KNEC Diploma in Electrical and Electronic Engineering (Power Option), Module III
Paper codes: 2521/305 and 2601/305 · Session: October/November 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 contains eight 20-mark questions: answer any three from Section A and any two from Section B.
SECTION A: ELECTRICAL POWER SYSTEMS
Question 1 (20 marks)
(a) Explain galloping and high-frequency vibrations as applied to overhead-line conductors. (4 marks)
Answer (a).
(i) Galloping: A low-frequency, large-amplitude oscillation of an overhead conductor, commonly caused when wind acts on an ice-coated or otherwise uneven conductor. The asymmetric shape produces changing aerodynamic lift, so the conductor moves vertically and laterally. Severe galloping can make phases clash and can damage conductors or supports.
(ii) High-frequency vibration: Small-amplitude, rapid oscillation, often called aeolian vibration, caused by alternating vortices shed by a steady crosswind. It is most damaging near clamps and supports, where repeated bending can cause strand fatigue. Vibration dampers are fitted to dissipate the energy.
(b)(i) Explain the effect of conductor surface on corona formation. (2 marks)
Answer (b)(i). A rough, dirty, stranded or damaged surface has local projections that concentrate the electric field. This increases the peak surface stress and causes corona to start at a lower line voltage. A smooth, clean, polished conductor has a more uniform field and a higher corona inception voltage. The effect is represented by a surface irregularity factor in corona calculations.
(b)(ii) A 132 kV transmission line has a conductor diameter of 2 cm. Corona occurs when the line voltage exceeds 210 kV rms at a potential gradient of 40 kV/cm. Taking the air dielectric strength as 30 kV/cm, air-density factor and irregularity factor as unity, determine (I) critical disruptive voltage per phase and (II) conductor spacing. (6 marks)
Answer (b)(ii). The question does not specify whether the quoted field gradients are peak or rms, and it calls the line 132 kV while giving a 210 kV corona threshold. Use 210 kV as the stated corona-onset line voltage. For the main answer, treat the 40 kV/cm and 30 kV/cm gradients as peak values, as is standard when air dielectric strength is quoted; convert phase voltage to peak for the spacing calculation. The nominal 132 kV label is not used in that calculation.
Conductor radius is r=2/2=1 cm. At the stated onset, the phase voltage is:
Vph,rms=(210)/(√(3))=121.24 kV
Convert this to peak and use the peak surface-gradient relation g=Vph,peak/[rln(D/r)]:
ln((D)/(r))=(√(2) Vph,rms)/(gr)=(√(2)×121.24)/(40×1)=4.287
D=r e4.287=1× e4.287≈72.8 cm
The disruptive critical voltage uses the air-strength gradient of 30 kV/cm peak:
Vd,ph,rms=(30)/(√(2)) rln((D)/(r))=(30)/(√(2))×1×4.287≈90.9 kV per phase
Answers: (I) V d≈90.9 kV rms per phase; (II) D≈72.8 cm. If the examiner intended both quoted gradients as rms values, the alternative spacing is about 20.7 cm; the paper does not state the convention.
(c) Two towers of heights 40 m and 100 m support a conductor across a 600 m span. The conductor tension is 2000 kg and its weight is 2 kg/m. Determine (i) the distance from the lower support to the lowest point and (ii) the sag at the lower support. (8 marks)
Answer (c). Treat the stated masses as kgf, as is customary in this sag approximation. Let x be the horizontal distance from the lower support to the lowest point. The supports differ in height by h=100−40=60 m. With weight w=2 kgf/m and tension T=2000 kgf:
h=(w)/(2T)[(L−x)2−x2]=(w)/(2T)(L2−2Lx)
60=(2)/(4000)(6002−2(600)x) ⇒ x=200 m
The sag at the lower support is:
flow=(wx2)/(2T)=(2(200)2)/(2(2000))=20 m
Answers: Lowest point is 200 m horizontally from the lower support; sag at the lower support is 20 m.
