KNEC Diploma in Electrical and Electronic Engineering (Power Option), Module III
Paper codes: 2521/305 and 2601/305 · Session: June/July 2024 · Time: 3 hours
Revision note: Attempt the paper before reading the solutions. These independently prepared answers support revision and are not an official KNEC marking scheme. The paper has eight questions in two sections: answer any three from Section A and any two from Section B.
SECTION A: ELECTRICAL POWER SYSTEMS
Question 1 (20 marks)
(a) With reference to overhead-line conductors, state two (i) factors affecting conductor vibrations; and (ii) requirements of high-frequency vibration dampers. (4 marks)
Answer (a)(i). Factors include wind speed and direction, conductor diameter and mass, span length, conductor tension, surface roughness and the amount of damping at the supports.
Answer (a)(ii). A damper should absorb energy over the range of vibration frequencies expected for the conductor and should be positioned and clamped so it does not damage or abrade the strands. It should also withstand weather and remain mechanically secure in service.
(b)(i) With reference to corona in overhead lines, explain (I) visual critical voltage and (II) critical disruptive voltage. (b)(ii) A 220 kV, 50 Hz, three-phase transmission line has conductors of 2 cm radius spaced 3 m apart in equilateral triangular formation. Air temperature is 50 °C, atmospheric pressure is 76 cm Hg and irregularity factor is 0.8. Determine the air-density factor, critical disruptive voltage and corona power loss per kilometre. (11 marks total for part (b))
Answer (b)(i). The critical disruptive voltage is the phase-to-neutral rms voltage at which the electric field at the conductor surface first becomes sufficient to ionise the surrounding air and start corona. The visual critical voltage is the higher voltage at which the corona glow first becomes visible to the eye.
Answer (b)(ii). Use r = 2 cm, D = 300 cm, m0 = 0.8 and phase voltage Vph = 220/√3 kV.
δ = 3.92b/(273 + t) = 3.92 × 76/(273 + 50) = 0.92235
Vph = 220/√3 = 127.02 kV rms
Vd = 21.1m0δr ln(D/r) = 21.1 × 0.8 × 0.92235 × 2 × ln(300/2) = 156.02 kV per phase
Because Vph is below Vd, disruptive corona does not start under the stated fair-weather conditions. Peek’s corona-loss expression therefore does not apply and the predicted loss is zero.
Vph = 127.02 kV < Vd = 156.02 kV ⇒ Pc = 0 kW/km per phase
(c) A 132 kV transmission line has a conductor weight of 700 kg/km and a span length of 300 m. Ultimate strength is 4000 kg with a safety factor of 2. Determine (i) working tension; and (ii) sag. (5 marks)
Answer (c). Treat the stated conductor loads as kgf, as is customary in this sag calculation. For supports at equal level and shallow sag, f = wL²/(8T).
Twork = 4000/2 = 2000 kgf
w = 700 kgf/km = 0.7 kgf/m
f = wL²/(8Twork) = 0.7 × 300²/(8 × 2000) = 3.94 m
Question 2 (20 marks)
(a) State two merits of an expulsion-type arrester. (2 marks)
Answer (a). It is relatively simple and inexpensive, and its arc-quenching action can interrupt the power-frequency follow current after a surge has been discharged to earth.
(b) With the aid of a circuit diagram and volt-current characteristic curve, describe the operation of a surge diverter. (6 marks)
Answer (b). A surge diverter is connected between the line and earth. At normal voltage it presents a very high impedance. When an overvoltage exceeds its operating level, the gap or nonlinear element conducts and diverts surge current to earth. Its nonlinear resistance limits the voltage across protected equipment. After the surge, the element returns to high impedance and any follow current is interrupted.
Line ────────┬──────────────── protected equipment
│
[spark gap /]
[nonlinear ] surge diverter
[resistor ]
│
Earth
V
│ /
│ _/ steep rise after clamping region
│ __/
│___________________/________________ I
low leakage surge current diverted
(c) A 110 kV, three-phase, 50 Hz transmission line 120 km long delivers 50 MW at 0.8 power factor lagging. Per-phase, per-kilometre line resistance is 0.4 Ω and reactance is 0.6 Ω. Capacitive admittance is 3 × 10−6 S/km/phase. Using the nominal-T method, determine (i) receiving-end current; and (ii) sending-end current. (12 marks)
Answer (c). Take receiving-end phase voltage as the reference. The total series impedance is split equally around the shunt admittance.
