KNEC Diploma in Electrical and Electronic Engineering (Power Option), Module III
Paper codes: 2521/305 and 2601/305 · Session: March/April 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) Explain the following mechanical vibrations in overhead lines: (i) high frequency; (ii) galloping. (4 marks)
Answer (a)(i): High-frequency vibration. This is the small-amplitude, high-frequency oscillation of a conductor caused mainly by steady cross-wind and alternating vortices shed from the conductor. Repeated bending near a clamp can cause fatigue damage and broken strands.
Answer (a)(ii): Galloping. This is a low-frequency, large-amplitude motion of an overhead conductor, often caused when wind acts on an ice-coated or otherwise asymmetrical conductor. It can make phases clash and can damage conductors, insulators and supports.
(b) Explain the use of bundled conductors in reducing corona on overhead lines. (2 marks)
Answer (b). Several sub-conductors in each phase act electrically like one conductor with a larger effective radius. This lowers the surface electric-field gradient and raises the voltage at which corona begins, reducing corona loss, audible noise and radio interference.
(c) A 110 kV, 50 Hz, 200 km three-phase overhead line has a conductor diameter of 2 cm. The conductors are symmetrically spaced 3 m apart. Air temperature is 30 °C and barometric pressure is 74 cm of mercury. Determine (i) air-density factor; (ii) critical disruptive voltage; and (iii) corona power loss per phase. (8 marks)
Answer (c). Use the standard Peek roughness factor m0 = 0.85 for an ordinary stranded conductor because the paper does not specify a polished surface. The radius is 1 cm and the phase voltage is 110/√3 kV.
δ = 3.92b/(273 + t) = 3.92 × 74/(273 + 30) = 0.9574
Vph = 110/√3 = 63.51 kV rms
Vd = 21.1m0δr ln(D/r) = 21.1 × 0.85 × 0.9574 × 1 × ln(300/1) = 97.94 kV per phase
Since 63.51 kV is below the disruptive voltage, the conductor is below corona inception under these assumptions. Therefore Peek’s corona-loss equation is not activated. Even taking m0 = 1 for a polished conductor gives Vd ≈ 115.22 kV per phase, so the conclusion is unchanged.
Pc,km = [242.4(f + 25)/δ]√(r/D)(Vph − Vd)² × 10⁻⁵ kW/km per phase, for Vph > Vd
Vph = 63.51 kV < Vd = 97.94 kV ⇒ Pc = 0 kW/km per phase
Pc,total per phase = 200 km × 0 = 0 kW
(d) With the aid of a labelled diagram, derive an expression for sag of a conductor supported by equal levels of support. (6 marks)
Answer (d). Let the span be L, conductor weight be w newtons per metre of horizontal span, horizontal tension at the lowest point be H, and maximum sag be f. For a shallow conductor curve, take the lowest point O as the origin and x horizontally.
Support A Support B
●-----------------------------------●
\ /
\ ↓ w /
\ O /
\___________ lowest __________/
<----------- L/2 ----------->
↑ f to support level
At a point a horizontal distance x from O, the vertical component of tension balances the conductor weight wx. Since the horizontal tension is H, the curve slope is dy/dx ≈ wx/H. Integrating from O, where y = 0, gives the parabolic sag curve:
dy/dx = wx/H
y(x) = wx²/(2H)
The support is L/2 from the lowest point, so:
f = w(L/2)²/(2H) = wL²/(8H)
For a more exact catenary model, the corresponding mid-span sag is H/w[cosh(wL/(2H)) − 1]. The parabolic result is the usual engineering approximation for small sag.
Question 2 (20 marks)
(a) State two (i) internal causes of overvoltages; and (ii) merits of overhead ground wires. (4 marks)
Answer (a)(i). Examples of internal overvoltage causes are switching operations (including energising or interrupting an inductive load) and sudden load rejection. Arcing ground and resonance are other examples.
Answer (a)(ii). Overhead ground wires intercept direct lightning strokes and conduct the lightning current safely to earth. They reduce the chance of a line conductor suffering a direct stroke and lower lightning-induced flashovers, improving supply reliability.
