KNEC Electrical Power Systems and Electromagnetic Field Theory Past Paper with Answers — June/July 2023

KNEC Diploma in Electrical and Electronic Engineering (Power Option), Module III

Paper: Electrical Power Systems and Electromagnetic Field Theory | Paper codes: 2521/305 and 2601/305 | Session: June/July 2023

Revision note: Attempt the paper before reading the solutions. These independently prepared answers are for revision and are not an official KNEC marking scheme. Use the examination instructions: 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) Distinguish between primary and back-up protection with respect to power systems. (ii) Describe the pilot power system protection scheme as used in power systems. (4 marks)

(b) With the aid of a labelled schematic circuit diagram, describe the Merz-Price three-phase transformer protection scheme. (8 marks)

(c) A star-connected, three-phase, 10 MVA, 6.6 kV generator has a per-phase reactance of 10%. It is protected by the Merz-Price circulating-current method, set to operate at fault currents greater than 175 A. Determine: (i) full-load current, I_L; (ii) reactance per phase, X; (iii) earthing resistance, r, when 10% of the generator winding remains unprotected. (8 marks)

Answer

(a)(i) Primary and back-up protection. Primary protection is the fast, selective protection assigned to a defined equipment zone. It should detect and isolate a fault in that zone. Back-up protection operates if the primary relay, circuit breaker, or its trip circuit fails; it normally has a time delay and may disconnect a wider part of the system.

(a)(ii) Pilot protection. Protective relays at the ends of a line or equipment zone exchange information over a pilot wire or communication channel. The relays compare the currents or directional indications at both ends. For a healthy line or an external fault, the through-current pattern is balanced and the relays restrain. For an internal fault, the end currents indicate fault current entering the protected zone, so the relays trip the circuit breakers at both ends.

(b) Merz-Price transformer differential protection. Current transformers (CTs) are placed on the high- and low-voltage sides. Their ratios and connections compensate for the transformer turns ratio and, where needed, its phase displacement. The CT secondary currents are connected in opposition to a differential relay for each phase.

HV supply ── CB ── CTs ──┌────────────────────┐── CTs ── CB ── LV load
                         │ 3-phase transformer │
                         └────────────────────┘

CT secondary currents on each phase ── connected in opposition ── differential relay
                                                            └── trips both side breakers

During normal load or a fault outside the protected zone, the compensated CT secondary currents are approximately equal, so their difference through the relay is near zero. An internal winding or phase fault creates a differential current. The relay then operates and trips the breakers on both sides, isolating the transformer.

(c)(i) Full-load current.

I_L = S / (√3 V_L) = 10,000,000 / (√3 × 6,600) = 874.8 A ≈ 875 A

(c)(ii) Reactance per phase. The base impedance per phase is V_L²/S for a three-phase machine.

Z_base = 6,600² / 10,000,000 = 4.356 Ω

X = 0.10 Z_base = 0.4356 Ω per phase

(c)(iii) Earthing resistance. With 10% of the phase winding unprotected, the voltage driving an earth fault at the protected boundary is 10% of the phase voltage. Using the usual simplified setting calculation and neglecting the small winding reactance over that 10% length:

V_fault = 0.10 × (6,600 / √3) = 381.1 V

r = V_fault / I_set = 381.1 / 175 = 2.18 Ω

Question 2 (20 marks)

(a) Describe aeoline vibrations with respect to overhead transmission lines. (4 marks)

(b) A transmission line has a span of 150 m between level supports. The conductor has a cross-sectional area of 2 cm². Its tension is 2,000 kg. The conductor material has a specific gravity of 9.9 g/cm³ and the wind pressure is 1.5 kg/m of line. Determine: (i) conductor weight per metre before wind pressure; (ii) resultant weight per metre due to wind pressure; (iii) sag. (6 marks)

(c) State three: (i) factors affecting corona; (ii) disadvantages of corona on power transmission lines. (6 marks)

(d) A three-phase, 220 kV, 50 Hz transmission line has conductors of radius 1.5 cm spaced 2.0 m apart. The irregularity factor is 0.85; the temperature is 40°C and the pressure is 76 cm Hg. Determine: (i) air-density factor, δ; (ii) critical disruptive voltage, V_c. (4 marks)

Answer

(a) Aeoline (aeolian) vibration. Wind flowing past a conductor sheds alternating vortices, producing small-amplitude, relatively high-frequency vibrations across the wind direction. Repeated bending at suspension clamps can cause fatigue and strand failure. Stockbridge-type dampers are fitted near supports to reduce the vibration.