IR = P/(√3VLLpf) = 50 × 10⁶/(√3 × 110 × 10³ × 0.8) = 328.04∠−36.87° A
Z = 120(0.4 + j0.6) = 48 + j72 Ω per phase
Y = j(3 × 10⁻⁶ × 120) = j3.60 × 10⁻⁴ S per phase
VR,ph = 110/√3 = 63.509 kV
Vm = VR,ph + IRZ/2 = 76.893 + j4.724 kV
Ic = YVm = −1.701 + j27.681 A
IS = IR + Ic = 260.731 − j169.143 A = 310.79∠−32.97° A
IR = 328.04∠−36.87° A; IS ≈ 310.79∠−32.97° A
Question 3 (20 marks)
(a) Explain each of the following with reference to surges: (i) arcing ground; and (ii) resonance. (4 marks)
Answer (a)(i). Arcing ground is an intermittent earth fault, commonly on an ungrounded or weakly grounded system, in which the arc repeatedly extinguishes and restrikes. Each restrike can produce a transient overvoltage on the healthy phases.
Answer (a)(ii). Resonance occurs when inductive and capacitive reactances in a system interact at or near a natural frequency. Energy exchanged between L and C can produce abnormally high voltage or current, including damaging surge-like overvoltages.
(b) A 120 kV surge travels along a cable towards its junction with an overhead line. The cable has L = 0.3 mH/km and C = 0.5 μF/km; the overhead line has L = 2 mH/km and C = 0.015 μF/km. Determine (i) cable surge impedance; (ii) line surge impedance; (iii) transmitted surge voltage; and (iv) transmitted surge current. (8 marks)
Answer (b). The incident wave travels from the cable into the higher-impedance overhead line. For a lossless line, Z₀ = √(L/C).
Zcable = √[(0.3 × 10⁻³)/(0.5 × 10⁻⁶)] = 24.49 Ω
Zline = √[(2 × 10⁻³)/(0.015 × 10⁻⁶)] = 365.15 Ω
Vt = [2Zline/(Zcable + Zline)]Vi = 224.91 kV
It = Vt/Zline = 224.91 × 10³/365.15 = 615.95 A
(c) A 60 MVA, 11 kV three-phase alternator has fault currents of 1000 A for a three-phase fault, 1500 A for a line-to-line fault and 2500 A for a line-to-ground fault. The generator neutral is solidly grounded. Determine (i) positive-sequence reactance; (ii) negative-sequence reactance; and (iii) zero-sequence reactance. (8 marks)
Answer (c). Assume the prefault phase voltage equals its rated value. Use the fault-current relations for bolted faults.
Vph = 11/√3 = 6.351 kV
X1 = Vph/I3φ = 6350.85/1000 = 6.351 Ω
ILL = √3Vph/(X1 + X2)
X2 = (√3 × 6350.85/1500) − 6.351 = 0.982 Ω
ILG = 3Vph/(X1 + X2 + X0)
X0 = 3 × 6350.85/2500 − 6.351 − 0.982 = 0.288 Ω
The corresponding reactances on the 60 MVA, 11 kV base (Zbase = 2.0167 Ω) are:
X1 = 3.149 pu; X2 = 0.487 pu; X0 = 0.143 pu
Question 4 (20 marks)
(a) With reference to protective relays, explain (i) pick-up current; and (ii) plug-setting multiplier. (4 marks)
Answer (a)(i). Pick-up current is the minimum current at which a relay begins to operate or close its contacts.
Answer (a)(ii). Plug-setting multiplier (PSM) is the actual current through the relay divided by the relay’s pick-up current, using secondary quantities at the relay.
PSM = relay current / relay pick-up current
(b) With the aid of a labelled diagram, describe the operation of a balanced-beam relay. (6 marks)
Answer (b). Under normal current the beam remains balanced by its spring or counterweight and the trip contact stays open. When current in the operating coil exceeds pick-up, the electromagnet attracts its armature and unbalances the beam. The moving contact closes the trip circuit, energising the circuit-breaker trip coil.
CT secondary
│
[operating coil] ── attracts armature
│ ↓
└────── pivoted beam ──────[restraining spring]
│
trip contact ─── trip coil / breaker
(c) State two types of relays used in transformer protection. (2 marks)
Answer (c). Buchholz relay and differential relay. Overcurrent and earth-fault relays are also commonly used for backup protection.
(d) A star-connected, 20 MVA, 6.6 kV three-phase alternator is protected by a Merz–Price circulating-current scheme using 1000/10 A CTs. The star point is earthed through 8 Ω. If the relay minimum operating current is 0.8 A, determine the percentage of stator winding that is unprotected. (8 marks)
Answer (d). The CT ratio is 100:1, so the relay’s 0.8 A pick-up corresponds to 80 A primary. At a fraction x of the phase winding measured from the neutral, the phase-to-earth voltage is approximately xVph, and the earth-fault current through the neutral resistor is xVph/Rn. The portion nearest the neutral remains below relay pick-up.
