# Chapter 6 : Direct Current Power Transmission¶

## Example 6.1 Page No : 288¶

In [1]:
import math
from sympy import Symbol

# GIVEN DATA
K_1 = 2.5 ; # Factor
K_2 = 1.7 ; # Factor

# CALCULATIONS
# For case (b)
I_d = Symbol('I_d') ; # math.since P_loss(dc) = P_loss (ac)
I_L = Symbol('I_L') ; # i.e 2*I_d**2*R_dc = 3*I_L**2*R_ac
I_d = math.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 = Symbol('V_d') ; # Defining a ploynomial V_d
E_p = Symbol('E_p') ; # math.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 = Symbol('dc_i') ; # dc_i = dc insulation level
ac_i = Symbol('ac_i') ; # ac_i = ac insulation level
dc_i = ins_lvl_1 * ac_i ;

# DISPLAY RESULTS
print ("EXAMPLE : 6.1 : SOLUTION :-") ;
print " a) Line-to-line dc voltage of V_d in terms of line-to-neutral voltage E_p , V_d = ", V_d ;
print " b) The dc line current I_d in terms of ac line current I_L , I_d = ", I_d1 ;
print " c Ratio of dc insulation level to ac insulation level =  %.3f"% (dc_i/ac_i) ;
print "  or dc insulation level = " , dc_i ;

EXAMPLE : 6.1 : SOLUTION :-
a) Line-to-line dc voltage of V_d in terms of line-to-neutral voltage E_p , V_d =  2.45*E_p
b) The dc line current I_d in terms of ac line current I_L , I_d =  1.225*I_L
c Ratio of dc insulation level to ac insulation level =  0.833
or dc insulation level =  0.833*ac_i


## Example 6.2 Page No : 290¶

In [2]:
import math
from sympy import Symbol

# GIVEN DATA
K = 3 ; # factor

# CALCULATIONS
# For case (a)
V_d = Symbol('V_d') ; # defining a polynomial
E_p = Symbol('E_p') ;
V_d = K*2*E_p ; # From equ 6.18

# For case (b)
P_dc = Symbol('P_dc') ;
P_ac = Symbol('P_ac') ;
P_dc = 2*P_ac ;

# For case (c)
P_ld = Symbol('P_ld') ; # P_loss(dc)
P_la = Symbol('P_la') ; # P_loss(ac)
P_ld = (2./3)*P_la ;

# DISPLAY RESULTS
print ("EXAMPLE : 6.2 : SOLUTION :-") ;
print " a) Maximum operating V_d in terms of voltage E_p , V_d = ", V_d ;
print " b) Maximum power transmission capability ratio,i.e,ratio of P_dc to P_ac , P_dc/P_ac = %.3f"%(P_dc/P_ac) ;
print " or P_dc = ", P_dc ;
print " c) Ratio of total I**2*R losses , i.e ,Ratio of P_lossdc) to P_lossac),which accompany maximum power flow = %.3f"%(P_ld/P_la) ;
print " or P_lossdc = ", P_ld ;

EXAMPLE : 6.2 : SOLUTION :-
a) Maximum operating V_d in terms of voltage E_p , V_d =  6*E_p
b) Maximum power transmission capability ratio,i.e,ratio of P_dc to P_ac , P_dc/P_ac = 2.000
or P_dc =  2*P_ac
c) Ratio of total I**2*R losses , i.e ,Ratio of P_lossdc) to P_lossac),which accompany maximum power flow = 0.667
or P_lossdc =  0.666666666666667*P_la


## Example 6.3 Page No : 295¶

In [3]:
import math

# 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
print " a) Three-phase kilovolt-ampere rating , S_B = %d kVA "%(S_B) ;
print " b) Wye-side kilovolt rating , E_L-N = %.4f kV "%(E_LN) ;

 a) Three-phase kilovolt-ampere rating , S_B = 209400 kVA
b) Wye-side kilovolt rating , E_L-N = 53.4188 kV


