KNEC Diploma in Electrical and Electronic Engineering (Power Option), Module III
Past paper: Electrical Power Systems and Electromagnetic Field Theory | Paper codes: 2521/305 and 2601/305 | Session: October/November 2023 | Time: 3 hours
Revision note: Attempt the paper before reading the solutions. These independently prepared answers are for revision and are not an official KNEC marking scheme. The paper has eight questions in two sections. The examination instructions say to answer any THREE questions from Section A and any TWO from Section B; each question carries 20 marks.
SECTION A: ELECTRICAL POWER SYSTEMS
Question 1 (20 marks)
(a) (i) State two demerits of conductor vibrations in overhead lines. (ii) With the aid of a labelled diagram, explain the method of using armour rods to reduce conductor vibrations. (5 marks)
Answer (a)(i). Conductor vibration can cause fatigue cracks and broken strands, especially near suspension clamps. It can also abrade the conductor or loosen fittings and damage insulators.
Answer (a)(ii). Armour rods are preformed helical metal rods wrapped tightly around the conductor on both sides of the suspension clamp. They increase the conductor’s effective diameter and distribute bending and contact stresses over a longer length, protecting strands from clamp wear and reducing vibration damage.
Labelled schematic (not to scale):
suspension insulator
│
[suspension clamp]
│
span conductor ────────(((( armour rods ))))──────── span conductor
helical rods cover the clamp zone
(b) A three-phase 230 kV line consists of 24 mm diameter conductors symmetrically spaced 2 m apart. The line operates at 30°C and a pressure of 76 cm Hg. The irregularity factors for critical and visual corona are 0.82 and 0.72 respectively. Determine (i) the air-density factor; (ii) the critical disruptive voltage per phase; and (iii) the visual critical voltage per phase. (7 marks)
Answer (b). Use the standard Peek equations with conductor radius and phase spacing in centimetres. These voltages are rms values.
r = 24/2 = 12 mm = 1.2 cm; D = 2 m = 200 cm; m0 = 0.82; mv = 0.72
δ = (3.92 × b)/(273 + t) = (3.92 × 76)/(273 + 30) = 0.9832
ln(D/r) = ln(200/1.2) = 5.116
Vd = 21.1 m0 δ r ln(D/r) = 21.1 × 0.82 × 0.9832 × 1.2 × 5.116 = 104.4 kV per phase
Vv = 21.1 mv δ r (1 + 0.3/√(δr)) ln(D/r) = 117.0 kV per phase
The corresponding line-to-line values are approximately 180.9 kV for disruptive corona and 202.7 kV for visual corona.
(c) A transmission-line conductor has a diameter of 20 mm and weighs 0.9 kg/m. Its span is 300 m and the wind pressure is 50 kg/m². An ice coating 15 mm thick covers the conductor. Given an ultimate strength of 9000 kg, a safety factor of 3 and an ice density of 920 kg/m³, determine (i) the wind load per metre; (ii) the ice load per metre; and (iii) the sag. (8 marks)
Answer (c). Treat the stated kilogram loads as kgf, as is customary in this transmission-line calculation. The ice surrounds the conductor, so its outer diameter includes twice the radial ice thickness.
Do = 20 mm + 2(15 mm) = 50 mm = 0.05 m
wwind = 50 × 0.05 = 2.50 kgf/m
wice = 920 × (π/4)(0.05² − 0.02²) = 1.517 kgf/m
wvertical = 0.9 + 1.517 = 2.417 kgf/m
wresultant = √(2.50² + 2.417²) = 3.478 kgf/m; Tallow = 9000/3 = 3000 kgf
fresultant = wL²/(8T) = (3.478 × 300²)/(8 × 3000) = 13.04 m
The resultant sag is about 13.04 m in the direction of the combined wind and weight load. Its vertical component is about 9.07 m.
Question 2 (20 marks)
(a) (i) State two requirements of a direct lightning-stroke protection system. (ii) A voltage surge with a crest value of 4000 kV travels on an 800 kV line fitted with a protective arrester rated at 1800 kV. The surge impedance of the line is 400 Ω. Determine the transmitted current, reflected voltage, reflection coefficient and transmission coefficient. (8 marks)
Answer (a)(i). The system must intercept the stroke using correctly positioned air terminals or shield wires so that the protected equipment lies within the protected zone. It must also provide a continuous, low-impedance path through down-conductors and earth electrodes to discharge the lightning current safely.