Iprimary,pick-up = 0.8 × (1000/10) = 80 A
Vph = 6.6/√3 = 3.811 kV
If(at fraction x) = xVph/Rn
x = Iprimary,pick-upRn/Vph = (80 × 8)/3810.5 = 0.16796
Unprotected winding = 0.16796 × 100% ≈ 16.8%
Question 5 (20 marks)
(a) State two assumptions made in stability studies of a power system. (2 marks)
Answer (a). Common assumptions are that mechanical input power remains constant over the short transient and that damping and losses are neglected in a first-swing study. A balanced network and constant generator internal voltage are also often assumed.
(b) A generator delivers 1.0 pu power to an infinite bus through a purely reactive network. Its maximum transferable power is 2 pu. A three-phase terminal fault reduces output to zero; when cleared, the original network is restored. The maximum rotor swing is δmax = 110°. Sketch the equal-area criterion plot and determine the critical clearing angle. (11 marks)
Answer (b). Before the fault, Pm = 1 pu and Pmax = 2 pu. The initial angle satisfies Pm = Pmax sin δ₀. During the fault Pe = 0, so the rotor accelerates. At the stability limit, accelerating and decelerating areas are equal.
Power P
2 ┤ post-fault Pe = 2 sin δ
│ /¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 1 ┼───────●──── A₁ ────────●──────── Pm = 1 pu
│ δ₀=30° δc A₂
0 ┼────── fault Pe=0 ────────────────●──── δ
δmax=110°
δ₀ = sin⁻¹(Pm/Pmax) = sin⁻¹(1/2) = 30°
A₁ = Pm(δc − δ₀) = δc − δ₀
A₂ = ∫δcδmax(2 sin δ − 1)dδ = 2(cos δc − cos δmax) − (δmax − δc)
A₁ = A₂ ⇒ cos δc = cos δmax + (δmax − δ₀)/2
δmax = 110° = 1.9199 rad; δ0 = 30° = 0.5236 rad
cos δc = cos(1.9199) + (1.9199 − 0.5236)/2 = 0.3561
δc = cos⁻¹(0.3561) = 69.14°
(c) Derive the swing equation of a synchronous machine connected to an infinite bus bar. (7 marks)
Answer (c). Neglect damping. The rotor’s accelerating torque is the difference between mechanical input torque and electromagnetic torque. Newton’s rotational equation gives Jd²θm/dt² = Tm − Te. Multiplying by synchronous angular speed converts accelerating torque to accelerating power.
Pa = Pm − Pe = ωs(Tm − Te)
Jd²θm/dt² = (Pm − Pe)/ωs
Define H as stored kinetic energy at synchronous speed in seconds on the machine MVA base and δ as rotor electrical angle relative to the infinite bus. Expressing the rotor equation in per-unit gives the standard swing equation:
(2H/ωs)d²δ/dt² = Pm − Pe
d²δ/dt² = (ωs/(2H))(Pm − Pe)
SECTION B: ELECTROMAGNETIC FIELD THEORY
Question 6 (20 marks)
(a) State two (i) types of electromagnetic radiations; and (ii) differences between progressive and stationary waves. (6 marks)
Answer (a)(i). Examples of electromagnetic radiation types include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.
Answer (a)(ii). A progressive wave travels through space and carries energy; its phase changes with position and it has no fixed nodes. A stationary wave is formed by superposition of equal waves travelling in opposite directions; its nodes and antinodes stay fixed and it has no net average energy transport along the medium.
(b) With the aid of a labelled diagram, describe the construction of a Geiger–Müller tube for detection of electromagnetic radiation. (6 marks)
Answer (b). A Geiger–Müller tube is a sealed gas-filled detector for ionising radiation such as X-rays and gamma rays. A thin central anode wire is insulated from a cylindrical metal cathode. The tube contains a low-pressure inert gas and a small amount of quenching gas. A thin mica window allows incident radiation to enter. A high-voltage supply is applied between anode and cathode through a large series resistor; pulses from the anode are sent to an amplifier or counter.
Incident ionising radiation
↓
thin mica window
┌───────────────────┐
│ low-pressure gas │ metal cylinder = cathode (−)
│ │ │
│ thin anode wire │──────── high voltage (+)
└────────┬──────────┘
└──────── series resistor ── pulse counter
(c) A parallel-plate capacitor has square metal plates each 100 cm on a side and 4 mm apart. The space is filled with a dielectric of relative permittivity 4. A potential difference of 320 V is maintained. Determine (i) capacitance; (ii) charge; (iii) electric flux density; and (iv) potential gradient. (8 marks)
Answer (c). Plate side = 1 m, area A = 1 m², separation d = 0.004 m and ε = ε₀εr.