(b) An overhead line has inductance and capacitance per kilometre of 1.5 mH and 0.08 μF. It is connected in series with a cable having inductance 0.3 mH/km and capacitance 0.4 μF/km. A 200 kV surge travels from the line towards the line-cable junction. Determine (i) each surge impedance; (ii) transmitted voltage; and (iii) transmitted current. (8 marks)
Answer (b). For a lossless line, surge impedance is √(L/C). At the junction, use the voltage-wave transmission coefficient from line impedance Z1 into cable impedance Z2.
Z1 = √[(1.5 × 10−3)/(0.08 × 10−6)] = 136.93 Ω
Z2 = √[(0.3 × 10−3)/(0.4 × 10−6)] = 27.39 Ω
Incident surge Vᵢ → line (Z₁) ──●── cable (Z₂) → transmitted wave Vₜ
Vt = [2Z2/(Z1 + Z2)]Vi = [2 × 27.39/(136.93 + 27.39)] × 200 = 66.67 kV
It = Vt/Z2 = 66.67 × 10³/27.39 = 2.434 kA
(c) A short three-phase transmission line delivers 6000 kW at 33 kV and 0.85 power factor lagging. The resistance and reactance of each conductor are 6 Ω and 10 Ω. Determine (i) sending-end current; and (ii) sending-end voltage per phase. (8 marks)
Answer (c). Neglect the line shunt capacitance, as appropriate for a short line. Take the receiving-end phase voltage as the reference.
IR = P/(√3VLLpf) = 6 × 10⁶/(√3 × 33 × 10³ × 0.85) = 123.50 A
φ = cos−1(0.85) = 31.79°; IS = IR = 123.50∠−31.79° A
VR,ph = 33/√3 = 19.053∠0° kV
Z = 6 + j10 = 11.662∠59.04° Ω
VS,ph = VR,ph + IRZ = 20.344∠1.86° kV per phase
Thus the sending-end line-to-line magnitude, for reference, is approximately 35.24 kV.
Question 3 (20 marks)
(a) With reference to faults, explain (i) symmetrical faults; and (ii) negative-sequence components. (4 marks)
Answer (a)(i). A symmetrical fault affects all three phases equally, such as a balanced three-phase short circuit. The three fault currents have equal magnitudes and remain 120° apart, so only the positive-sequence network is involved in the ideal fault calculation.
Answer (a)(ii). Negative-sequence components are three equal-magnitude phasors displaced by 120° in the reverse phase order. In a machine they produce a magnetic field rotating opposite to the normal positive-sequence field.
(b) With the aid of a labelled diagram, describe the operation of an expulsion-type arrester. (6 marks)
Answer (b). The arrester is connected between the line and earth. During a surge, the series spark gap breaks down and current flows through the arrester to earth. The arc heats the fibre-lined expulsion tube, generating gas that builds pressure and is expelled through the vent. The gas cools and de-ionises the arc path; at the next current zero the arc is extinguished and the power-frequency follow current is interrupted.
Line ── series spark gap ── arc in fibre-lined tube
│
gas pressure / vent
│
Earth
(c) A 30 MVA, 15 kV alternator has a solidly grounded neutral. Its positive-, negative- and zero-sequence reactances are 0.5 pu, 0.4 pu and 0.1 pu. A single-line-to-ground fault occurs at the terminals of the unloaded alternator. Before the fault, line-to-neutral voltage is 1 + j0 pu. Sketch the sequence-network connection and find (i) line-to-ground fault current in pu; and (ii) fault current in amperes. (10 marks)
Answer (c). For a solid single-line-to-ground fault, the positive-, negative- and zero-sequence networks are connected in series. Their sequence currents are equal; the phase fault current is three times the sequence current.