Answer (a)(ii). Assume the arrester clamps the transmitted crest voltage to 1800 kV. Take the incident wave as positive.
Vi = 4000 kV; Vt = 1800 kV; Z0 = 400 Ω
Vr = Vt − Vi = 1800 − 4000 = −2200 kV
It = Vt/Z0 = 1.8 × 106/400 = 4500 A = 4.50 kA
Γ = Vr/Vi = −2200/4000 = −0.55; τ = Vt/Vi = 0.45
The negative reflection means the reflected voltage reverses polarity. The stated 800 kV is the line’s nominal operating voltage; it is not substituted for the surge crest in this wave calculation.
(b) Explain each of the following with respect to overhead transmission lines: (i) short; (ii) medium. (4 marks)
Answer (b). In the usual textbook classification, a short line is roughly up to 80 km; its shunt capacitance is small enough to neglect, so the model uses series resistance and reactance. A medium line is roughly 80–250 km; its shunt capacitance must be included, usually as a lumped nominal-T or nominal-π model. Length boundaries vary slightly between textbooks.
(c) A 66 kV, 120 km, 50 Hz three-phase line delivers 30 MW at 0.8 power factor lagging to a balanced load. Its resistance is 0.2 Ω/km, inductive reactance 0.35 Ω/km and capacitive susceptance 0.5 × 10⁻⁴ S/km. Using the nominal-T method, determine (i) the voltage across the capacitor; (ii) the current through the capacitor; and (iii) the sending-end current. (8 marks)
Answer (c). Work per phase. The receiving-end voltage is the reference phasor, and the full shunt susceptance is placed at the midpoint in the nominal-T model.
VR = 66/√3 = 38.105 kV per phase; φ = cos−1(0.8) = 36.87°
IR = (30 × 106)/(√3 × 66 × 103 × 0.8) ∠−36.87° = 328.04∠−36.87° A
Z = (0.2 + j0.35) × 120 = 24 + j42 Ω; Y = j(0.5 × 10−4 × 120) = j0.006 S
VC = VR + IR(Z/2) = 45.388 + j3.149 kV = 45.497∠3.97° kV per phase
|VC,LL| = √3 |VC| = 78.80 kV
IC = YVC = −18.895 + j272.326 A = 272.98∠93.97° A
IS = IR + IC = 243.537 + j75.502 A = 254.97∠17.22° A
The capacitor voltage is 45.50 kV phase-to-neutral (about 78.80 kV line-to-line for the balanced system).
Question 3 (20 marks)
(a) State three (i) demerits of the rod-gap arrester; and (ii) effects of arcing ground. (6 marks)
Answer (a)(i): demerits of a rod-gap arrester.
- Its spark-over voltage is affected by atmospheric conditions, electrode condition and gap spacing, so protection is not highly consistent.
- After a surge flashover, power-frequency follow current can continue; the rod gap does not extinguish it reliably by itself.
- The arc can damage or erode the electrodes, and the simple gap gives limited coordination and protection compared with a modern arrester.
Answer (a)(ii): effects of arcing ground.
- Intermittent restriking produces transient overvoltages that can rise well above normal phase-to-earth voltage.
- The overvoltages can puncture insulation and cause repeated earth faults or equipment damage.
- The neutral shifts, raising the healthy phases’ earth voltage and increasing insulation stress.
(b) Distinguish between positive and negative symmetrical components of an alternator voltage. (4 marks)
Answer (b). Positive-sequence voltages have equal magnitudes, are separated by 120°, and follow the normal phase order R–Y–B. Negative-sequence voltages also have equal magnitudes and 120° separation, but their phase order is reversed, R–B–Y.
