Objective Questions on Electrical Transmission | 2

  1. For a 50 Hz 300 km long transmission line VS = 145∠4.9° kV, I S = 0.164∠-36.9° kA. What is the sending end power?

    Sending end power factor = cos(4.9°+36.9°) = cos41.8° = 0.745 lagging
    VSline = 145√3
    Sending end power = √ VSline IScosΦ = √3X145√3X0.164X0.745 = 53.2 MW.

  2. A 275 kV transmission line has the following line constants. A = 0.85∠5°, B = 200∠75°
    Determine the power at unity power factor that can be received at if the voltage profile at each end is to be maintained at 275 kV. Assume δ = 22°.

    Given VS = VR =275 kV, α = 5°, β = 75°
    PR = [VS X VR] / [B X cos(β-δ) − A/B X V2Rcos(β-α)]
    ∴ PR = 275X275 / 200 cos(75°−22°) − 0.85 / 200X2752cos(75°−5°) = 227.6-109.9 = 117.7 MW

  3. Find the load angle in the above question if power demanded by load is 150 MW?

    Given VS = VR = 275 kV
    PD = PR = 150 MW
    PR = VS X VR/B X cos(β-δ)−A/B X V2Rcos(β-α)
    ∴ 150 = 275X275/200 cos(75° − δ°) − 0.85/200X2752cos(75°−5°)
    150 = 378cos(75°&minusδ)-110
    ∴ δ = 28.46°

  4. What is the % regulation of a balanced 3Φ transmission line at no load, if its sending end voltage is 82.6 kV, receiving end voltage is 76.2 kV. Given transmission line parameters are : A = 0.992, B = 55.5, C = 0.000315, D = 0.0175

    VS⃗ = A⃗ VR⃗ + B⃗ IR
    At no load, IR⃗ = 0
    ∴ VS⃗ = A⃗ VR0
    VR0⃗ = 82.6/0.992 = 83.27kV
    %regulation = (83.27-76.2)/76.2 X100% = 9.25%

  5. A 200 km long 3Φ transmission line having sending end voltage of 44 kV and electric current of 178 A. If the line supplies load of 20 MW at 0.9 p.f. then determine the transmission efficiency?

    Sending end power = 3 VsIscosΦs
    = 3 X 44 X 103 X 178 X 0.9 = 21.2 MW
    Transmission efficiency = (20/21.2) X 100 % = 94 %

  6. A 3Φ 100 km long transmission line is supplied by load of 20 MW at 0.9 p.f. lagging at 66 kV at receiving end. Calculate the load current?

    For a long transmission line load current, IL = Power/(√3 Voltage X load p.f)
    or IL = 20 X 106/(√3X66X103X0.9) = 195 A

  7. A 3Φ line delivers 3600 kW at a p.f. 0.8 lagging to a load. If the line voltage at receiving end is 32 kV, determine line current.

    Receiving end voltage per phase = 32/√3 = 18.48 kV
    Line current, Ir = (power deliverd/phase) / VrXcosΦr [Where, Vr is line voltage at receiving end and cosΦr is load power factor.]
    = 1200X103 / (18.48X103X0.8) = 81.2 A

  8. In overhead transmission lines the effect of capacitance can be neglected when the length of transmission line is

    Effect of line capacitance to produce a charging electric current in quadrature with voltage. Transmission lines having the length less than 80 km operating at a voltage less than 20KV fall in category of short transmission. Due to small length lower voltage capacitance effects are small. Hence the effect of parameter C can be neglected. Hence performance of a short transmission line depends upon the parameters resistance and inductance.

  9. A single phase transmission line of impedance j0.8 ohm supplies a resistive load of 500 A at 300V. The sending end power factor is

    VS = VR+IZS

  10. A 220 kV, 20 km long, 3-phase transmission line has the following A, B, C, D constants: A = D = 0.96, C = 0.037, B = 25. Its charging electric current will be

    IS = CVR+DIR
    Here, VR = 220/√3
    Let, IR = 0, ∴ IS = 0.05X220 / √3 = 11 / √3A

  11. An 11 kV, 3-phase transmission line has a resistance of 1.5 Ω and reactance of 4 Ω per phase. Calculate the percentage regulation and efficiency of the line when a total load of 5000 kVA at 0.8 lagging power factor is supplied at 11 kV at the distant end.

