Submitted by : Reza Ashraf (Rabbi), B.Sc. in EEE, 30 | B | Day, SUB.

Book – Principle of power system (V.K. Mehta)/Chapter 10 –

Performance of transmission lines/Page – 229/Q:2

A single phase 11 Kv line with a length of 15Km is to transmit 500kVA. The inductive reactance of the line is 0.5Ω/Km & the resistance is 0.3Ω/Km. Calculate the efficiency and regulation of the line for 0.8 lagging power factor.

Percentage of efficiency={V _RI_R Cos ф_R/ (V _RI_R Cos ф_R + I²R)} ×100%

 Total R = 0.3 × 15km = 4.5 Ω

 Total X_L = 0.5 × 15 = 7.5 Ω

V_R = 11 10³/ √3 = 6350.8 V

I_R = (500 × 10³ )/(√3 × 11 ×10³ × 0.8) =32.8 A

I²R = (32.8)² 1.5 = 1613.76 W

Z = 1.5 + j7.5

Cos ф_R = 0.8            Sinф_R = 0.6

V_S = √{(V_RCos ф_R+IR)² + (V_R Sinф_R+IX₂)²}

=√{27334493.5 + 16455029.99}

     =6617.36 V

        I.            % of transmission efficiency = V _RI_RCos ф_R/ V _RI_RCos ф_R + I²R ×100%

                                                                =6350.8 32.8 0.8 / 6350.8 32.8 0.8 + 1613.76

                                                                =166644.992/166644.99 + 1613.76 100

                                                                =99.04%

      II.            %Regulation efficiency = V_S – V_R / V_R  100

                                                =6617.36 – 6350.8 / 6350.8 ×100

                                                =4.2 %

(Answer)

Book – Principle of power system (V.K. Mehta)/Chapter 10 –

Performance of transmission lines/Page – 240/Q:1

A (medium) single phase transmission line 100Km long has the following constants Resistance/Km/phase = 0.15Ω.

Inductive reactance/Km/phase = 0.377Ω

Capacitive reactance/Km/phase = 31.87Ω

Receiving end voltage = 132kV

Assuming that the total capacitance of the line is localized at the receiving end along, determine:

(1)Sending end current,(2) Line value of sending end voltage(3) Regulation

(4) Sending end power factor

Solution:    

Total Resistance /Phase R = 15Ω

Total Reactance /Phase X_L = 37.7Ω

Capacitive Reactance / Phase X_C = 3.187Ω

Capacitive Admittance / Phase  Y = 1/X_C = 3.13 ×10^-4 S

Receiving end line voltage V_R =132×10³ V

Load current I _R = 72× 10⁶/(132 ×10³× 0.8)

                                  = 681.82 A

Taking receiving end voltage as the reference phasor, we have,

V_R = V_R + j0 = 132 ×10³ V

I_R = I_R (0.8 – j 0.6)

    = 681.82 (0.8- j 0.6 

   = 545.46 – j 409.09            

    = 681.82<-36.87

Capacitive current I_C = j YV_R

                                            = j 3.13 ×10^-4 × 132 ×10³

                                            = j 41.32

        I.            I_S = I_R + I_C

    = 545.46 – j 409.06 + j 41.32

    = 545.46 – j 367.77

    = 545.46<- 33.99

    Magnitude of I_S = √{(545.46)² + (367.77)²}

                       = 657.86 A

      II.            V_S = V_R + I_SZ

      = 132 ×10³ + {(657.86<-33.99) (40.57<68.30)}

      = 132 ×10³ + 26689.38<34.31

      = 132 ×10³ + 22045.43 + j 15044

      = 154045.43 + j 15044

      =154778.28<5.58

    III.            % Regulation = {(V_S – V_R)/ V_R} ×100

                                           = {(154.77 – 132)/132} × 100

                                           = 17.3%

    IV.            Ѳ₁ = 33.99

              Ѳ₂ = 5.58

              Ф_s  = Ѳ₁  + Ѳ₂= 33.99+5.58

                             = 39.57

Sending end power factor   Cos ф_s = Cos 39.57

        = 0.77 lagging                                                        (Answer)

Book – Principle of power system (V.K. Mehta)/Chapter 10 –

Performance of transmission lines/Page – 240/Q:4

A 3-phase, 50Hz, 100Km transmission has the following constants:

Resistance/Km/phase = 0.1Ω.

