# Appendix C : Review of Basics¶

## Example C.1 Page No : 779¶

In [2]:
import math
from numpy import conj,exp

# GIVEN DATA
z = 100 * exp(60*1j*math.pi/180) ; # Impedance of transmission line in Ω
v1 = 73034.8 * exp(30*1j*math.pi/180) ; # Bus voltages in V
v2 = 66395.3 * exp(20*1j*math.pi/180) ; # Bus voltages in V

# CALCULATIONS
# For case (a)
S_12 = v1 * ( conj(v1) - conj(v2) )/( conj(z) ) ; # Complex power per phase in VA

# For case (b)
P_12 = S_12.real ; # Active power per phase in W

# For case (c)
Q_12 = S_12.imag ; # Reactive power per phase in vars

# DISPLAY RESULTS
print ("EXAMPLE : C.1 : SOLUTION :-") ;
print " a) Complex power per phase that is being transmitted from bus 1 to bus 2 , S12 = %.2f<%.2f VA "%(abs(S_12)\
, math.atan2(S_12.imag,S_12.real)*180/math.pi) ;
print " b) Active power per phase that is being transmitted , P12 = %.2f W "%(P_12) ;
print " b) Reactive power per phase that is being transmitted , Q12 = %.2f vars "%(Q_12) ;

# rounding off error

EXAMPLE : C.1 : SOLUTION :-
a) Complex power per phase that is being transmitted from bus 1 to bus 2 , S12 = 10104766.69<3.56 VA
b) Active power per phase that is being transmitted , P12 = 10085280.57 W
b) Reactive power per phase that is being transmitted , Q12 = 627236.52 vars


## Example C.2 Page No : 791¶

In [3]:
import math

# GIVEN DATA
X_pu = 12./100 ; # Leakage reactancw in pu
kV_B_HV = 345 ; # HV side ratings in Y kV
kV_B_LV = 34.5 ; # LV side ratings in Y kV
MVA_B = 20. ; # selected Base on HV side in MVA

# CALCULATIONS
# For case (a)
X_pu = 12./100 ; # reactancw of transformer in pu

# For case (b)
Z_B_HV = (kV_B_HV)**2/MVA_B ; # HV side base impedance in Ω

# For case (c)
Z_B_LV = (kV_B_LV)**2/MVA_B ; # LV side base impedance in Ω

# For case (d)
X_HV = X_pu * Z_B_HV ; # reactancw referred to HV side in Ω

# For case (e)
X_LV = X_pu * Z_B_LV ; # reactancw referred to LV side in Ω
n = (kV_B_HV/math.sqrt(3))/(kV_B_LV/math.sqrt(3)) ; # Turns ratio of winding
X_LV1 = X_HV/n**2 ; # From equ C.89

# DISPLAY RESULTS
print ("EXAMPLE : C.2 : SOLUTION :-") ;
print " a) reactancw of transformer in pu , X_pu = %.2f pu "%(X_pu) ;
print " b) High-voltage side base impedance , Z_B_HV = %.2f Ω "%(Z_B_HV) ;
print " c) Low-voltage side base impedance , Z_B_LV = %.4f Ω "%(Z_B_LV) ;
print " d) Transformer reactancw referred to High-voltage side , X_HV = %.2f Ω "%(X_HV) ;
print " e) Transformer reactancw referred to Low-voltage side , X_LV = %.4f Ω "%(X_LV) ;
print "     or) From another equation C.89 ," ;
print "     Transformer reactancw referred to Low-voltage side , X_LV = %.4f Ω "%(X_LV1) ;

EXAMPLE : C.2 : SOLUTION :-
a) reactancw of transformer in pu , X_pu = 0.12 pu
b) High-voltage side base impedance , Z_B_HV = 5951.25 Ω
c) Low-voltage side base impedance , Z_B_LV = 59.5125 Ω
d) Transformer reactancw referred to High-voltage side , X_HV = 714.15 Ω
e) Transformer reactancw referred to Low-voltage side , X_LV = 7.1415 Ω
or) From another equation C.89 ,
Transformer reactancw referred to Low-voltage side , X_LV = 7.1415 Ω