Question 2 (20 marks)
(a) With reference to an overhead transmission line, explain the effect of surges on (i) an open-circuited line and (ii) arcing ground. (4 marks)
Answer (a).
(i) Open circuit: At an open end, a travelling voltage wave is reflected with the same polarity. The incident and reflected waves can add at the termination, producing an ideal voltage as high as twice the incident value. This can overstress insulation and cause flashover.
(ii) Arcing ground: An intermittent arc from a phase to earth repeatedly extinguishes and restrikes. The resulting abrupt changes in current and charge can create high transient overvoltages on the healthy phases and at the fault point, risking insulation breakdown and further faults.
(b) A 220 kV surge travels on a cable of surge impedance 60 Ω toward a junction with an overhead line of surge impedance 500 Ω. Find the transmitted and reflected voltages. (2 marks)
Answer (b). Let Z1=60 Ω, Z2=500 Ω, and incident voltage Vi=220 kV. At a junction:
τV=(2Z2)/(Z1+Z2)=(1000)/(560)=1.7857, Vt=τVVi=392.86 kV
ΓV=(Z2−Z1)/(Z1+Z2)=(440)/(560)=0.7857, Vr=ΓVVi=172.86 kV
Answers: Transmitted voltage 392.9 kV; reflected voltage +172.9 kV (same polarity as the incident wave).
(c) Explain the effect of power factor on the efficiency of an overhead transmission line. (2 marks)
Answer (c). For a given real power and line voltage, a lower power factor requires a larger current. The increased I2R loss and voltage drop reduce efficiency. Raising the power factor toward unity reduces current and line losses, improving efficiency.
(d) A 66 kV, 50 Hz, 200 km, three-phase line supplies 20 MW at 0.8 power factor lagging. Per phase per kilometre, R=0.2 Ω, X=0.4 Ω, and susceptance =10×10−6 S. Using the nominal-π method, find (i) receiving-end current and (ii) sending-end phase voltage. (12 marks)
Answer (d). Take the receiving-end phase voltage as the reference. The total per-phase line constants are:
VR=(66)/(√(3))=38.105 kV, Z=200(0.2+j0.4)=40+j80 Ω, Y=j(200)(10−5)=j0.002 S
The receiving-end load current is:
IR=(20×106)/(√(3)(66×103)(0.8))∠[−cos−1(0.8)]=218.69∠−36.87circ A
In the nominal-π model, half of the shunt admittance is at the receiving end. Its charging current is:
IcR=(Y)/(2)VR=j0.001(38.105×103)=j38.105 A
Thus the series current and sending-end voltage are:
Iseries=IR+IcR=174.95−j93.11=198.18∠−28.0circ A
VS=VR+ZIseries=(52.55+j10.27) kV=53.55∠11.06circ kV per phase
Answers: Receiving-end load current 218.69∠−36.87circ A; sending-end phase voltage 53.55∠11.06circ kV. The series current in the nominal-π model is approximately 198.18∠−28.0circ A.
Question 3 (20 marks)
(a) State two demerits of a valve-type arrester. (2 marks)
Answer (a). It is comparatively bulky and costly, and its series spark gaps and valve resistors need careful maintenance and can be affected by moisture or contamination.
(b) With a labelled diagram, describe the operation of an expulsion-type arrester. (6 marks)
Answer (b). A typical expulsion arrester has a line terminal, an external series gap, a fibre-lined insulating tube, an internal gap/electrode and an earth terminal. Connect it between the line and earth:
Protected line
│
External series gap
│
┌───────────┐
│ Fibre-lined│ ← insulating arc tube
│ expulsion │
│ tube │
└─────┬─────┘
Internal gap /
electrode
│
Earth
At normal voltage the gaps prevent significant current. A surge causes the gaps to break down, providing a path from line to earth and limiting the voltage across protected equipment. The resulting power-frequency follow-current arc heats and decomposes the fibre lining, producing gas. Gas pressure drives the ionised products out through the vent, lengthens and de-ionises the arc, and extinguishes the follow current at a current zero. The arrester then returns to its insulating state.
