.2/Example6_2.sce @@ -0,0 +1,36 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+K = 3 ; // factor
+
+// CALCULATIONS
+// For case (a)
+V_d = poly(0,'V_d') ; // defining a polynomial
+E_p = poly(0,'E_p') ;
+V_d = K*2*E_p ; // From equ 6.18
+
+// For case (b)
+P_dc = poly(0,'P_dc') ;
+P_ac = poly(0,'P_ac') ;
+P_dc = 2*P_ac ;
+
+// For case (c)
+P_ld = poly(0,'P_ld') ; // P_loss(dc)
+P_la = poly(0,'P_la') ; // P_loss(ac)
+P_ld = (2/3)*P_la ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.2 : SOLUTION :-") ;
+printf("\n (a) Maximum operating V_d in terms of voltage E_p , V_d = \n") ; disp(V_d) ;
+printf("\n (b) Maximum power transmission capability ratio,i.e,ratio of P_dc to P_ac , P_dc/P_ac = \n") ; disp(P_dc/P_ac) ;
+printf("\n (or) P_dc = \n") ; disp(P_dc) ;
+printf("\n (c) Ratio of total I^2*R losses , i.e ,Ratio of P_loss(dc) to P_loss(ac),which accompany maximum power flow = \n") ; disp(P_ld/P_la) ;
+printf("\n (or) P_loss(dc) = \n") ; disp(P_ld) ;
diff --git a/1820/CH6/EX6.3/Example6_3.sce b/1820/CH6/EX6.3/Example6_3.sce new file mode 100755 index 000000000..1e22f6b1c --- /dev/null +++ b/1820/CH6/EX6.3/Example6_3.sce @@ -0,0 +1,28 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_d0 = 125 ; // voltage rating of bridge rectifier in kV
+V_dr0 = V_d0 ; // Max continuos no-load direct voltage in kV
+I = 1600 ; // current rating of bridge rectifier in A
+I_d = I ; // Max continuous current in A
+
+// CALCULATIONS
+// For case (a)
+S_B = 1.047 * V_d0 * I_d ; // 3-phase kVA rating of rectifier transformer
+
+// For case (b)
+// SINCE V_d0 = 2.34*E_LN
+E_LN = V_d0/2.34 ; // Wye side kV rating
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.3 : SOLUTION :-") ;
+printf("\n (a) Three-phase kilovolt-ampere rating , S_B = %d kVA \n",S_B) ;
+printf("\n (b) Wye-side kilovolt rating , E_L-N = %.4f kV \n",E_LN) ;
diff --git a/1820/CH6/EX6.4/Example6_4.sce b/1820/CH6/EX6.4/Example6_4.sce new file mode 100755 index 000000000..72b982f6c --- /dev/null +++ b/1820/CH6/EX6.4/Example6_4.sce @@ -0,0 +1,49 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+E_LN = 53.418803 ; // Wye-side kV rating . From exa 6.3
+I = 1600 ; // current rating of bridge rectifier in A
+I_d = I ; // Max continuous current in A
+X_tr = 0.10 ; // impedance of rectifier transformer in pu Ω
+
+// For case (a)
+sc_MVA1 = 4000 ; // short-ckt MVA
+
+// For case (b)
+sc_MVA2 = 2500 ; // short-ckt MVA
+
+// For case (c)
+sc_MVA3 = 1000 ; // short-ckt MVA
+
+// CALCULATIONS
+nom_kV = sqrt(3) * E_LN ; // Nominal kV_L-L
+I_1ph = sqrt(2/3) * I_d ; // rms value of wye-side phase current
+E_LN1 = E_LN * 10^3 ; // Wye-side rating in kV
+X_B = (E_LN1/I_1ph) ; // Associated reactance base in Ω
+
+// For case (a)
+X_sys1 = nom_kV^2/sc_MVA1 ; // system reactance in Ω
+X_tra = X_tr * X_B ; // Reactance of rectifier transformer
+X_C = X_sys1 + X_tra ; // Commutating reactance in Ω
+
+// For case (b)
+X_sys2 = nom_kV^2/sc_MVA2 ; // system reactance in Ω
+X_C2 = X_sys2 + X_tra ; // Commutating reactance in Ω
+
+// For case (b) When breaker 1 & 2 are open
+X_sys3 = nom_kV^2/sc_MVA3 ; // system reactance in Ω
+X_C3 = X_sys3 + X_tra ; // Commutating reactance in Ω
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.4 : SOLUTION :-") ;
+printf("\n (a) Commutating reactance When all three breakers are closed, X_C = %.4f Ω \n",X_C) ;
+printf("\n (b) Commutating reactance When breaker 1 is open, X_C = %.4f Ω \n",X_C2) ;
+printf("\n (c) Commutating reactance When breakers 1 and 2 are open, X_C = %.4f Ω \n",X_C3) ;
diff --git a/1820/CH6/EX6.5/Example6_5.sce b/1820/CH6/EX6.5/Example6_5.sce new file mode 100755 index 000000000..919af5d8a --- /dev/null +++ b/1820/CH6/EX6.5/Example6_5.sce @@ -0,0 +1,45 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_C = 6.2292017 ; // commutating reactance when all 3 breakers are closed
+E_LN = 53.418803 * 10^3 ; // Wye-side volt rating