## Example C.3 Page No : 792¶

In [4]:
import math

# GIVEN DATA
X_pu = 12./100 ; # Leakage reactancw in pu
kV_B_HV = 345. ; # HV side ratings in Y kV
kV_B_LV = 34.5 ; # LV side ratings in Δ kV
MVA_B = 20. ; # Base on HV side in MVA

# CALCULATIONS
# For case (a)
n = ( kV_B_HV/math.sqrt(3) )/kV_B_LV ; # Turns ratio of windings

# For case (b)
Z_B_HV = (kV_B_HV)**2/MVA_B ; # HV side base impedance in Ω
X_HV = X_pu * Z_B_HV ; # reactancw referred to HV side in Ω
X_LV = X_HV/(n**2) ; # transformer reactancw referred to delta LV side in Ω

# For case (c)
Z_dt = X_LV ;
Z_Y = Z_dt/3 ; # reactancw of equi wye connection
Z_B_LV = kV_B_LV**2/MVA_B ; # LV side base impedance in Ω
X_pu1 = Z_Y/Z_B_LV ; # reactancw in pu referred to LV side

# Alternative method For case (c)
n1 = kV_B_HV/kV_B_LV ; # Turns ratio if line-to-line voltages are used
X_LV1 = X_HV/(n1**2) ; # reactancw referred to LV side in Ω
X_pu2 = X_LV1/Z_B_LV ; # reactancw in pu referred to LV side

# DISPLAY RESULTS
print ("EXAMPLE : C.3 : SOLUTION :-") ;
print " a) Turns ratio of windings , n = %.4f "%(n) ;
print " b) Transformer reactancw referred to LV side in ohms ,X_LV = %.4f Ω "%(X_LV) ;
print " c) Transformer reactancw referred to LV side in per units ,X_pu = %.2f pu "%(X_pu1) ;
print "    or) From another equation if line-to-line voltages are used ," ;
print "     Transformer reactancw referred to LV side in per units ,X_pu = %.2f pu "%(X_pu2) ;

EXAMPLE : C.3 : SOLUTION :-
a) Turns ratio of windings , n = 5.7735
b) Transformer reactancw referred to LV side in ohms ,X_LV = 21.4245 Ω
c) Transformer reactancw referred to LV side in per units ,X_pu = 0.12 pu
or) From another equation if line-to-line voltages are used ,
Transformer reactancw referred to LV side in per units ,X_pu = 0.12 pu


## Example C.4 Page No : 794¶

In [3]:
import math
from numpy import exp,degrees,arctan2

# GIVEN DATA
I_1 = 1000. ; # Physical current in A for 2.4 kV circuit
Z_pu = 0.04 ; # Leakage reactancw in pu
I_pu = 2.08*exp(1j*(-90)*math.pi/180) ; # Generator supply for pure inductive load
kVA_Bg1 = 6000. ; # Rated kVA values for T1
kVA_Bg2 = 4000. ; # Rated kVA values for T2
N2 = 2.4 ; # N2 = V2 in Y kV ,refer fig C.4
N1 = 24. ; # N1 = V1 in Y kV ,refer fig C.4
N3 = 24. ; # N3 = V3 = N1 in Y kV ,refer fig C.4
N4 = 12. ; # N4 = V4 in Y kV ,refer fig C.4

# CALCULATIONS
# For case (a)
kVA_B = 2080. ; # arbitrarily selected kVA values for all 3 ckt

# For case (b)
n1 = N2/N1 ; # Turns ratio of transformer T1 & T2 i.e N2/N1
n2 = N3/N4 ; # Turns ratio N1'/N2'
kV_BL_L1 = 2.5 ; # arbitrarily selected Base voltage for 2.4 kV ckt in kV
kV_BL_L2 = kV_BL_L1/n1 ; # arbitrarily selected Base voltage for 24 kV ckt in kV
kV_BL_L3 = kV_BL_L2/n2 ; # arbitrarily selected Base voltage for 12 kV ckt in kV