(c)(i) State two merits of the symmetrical-component method in fault analysis. (2 marks)
Answer (c)(i). It converts an unbalanced three-phase set into positive-, negative- and zero-sequence sets, making an unbalanced fault easier to analyse with balanced sequence networks. The sequence networks can be interconnected according to the fault type, simplifying calculation of fault currents and voltages.
(c)(ii) Three inductive reactances of 2 Ω, 3 Ω and 4 Ω are connected in delta across a three-phase supply. Given ER=10∠0circ V, EY=10∠−90circ V, and EB=10∠120circ V, determine (I) phase currents, (II) positive-sequence current in the red phase and (III) zero-sequence current in the yellow phase. (10 marks)
Answer (c)(ii). Interpret the figure as delta branches RY:j2 Ω, YB:j3 Ω, and BR:j4 Ω. Treat the given ER,EY,EB as phase-to-neutral phasors and take each branch voltage as the difference between its terminal phase voltages. Branch currents are:
VRY=ER−EY=10+j10 V, IRY=(VRY)/(j2)=5−j5 A
VYB=EY−EB=5−j18.66 V, IYB=(VYB)/(j3)=−6.22−j1.67 A
VBR=EB−ER=−15+j8.66 V, IBR=(VBR)/(j4)=2.17+j3.75 A
Using the directions R→ Y, Y→ B, and B→ R, line (phase) currents are:
IR=IRY−IBR=2.835−j8.750=9.20∠−72.05circ A
IY=IYB−IRY=−11.220+j3.333=11.71∠163.45circ A
IB=IBR−IYB=8.385+j5.417=9.98∠32.86circ A
Let a=1∠120circ. The positive-sequence red-phase component is:
IR1=(IR+aIY+a2IB)/(3)=2.019−j10.034=10.24∠−78.62circ A
For a three-wire delta load, the three line currents sum to zero, so the zero-sequence component in every line phase is zero:
IY0=(IR+IY+IB)/(3)=0 A
Answers: IR=9.20∠−72.05circ A, IY=11.71∠163.45circ A, IB=9.98∠32.86circ A; IR1=10.24∠−78.62circ A; IY0=0 A. These results state the phase-voltage and branch-direction interpretation used because the printed figure is not fully explicit.
Question 4 (20 marks)
(a) State two merits of solid-state relays. (2 marks)
Answer (a). They operate quickly and have no moving contacts, so they are reliable and need little mechanical maintenance. They are also compact and can be designed with sensitive, repeatable operating characteristics.
(b) With a labelled diagram, describe the operation of an induction-time distance impedance relay. (8 marks)
Answer (b). An impedance distance relay measures the apparent impedance Zapp=V/I between the relay location and the fault. A current coil supplies operating torque; a voltage coil supplies restraining torque. The induction element’s contacts are connected to the circuit-breaker trip circuit.
Protected line ── CT ───────────────→
│ current
▼
┌────────────┐ trip contacts
VT across ──→│ induction │───────────→ CB trip coil
line voltage │ distance │
│ relay │
└────────────┘
│
Earth
During normal load or a fault outside the set reach, the measured V/I is above the relay’s reach setting, so the restraining action keeps the contacts open. For a fault within reach, voltage at the relay falls and current rises; the apparent impedance drops below the setting. Operating torque then exceeds restraining torque, the induction disc moves and closes the trip contacts, tripping the circuit breaker. The selected time characteristic/time dial coordinates the trip with other protection. A plain impedance element is not inherently directional, so directional supervision may be added where required.
(c)(i) State two merits of the Petersen-coil earthing system. (2 marks)
Answer (c)(i). The coil can be tuned to cancel most of the network’s capacitive earth-fault current, reducing arc current and the chance of a sustained arcing fault. A transient earth fault may self-extinguish, allowing continued operation long enough to locate the fault.