+V_d0 = 125 * 10^3 ; // voltage rating of bridge rectifier in V
+V_dr0 = V_d0 ; // Max continuos no-load direct voltage in V
+I = 1600 ; // current rating of bridge rectifier in A
+I_d = I ; // Max continuous current
+nom_kV = sqrt(3) * E_LN ; // Nominal kV_L-L
+X_tr = 0.10 ; //impedance of rectifier transformer in pu Ω
+alpha = 0 ; // delay angle α = 0 degree
+
+// CALCULATIONS
+// For case (a)
+E_m = sqrt(2) * E_LN ;
+u = acosd(1 - (2*X_C*I_d)/(sqrt(3)*E_m)); // overlap angle when delay angle α = 0 degree
+
+// For case (b)
+R_C = (3/%pi) * X_C ; // Equ commutation resistance per phase
+V_d = V_d0 * cosd(alpha) - R_C * I_d ; // dc voltage of rectifier in V
+
+// For case (c)
+cos_theta = V_d/V_d0 ; // Displacement or power factor of rectifier
+
+// For case (d)
+Q_r = V_d * I_d * tand( acosd(cos_theta) ) ; // magnetizing var I/P
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.5 : SOLUTION :-") ;
+printf("\n (a) Overlap angle u of rectifier, u = %.2f degree\n",u) ;
+printf("\n (b) The dc voltage V_dr of rectifier, V_dr = %.2f V \n",V_d) ;
+printf("\n (c) Displacement factor of rectifier, cosθ = %.3f \n",cos_theta) ;
+printf("\n and θ = %.1f degree \n ",acosd(cos_theta)) ;
+printf("\n (d) Magnetizing var input to rectifier, Q_r = %.4e var \n",Q_r) ;
+
+printf("\n NOTE : In case(d) 7.6546e+07 var is same as 7.6546*10^7 var = 76.546 Mvar \n") ;
diff --git a/1820/CH6/EX6.6/Example6_6.sce b/1820/CH6/EX6.6/Example6_6.sce new file mode 100755 index 000000000..179317862 --- /dev/null +++ b/1820/CH6/EX6.6/Example6_6.sce @@ -0,0 +1,42 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+I_d = 1600 ; // Max continuous dc current in A
+V_d0 = 125 * 10^3 ; // voltage rating of bridge rectifier in V
+V_d = 100 * 10^3 ; // dc voltage of rectifier in V
+X_C = 6.2292017 ; // commutating reactance when all 3 breakers are closed
+
+// CALCULATIONS
+// For case (a)
+R_C = (3/%pi) * X_C ;
+cos_alpha = (V_d + R_C*I_d)/V_d0 ; // Firing angle α
+alpha = acosd(cos_alpha) ;
+
+// For case (b)
+// V_d = (1/2)*V_d0*(cos_alpha + cos_delta)
+cos_delta = (2 * V_d/V_d0) - cos_alpha ;
+delta = acosd(cos_delta) ;
+u = delta - alpha ; // Overlap angle u in degree
+
+// For case (c)
+cos_theta = V_d/V_d0 ; // power factor
+theta = acosd(cos_theta) ;
+
+// For case (d)
+Q_r = V_d * I_d * tand(theta) ; // magnetizing var I/P
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.6 : SOLUTION :-") ;
+printf("\n (a) Firing angle α of rectifier, α = %.2f degree\n",alpha) ;
+printf("\n (b) Overlap angle u of rectifier, u = %.2f degree\n",u) ;
+printf("\n (c) Power factor , cosθ = %.2f \n",cos_theta) ;
+printf("\n and θ = %.2f degree \n ",theta) ;
+printf("\n (d) Magnetizing var input , Q_r = %.2e var \n",Q_r) ;
diff --git a/1820/CH6/EX6.7/Example6_7.sce b/1820/CH6/EX6.7/Example6_7.sce new file mode 100755 index 000000000..6200c30ca --- /dev/null +++ b/1820/CH6/EX6.7/Example6_7.sce @@ -0,0 +1,32 @@ +// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.7 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_C = 12.649731 ; // commutating reactance when 2 breakers are open
+alpha = 0 ;
+I_d = 1600 ; // DC current in A
+E_LN = 53.4188 * 10^3 ; // Wye-side rating in V
+V_d0 = 125 * 10^3 ; // voltage rating of bridge rectifier in V
+
+// CALCULATIONS
+// For case (a)
+E_m = sqrt(2) * E_LN ;
+u = acosd(1 - (2 * X_C * I_d)/(sqrt(3) * E_m)) ; // overlap angle u = δ
+
+// For case (b)
+// since rectifier operates in first mode i.e doesn't operate in second mode
+R_C = (3/%pi) * X_C ;
+V_dr = ( V_d0 * cosd(alpha) ) - (R_C*I_d) ; // dc voltage of rectifier in V
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.7 : SOLUTION :-") ;
+printf("\n (a) u = %.1f degree \n",u) ;
+printf("\n since u < 60 degree . The rectifier operates at FIRST mode , the normal operating mode \n") ;
+printf("\n (b) When dc current is 1600 A , V_dr = %.2f V \n",V_dr) ;
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