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A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 μF/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end and energized with 400 kV at the sending end, the receiving end line voltage end line voltage in kV (round off to two decimal places) will be _____________.
    Correct answer is '418.85KV'. Can you explain this answer?
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    A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line...
    VS = 400 KV
    l = 300 km
    L1 = 1 mH / km / phase
    C1 = 0.01 μF / km / phase
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    Most Upvoted Answer
    A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line...
    ΜF/km per phase. The resistance of each phase is negligible. Find the sending-end voltage, receiving-end voltage, and the voltage regulation if the load at the receiving end is 800 MW at 0.8 power factor lagging.

    Solution:

    Given parameters are:

    Line length, l = 300 km
    Line inductance, L = 1 mH/km per phase
    Line capacitance, C = 0.01 μF/km per phase
    Load at the receiving end, S = 800 MW
    Power factor, cos(φ) = 0.8 lagging
    Frequency, f = 50 Hz

    First, we need to calculate the line parameters per phase for the entire transmission line.

    Inductance per phase, L' = L × l = 1 × 300 = 300 mH
    Capacitance per phase, C' = C × l = 0.01 × 300 = 3 μF

    Next, we can calculate the characteristic impedance of the transmission line per phase as follows:

    Zc = √(L'/C') = √(300/3 × 10^-6) = 1000 Ω

    The sending-end voltage can be calculated using the following formula:

    Vs = Vr + Iline × Zc

    where Vr is the receiving-end voltage and Iline is the current in the line.

    Since the load is inductive, the power factor is lagging, and we need to use the following formula to calculate the current in the line:

    cos(φ) = P/S = 0.8
    sin(φ) = √(1-cos²(φ)) = 0.6
    Q = S × sin(φ) = 800 × 0.6 = 480 MVAr
    Iline = Q/Vr = 480/(Vr × 10^6)

    Substituting the values in the sending-end voltage formula, we get:

    Vs = Vr + (480/Vr) × 1000

    To find the receiving-end voltage and the voltage regulation, we need to solve the above equation using an iterative method.

    Assuming the receiving-end voltage, Vr = 400 kV, we get:

    Vs = 400 + (480/400) × 1000 = 1520 kV

    The voltage drop in the line is given by:

    Vdrop = Iline × Zline = Iline × Zc

    where Zline is the total impedance of the transmission line per phase. Since the resistance of the line is negligible, the total impedance is equal to the characteristic impedance, i.e., Zline = Zc.

    The voltage regulation is given by:

    Regulation = (Vs - Vr)/Vr × 100%

    Substituting the values, we get:

    Regulation = (1520 - 400)/400 × 100% = 280%

    Since the voltage regulation is very high, we need to increase the voltage at the sending end to compensate for the voltage drop in the line. Assuming a new receiving-end voltage of Vr = 420 kV, we repeat the above calculations:

    Iline = 480/(420 × 10^6) = 1.14 kA
    Vs = 420 + (1.14 × 1000) = 1534 kV
    Regulation = (1534 -
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    Community Answer
    A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line...
    Consider a 300 Km power transmission line which is working with 50 Hz
    frequency. The source side voltage considered as 1 pu and theta is 18
    degree
    a)
    Find the mid-point voltage magnitude.
    b)
    Find the reactive power absorbed at each ending points based on the P0.
    c)
    Find the X value in order to maintain the mid-point voltage equal to
    unity

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    it depends on the length of the conductor the capacitance of the line is proportional to the length of the transmission line their effect is negligible on the performance of short having a length less than 80 km and low voltage transmission accidents of the transmission line along with the conductances forms the shunted mittens the conductance and the transmission line is because of the leakage over the surface of the conductor considered a line consisting of two conductors and be each of radius are the distance between the conductors being Des shown in the diagram below minus the potential difference between the conductors and via's work QA charge on conductor QB charge on conductor vvab pencil difference between conductor and the Epsilon minus absolute primitivity QA plus QV = 0 so that QA equals QB - equals DBA equals data equals DB equals our substituting these values and voltage equation we get the capacitance between the conductors is cab is referred to as lying to line capacitance if the two conductors are in VR oppositely charge then the potential difference between them is zero then the potential of each conductor is given by one half bath the capacitance between each conductor and point of zero potential and is capacitive CN is called the capacitance to neut or capacitance to ground capacitance cab is the combination of two equal capacity and VN series thus capacitance to neutral is twice the capacitance between the conductors IE CN equals to Cave the absolute primitivity Epsilon is given by Epsilon equals epsilono Epsilon are where epsilano is the permittivity of the free space and Epsilon or is the relative primitivity of the medium prayer capacitance reactants between one conductor and neutral capacitance of the symmetrical three phase line let a balanced system of voltage be applied to a symmetrical three-phase line shown below the phasor diagram of the three phase line with equilateral spacing is shown below take the voltage of conductor to neutral as a reference phaser the potential difference between conductor and we can be written the similarly potential difference between conductors and sea is on adding equations one and two we get also combining equation three and four from equation 6 and 7 the line to neutral capacitance the capacitance of symmetrical three phase line is same as that of the two wire line Related: Capacitance of Transmission Lines?

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    A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 μF/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end and energized with 400 kV at the sending end, the receiving end line voltage end line voltage in kV (round off to two decimal places) will be _____________.Correct answer is '418.85KV'. Can you explain this answer?
    Question Description
    A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 μF/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end and energized with 400 kV at the sending end, the receiving end line voltage end line voltage in kV (round off to two decimal places) will be _____________.Correct answer is '418.85KV'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 μF/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end and energized with 400 kV at the sending end, the receiving end line voltage end line voltage in kV (round off to two decimal places) will be _____________.Correct answer is '418.85KV'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 μF/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end and energized with 400 kV at the sending end, the receiving end line voltage end line voltage in kV (round off to two decimal places) will be _____________.Correct answer is '418.85KV'. Can you explain this answer?.
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