(c)(ii) A 240 kV, three-phase, 50 Hz, 300 km line has capacitance to earth of 0.05 μF/km per phase. For Petersen-coil earthing, determine (I) coil inductance, (II) coil current and (III) kVA rating. (8 marks)
Answer (c)(ii). Total capacitance per phase is C=0.05×300=15 μF. At resonance, the neutral coil compensates the total three-phase capacitive earth-fault current, so 1/(ω L)=3ω C:
L=(1)/(3ω2C), ω=2π(50)=314.16 rad/s
L=(1)/(3(314.16)2(15×10−6))=0.2252 H
Phase voltage and tuned coil current are:
Vph=(240)/(√(3))=138.56 kV, IL=3ω C Vph=1.959 kA
S=VphIL=138.56×1.959≈271.4 MVA=271,400 kVA
Answers: L≈0.225 H, IL≈1.959 kA, and coil rating ≈271,400 kVA on the assumed full-compensation basis.
Question 5 (20 marks)
(a) Explain dynamic stability and steady-state stability in power systems. (4 marks)
Answer (a).
(i) Dynamic stability: The ability of a power system to remain in synchronism, or regain synchronism, after a disturbance when the time response of generators, excitation systems, governors and other controls is considered.
(ii) Steady-state stability: The ability of a power system to maintain synchronism for small, gradual changes in load or operating conditions without losing its stable operating point.
(b) Explain how (i) braking resistors, (ii) single-pole switching and (iii) a fast-acting automatic voltage regulator improve power-system stability. (6 marks)
Answer (b).
(i) Braking resistors: Switched across generator terminals during or after a disturbance, they absorb electrical energy and reduce the accelerating power of the rotor. This reduces rotor-angle swings and helps it remain in synchronism.
(ii) Single-pole switching: For a single-phase-to-earth fault, only the faulty phase is opened. The healthy phases continue to transfer power, so the disturbance is smaller than a three-pole interruption. The faulty phase can then be reclosed after the arc de-ionises.
(iii) Fast-acting AVR: Rapidly increases field excitation when voltage falls. This raises the generator internal emf and synchronising power, improves voltage recovery and helps arrest rotor-angle swings.
(c) A cylindrical-rotor generator has delivered power 0.5 p.u., no-load voltage 1.5 p.u., infinite-bus voltage 1 p.u., line reactance 0.5 p.u. and inertia constant 5 MW·s/MVA. Generator reactance is 1 p.u. A three-phase fault occurs on the line. Sketch the arrangement and determine (I) steady-state torque angle and (II) critical clearing angle. (10 marks)
Answer (c). The paper does not specify the fault location or the post-fault network configuration. For a determinate equal-area calculation, assume a solid three-phase fault makes electrical power transfer zero during the fault (Pe,fault=0), and clearing restores the original generator–line–infinite-bus path.
Generator internal emf E′
│
jXd′ = j1.0 p.u.
│
├──── line, jXL = j0.5 p.u. ──── Infinite bus, V = 1.0 p.u.
│
└── 3-phase line fault (assumed Pe = 0 during fault)
Total transfer reactance after clearing is:
Xeq=Xd‘+XL=1.0+0.5=1.5 p.u., Pmax=(E’V)/(Xeq)=(1.5(1.0))/(1.5)=1.0 p.u.
Before the fault, Pm=Pe=0.5 p.u., so the steady-state torque (power) angle is:
δ0=sin−1((Pm)/(Pmax))=sin−1(0.5)=30circ
With the original network restored, the unstable equilibrium angle is δu=180circ−δ0=150circ. For a solid fault with zero electrical output, the equal-area criterion gives:
cosδc=cosδu+(Pm)/(Pmax)(δu−δ0)
cosδc=cos150circ+0.5(150circ−30circ)(π)/(180)=−0.8660+1.0472=0.1812
δc=cos−1(0.1812)≈79.6circ
Answers under the stated fault assumption: steady-state torque angle 30circ; critical clearing angle 79.6circ. The inertia constant is used to calculate a critical clearing time, but is not needed for this angle calculation.
