Line with length of approximately between 50 and 150 km and line voltage of between 20kV and 100kV is considered Medium Transmission Line. The capacitance effect is actually uniformly distributed over the entire line. For the purpose of calculation, it is considered that the distributed capacitance is divided and lumped in the form of capacitors shunted across the line. The localised capacitance methods known as (i) Nominal PI method (ii) Nominal T methods and (iii) End Condensor method are used.

REGULATION AND EFFICIENCY OF MEDIUM TRANSMISSION LINE USING NOMINAL PI METHOD

Demonstrating below is a typical example of using Nominal PI Method to find voltage regulation and efficiency of a line .

 

 

Derive equations.

A 3-phase line delivers P3 MW at lagging power factor pf at receiving end line-to-line voltage of V_L has line impedance of Z=R + jX Ohm, admittance Y=1/Xc Ohm

V_R =V_L/1.732 /_0°

Vs = V_R + I_L Z = V_R + I_L (R + jX) = = |Vs | /_ ø _vs

I_R= P3 / [(1.732 x V_L x pf] = 1/3 x P3 /[(V_L/1.732) x pf ] /_- cos ^-1 pf

I_c1= V_R/Xc =Y/2 x V_R = 1/2 (Y V_R) /_90°

Ic2 = Vs/Xc =Y/2 x Vs = 1/2 x ( |Y| /_90° x |Vs | /_ø _vs ) = 1/2 x |Y| |Vs | /_(90° + ø _vs)

I_L = I_R + I_c1 = |I_L| /_ø_IL

I_s = I_L + I_c2 = |Is | /_ ø _Is

=>sending end power factor = cos (ø _vs + ø _Is )

=>voltage regulation = (V_R -Vs)/V_R x 100%

=>efficiency of the line = Pd /[Pd + lossess)

(d) efficiency of transmission line = Power delivered /[Power delivered + Losses] = [1/3 x 15MW]/ [(1/3 x 15MW) + (87.88²A x 10 Ohm x 10^-6 ]=98.5%

Derive equations.

A 3-phase line delivers P3 MW at lagging power factor pf at receiving end line-to-line voltage of VL has line impedance of Z=R + jX Ohm, admittance Y=1/Xc Ohm

V_R =V_L/1.732 /_0°

Vc = V_R + I_R Z /2 = |Vc| /_ø _vc

Vs = V_c + I_s Z /2 =|Vs| /_ø _vs

I_R= P3 / [(1.732 x V_L x pf] = 1/3 x P3 /[(V_L/1.732) x pf ] /_- cos ^-1 pf

I_c= V_c/Xc =Y x V_c = =|Ic| /_ø _ic

I_s = I_R + I_c = |Is| /_ø _is

=>sending end power factor = cos (ø _vs + ø _Is )

=>voltage regulation = (V_R -Vs)/V_R x 100%

=>efficiency of the line = Pd /[Pd + lossess)

 

Typical example

Using same electrical data above but solve using nominal T-method.

Find
(a) sending end voltage Vs
(b) sending end Is
(c)sending end power factor
(d) efficiency of transmission line

Answers

R=0.1 Ohm/km x 100km=10 Ohm
XL=0.2 Ohm/km x 100km = 20 Ohm
Y=0.005 mMho/ km x 100km= 0.5 mMho

Let us choose Red phase for analysis
Set receiving end voltage Vr as reference
Vr= 110kV/1.732 /_0° =63.51/_0° kV

Load current Ir =(1/3 x 15MW) /[63.51/_0° kV x 0.82] =96 /_-34.9° Amps where ø =cos ^-1 0.82=34.9°
Half of line impedance,Z/2=1/2 [R+jX_L]=1/2 [10 + j20] =11.18 /_63.4° Ohm

Vc=[Ir x Z/2] + Vr =64.45 /_0.5°

Capacitor current I_c = Vc/Xc=Y x Vc = 0.5 /_90° mMho x 64.45 /_0.5° kV =32.22/_90.5°
Transmission line current I_L = Ir + Ic = 81.67 /_16.14° Amps