(c) The unbalanced currents in a three-phase system are IR = 200∠40° A, IY = 80∠320° A and IB = 50∠180° A. Determine the red-phase symmetrical sequence components. (10 marks)
Answer (c). Let a = 1∠120°. For the red phase, the zero-, positive- and negative-sequence components are:
IR0 = (IR + IY + IB)/3
IR1 = (IR + aIY + a²IB)/3
IR2 = (IR + a²IY + aIB)/3
IR0 = 54.831 + j25.712 A = 60.56∠25.12° A
IR1 = 64.034 + j83.548 A = 105.26∠52.53° A
IR2 = 34.344 + j19.298 A = 39.39∠29.33° A
Question 4 (20 marks)
(a) With reference to power-system protection, explain pilot schemes. (2 marks)
Answer (a). A pilot scheme uses a communication link between the ends of a protected line to exchange or compare current/phase information. If the comparison indicates a fault inside the protected section, relays at the ends trip their circuit breakers; external faults remain stable.
(b) With the aid of a labelled circuit diagram, explain the operation of the definite-distance relay. (7 marks)
Answer (b). A current transformer supplies the relay current input and a potential transformer supplies its voltage input. The relay measures the apparent impedance Zapp = V/I. If it falls below the preset reach, the fault is within the protected distance; the definite-distance element then trips after its fixed operating time.
Source ── CT ── CB ───────── protected line ───────── fault
│ ▲
│ └────────────── trip coil
└── I input ─┐
Bus ── PT ── V input ─┴── [definite-distance relay: compare |V/I| with Zset]
relay trip contact ── DC control supply ── CB trip coil
The relay reach is set by the line impedance to be protected. Its operating time is definite for faults within that reach, rather than increasing continuously with distance.
(c) State three merits of ring-feeder busbar protection. (3 marks)
Answer (c).
- A circuit breaker can be isolated for inspection or maintenance while the other ring path keeps circuits supplied.
- A bus or circuit fault can be isolated by opening the adjacent breakers, leaving the remainder of the ring in service.
- The arrangement gives flexible switching and good continuity of supply because each circuit can be fed from either direction.
(d) A 20 MVA, 6.6 kV, three-phase star-connected alternator is protected using the Merz-Price scheme. The CT ratio is 1000/10, the relay minimum operating current is 0.9 A, and neutral-earthing resistance is 8 Ω. Determine (i) the percentage of unprotected winding; and (ii) the resistance required to protect 90% of the stator winding. (8 marks)
Answer (d). Convert the relay pickup to its equivalent primary current and compare it with the earth-fault current available at a fault a fraction x of the phase winding from the neutral.
Vph = 6600/√3 = 3810.5 V; Ipickup,pri = 0.9 × (1000/10) = 90 A
x = Ipickup,priRn/Vph = (90 × 8)/3810.5 = 0.18895
Thus about 18.9% of each phase winding near the neutral is unprotected, so this setting protects approximately 81.1%.
For 90% coverage, x = 0.10: Rn,limit = (0.10 × 3810.5)/90 = 4.23 Ω
For at least 90% coverage, the neutral resistance must be no greater than about 4.23 Ω under these assumptions. A lower resistance improves sensitivity; the question’s wording “minimum resistance” is interpreted as the required limiting value.
Question 5 (20 marks)
(a) (i) With reference to power-system stability, explain (I) steady-state stability limit and (II) transient stability. (ii) State three assumptions made during power-system stability analysis. (7 marks)
Answer (a)(i). The steady-state stability limit is the greatest power that can be transferred without loss of synchronism when the system changes gradually and remains close to its operating point. Transient stability is the ability of synchronous machines to remain in synchronism after a large, sudden disturbance such as a short circuit, switching event or sudden loss of generation.
Answer (a)(ii). Common classical-analysis assumptions include:
- A synchronous generator is represented by a constant internal emf behind its transient reactance during the first swing.
- Mechanical input power is constant over the short disturbance interval.
- Armature resistance and damping/governor effects are neglected, and the infinite-bus voltage is treated as constant.
(b) Figure 1 in the paper shows a generator connected to an infinite bus. The generator delivers real power 0.8 p.u. at 0.8 power factor lagging. The generator transient reactance is 0.25 p.u., transformer reactance is 0.2 p.u., and each of two parallel transmission lines has reactance 0.4 p.u. (i) Draw the reactance diagram. (ii) Determine the supply current, equivalent impedance and power angle. (10 marks)
Answer (b). First combine the two transmission lines in parallel.