# For case (c)
Z_B1 = (kV_BL_L1)**(2) * 1000/(kVA_B) ; # Base impedance in Ω for 2.4 kV ckt
Z_B2 = (kV_BL_L2)**(2) * 1000/(kVA_B) ; # Base impedance in Ω for 24 kV ckt
Z_B3 = (kV_BL_L3)**(2) * 1000/(kVA_B) ; # Base impedance in Ω for 12 kV ckt

# For case (d)
I_B1 = kVA_B/(math.sqrt(3)*kV_BL_L1) ; # Base current in A for 2.4 kV ckt
I_B2 = kVA_B/(math.sqrt(3)*kV_BL_L2) ; # Base current in A for 24 kV ckt
I_B3 = kVA_B/(math.sqrt(3)*kV_BL_L3) ; # Base current in A for 12 kV ckt

# For case (e)
I_2 = (n1) * I_1 ; # Physical current in A for 24 kV circuit
I_4 = (n2) * I_2 ; # Physical current in A for 12 kV circuit

# For case (f)
I_pu_3ckt = abs(I_pu) ; # per-unit current values for all 3-ckt

# For case (g)
kV_B1 = N2 ; # Given voltage in kV
kV_B2 = N4 ; # Given voltage in kV
Z_pu_T1 = (1j)*Z_pu*(kVA_B/kVA_Bg1)*(kV_B1/kV_BL_L1)**(2) ; # New reactancw of T1
Z_pu_T2 = (1j)*Z_pu*(kVA_B/kVA_Bg2)*(kV_B2/kV_BL_L3)**(2) ; # New reactancw of T2

# For case (h)
V1 = kV_B1/kV_BL_L1 ; # voltage in pu at bus 1
V2 = V1 - I_pu * (Z_pu_T1) ; # voltage in pu at bus 2
V4 = V2 - I_pu * (Z_pu_T2) ; # voltage in pu at bus 3

# For case (i)
S1 = V1 * abs(I_pu) ; # Apparent power value at bus 1 in pu
S2 = V2 * abs(I_pu) ; # Apparent power value at bus 2 in pu
S4 = V4 * abs(I_pu) ; # Apparent power value at bus 4 in pu

