To find the transmitted voltage when a surge encounters a capacitance, we can use the following formula:

[ V_t = V_i times frac{Z_0}{Z_0 + Z_L} ]

Where:

– ( V_t ) = transmitted voltage

– ( V_i ) = incident voltage (240 kV)

– ( Z_0 ) = surge impedance (350 ohms)

– ( Z_L ) = load impedance

The load impedance ( Z_L ) for a capacitive load can be calculated using the formula:

[ Z_L = frac{1}{jomega C} ]

Where:

– ( C = 3000 , text{pF} = 3000 times 10^{-12} , text{F} )

– ( omega = 2pi f ), and for a 2 μs pulse, we can approximate ( f ) using the rise time, but for surge calculations, we can treat it for a high-frequency response effectively causing a near short circuit (ideally Z_L → 0), making it mostly capacitance impedance.

However, for simplification and practical scenarios in surge calculations, we will approximate ( Z_L ) as a very low impedance due to capacitance, and thus the transmitted voltage can be calculated as:

Using the approximation: ( Z_L approx 0 ), the formula simplifies to:

[ V_t = V

To find the no-load line voltage at the receiving end of a medium transmission line, you can use the formula:

[ V_R = V_S – I_S cdot Z ]

where:

– ( V_R ) is the receiving end voltage,

– ( V_S ) is the sending end voltage,

– ( I_S ) is the sending end current,

– ( Z ) is the line impedance.

However, for a no-load condition, the current ( I_S ) can be assumed to be very small, and the shunt admittance can be neglected for this calculation. Thus, the no-load receiving end voltage (( V_R )) can be approximated to be equal to the sending end voltage (( V_S )).

Given:

– Sending end voltage, ( V_S = 143 ) kV,

– Line impedance, ( Z = 101.24angle74° ) Ω,

– Shunt admittance ( Y = 7.38 times 10^{-4}angle90° ) Ω(^{-1}) (not needed for this approximation).

For the no-load condition:

[ V_R approx V_S ]

Hence, the no-load line voltage at the receiving end would be approximately:

143 kV.

To calculate the power factor angle at the reduced voltage regulation of 50%, we can follow these steps:

1. Calculate the apparent power (S):

[

P = 5 , text{MW} quad text{power factor (pf)} = 0.8

]

[

S = frac{P}{pf} = frac{5 , text{MW}}{0.8} = 6.25 , text{MVA}

]

2. Calculate the current (I):

Using the receiving end voltage (Vr = 10 kV):

[

S = V_r cdot I implies I = frac{S}{V_r} = frac{6.25 times 10^6 , text{VA}}{10 times 10^3 , text{V}} = 625 , text{A}

]

3. Calculate the voltage drop (V_drop) across the line:

[

V_{drop} = I cdot (R + jX) = 625 , text{A} cdot (0.39 , Omega + j3.96 , Omega)

]

[

V_{drop} = 625 cdot 0.39 + j(625