E′ ── j0.25 ── j0.20 ──┬── j0.40 ──┬── V∞ = 1∠0° p.u.
└── j0.40 ──┘
Xline,eq = 0.4 ∥ 0.4 = 0.20 p.u.; Xeq = 0.25 + 0.20 + 0.20 = 0.65 p.u.
S = P + jQ = 0.8 + j(0.8 tan(cos−10.8)) = 0.8 + j0.6 p.u.
I = S*/V* = 0.8 − j0.6 = 1∠−36.87° p.u.
Zeq = j0.65 p.u.
E′ = V∞ + jXeqI = 1 + j0.65(0.8 − j0.6) = 1.39 + j0.52 = 1.484∠20.51° p.u.
The supply current is 1.00∠−36.87° p.u., the equivalent transfer impedance is j0.65 p.u., and the generator power angle is δ = 20.51°.
(c) State three merits of neutral earthing in a power system. (3 marks)
Answer (c). Neutral earthing provides a defined earth-fault return path so protective relays can detect and clear faults; it limits neutral displacement and transient overvoltages on healthy phases; and it reduces the duration and damage of earth faults, improving safety and continuity.
SECTION B: ELECTROMAGNETIC FIELD THEORY
Answer TWO questions from this section.
Question 6 (20 marks)
(a) (i) With reference to electromagnetic waves, explain wave polarization and skin depth. (ii) State two characteristics of electromagnetic detectors. (6 marks)
Answer (a)(i). Wave polarization describes the orientation and time-varying path traced by the electric-field vector at a fixed point as the wave passes; it may be linear, circular or elliptical. Skin depth is the distance into a conducting medium at which the field amplitude falls to 1/e (about 36.8%) of its surface value. It equals 1/α when α is the attenuation constant.
Answer (a)(ii). A useful electromagnetic detector has adequate sensitivity to resolve the field level being measured and a known, calibrated frequency response. Its reading can also depend on probe orientation and wave polarization, so those conditions must be stated when comparing readings.
(b) A lossless medium has intrinsic impedance η = 60π Ω and relative permeability μr = 1. An electromagnetic wave has magnetic-field intensity H = −0.1 cos(ωt − z) ax + 0.5 sin(ωt − z) ay A/m and propagates through the medium. Determine (i) the relative permittivity; and (ii) the angular frequency ω. (6 marks)
Answer (b). The intrinsic impedance of a lossless medium is η = η0√(μr/εr). Also, the phase term ωt − z gives β = 1 rad/m.
60π = 120π√(1/εr) ⇒ εr = 4
v = c/√(μrεr) = 3 × 108/2 = 1.5 × 108 m/s
ω = βv = 1 × 1.5 × 108 = 1.5 × 108 rad/s
(c) (i) State Gauss’s law. (ii) Charges −4 μC and +5 μC are located at (2, −1, 3) and (0, 4, −2), respectively. Taking zero potential at infinity, determine the potential at (1, 0, 1). (8 marks)
Answer (c)(i). Gauss’s law states that the total electric flux through a closed surface equals the charge enclosed by that surface.
∯S D · dS = Qenclosed = ∭V ρv dV
Answer (c)(ii). The distances from the two charges to P(1, 0, 1) are √6 m and √26 m.
V(P) = (1/(4πε0))[(−4 × 10−6/√6) + (5 × 10−6/√26)]
V(P) = −5.864 × 103 V ≈ −5.86 kV
Question 7 (20 marks)
(a) With reference to magnetostatic fields, state the Biot–Savart law and give its expression. (3 marks)
Answer (a). The Biot–Savart law gives the magnetic field produced at a point by a steady current element. Each current element contributes in proportion to current and element length, inversely with the square of the distance, and in the direction of dℓ × aR.
dB = (μ0I/4π)(dℓ × aR)/R²; B = (μ0I/4π) ∫(dℓ × aR)/R²
(b) A charge distribution has spherical volume density ρv = ρ0r/R for 0 ≤ r ≤ R, and ρv = 0 for r > R. Using Gauss’s law, determine the electric-field intensity for (i) r < R and (ii) r > R. (9 marks)
Answer (b). By spherical symmetry, E is radial and constant over a Gaussian sphere of radius r.
