01․ The voltage and current instantaneous values are given as 5 sin (ωt + 30°) pu and 2 sin (ωt - 15°) pu respectively. Find the per unit active power?

The voltage and current instantaneous values are given as
V = V

_{m}sin (ωt + θ_{1}) pu I = I_{m}sin (ωt + θ_{2}) pu Apparent power in pu (S_{pu}) = (V_{m}*I_{m})/2 Active power in pu (P_{pu}) = (V_{m}*I_{m})/2*cos (θ_{1}-θ_{2}) Reactive power (Q_{pu}) = (V_{m}*I_{m})/2*sin (θ_{1}-θ_{2}) From given data, Active power in pu (P_{pu}) = 5*2/2 * cos (30+15) = 3.535 pu02․ The angular moment of 10 MVA generator is 0.2. Find the new value of angular moment for 20 MVA base?

Angular moment (M) in pu = M/S

_{b}S_{b}= Base MVA M_{pun}= M_{puo}* (MVA)_{bo}/(MVA)_{bn}Here, "n" represents the new values "o" represents the old values M_{pun}= 0.2*10/20 = 0.1 pu03․ The per unit impedance Z(pu) is given by

Z

_{pu}= Z_{Ω}* (MVA)_{b}/(KV)²_{b}Where, Z_{pu}= Per unit impedance Z_{Ω}= Impedance in Ω (MVA)_{b}= Base MVA (KV)_{b}= Base voltage04․ The per unit impedance of a circuit element is 0.30. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element will be

Z

_{pu}= Z_{Ω}* (MVA)_{b}/(kV)²_{b}Where, Z_{pu}= Per unit impedance Z_{Ω}= Impedance in Ω (MVA)_{b}= Base MVA (kV)_{b}= Base voltage Z_{pun}= Z_{puo}* (MVA)_{bn}/(MVA)_{bo}* (kV)²_{bo}/(kV)²_{bn}Here, "n" represents the new values "o" represents the old values Z_{pun}= 0.30* 1/2 *4 = 0.60 pu05․ The per unit value of a 2 ohm resistor at 100 MVA and 10 kV base voltage is

Z

_{pu}= Z_{Ω}* (MVA)_{b}/(KV)²_{b}Where, Z_{pu}= Per unit impedance Z_{Ω}= Impedance in Ω (MVA)_{b}= Base MVA (KV)_{b}= Base voltage Z_{pu}= 2 * 100/10² = 2 pu06․ For a 500 Hz frequency excitation, a 100 km long power line will be modelled as

Classification of transmission lines:
Short transmission line = Length(L) < 80 km and product of length(L) and frequency(f) L.f< 4000
Medium transmission line = 80 km ≤ L ≤ 200 Km and 4000 ≤ L.f ≤ 10000
Long transmission line = L > 200 km and L.f >10000

07․ The charging current of a 400 kV line is more than that of a 220 kV line of the same length?

Charging current I

_{c}= V/X_{c}Where, V = Voltage X_{c}= Charging reactance = 1/(2πfl) l = length of the transmission line Charging current I_{c}= V*2πfl I_{c}∝ V Therefore, the given statement is correct statement.08․ If in a line, resistance and reactance are found to be equal and regulation is zero, then load will have

Voltage regulation= (Vs - Vr)/Vr = Ir(R cosφ ± X sinφ)/Vr
For lagging load:
Voltage regulation = Ir(R cosφ + X sinφ)/Vr
For leading load:
Voltage regulation = Ir(R cosφ - X sinφ)/Vr
For unity load power factor:
Voltage regulation = Ir(R cosφ)/Vr
Therefore, zero voltage regulation is possible only for leading power factor.
From given data
Cosφ = X/Z = 1/√2 = 0.707 leading

09․ The voltage regulation of a line at full load 0.8 power factor is 11.
A long transmission line under no load conditions, for a good voltage profile needs

Under no load conditions or light load conditions of a transmission line, receiving end voltage is greater than sending ending voltage will occur in case of medium and transmission lines. This is called ferranti effect. This is due to capacitance effect is dominating compare to inductance of transmission line under no load only.
To over come this problem, shunt reactors are connected at receiving end.

10․ The load current in short circuit calculations are neglected because

The load current in short circuit calculations are neglected because
1. Short circuit currents are much larger than load currents
2. Short circuit currents are greatly out of phase with load currents
Therefore, both statements are correct statements.

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