(b)(i) Conductor weight per metre.

Volume per metre = 2 cm² × 100 cm = 200 cm³

Mass per metre = 200 × 9.9 = 1,980 g = 1.98 kg/m

Thus the vertical conductor load is approximately 1.98 kgf/m (the paper gives the tension and wind load in kg units, treated here as kgf).

(b)(ii) Resultant load per metre with wind. The wind load acts horizontally, at right angles to the conductor’s vertical weight.

w_r = √(1.98² + 1.50²) = 2.484 kgf/m

(b)(iii) Sag. For a level span and approximately parabolic conductor curve, use the resultant load.

d = w_rL² / (8T) = 2.484 × 150² / (8 × 2,000) = 3.49 m

(c)(i) Factors affecting corona

• Conductor surface condition and irregularity; rough, dirty, or stranded surfaces increase local electric stress.
• Conductor radius and spacing; larger radius and greater spacing reduce surface electric stress.
• Operating voltage, air density (pressure and temperature), humidity, and weather conditions.

(c)(ii) Disadvantages of corona

• Power loss and reduced transmission efficiency.
• Radio/telephone interference and audible noise.
• Ozone and chemical action that may corrode or damage nearby materials.

(d)(i) Air-density factor. With pressure b in cm Hg and temperature t in °C:

δ = 3.92b / (273 + t) = (3.92 × 76) / (273 + 40) = 0.9518

(d)(ii) Critical disruptive voltage. Using Peek’s RMS phase-voltage expression, with irregularity factor m₀ = 0.85, conductor radius r = 1.5 cm and spacing D = 200 cm:

V_c,ph = 21.1 m₀ δ r ln(D/r) kV

V_c,ph = 21.1 × 0.85 × 0.9518 × 1.5 × ln(200/1.5) = 125.3 kV per phase (r.m.s.)

Equivalent line-to-line value = √3 × 125.3 = 217.0 kV

Question 3 (20 marks)

(a) State two causes of surges in transmission lines. (2 marks)

(b) Figure 1 shows a surge-diverter circuit comprising a spark gap and a non-linear resistor. Explain its operation. (4 marks)

Power line ────────────────┬────────────────
                           │
                        [spark gap]
                           │
                    [non-linear resistor]
                           │
                         earth

(c) Table 1 shows classes of transmission lines and their parameters. Complete the table. (6 marks)

Classification	Typical length	Typical voltage
Short	Less than about 80 km	Below about 20 kV
Medium	About 80–250 km	About 20–100 kV
Long	More than about 250 km	Above about 100 kV

(d) A short three-phase transmission line has an impedance of (6 + j8) Ω per phase. The sending- and receiving-end line voltages are 120 kV and 110 kV respectively. The receiving-end load power factor is 0.9 lagging. Determine: (i) receiving-end phase voltage, V_R; (ii) sending-end phase voltage, V_S; (iii) load current, I; (iv) output power, P_o, in kW. (8 marks)

Answer

(a) Lightning strokes to or near a line, and switching operations such as energising or interrupting lines, transformers, or large loads, can produce travelling overvoltage surges.

(b) At normal system voltage the spark gap is non-conducting. When a surge raises the line voltage above the gap’s breakdown value, the gap conducts. The non-linear resistor then provides a low-resistance path to earth and limits the surge voltage/current at the protected equipment. When the surge has passed, the resistor’s resistance rises and the gap de-ionises so normal operation resumes.

(c) One commonly used textbook classification is shown above; boundary values vary slightly between references. The key distinction is that short lines use a lumped series-impedance model, medium lines include shunt capacitance in a lumped model, and long lines require distributed-parameter analysis.

(d) Treat the stated sending and receiving voltages as line-to-line r.m.s. values. Let the receiving phase voltage be the reference. The current lags it by φ = cos⁻¹(0.9) = 25.84°. Use kV and kA consistently with Ω.