Positive: E₁ = 1 pu ── jX₁ ──┐
Negative: jX₂ ─────┼── fault ── ground
Zero: jX₀ ─────┘
I₁ = I₂ = I₀
I₁ = I₂ = I₀ = E₁/(jX₁ + jX₂ + jX₀) = 1/[j(0.5 + 0.4 + 0.1)] = −j1.0 pu
If = 3I₀ = −j3.0 pu = 3.0∠−90° pu
Ibase = Sbase/(√3VLL,base) = 30 × 10⁶/(√3 × 15 × 10³) = 1154.7 A
|If| = 3 × 1154.7 = 3464 A
Question 4 (20 marks)
(a)(i) Differentiate between unit and non-unit protection. (4 marks)
Answer (a)(i). Unit protection compares electrical quantities at the boundaries of a defined zone, usually currents measured by CTs at each end. It trips for an internal fault in that zone and remains stable for external faults. Non-unit protection does not directly compare both ends of a bounded zone; examples include overcurrent and distance protection, which use local measurements and coordination/time grading and may also provide backup coverage.
(a)(ii) State two demerits of the Merz–Price protection scheme. (2 marks)
Answer (a)(ii). It requires CTs at both ends and pilot wiring, increasing installation cost and complexity. CT ratio errors or saturation during heavy through-faults can also produce spill current and risk maloperation.
(b)(i) With the aid of a labelled schematic diagram, describe the operation of the voltage-balance differential relay. (5 marks)
Answer (b)(i). Current transformers at the two ends of the protected section feed pilot circuits arranged so their secondary voltages oppose. With normal load or an external fault, the CT secondary voltages are substantially balanced, so the operating relay receives negligible voltage and does not trip. An internal fault makes the end currents unequal; the resulting voltage difference drives current through the relay operating coil and trips the circuit breakers at the line ends.
Bus A CT₁ protected line CT₂ Bus B
────────[ ]─────────────── fault ────────[ ]────────
╲ ╱
╲──── pilot wires, opposing ───╱
[voltage relay]
│
trip both ends
(b)(ii) State three merits of Translay protection systems for transmission lines. (4 marks)
Answer (b)(ii). The Translay arrangement uses only two pilot wires, reducing wiring cost. Its pilot circuit current and CT burden are comparatively small, which is useful on longer protected sections. It provides fast, selective protection for faults within the protected line section and is stable for external faults when correctly set.
(c) Figure 1 shows a transmission line fitted with a Petersen arc-suppression coil. Derive an expression for the coil inductance L. (5 marks)
Answer (c). Let C₀ be the line-to-earth capacitance per phase and ω = 2πf. During a single-line-to-ground fault, the three phase capacitances contribute a total charging current of approximately 3ωC₀Vph. The neutral coil supplies an opposing current Vph/(ωL). At resonance these magnitudes are equal:
3ωC₀Vph = Vph/(ωL)
L = 1/(3ω²C₀) = 1/(12π²f²C₀)
If CT denotes the sum of the three phase-to-earth capacitances, the same relation is L = 1/(ω²CT).
Question 5 (20 marks)
(a) Explain (i) equal-area criterion; and (ii) critical clearing angle as used in power-system stability. (4 marks)
Answer (a)(i). The equal-area criterion is a graphical test of first-swing transient stability. The rotor remains in synchronism after a disturbance when the area of accelerating energy gained during the fault is no greater than the available decelerating-energy area after clearing.
Answer (a)(ii). The critical clearing angle is the largest rotor angle at which the fault can be cleared while preserving synchronism. At the stability limit, the accelerating and decelerating areas are just equal.
(b) Explain three methods used to improve the transient stability of a power system. (6 marks)
Answer (b). (1) Clear faults rapidly with high-speed relays and circuit breakers; this limits the time and rotor-angle increase during acceleration. (2) Reduce transfer reactance by adding parallel transmission paths or series compensation, increasing the post-fault maximum power transfer. (3) Use fast excitation and voltage regulation to raise generator internal emf and electrical output after the disturbance. High-speed reclosing and dynamic braking are additional methods.