(b) sending end Is =Ic + Ir = 32.22/_90.5° + 96 /_-34.9° = 81.67 /_-16.14° Amps

(a) Sending end voltage Vs= I_s Z/2 + Vc =[ 81.67 /_-16.14° Amps] [11.18 /_63.4° Ohm] + 64,450 /_0.5° V= 65.08 /_1.09° kV
Vs (line-to-line)= 1.732 x 65.08 kV =112.72 kV

(c) sending end power factor = cos (16.14° + 1.09°) =cos 17.23 ° =0.96

(d) efficiency of transmission line = Power delivered /[Power delivered + Losses]
= [1/3 x 15MW]/ [(1/3 x 15MW) + (81.67²A x 5 Ohm x 10^-6 ) +( 96²A x 5 Ohm x 10^-6)]= 5MW /[5MW + 0.033 MW + 0.046 MW] =98.4%

 

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REGULATION AND EFFICIENCY OF MEDIUM TRANSMISSION LINE USING END CONDENSOR METHOD Demonstrating here is a typical example of End Condensor Method
to find voltage regulation and efficiency of a line.

 

 

Example.

A 3-phase line delivers P₃ MW at lagging power factor of pf at receiving end line-line V_L kV has following constants.

R Ohm/phase
X_L Ohm/phase
X_C Ohm/phase -->Susceptance, Y= 1/X_Cmho
Line impedance, Z= R + jX Ohm

θ_R= cos ^-1 pf

Let receiving end voltage of phase R to neutral be the reference voltage,V_R =1/1.732 x V_L /_0° kV =| V_R| /_0° kV

Load current, I_R= P₃ /[ 1.732 x V_L x pf] /_- cos ^-1 pf = |I_R| /_-cos ^-1 pf

Charging current, I_C= V_R/X_C = Y /_90° x V_R/_0° = [ Y V_R ] /_90° A

Transmission line current, I_S=[ I_R + I_C ] = |Is| /_ø _Is A

Sending end voltage, V_s= I_S Z + V_R =|Vs| /_ø_Vs kV

Sending end voltage (line-line) = 1.732 x V_s kV

=>Sending end power factor= cos (ø_Vs + ø_Is) = cos ø_s

=> Voltage regulation = {V_s - V_R}/V_R x 100%

=> Efficiency of the line = Pd/{Pd + losses} x 100% where Pd: delivered power per line., Losses= I² x R

 

Typical example

Using same electrical data above but solve using End Condensor method.

Find
(a) sending end voltage Vs
(b) sending end Is
(c)sending end power factor
(d) efficiency of transmission line

Answers

R=0.1 Ohm/km x 100km=10 Ohm
XL=0.2 Ohm/km x 100km = 20 Ohm
Y=0.005 mMho/ km x 100km= 0.5 mMho

Let us choose Red phase for analysis
Set receiving end voltage Vr as reference
Vr=Vc= 110kV/1.732 /_0° =63.51/_0° kV

Load current Ir =(1/3 x 15MW) /[63.51/_0° kV x 0.82] =96 /_-34.9° Amps where ø =cos ^-1 0.82=34.9°
Line impedance,Z=[R+jX_L]=10 + j20 =22.36 /_63.4° Ohm

Capacitor current I_c = Vc/Xc=Y x Vc = 0.5 /_90° mMho x 63.51/_0° kV = 31.76 /_90°
Transmission line current I_L = Ir + Ic = 81.67 /_16.14° Amps

(b) sending end Is =Ic + Ir = 31.76 /_90° + 96 /_-34.9° = 82.07 /_-16.4° Amps

(a) Sending end voltage Vs= I_s Z + Vc =[ 82.07 /_-16.4° Amps] [22.36 /_63.4° Ohm] + 63,510 /_0° V= 64.78 /_1.09° kV
Vs (line-to-line)= 1.732 x 64.78 kV =112.2 kV

(c) sending end power factor = cos (16.4° + 1.09°) =cos 17.49 ° =0.95

(d) efficiency of transmission line = Power delivered /[Power delivered + Losses]
= [1/3 x 15MW]/ [(1/3 x 15MW) + (82.07²A x 10 Ohm x 10^-6 ) ] = 5/[5 + 0.067] = 98.6%