# DISPLAY RESULTS
print ("EXAMPLE : C.3 : SOLUTION :-") ;
print " a) Base kilovoltampere value for all 3-circuits is , kVA_B = %.1f kVA "%(kVA_B) ;
print " b) Base line-to-line kilovolt value for 2.4 kV circuit , kV_BL_L = %.1f kV "%(kV_BL_L1) ;
print "     Base line-to-line kilovolt value for 24 kV circuit , kV_BL_L = %.1f kV "%(kV_BL_L2) ;
print "     Base line-to-line kilovolt value for 24 kV circuit , kV_BL_L = %.1f kV "%(kV_BL_L3) ;
print " c) Base impedance value of 2.4 kV circuit , Z_B = %.3f Ω "%(Z_B1) ;
print "     Base impedance value of 24 kV circuit , Z_B = %.1f Ω "%(Z_B2) ;
print "     Base impedance value of 12.5 kV circuit , Z_B = %.1f Ω "%(Z_B3) ;
print " d) Base current value of 2.4 kV circuit , I_B = %d A "%(I_B1) ;
print "     Base current value of 24 kV circuit , I_B = %d A "%(I_B2) ;
print "     Base current value of 2.4 kV circuit , I_B = %d A "%(I_B3) ;
print " e) Physical current of 2.4 kV circuit , I = %.f A "%(I_1) ;
print "     Physical current of 24 kV circuit , I = %.f A "%(I_2) ;
print "     Physical current of 12 kV circuit , I = %.f A "%(I_4) ;
print " f) Per unit current values for all 3 circuits , I_pu = %.2f pu "%(I_pu_3ckt) ;
print " g) New transformer reactancw of T1 , Z_pu_T1 =  j%.4f pu "%(abs(Z_pu_T1.real)) ;
print "     New transformer reactancw of T2 , Z_pu_T2 =  j%.4f pu "%(abs(Z_pu_T2)) ;
print " h) Per unit voltage value at bus 1 ,V1 = %.2f<%.1f pu "%(abs(V1),degrees(arctan2(V1.imag,V1.real))) ;
print "     Per unit voltage value at bus 2 ,V2 = %.4f<%.1f pu "%(abs(V2),degrees(arctan2(V2.imag,V2.real))) ;
print "     Per unit voltage value at bus 4 ,V4 = %.4f<%.1f pu "%(abs(V4),degrees(arctan2(V4.imag,V4.real))) ;
print " i) Per-unit apparent power value at bus 1 , S1 = %.2f pu "%(S1.real) ;
print "     Per-unit apparent power value at bus 2 , S2 = %.4f pu "%(S2) ;
print "     Per-unit apparent power value at bus 4 , S4 = %.4f pu "%(S4) ;
print " j TABLE C.2 " ;
print "     Results Of Example C.4 " ;
print "     ___________________________________________________________________________________" ;
print "     QUANTITY      \t  2.4-kV circuit   \t  24-kV circuit   \t  12-kV circuit   ";
print "     ___________________________________________________________________________________" ;
print "     kVA_B3-Φ)    \t  %d kVA           \t  %d kVA          \t  %d kVA "%(kVA_B,kVA_B,kVA_B) ;
print "     kV_BL-L)     \t  %.1f kV          \t  %d kV           \t  %.1f kV "%(kV_BL_L1,kV_BL_L2,kV_BL_L3) ;
print "     Z_B           \t  %.3f Ω           \t  %.1f Ω          \t  %.1f Ω "%(Z_B1,Z_B2,Z_B3) ;
print "     I_B           \t  %d A             \t  %d A            \t  %d A "%(I_B1,I_B2,I_B3) ;
print "     I_physical    \t  %d A             \t  %.f A           \t  %.f A "%(I_1,I_2,I_4) ;
print "     I_pu          \t  %.2f pu          \t  %.2f pu         \t  %.2f pu "%(I_pu_3ckt,I_pu_3ckt,I_pu_3ckt) ;
print "     V_pu          \t  %.2f pu          \t  %.4f pu         \t  %.4f pu "%(abs(V1),abs(V2),abs(V4)) ;
print "     S_pu          \t  %.2f pu          \t  %.4f pu         \t  %.4f pu "%(S1,S2,S4) ;
print "     ___________________________________________________________________________________" ;

EXAMPLE : C.3 : SOLUTION :-
a) Base kilovoltampere value for all 3-circuits is , kVA_B = 2080.0 kVA
b) Base line-to-line kilovolt value for 2.4 kV circuit , kV_BL_L = 2.5 kV
Base line-to-line kilovolt value for 24 kV circuit , kV_BL_L = 25.0 kV
Base line-to-line kilovolt value for 24 kV circuit , kV_BL_L = 12.5 kV
c) Base impedance value of 2.4 kV circuit , Z_B = 3.005 Ω
Base impedance value of 24 kV circuit , Z_B = 300.5 Ω
Base impedance value of 12.5 kV circuit , Z_B = 75.1 Ω
d) Base current value of 2.4 kV circuit , I_B = 480 A
Base current value of 24 kV circuit , I_B = 48 A
Base current value of 2.4 kV circuit , I_B = 96 A
e) Physical current of 2.4 kV circuit , I = 1000 A
Physical current of 24 kV circuit , I = 100 A
Physical current of 12 kV circuit , I = 200 A
f) Per unit current values for all 3 circuits , I_pu = 2.08 pu
g) New transformer reactancw of T1 , Z_pu_T1 =  j0.0000 pu
New transformer reactancw of T2 , Z_pu_T2 =  j0.0192 pu
h) Per unit voltage value at bus 1 ,V1 = 0.96<0.0 pu
Per unit voltage value at bus 2 ,V2 = 0.9334<-0.0 pu
Per unit voltage value at bus 4 ,V4 = 0.8935<-0.0 pu
i) Per-unit apparent power value at bus 1 , S1 = 2.00 pu
Per-unit apparent power value at bus 2 , S2 = 1.9415 pu
Per-unit apparent power value at bus 4 , S4 = 1.8586 pu
j TABLE C.2
Results Of Example C.4
___________________________________________________________________________________
QUANTITY      	  2.4-kV circuit   	  24-kV circuit   	  12-kV circuit
___________________________________________________________________________________
kVA_B3-Φ)    	  2080 kVA           	  2080 kVA          	  2080 kVA
kV_BL-L)     	  2.5 kV          	  25 kV           	  12.5 kV
Z_B           	  3.005 Ω           	  300.5 Ω          	  75.1 Ω
I_B           	  480 A             	  48 A            	  96 A
I_physical    	  1000 A             	  100 A           	  200 A
I_pu          	  2.08 pu          	  2.08 pu         	  2.08 pu
V_pu          	  0.96 pu          	  0.9334 pu         	  0.8935 pu
S_pu          	  2.00 pu          	  1.9415 pu         	  1.8586 pu
___________________________________________________________________________________