(c) A 50 Hz, 4-pole, 500 MVA, 22 kV generator delivers rated MVA at 0.8 power factor lagging. A fault suddenly reduces its electrical power output by 40%. Neglect losses and assume constant shaft input. Determine (i) power before fault Pm; (ii) power during fault Pe; (iii) angular speed ω; and (iv) accelerating torque Ta. (10 marks)
Answer (c). Before the fault, real power is the rated apparent power multiplied by power factor. With losses neglected, this is the mechanical input Pm. The fault reduces electrical output to 60% of its pre-fault value.
Pm = 500 × 0.8 = 400 MW
Pe = (1 − 0.40) × 400 = 240 MW
ns = 120f/P = 120 × 50/4 = 1500 r/min
ωm = 2πns/60 = 157.08 rad/s
Ta = (Pm − Pe)/ωm = 160 × 10⁶/157.08 = 1.019 × 10⁶ N·m
SECTION B: ELECTROMAGNETIC FIELD THEORY
Question 6 (20 marks)
(a) State two (i) sources of electromagnetic radiations; and (ii) applications of electromagnetic waves. (4 marks)
Answer (a). Sources include accelerating electric charges, such as oscillating currents in an antenna, and thermal emission from hot matter. Applications include radio and mobile communications, radar and satellite links, medical imaging, and infrared heating or sensing.
(b) In free space, the magnetic-field intensity is H = 0.1 cos(2 × 10⁸t − kx) ay A/m. Determine (i) periodic time T; (ii) wavelength λ; (iii) phase constant β; and (iv) time for the wave to travel λ/8. (8 marks)
Answer (b). The angular frequency is ω = 2 × 10⁸ rad/s. In free space, wave speed is c ≈ 3 × 10⁸ m/s.
T = 2π/ω = 2π/(2 × 10⁸) = 3.142 × 10⁻⁸ s = 31.42 ns
β = ω/c = (2 × 10⁸)/(3 × 10⁸) = 0.6667 rad/m
λ = 2π/β = 9.425 m
tλ/8 = (λ/8)/c = T/8 = 3.927 × 10⁻⁹ s = 3.93 ns
(c)(i) State Ampère’s circuital law. (3 marks)
Answer (c)(i). The circulation of magnetic-field intensity around a closed path equals the enclosed conduction current for magnetostatic conditions. The general Maxwell–Ampère form also includes displacement current.
∮C H · dl = Ienc (magnetostatic form)
∮C H · dl = Ienc + d/dt ∫S D · dS (general form)
(c)(ii) A thin ring of radius 5 cm is centred at (0, 0, 1 cm) in the plane z = 1 cm and carries a 50 mA loop current. The paper gives the axial magnetic-field expression H = Ir²/[2(r² + z²)3/2] az. Determine the field at (0, 0, −1 cm) and (0, 0, 10 cm). The scan labels the requested quantity “electric field intensity,” although the supplied equation is for H. (5 marks)
Answer (c)(ii) — source wording note. The given expression determines magnetic field H, not electric field E. A static current loop does not by itself determine E, so the values below evaluate the supplied H expression; the paper likely intended to ask for magnetic-field intensity. Distances z in the expression are measured from the ring centre.
a = 0.05 m; I = 0.05 A
At (0, 0, −0.01 m): z = −0.01 − 0.01 = −0.02 m
Hz = Ia²/[2(a² + z²)3/2] = 0.4002 A/m
At (0, 0, 0.10 m): z = 0.10 − 0.01 = 0.09 m
Hz = Ia²/[2(a² + z²)3/2] = 0.05727 A/m
Both field vectors point along +az for the current sense represented by the printed axial-field expression.
Question 7 (20 marks)
(a) Distinguish between lossy and lossless media. (4 marks)
Answer (a). A lossy medium dissipates some wave energy, usually through finite conductivity or dielectric loss; its wave amplitude decreases with distance and its attenuation constant α is positive. An ideal lossless medium has no material power loss (σ = 0 and no dielectric loss); α = 0, so a uniform plane wave does not decay in amplitude as it propagates.