SECTION B: ELECTROMAGNETIC FIELD THEORY
Question 6 (20 marks)
(a) State two (i) properties of electromagnetic waves and (ii) modes of photometric detection and analysis. (4 marks)
Answer (a).
(i) Electromagnetic-wave properties: (1) The electric field, magnetic field and direction of propagation are mutually perpendicular in a uniform plane wave. (2) Electromagnetic waves transport energy and can propagate through a vacuum; in free space their speed is c≈3.0×108 m/s.
(ii) Photometric detection modes: (1) Visual photometry, in which an observer compares brightness or colour with a standard. (2) Photoelectric photometry, in which a photocell/photodiode converts incident light into an electrical signal that is measured against a calibrated reference.
(b) With a labelled diagram, describe the construction of a diffusion cloud-chamber photometric detector. (6 marks)
Answer (b). A diffusion cloud chamber has a transparent, closed chamber with an alcohol-soaked felt or wick near its warm top, a cold black metal base cooled by dry ice, and a particle source positioned near the base. A lamp directs a thin beam of light across the sensitive layer; the tracks can be observed through the transparent wall or recorded by a camera.
Warm top / lid
┌─────────────────────┐
│ Felt or wick soaked │ ← alcohol vapour source
│ with alcohol │
│ │
│ Saturated vapour │
Lamp ──→ thin light beam │
│ Sensitive, │
│ supersaturated layer│ ← particle tracks seen here
│ • radioactive │
│ source near base │
├─────────────────────┤
│ Cold black metal │ ← droplets stand out
│ base plate │
└─────────────────────┘
Dry ice
Alcohol evaporates at the warm top and diffuses downward. Near the cold base it becomes supersaturated, forming a sensitive layer. A charged particle ionises the vapour along its path; alcohol condenses on those ions as tiny droplets. Side lighting makes the droplet track visible against the dark base. Thus the chamber detects the particle through the light scattered by the condensation track.
(c) A 1.8 GHz wave propagates in a medium with μr=1.8, εr=30 and σ=3.0 S/m. Find (i) attenuation constant, (ii) propagation constant and (iii) skin depth. (10 marks)
Answer (c). Use μ=1.8μ0, ε=30ε0, f=1.8×109 Hz, and ω=2π f. The loss tangent is:
tanδm=(σ)/(ωε)=(3)/(2π(1.8×109)(30ε0))≈0.999
For a lossy medium:
α=ω√((με)/(2)[√(1+((σ)/(ωε))2)−1])≈126.0 Np/m
β=ω√((με)/(2)[√(1+((σ)/(ωε))2)+1])≈304.5 rad/m
Therefore:
γ=α+jβ=(126.0+j304.5) m−1
δs=(1)/(α)=(1)/(126.0)=7.94×10−3 m=7.94 mm
Answers: α≈126.0 Np/m; γ≈(126.0+j304.5) m−1; skin depth δs≈7.94 mm.
Question 7 (20 marks)
(a) Explain (i) electric field intensity and (ii) electric flux density in an electrostatic field. (4 marks)
Answer (a).
(i) Electric field intensity, E: Force experienced per unit positive test charge at a point in the field. Its SI unit is N/C, equivalently V/m.
E=( F)/(q)
(ii) Electric flux density, D: Electric flux passing normally through unit area. Its SI unit is C/m². In a linear, isotropic dielectric, D=ε E.
(b) State (i) Gauss’s divergence theorem in integral form and (ii) two methods of electromagnetic shielding. (5 marks)
Answer (b).