## Example 6.4 Page No : 296¶

In [4]:
import math

# 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 = math.sqrt(3) * E_LN ; # Nominal kV_L-L
I_1ph = math.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 reactancw base in Ω

# For case (a)
X_sys1 = nom_kV**2/sc_MVA1 ; # system reactancw in Ω
X_tra = X_tr * X_B ; # reactancw of rectifier transformer
X_C = X_sys1 + X_tra ; # Commutating reactancw in Ω

# For case (b)
X_sys2 = nom_kV**2/sc_MVA2 ; # system reactancw in Ω
X_C2 = X_sys2 + X_tra ; # Commutating reactancw in Ω

# For case (b) When breaker 1 & 2 are open
X_sys3 = nom_kV**2/sc_MVA3 ; # system reactancw in Ω
X_C3 = X_sys3 + X_tra ; # Commutating reactancw in Ω

# DISPLAY RESULTS
print ("EXAMPLE : 6.4 : SOLUTION :-") ;
print " a) Commutating reactancw When all three breakers are closed, X_C = %.4f Ω "%X_C ;
print " b) Commutating reactancw When breaker 1 is open, X_C = %.4f Ω "%X_C2 ;
print " c) Commutating reactancw When breakers 1 and 2 are open, X_C = %.4f Ω "%(X_C3) ;

EXAMPLE : 6.4 : SOLUTION :-
a) Commutating reactancw When all three breakers are closed, X_C = 6.2292 Ω
b) Commutating reactancw When breaker 1 is open, X_C = 7.5133 Ω
c) Commutating reactancw When breakers 1 and 2 are open, X_C = 12.6497 Ω


## Example 6.5 Page No : 298¶

In [5]:
import math

# GIVEN DATA
X_C = 6.2292017 ; # commutating reactancw 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 = math.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 = math.sqrt(2) * E_LN ;
u = math.acos((1 - (2*X_C*I_d))/(math.sqrt(3)*E_m)); # overlap angle when delay angle α = 0 degree

# For case (b)
R_C = (3/math.pi) * X_C ; # Equ commutation resistance per phase
V_d = V_d0 * math.cos(math.radians(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 * math.degrees(math.atan( math.acos(math.radians(cos_theta))) ) ; # magnetizing var I/P

# DISPLAY RESULTS
print ("EXAMPLE : 6.5 : SOLUTION :-") ;
print " a) Overlap angle u of rectifier, u = %.2f degree"%u ;
print " b) The dc voltage V_dr of rectifier, V_dr = %.2f V "%V_d ;
print " c) Displacement factor of rectifier, math.cosθ = %.3f  "%cos_theta ;
print "     and   θ = %.1f degree  "%math.acos(cos_theta) ;
print " d) Magnetizing var input to rectifier, Q_r = %.4e var "%Q_r ;

print " NOTE : In cased 7.6546e+07 var is same as 7.6546*10**7 var = 76.546 Mvar " ;

EXAMPLE : 6.5 : SOLUTION :-
a) Overlap angle u of rectifier, u = 1.72 degree
b) The dc voltage V_dr of rectifier, V_dr = 115482.48 V
c) Displacement factor of rectifier, math.cosθ = 0.924
and   θ = 0.4 degree
d) Magnetizing var input to rectifier, Q_r = 1.0578e+10 var
NOTE : In cased 7.6546e+07 var is same as 7.6546*10**7 var = 76.546 Mvar


## Example 6.6 Page No : 299¶

In [8]:
import math

# 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 reactancw when all 3 breakers are closed

# CALCULATIONS
# For case (a)
R_C = (3/math.pi) * X_C ;
cos_alpha = (V_d + R_C*I_d)/V_d0 ; # Firing angle α
alpha = math.acos(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 = math.acos(cos_delta) ;
u = delta - alpha ; # Overlap angle u in degree