V_Rφ = 110 / √3 = 63.51 kV;   V_Sφ = 120 / √3 = 69.28 kV

V_Sφ = V_Rφ + I∠(−25.84°)(6 + j8)

69.282² = 63.509² + 100I² + 2(63.509)I[6(0.9) + 8 sin(25.84°)]

I = 0.6426 kA = 642.6 A

P_o = √3 V_R,LLI cosφ = √3 × 110 × 0.6426 × 0.9 = 110.2 MW = 110,200 kW

SECTION B: ELECTROMAGNETIC FIELD THEORY

Question 4 (20 marks)

(a) With the aid of circuit diagrams, describe each of the following types of symmetrical faults: (i) line-to-line; (ii) double-line-to-ground. (6 marks)

(b) Figure 2 shows a phasor diagram of a three-phase system. Draw the corresponding: (i) positive-sequence current; (ii) negative-sequence current; (iii) zero-sequence current. (6 marks)

(c) A three-phase, four-wire system has the following currents under abnormal loading conditions: I_R = 100∠30° A, I_Y = 50∠300° A, I_B = 30∠180° A. Determine: (i) positive-sequence current in the R line; (ii) negative-sequence current in the R line; (iii) zero-sequence current in the R line; (iv) current in the neutral wire. (8 marks)

Answer

(a) Fault arrangements. The paper labels these “symmetrical faults”, but line-to-line and double-line-to-ground faults are technically unsymmetrical faults. The descriptions below address the two fault types actually listed.

Line-to-line fault (two phases joined; no earth path):
R ────────────────┐
                  ├── fault impedance ──┐
Y ────────────────┘                      │   B remains unfaulted
                                         └──

Double-line-to-ground fault (two phases joined and earthed):
R ────────────────┐
                  ├── fault impedance ── earth
Y ────────────────┘
B ─────────────────── remains unfaulted

In a line-to-line fault, two phase conductors touch through a fault impedance without a direct earth connection. In a double-line-to-ground fault, two phase conductors are connected together and to ground, so ground carries part of the fault return current.

(b) Sequence-current diagrams. The supplied diagram is the positive R-Y-B sequence: R is the reference, Y lags R by 120°, and B leads R by 120°. For negative sequence the phase order is reversed. Zero-sequence currents are equal and in phase.

Positive sequence (RYB):                 Negative sequence (RBY):
              I_R (0°)                              I_R (0°)
                  ↑                                     ↑
       I_B ↙             ↘ I_Y              I_Y ↙             ↘ I_B

Zero sequence:
        I_R, I_Y, I_B all point in the same direction and have equal phase.

Positive sequence: I_R = I∠0°, I_Y = I∠−120°, I_B = I∠+120°

Negative sequence: I_R = I∠0°, I_B = I∠−120°, I_Y = I∠+120°

Zero sequence: I_R = I_Y = I_B = I∠0°

(c) Symmetrical components. For R-Y-B phase order, let a = 1∠120° = −0.5 + j0.866. Then:

I₀ = (I_R + I_Y + I_B) / 3

I₁ = (I_R + aI_Y + a²I_B) / 3;   I₂ = (I_R + a²I_Y + aI_B) / 3

I_0,R = 27.20 + j2.23 = 27.29∠4.69° A

I_1,R = 42.20 + j39.76 = 57.98∠43.29° A

I_2,R = 17.20 + j8.01 = 18.97∠24.96° A

The neutral current is the phasor sum of the three line currents (its return direction is opposite to the line-current sum):

I_N = I_R + I_Y + I_B = 81.60 + j6.70 = 81.88∠4.69° A

Question 5 (20 marks)

(a) (i) Define transient stability and critical clearing angle. (ii) State three factors affecting the transient stability of a generator. (7 marks)

(b) A 50 Hz synchronous generator has an internal voltage of 1.2 per unit, inertia constant H = 5.2 MJ/MVA and reactance of 0.4 per unit. It is connected to an infinite bus through a double-circuit line of reactance 0.35 per unit each. The generator delivers 0.8 per unit power and the infinite-bus voltage is 1.0 per unit. Determine: (i) total system reactance X; (ii) maximum power transferred to the busbar, P_max; (iii) steady-state operating angle, δ. (6 marks)

(c) With the aid of a diagram, describe the dark-lamp method of synchronising two generators. (7 marks)

Answer

(a)(i) Transient stability is the ability of a power system to maintain synchronism after a large, sudden disturbance such as a short circuit, loss of generation, or switching event. The critical clearing angle is the largest rotor angle reached at the instant a fault is cleared for which the generator can still remain in synchronism.

(a)(ii) Factors include the pre-fault loading/power angle; fault type and location and the resulting power transfer during the fault; fault-clearing time and the post-fault network; and generator inertia. Any three well-explained factors earn the requested points.

(b)(i) Total reactance. The two identical line circuits are in parallel.