(i) Gauss divergence theorem: The outward flux of a vector field through a closed surface equals the volume integral of the field’s divergence within the surface:
∯S A· d S=∭V(∇· A) dV
(ii) Shielding methods: (1) Enclose the source or protected equipment in a conducting screen or Faraday cage to redistribute electric charges and reduce electric-field coupling. (2) Use a high-permeability magnetic shield, or a conductive enclosure that supports induced eddy currents, to reduce magnetic-field coupling.
(c)(i) Explain an electric dipole as applied in electrodynamics. (2 marks)
Answer (c)(i). An electric dipole consists of two equal and opposite charges separated by a small distance. Its dipole moment has magnitude p=qd and direction from the negative charge to the positive charge.
(c)(ii) A metallic sphere of radius 10 cm has uniform surface charge density 10 nC/m2. Find (I) surface potential, (II) total charge and (III) stored energy. (9 marks)
Answer (c)(ii). Use R=0.10 m, σs=10×10−9 C/m2, and ε0=8.854×10−12 F/m. The total charge is:
Q=σs(4π R2)=10−8×4π(0.1)2=1.257×10−9 C=1.257 nC
The potential on an isolated charged conducting sphere is:
Vs=(Q)/(4πε0R)=(1.257×10−9)/(4π(8.854×10−12)(0.1))≈112.94 V
The electrostatic energy stored is:
W=(1)/(2)QVs=(1)/(2)(1.257×10−9)(112.94)=7.10×10−8 J
Answers: Vs≈112.94 V, Q≈1.257 nC, W≈7.10×10−8 J.
Question 8 (20 marks)
(a) State the characteristics of conductivity σ, permittivity ε, and permeability μ for (i) lossy media and (ii) good conductors in plane-wave propagation. (4 marks)
Answer (a).
(i) Lossy medium: Conductivity is finite and non-zero, so conduction current causes appreciable attenuation. The ratio σ/(ωε) is not negligible; it may be comparable with unity. Permittivity and permeability are finite material parameters that govern electric and magnetic energy storage and wave speed.
(ii) Good conductor: Conductivity is very large compared with the displacement-current term, σ≫ωε. Conduction current dominates, fields are strongly attenuated and skin depth is small. Permittivity has little effect on conduction compared with σ; permeability remains the medium’s magnetic parameter and may differ from μ0 for magnetic conductors.
(b) State Maxwell’s equations in integral form, citing the applicable law. (8 marks)
Answer (b).
Gauss’s law for electricity: Electric flux through a closed surface equals enclosed charge.
∯S D· d S=Qenc=∭Vρv dV
Gauss’s law for magnetism: Net magnetic flux through any closed surface is zero; there are no isolated magnetic monopoles.
∯S B· d S=0
Faraday’s law of electromagnetic induction: The circulation of electric field around a closed path equals the negative rate of change of magnetic flux through the surface.
∮C E· d l=−(d)/(dt)∬S B· d S
Ampère–Maxwell law: Magnetic-field circulation equals the enclosed conduction current plus the displacement current.
∮C H· d l=Ienc+(d)/(dt)∬S D· d S
(c) A parallel-plate capacitor has plate area 10 cm2, spacing 0.2 cm, dielectric εr=2, conductivity σ=4×10−5 S/m, and applied potential difference 120 V. Find (i) electric field intensity, (ii) current density, (iii) current and (iv) power density. (8 marks)
Answer (c). Convert area and plate spacing to SI units: A=10×10−4=10−3 m2, d=0.2 cm=2×10−3 m. For the stated steady applied voltage:
E=(V)/(d)=(120)/(2×10−3)=6.0×104 V/m=60 kV/m
Using Ohm’s law in point form, J=σ E:
J=(4×10−5)(6.0×104)=2.4 A/m2
I=JA=2.4(10−3)=2.4×10−3 A=2.4 mA
pv= J· E=σ E2=(4×10−5)(6.0×104)2=1.44×105 W/m3=144 kW/m3
Answers: E=60 kV/m; J=2.4 A/m2; I=2.4 mA; power density pv=144 kW/m3.