C = ε₀εrA/d = (8.854 × 10⁻¹² × 4 × 1)/0.004 = 8.854 × 10⁻⁹ F = 8.854 nF
Q = CV = 8.854 × 10⁻⁹ × 320 = 2.833 × 10⁻⁶ C = 2.833 μC
D = Q/A = 2.833 × 10⁻⁶ C/m²
E = V/d = 320/0.004 = 8.00 × 10⁴ V/m = 80.0 kV/m
Question 7 (20 marks)
(a) State the four Maxwell equations in differential form. (4 marks)
Answer (a).
∇ · D = ρv
∇ · B = 0
∇ × E = −∂B/∂t
∇ × H = J + ∂D/∂t
(b) In free space, E = 50 cos(10⁸t + βx) ay V/m. Determine (i) direction of wave propagation; (ii) phase constant β; (iii) wavelength λ; and (iv) time taken for the wave to travel λ/2. (8 marks)
Answer (b). A constant phase in ωt + βx moves toward decreasing x, so the wave propagates in −x. In free space c = 3 × 10⁸ m/s.
Direction = −ax
β = ω/c = 10⁸/(3 × 10⁸) = 0.3333 rad/m
λ = 2π/β = 18.85 m
tλ/2 = (λ/2)/c = π/ω = 3.142 × 10⁻⁸ s = 31.42 ns
(c) A transverse electromagnetic wave of frequency 14 GHz propagates in a homogeneous medium with μr = 1. Its field phase term is ej(ωt − 280πy). Assuming c = 3 × 10⁸ m/s, determine (i) phase constant; (ii) wavelength; (iii) wave velocity; and (iv) relative permittivity. (8 marks)
Answer (c). The spatial phase coefficient gives β = 280π rad/m; the wave travels in +y. Its angular frequency is ω = 2πf.
β = 280π = 879.65 rad/m
λ = 2π/β = 1/140 = 7.143 × 10⁻³ m
v = ω/β = 2π × 14 × 10⁹/(280π) = 1.00 × 10⁸ m/s
εr = (c/v)²/μr = (3 × 10⁸/1 × 10⁸)²/1 = 9
Question 8 (20 marks)
(a)(i) With reference to energy and momentum in an electromagnetic field, define skin effect. (a)(ii) State Poynting’s theorem. (4 marks)
Answer (a)(i). Skin effect is the concentration of alternating current and electromagnetic field near the surface of a conductor. The field decays with depth, reducing the effective conducting area at higher frequency.
Answer (a)(ii). Poynting’s theorem is electromagnetic energy conservation: outward electromagnetic power flow plus the rate of increase of stored field energy and power dissipated in the volume sums to zero.
∇ · (E × H) + ∂u/∂t + J · E = 0, u = ½(E · D + B · H)
∮S(E × H) · dS = −d/dt ∫Vu dV − ∫VJ · E dV
(b) Describe (i) diamagnetic; and (ii) paramagnetic materials. (4 marks)
Answer (b)(i). A diamagnetic material develops a small magnetisation opposing the applied field. Its susceptibility is slightly negative and its relative permeability is slightly below one; it is weakly repelled by a magnetic field.
Answer (b)(ii). A paramagnetic material has a small positive susceptibility and relative permeability slightly above one. Its magnetic dipoles tend to align with the applied field, so it is weakly attracted; the magnetisation largely disappears when the field is removed.
(c) Charges Q₁ = 200 nC and Q₂ = 100 nC are at A(3, 2, −1) and B(−1, −1, 4). A charge Q₃ = 500 nC is at C(0, 3, 1). Determine (i) distance vectors RAC and RBC; (ii) force on Q₃ due to Q₁; and (iii) force on Q₃ due to Q₂. (12 marks)
Answer (c). Treat the coordinates as metres. For each force, use Coulomb’s law with displacement from the source charge to C.
R⃗AC = r⃗C − r⃗A = (−3ax + ay + 2az) m
RAC = √14 = 3.742 m
R⃗BC = r⃗C − r⃗B = (ax + 4ay − 3az) m
RBC = √26 = 5.099 m
F⃗13 = (1/4πε₀)(Q₁Q₃/RAC³)R⃗AC
F⃗13 = (−5.147ax + 1.716ay + 3.431az) × 10⁻⁵ N
|F⃗13| = 6.420 × 10⁻⁵ N
F⃗23 = (1/4πε₀)(Q₂Q₃/RBC³)R⃗BC
F⃗23 = (3.390ax + 13.558ay − 10.169az) × 10⁻⁶ N
|F⃗23| = 1.728 × 10⁻⁵ N
Revision tip: For every fault and field calculation, state the reference direction, units and assumptions before substituting the supplied values.