X_line,eq = 0.35 ∥ 0.35 = 0.175 p.u.

X_total = 0.40 + 0.175 = 0.575 p.u.

(b)(ii) Maximum power transfer. For the lossless generator–infinite-bus model P = EV sinδ / X, so:

P_max = EV / X = (1.2 × 1.0) / 0.575 = 2.087 p.u.

(b)(iii) Steady-state operating angle.

0.8 = 2.087 sinδ   ⇒   δ = sin⁻¹(0.8 / 2.087) = 22.54°

The given frequency and inertia constant are not needed for these three requested steady-state quantities.

(c) Dark-lamp synchronising method. Three lamps are connected across the open contacts of the three corresponding circuit-breaker poles between the incoming generator and live busbars.

Live busbar R ─────o/ o───── Generator R
                   │ L1 │
Live busbar Y ─────o/ o───── Generator Y
                   │ L2 │
Live busbar B ─────o/ o───── Generator B
                   │ L3 │

First verify the phase sequence. Adjust the incoming generator’s excitation until its voltage is close to the busbar voltage, then adjust its speed until its frequency matches. With the correct phase sequence, all three lamps brighten and darken together as the phase angle changes. Close the breaker when all lamps are simultaneously dark, indicating that corresponding phase voltages are momentarily equal and in phase. The voltage, frequency, and phase sequence must be checked before closing.

Question 6 (20 marks)

(a) (i) State Coulomb’s law. (ii) Figure 3 shows three charges: Q₁ = 5 nC at the origin; Q₂ = 6 nC at x = 0.3 m; Q₃ = −3 nC at y = −0.1 m. Determine the magnitude of the force on Q₁ due to (I) Q₂; (II) Q₃; (III) both Q₂ and Q₃. (8 marks)

(b) With the aid of a labelled diagram, describe the determination of the hysteresis loop using the method of reversal. (7 marks)

(c) (i) Define magnetic flux density. (ii) A magnetic material has relative permeability 3 and is placed in a field strength of 8 A/m. Determine: (I) magnetic flux density, B; (II) magnetization. (5 marks)

Answer

(a)(i) Coulomb’s law. The electrostatic force between two point charges is proportional to the product of their charges and inversely proportional to the square of their separation. It acts along the line joining them; like charges repel and unlike charges attract.

F = (1 / 4πε) |Q₁Q₂| / r²;   in free space, 1 / (4πε₀) ≈ 8.99 × 10⁹ N·m²/C²

(a)(ii) Forces on Q₁. Since Q₁ and Q₂ are both positive, the force due to Q₂ acts in the −x direction. Q₃ is negative, so it attracts Q₁ in the −y direction.

F₁₂ = k|Q₁Q₂|/(0.3)² = 2.996 × 10⁻⁶ N = 3.00 μN

F₁₃ = k|Q₁Q₃|/(0.1)² = 1.348 × 10⁻⁵ N = 13.48 μN

|F_1,total| = √(F₁₂² + F₁₃²) = 1.381 × 10⁻⁵ N = 13.81 μN

(b) Method of reversal. Place a ring or specimen core in a magnetic circuit. A primary magnetising coil carries a controllable current and produces H. A secondary search coil is connected through a ballistic galvanometer or fluxmeter. Reverse the primary current between equal positive and negative values. The change of flux produces a galvanometer throw proportional to the change of flux linkage (a full reversal corresponds to a change from +B to −B). Repeat for successive magnetising-current values, convert current to H and the calibrated throws to B, and plot B against H to obtain the hysteresis loop.

DC supply ── reversing switch ── magnetising coil (N₁ turns)
                                      ║
                                 specimen ring/core
                                      ║
search coil (N₂ turns) ── ballistic galvanometer / fluxmeter

(c)(i) Magnetic flux density is magnetic flux per unit area normal to the field; its SI unit is the tesla (T).

(c)(ii) Flux density and magnetisation.