Qenc(r<R) = ∫0r(ρ0r′/R)4πr′²dr′ = πρ0r⁴/R
E(4πr²) = Qenc/ε0 ⇒ E = (ρ0r²/(4ε0R)) ar, r<R
Qtotal = πρ0R³
E = Qtotal/(4πε0r²) ar = (ρ0R³/(4ε0r²)) ar, r>R
(c) (i) List two characteristics of diamagnetic materials in magnetostatics. (ii) Define electric permittivity and electric-field strength with respect to electrodynamics. (4 marks)
Answer (c)(i). Diamagnetic materials have a small negative magnetic susceptibility and relative permeability slightly below one. Their induced magnetization opposes the applied magnetic field, so they are weakly repelled by it.
Answer (c)(ii). Electric permittivity ε is the material parameter relating electric flux density to electric-field strength. Electric-field strength E is force per unit positive test charge at a point.
D = εE; E = F/q
(d) A dielectric has relative permittivity εr = 5.7 and electric flux density D = 3.0 × 10⁻⁷ C/m². Determine (i) the electric susceptibility χe; and (ii) electric-field strength E. (4 marks)
Answer (d).
χe = εr − 1 = 5.7 − 1 = 4.7
E = D/(ε0εr) = (3.0 × 10−7)/(8.854 × 10−12 × 5.7) = 5.944 × 10³ V/m
Therefore, E ≈ 5.94 kV/m.
Question 8 (20 marks)
(a) Write Maxwell’s equations in differential form. (4 marks)
Answer (a).
∇ · D = ρv
∇ · B = 0
∇ × E = −∂B/∂t
∇ × H = J + ∂D/∂t
(b) An electromagnetic wave propagates through a medium with εr = 1, μr = 20 and conductivity σ = 3 S/m. The scanned field expression is E = e−z/3 sin(10⁸t − βz) ax V/m. Determine the loss tangent, attenuation factor α and intrinsic impedance η. (6 marks)
Answer (b). Take the printed conductivity as 3 S/m and ω = 10⁸ rad/s. The material properties give:
tan δ = σ/(ωε) = 3/(108 × ε0 × 1) = 3.388 × 10³
α = ω√[(με/2)(√(1 + tan²δ) − 1)] = 61.39 Np/m
η = √[jωμ/(σ + jωε)] = 20.47 + j20.46 Ω = 28.94∠44.99° Ω
Source-scan consistency note: the printed factor e−z/3 itself implies α = 1/3 Np/m when z is in metres, which conflicts with α = 61.39 Np/m from the printed εr, μr, σ and frequency. The scan therefore contains inconsistent data; both readings are shown so the discrepancy is not hidden.
(c) (i) State the energy-conservation theorem. (ii) The current density is J = (1/r³)(2 cos θ ar + sin θ aθ) A/m². Determine the current through (I) a hemispherical shell of radius 20 cm and (II) a spherical shell of radius 10 cm. (10 marks)
Answer (c)(i). Poynting’s theorem expresses conservation of electromagnetic energy: power flowing out of a volume plus the rate of increase of stored field energy and the power dissipated in the volume sums to zero.
∇ · (E × H) + ∂u/∂t + J · E = 0, where u = ½(E · D + B · H)
Answer (c)(ii). For a spherical surface, the area element is dS = r² sin θ dθ dφ ar. Only the radial component of J crosses the curved shell.
Ihemisphere = ∫02π∫0π/2(2 cos θ/r³)(r² sin θ)dθdφ = 2π/r
r = 0.20 m: Ihemisphere = 2π/0.20 = 10π = 31.42 A outward
Isphere = ∫02π∫0π(2 cos θ/r³)(r² sin θ)dθdφ = 0 A
The spherical-shell result is zero because the outward current through the northern half is cancelled by the inward current through the southern half.
Revision tip: Keep vector directions and units visible in each field calculation. When a source scan prints conflicting data, show the conflict and state which value comes from the stated material parameters.