    VR = (11X103)√3 = 6351 V
    Load electric current I= (5000X1000)/(3X6351)=262.43 A
    VS = VR + I R cos φR + I XL sin φR= 6351 + 262.43 X 1.5 X 0.8 + 262.43 X 4 X 0.6 = 7295.8 V
    % regulation =(7295.8-6351)/6351 X 100=14.88%.

  12. Estimate the distance over which a load of 15000 kW at a p.f. 0.8 lagging can be delivered by a 3-phase transmission line having conductors each of resistance 1Ω per kilometre. The voltage at the receiving end is to be 132 kV and the loss in the transmission is to be 5%.

    Line electric current I = (15000 X 103)/√ X 132 X 103 X 0.8 = 82 A. Line losses = 15000 X 0.05 = 750 kW = 3 X I2R.
    R=37.18Ω
    ∴ the length of the line is 37.18 km.

  13. A single phase overhead transmission line delivers 1100 kW at 33 kV at 0.8 p.f lagging. The total resistance and inductive reactance of the line are 10Ω and 15Ω respectively. Determine the sending end voltage.

    Total impedance Z = R+jXL = 10+j15
    Line electric current I = (1100 X 103) / (33000 X 0.8) = 41.67 A
    I = I(cosφR-jsinφR) = 41.67(0.8-j0.6)=33.33-j25.
    Sending end voltage = VR+IZ = 33000+(33.33-j25)(10+j15) = 33708.3+j250 = 33.7 kV

  14. When the load power factor in a three phase short transmission line is leading, the voltage regulation is

    Voltage regulation =
    ∴ When the load p.f. is leading that I R cos φr > I XL sinφr,then voltage regulation is positive.
    When the load p.f. is leading to an extent that I XLsinφr > I R cosφr,then voltage regulation is negative.

  15. The voltages at the two ends of a line are 66 KV &its reactance is 30Ω. The capacity of the line is

    P=V2⁄Z=662⁄30=145.2 MW

  16. The voltages at the two ends of a line are 66 KV & its reactance is 40Ω.The capacity of the line is-

    P=v2/R=(66)2/40=108.9 MW

  17. An 800 KV transmission line has a maximum power transfer capacity of P. If it is operated at 400 KV with series reactance unchanged, then the new maximum power transfer capacity is approximately

    Power transfer capacity proportional to v2 or, P1 ⁄ P2 = (V1 ⁄ V2)2 or, P2 = P*(400 ⁄ 800)2 = P ⁄ 4.

  18. A 800 KV transmission line is having per phase line inductor of 1.1 mH ⁄ Km & per phase line capacitance of 11.68 nF ⁄ Km. Ignore the length of the line, its ideal power transfer capability in MW is

    Parameter of the line, Inductance(L) = 1.1 mH ⁄ Km, Capacitance(C) = 11.68 nF ⁄ Km. Surge impedance of the line (Zs) = √(L ⁄ C) = √(1.1 X 10 − 3 ⁄ 11.68 X 10 − 9) = 306.88 Ω. Ideal power transfer capability = V2 ⁄ Zs = (800*103)2 ⁄ 306.88 = 2085MW.

  19. DC transmission has which following disadvantages

    In case of dc voltage transmission step up and step down of voltages are difficult because transformers don't work on dc. It is very difficult to get sufficiently high voltage in d.c. generation & it is not easy to step-up & step-down the d.c. voltage. D.c. transmission is very difficult for high voltage d.c. generation due to commutating problem.

  20. In which phenomenon sending end voltage is equal to receiving end voltage in transmission line?

    In case of long transmission line having high capacitance, sometimes receiving end voltage equals to or greater than sending end voltage, a phenomenon called Ferranti effect.