 Reactance/Km/phase = 0.5Ω

 Susceptance/Km/phase = 10^-5siemen

If the line supplies a load of 20MW at 0.9 p.f . lagging at 66kVat the receiving end, Calculate by using nominal π method:

(1)Sending end current (2)Line value of sending end voltage (3)Sending end power factor (4)Regulation

Solution:

Total R = 0.1 Ω×100 = 10Ω

Total X_L = 0.5Ω× 100 = 50Ω

Total Y = 10^-5 ×100 = 10^-3

p.f = 0.9

V = 66kv

P = 20MW

Z = R + j X_L

   = 10 + j 50 = 50.99 < 79^o

V_R = V_R + j 0 = 38105 V

I_R = I_R(Cos ф_R - j Sin ф_R)

      = 194 ( 0.9 – j 0.4)

      = 175 – j 78

      = 192 < -24^o

V_R = 66000/√3 = 38105 V

I_R = {20 × 10⁶/ (66000 × √3 × 0.9)}

    = 194 A

V₁= V_R + I_R Z/2

    = 38105 + (192 <-24^o × 26 < -24^o)

    =38105 + 4992 < -48^o

    = 38105 + 3340 – j 3710

    = 41445 – j 3710

    = 41610 < -5^o

I_C = jYV₁

     = j × 10^-3× (41445 – j 3710)

     = j 41.45 + 3.7

Impedance per phase, Z = R + j X_L

                                                   = 9.6 + j 30.46 = 31.9 < 73^o

Receiving end voltage, V_R = V_R + j 0

                                                        = 38105 V

Load current, I_R = I_R (Cos ф_R - j Sin ф_R)

                                   = 262 (0.8 – j 0.6)

                                   = 262 × 1< -37= 262 < -37= 209 – j 158

Cos ф_R = 0.8            Sinф_R = 0.6

Changing current, I_C = j YV₁

Now, V₁ = V_R + T_R Z/2

                   = 38105 + 262-37 15.9 < 73= 37105 = 4165.8 < 36

                   = 38105 + 3370 + j 2449 = 41475 + j 2449

I_C = j 2×10^-4 × (41475 + j 2449) = j 8.295 – 0.5

Sending current, I_S = I_R + I_C

                                         = 209 – j 158 + j 8.295 - 0.5

                                         = 108.5 – j 149. 7 = 257 <-36^o

I_S = I_R + I_C

    = 175 – j 78 + 3.7 + j 41.45

    = 178.7 – j 36.55 = 182 < -12^o

V_S = V₁ + I_S Z/2

     = 41445 – j 3710 + 182< -12^o× 26 <79^o

     = 41445 – j 3710 + 4732< 67^o

     = 41445 – j 3710 + 1849 + j 4356  = 43294 + j 646

V_S = (43294)² + (646)²= 43299 V                                                                              (Answer)

Book – Principle of power system (V.K. Mehta)/Chapter 10 –

Performance of transmission lines/Page – 240/Q:5

A 3-phase, overhead transmission line has the following constants:

Resistance/Km/phase = 10Ω., Inductive reactance/Km/phase = 35Ω

 Capacitance admittance/Km/phase = 3×10^-4siemen

If the line supplied a balanced load of 40000 kVA at 110kV and 0.8p.f . Lagging, Calculate :

(1)Sending end power factor.(2) % Regulation,(3) Transmission efficiency

Solution:

Given,

R = 10

X_L = 35

Y = 3 10-4  siemen

Cos ф_R = 0.8            Sin ф_R = 0.6

V_R = 110 × 10³/√3

      = 63508.5 V

I_R = 40000 × 10³/ 110 × 10³/√3 × 0.8

     = 262 A

Z = 10 + j 35 = 36.4 < 74

V_R = V_R + j O = 63508.5

I_R = I_R (Cos ф_R - j Sin ф_R)

     = 262(0.8 – j 0.6)

     = 209.8 – j 157.2

       = 262 < -36.84

V₁ = V_R + I_R Z/2

      = 63508.5 + (209.8 – j 157.2) (10 +j 38/2)

      = 63508.5 + 209.8 – j 157.2) (5 + j 17.5)

      = 63508.5 + [(262.2<-36.84) (18.2<74.05)     

      = 63508.5 + 4772.04 < 37.21

      = 63508.5 + ( 3800.5 + j 2885.8)

      = 67309 + j 2885.8

I_C = j YV₁

     = j 3 ×10^-4 (67309 + j 2885.8)

     = 20.19 + j 0.86

I_S = I_R + I_C

    = 209.8 – j 157.2 + 20.19 + j 0.86

    = 230 – j 156.34

    = 278 <-34.2

    = 278 <-34.12

V_S = V_R + I_S Z/2

      = 67309 + j 2885.8 + (230 – 156.340) (5 + j 17.5)

      = 67309 + j 2885.8 + (278<-34.2) (18.2 < 74.05)

      = 67309 + j 2885.8 +5059.6 + 39.85

      = 72368.6 + 2925.65

      = 72428 < 2.31 = 72428 < 2^o18”

Ѳ₁₌ angle between V_R­ & V_S

Ѳ₂= angle between V_R­ & I_S

Ф_s = Ѳ₁ + Ѳ₂

     = 2^o18” + 34^o12” = 36^o18”

(1)    Cos ф_s = 0.81 lagging

(2)    % Regulation = {(V_S – V_R) / V_R }×100 %

            = {(72428 – 63508.5) / 63508.5} × 100

            = 14.04%

(3)    Sending end power = 3V_SI_S Cos ф_s

       =3 ×72428 ×278 ×0.81

       =48.93 MW

         Now, Efficiency =( 40/48.93)×100

                                                =81%

(Answer)