Power is directly proportional

Section A
1. Power is directly proportional to V2, so if the voltage of a transmission line is double
power is expected to be increased by 4 times.
2. The formula for per unit impedance Zpu = ZΩ * (MVA)base / (KV)2base
3. Characteristic impedance is independent of the transmission line’s length.
4. The formula for voltage regulation of medium transmission line is (Vs - Vr) / Vr * 100
5. Protections schemes to be employed on transmission line are distance, earth fault and
overcurrent protection.
Section B
1. S = √ 3 * VL-L * IL............1
S = 500MVA
VL-L = 345kV
From equation 1
IL = S / √ 3 * VL-L
= (500 * 106) / √ 3 * (345 * 103)
= 836.74A
a) Load impedance per phase
Impedance per phase, Z = Vph / Iph
For delta connection phase voltage is equal to line voltage and line current is √ 3 * phase
current, thus, VRY = 345 < 00 kV
IRY = 836.74 / √ 3
= 483< -300 A
Impedance in RY phase, ZRY = VRY / IRY
= 345< 0 kV / 483< -300 A
= 237.67<30 Ω
Assuming a balanced load, Zry=Zyb = Zrb = 236.67<30 Ω
b) Load current and phase current
Phase currents
IRY = 483 < -30 A
IYB = 483 < -150 A

IBR = 483 < 90 A
Load current is equal to the line current which is equal to 1449.275<-30 A
c) Real and reactive power per phase
Real power, p = Vph * Iph * p.f
= 345 * 483 * 0.866
= 144.305 kW
Reactive power, Q = Vph * Iph * sin ∅
= 345 * 483 * sin 30
= 83.318kVar
d) Total real and reactive power by load
Total real power = 3 * real power per phase
= 3 * 144.305
= 432.915 kW.
Total reactive power = 3 * reactive power per phase
= 3 * 83.318
= 250 kVar
2. The main function of protection relay is to detect existence of a fault, its location and the
type of fault. Protection relay plays a role in tripping the circuit by operating the circuit
breaker, thus isolating the faulty section or system from the healthier one. This help in
reducing damage to equipment, animals and also human beings. The basic requirement of
a protection relay is to isolate the faulty system in the shortest time possible when a short
circuit is detected or when abnormal behavior is experienced.
Relays to be used in transformer protection are differential relay, Over-flux relay,
Buchholz relay, winding temperature trip relays, oil temperature trip and many more.
Relays to be used in transmission line protection are inverse time relays with definite
minimum time (IDMT), switched distance scheme. And finally, in busbars, overcurrent
relays, numerical relays, differential relays are widely used.
3. Ia = 10 < 0
Ib = 0
Ic = 10<120
a) Positive sequence current
I1 = 1/3 (Ia + aIb + a2Ic) where a = 1<1200 and a2 = 1<2400
= 1/3 (10 < 0 + 0 + 1<2400(10<120))
= 6.667<0
b) Negative sequence current
I2 = 1/3 (Ia + a2 * Ib + aIc) where a = 1<1200 and a2 = 1<2400

= 1/3 (10<0 + 0 + 1<120(10<120))
= 3.33<-60
c) Zero sequence current
Ia0 = Ib0 = Ic0 = 1/3 (Ia + Ib + Ic)
= 1/3 (10<00 + 0 + 10<1200)
= 3.33<60
d) The neutral current
In = Ia + Ib + Ic
= 10 <0 + 0 + 10<120
= 10<600 A
Section C
1. Industrial motors
a) The line current
Power of the load = 500kW
Power factor, cos ∅ = 0.6, thus ∅ = cos-1(0.6) = 530
P = √ 3 * VL * IL * cos ∅
IL = P / (√ 3 * VL * cos ∅)
=500kW / (√ 3 * 2300 * 0.6)
= 209.2 A
b) Voltage at the sending end
From ohms law
Voltage at the sending end, Vs = IL * (Zeq) + VR (voltage at the receiving end)
= 209.2 * (0.25 + j0.75) + 2300
= 2357.53 < 3.80 V
c) Power, reactive power and voltampere at the sending end
Real power, P = √ 3 * VS * IL * cos ∅
= √ 3 * 2357.53 * 209.2 * 0.6
= 512.543 kW
Reactive power, Q = √ 3 * VS * IL * Sin ∅
= √ 3 * 2357.53 * 209.2 * Sin 53
= 682.225 kVar
Voltampere, S = √ 3 * VS * IL
=√ 3 * 2357.53 * 209.2
= 854.24kVA