# For case (c)
cos_theta = V_d/V_d0 ; # power factor
theta = math.degrees(math.acos(cos_theta)) ;

# For case (d)
Q_r = V_d * I_d * math.tan(math.radians(theta)) ; # magnetizing var I/P

# DISPLAY RESULTS
print ("EXAMPLE : 6.6 : SOLUTION :-") ;
print " a) Firing angle α of rectifier, α = %.2f degree"%(alpha) ;
print " b) Overlap angle u of rectifier, u = %.2f degree"%(u) ;
print " c) Power factor , math.cosθ = %.2f  "%(cos_theta) ;
print "     and   θ = %.2f degree  "%(theta) ;
print " d) Magnetizing var input , Q_r = %.2e var "%(Q_r) ;

EXAMPLE : 6.6 : SOLUTION :-
a) Firing angle α of rectifier, α = 0.50 degree
b) Overlap angle u of rectifier, u = 0.26 degree
c) Power factor , math.cosθ = 0.80
and   θ = 36.87 degree
d) Magnetizing var input , Q_r = 1.20e+08 var


## Example 6.7 Page No : 301¶

In [10]:
import math

# GIVEN DATA
X_C = 12.649731 ; # commutating reactancw 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 = math.sqrt(2) * E_LN ;
u = math.acos(1 - (2 * X_C * I_d)/(math.sqrt(3) * E_m)) ; # overlap angle u = δ

# For case (b)
# math.since rectifier operates in first mode i.e doesn't operate in second mode
R_C = (3/math.pi) * X_C ;
V_dr = ( V_d0 * math.cos(math.radians(alpha)) ) - (R_C*I_d) ; # dc voltage of rectifier in V

# DISPLAY RESULTS
print ("EXAMPLE : 6.7 : SOLUTION :-") ;
print " a) u = %.1f degree "%(u) ;
print "  math.since u < 60 degree . The rectifier operates at FIRST mode , the normal operating mode ";
print " b) When dc current is 1600 A , V_dr = %.2f V "%(V_dr) ;

# note : rounding off error.

EXAMPLE : 6.7 : SOLUTION :-
a) u = 0.8 degree
math.since u < 60 degree . The rectifier operates at FIRST mode , the normal operating mode
b) When dc current is 1600 A , V_dr = 105672.63 V


## Example 6.10 Page No : 307¶

In [11]:
import math

# GIVEN DATA
X_C = 6.2292 ; # commutating reactancw 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/math.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)/math.cos(math.radians(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 = math.cos(math.radians(alpha)) - ( (X_c_pu * I_d_pu)/( (a/n) *E_1_pu) ) ;
delta = math.acos(cos_delta) ;
u = delta - alpha ;

# For case (d)
cos_theta = V_d/V_d0 ; # pf of rectifier
theta = math.acos(cos_theta) ;

# For case (e)
Q_r = V_d*I_d*math.degrees(math.atan(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
print ("EXAMPLE : 6.10 : SOLUTION :-") ;
print " a) Open circuit dc Voltage , V_d0 = %.2f V "%(V_d0);
print " b) Open circuit ac voltage on wye side of transformer , E = %.2f V "%(E);
print " c) Overlap angle , u = %.2f degree "%(u)
print " d) Power factor , math.cosθ = %.3f  "%(cos_theta);
print "     and   θ = %.2f degree  "%(theta);
print " e) Magnetizing var input to rectifier , Q_r = %.4e var "%(Q_r);
print " f) Number of 0.625 percent steps of buck required , No. of buck = %.f steps "%(no_buck);

EXAMPLE : 6.10 : SOLUTION :-
a) Open circuit dc Voltage , V_d0 = 109517.52 V
b) Open circuit ac voltage on wye side of transformer , E = 46802.36 V
c) Overlap angle , u = 0.56 degree
d) Power factor , math.cosθ = 0.913
and   θ = 0.42 degree
e) Magnetizing var input to rectifier , Q_r = 3.6451e+09 var
f) Number of 0.625 percent steps of buck required , No. of buck = 20 steps