(b) A lossy dielectric has intrinsic impedance η = 200∠30° Ω. A wave propagates with H = 10e−αx cos(ωt − x/2) ay A/m. Determine (i) electric-field amplitude E₀; (ii) attenuation factor α; and (iii) skin depth δ. (10 marks)
Answer (b). The phase constant is β = 0.5 rad/m. For a uniform plane wave in a medium with real μ, η = jωμ/(α + jβ), so the impedance angle satisfies tan(∠η) = α/β.
E₀ = |η|H₀ = 200 × 10 = 2000 V/m
α = β tan 30° = 0.5 × tan 30° = 0.2887 Np/m
δ = 1/α = 3.464 m
For the stated H direction and propagation in +x, E points in −az; with the ejωt phasor convention E leads H by 30°.
(c)(i) State Coulomb’s law between two point charges. (2 marks)
Answer (c)(i). The electrostatic force between two point charges acts along the line joining them, is proportional to the product of the charges, and is inversely proportional to the square of their separation.
F⃗₁₂ = [q₁q₂/(4πε₀R²)] aR
(c)(ii) Charges of 5 nC and −2 nC are at (2, 0, 4) and (−3, 0, 5) in free space. Determine (I) the force on a 1 nC charge at (1, −3, 7); and (II) the electric field at that point. (4 marks)
Answer (c)(ii). Treat the coordinates as metres. The displacement vectors from the source charges to the field point are R⃗₁ = (−1, −3, 3) m and R⃗₂ = (4, −3, 2) m.
R₁ = √19 m; R₂ = √29 m
E⃗ = (1/4πε₀)[(5 × 10⁻⁹R⃗₁/R₁³) + (−2 × 10⁻⁹R⃗₂/R₂³)]
E⃗ = (−1.003ax − 1.283ay + 1.398az) V/m
|E⃗| = 2.146 V/m
F⃗ = q₀E⃗ = (−1.003ax − 1.283ay + 1.398az) × 10⁻⁹ N
|F⃗| = 2.146 × 10⁻⁹ N
Question 8 (20 marks)
(a) Explain skin effect as applied in electromagnetic waves. (2 marks)
Answer (a). Skin effect is the tendency of alternating current and electromagnetic fields to concentrate near the surface of a conductor. The field amplitude falls with depth, so at higher frequency the effective conducting area is smaller and AC resistance is higher.
(b)(i) Explain electromagnetic shielding in energy momentum. (2 marks)
Answer (b)(i). Interpreting the printed wording as shielding against transfer of electromagnetic energy and momentum, a shield reduces the field and Poynting-energy flux reaching a protected region. It does this by reflecting incident energy, absorbing and dissipating part of it, and redirecting induced currents and fields at the boundary.
(b)(ii) State four methods of electromagnetic shielding. (4 marks)
Answer (b)(ii). Four methods are: (1) a bonded conductive enclosure or Faraday cage; (2) conductive sheet, foil or mesh around the protected equipment; (3) high-permeability magnetic material to divert low-frequency magnetic flux; and (4) lossy absorbing material or ferrite to absorb electromagnetic energy. Good bonding and filtered cable penetrations improve an enclosure’s effectiveness.
(c) In a non-magnetic medium, E = 5 sin(2π × 10⁷t − 0.8x) ay V/m. Determine (i) relative permittivity εr; (ii) intrinsic impedance η; and (iii) total power transferred. (12 marks)
Answer (c). The phase constant is β = 0.8 rad/m, angular frequency is ω = 2π × 10⁷ rad/s, and μ = μ₀. The field is a lossless travelling wave in +x. No cross-sectional area is provided, so the determinate power result is the average power flux per unit area; total power through an area A is SavA.
v = ω/β = (2π × 10⁷)/0.8 = 7.854 × 10⁷ m/s
εr = (c/v)² = (βc/ω)² = 14.57
η = η₀/√εr = 376.73/√14.57 = 98.70 Ω
Sav = E₀²/(2η) = 5²/(2 × 98.70) = 0.1267 W/m² in +ax
Ptotal = SavA = 0.1267A W, with A in m²
Revision tip: Keep the units and vector directions visible in field calculations. Where the scan’s wording conflicts with its supplied equation, identify the conflict and state exactly what the equation determines.