## Example C.5 Page No : 811¶

In [6]:
import math

# GIVEN DATA
D_ab = 6.8 ; # dismath.tance b/w conductors center-to-center in ft
D_bc = 5.5 ; # dismath.tance b/w conductors center-to-center in ft
D_ca = 4 ; # dismath.tance b/w conductors center-to-center in ft

# CALCULATIONS
# For case (a)
D_eq = (D_ab * D_bc * D_ca)**(1/3) ; # Equi spacing for pole top in ft
D_s = 0.01579 ; # GMR in ft From Table A.1
X_L = 0.1213 * math.log(D_eq/D_s) ; # Inductive reactancw in Ω/mi . From equ C.135

# For case (b)
X_a = 0.503 ; # Inductive reactancw in Ω/mi From Table A.1
X_d = 0.2026 ; # From Table A.8 for D_eq,by linear interpolation in Ω/mi
X_L1 = X_a + X_d ; # Inductive reactancw in Ω/mi

# DISPLAY RESULTS
print ("EXAMPLE : C.5 : SOLUTION :-") ;
print " a) Inductive reactancw umath.sing equation C.135 , X_L = %.4f Ω/mi "%(X_L );
print " b) Inductive reactancw umath.sing tables , X_L = %.4f Ω/mi "%(X_L1) ;

EXAMPLE : C.5 : SOLUTION :-
a) Inductive reactancw umath.sing equation C.135 , X_L = 0.5032 Ω/mi
b) Inductive reactancw umath.sing tables , X_L = 0.7056 Ω/mi


## Example C.6 Page No : 812¶

In [7]:
import math

# GIVEN DATA
D_ab = 6.8 ; # dismath.tance b/w conductors center-to-center in ft
D_bc = 5.5 ; # dismath.tance b/w conductors center-to-center in ft
D_ca = 4 ; # dismath.tance b/w conductors center-to-center in ft
l = 100 ; # Line length in miles

# CALCULATIONS
# For case (a)
D_m = (D_ab * D_bc * D_ca)**(1./3) ; # Equi spacing for pole top in ft
r = 0.522/(2 * 12) ; # feet
X_C = 0.06836 * math.log10 (D_m/r) ; # Shunt capacitive reactancw in MΩ*mi

# For case (b)
X_a = 0.1136 ; # Shunt capacitive reactancw in MΩ*mi , From table A.1
X_d = 0.049543 ; # Shunt capacitive reactancw spacing factor in MΩ*mi , From table A.9
X_C1 = X_a + X_d ; # Shunt capacitive reactancw in MΩ*mi
X_C2 = X_C1/l ; # Capacitive reactancw of 100 mi line in MΩ

# DISPLAY RESULTS
print ("EXAMPLE : C.6 : SOLUTION :-") ;
print " a) Shunt capacitive reactancw umath.sing equation C.156 , X_C = %.6f MΩ*mi "%(X_C) ;
print " b) Shunt capacitive reactancw umath.sing tables , X_C = %.6f MΩ*mi "%(X_C1) ;
print " c) Capacitive reactancw of total line , X_C = %.5e MΩ "%(X_C2) ;

EXAMPLE : C.6 : SOLUTION :-
a) Shunt capacitive reactancw umath.sing equation C.156 , X_C = 0.163211 MΩ*mi
b) Shunt capacitive reactancw umath.sing tables , X_C = 0.163143 MΩ*mi
c) Capacitive reactancw of total line , X_C = 1.63143e-03 MΩ