B = μ₀μ_rH = (4π × 10⁻⁷)(3)(8) = 3.016 × 10⁻⁵ T = 30.16 μT

M = χ_mH = (μ_r − 1)H = (3 − 1)(8) = 16 A/m

Question 7 (20 marks)

(a) (i) Define skin depth as used in electromagnetic waves. (ii) Draw a skin-depth-versus-frequency curve for a good conductor. (6 marks)

(b) A 4 GHz uniform plane wave with magnetic-field intensity 45 A/m propagates in air in the z-direction. Determine: (i) phase constant; (ii) wavelength; (iii) amplitude of electric-field intensity; (iv) velocity of propagation. (8 marks)

(c) In free space the potential is V = 2(x + 1)²(y + 2)²(z + 3)². At P(1, 2, 3), determine: (i) potential; (ii) electric field; (iii) electric flux density. (6 marks)

Answer

(a)(i) Skin depth. Skin depth, δ_s, is the distance into a conductor at which the amplitude of an electromagnetic field (and associated current density) has fallen to 1/e of its surface value.

δ_s = √(2 / (ωμσ)) = 1 / √(πfμσ) for a good conductor

(a)(ii) Curve. For fixed μ and σ, skin depth varies inversely with the square root of frequency. The curve therefore falls steeply at low frequency and gradually approaches zero as frequency increases.

skin depth, δs
↑
│╲
│ ╲
│  ╲___
│      ╲______
└────────────────→ frequency, f

(b) Plane wave in air. Treat 45 A/m as the magnetic-field amplitude. Use c ≈ 3.00 × 10⁸ m/s and the intrinsic impedance of air η₀ ≈ 376.7 Ω.

β = 2πf/c = 2π(4 × 10⁹)/(3 × 10⁸) = 83.78 rad/m

λ = c/f = (3 × 10⁸)/(4 × 10⁹) = 0.075 m

|E| = η₀|H| = 376.7 × 45 = 1.695 × 10⁴ V/m = 16.95 kV/m

v = c ≈ 3.00 × 10⁸ m/s

(c) Potential, electric field, and flux density. At P(1,2,3), the coordinate factors are x+1=2, y+2=4, z+3=6.

V(P) = 2(2)²(4)²(6)² = 4,608 V

∇V(P) = 4,608 a_x + 2,304 a_y + 1,536 a_z V/m

E(P) = −∇V = −4,608 a_x − 2,304 a_y − 1,536 a_z V/m

|E(P)| = 5,376 V/m

D(P) = ε₀E(P) = −4.08 × 10⁻⁸ a_x − 2.04 × 10⁻⁸ a_y − 1.36 × 10⁻⁸ a_z C/m²

Question 8 (20 marks)

(a) State two areas of application of electromagnetic waves in agriculture. (2 marks)

(b) A capacitor has square metal plates, each 80 cm on a side, separated by 3 mm. The dielectric between the plates has relative permittivity 3. A potential of 300 V is applied. Determine: (i) capacitance; (ii) charge; (iii) electric flux density; (iv) electric-field intensity. (8 marks)

(c) A steel conductor with relative permeability μ_r = 180 and conductivity σ = 4 × 10⁶ S/m carries a current i = 2 cos(10⁴πt) A. Determine: (i) frequency; (ii) depth of penetration of the wave. (6 marks)

(d) Sketch an electromagnetic wave propagating in the x-direction. (4 marks)

Answer

(a) Examples include satellite or aerial remote sensing of crop condition/vegetation and microwave or radar sensing of soil moisture. Any two relevant agricultural applications earn the marks.

(b) Parallel-plate capacitor. Assume the stated permittivity of 3 is relative permittivity, ε_r.

A = (0.80)² = 0.64 m²;   d = 3 mm = 0.003 m

C = ε₀ε_rA/d = (8.854 × 10⁻¹²)(3)(0.64)/(0.003) = 5.67 × 10⁻⁹ F = 5.67 nF

Q = CV = (5.67 × 10⁻⁹)(300) = 1.70 × 10⁻⁶ C = 1.70 μC

D = Q/A = 1.70 × 10⁻⁶ / 0.64 = 2.66 × 10⁻⁶ C/m²

E = V/d = 300/0.003 = 1.00 × 10⁵ V/m

(c) Skin effect in steel. Compare i = 2cos(10⁴πt) with I_mcos(ωt): ω = 10⁴π rad/s.

f = ω/(2π) = (10⁴π)/(2π) = 5,000 Hz

μ = μ₀μ_r = (4π × 10⁻⁷)(180) H/m

δ_s = √(2/(ωμσ)) = 2.653 × 10⁻⁴ m = 0.265 mm

(d) Electromagnetic wave in the +x-direction. The fields are transverse and mutually perpendicular. For example, E may point in +y and H in +z, so E × H points in +x.

                         E (+y)
                           ↑
Propagation (+x)  ─────────→
                           ⊙ H (+z, out of the page)