Revision material: This lesson transcribes the July 2019 paper and provides independently prepared worked answers for all questions in the KNEC Diploma in Electrical Engineering Module III Electrical Power Systems and Electromagnetic Field Theory July 2019 past paper. The scanned paper instructs candidates to answer three questions in Section A and two in Section B; all eight are answered here for revision. This is not an official KNEC marking scheme.
Section A: Electrical Power Systems
Question 1
(a) (3 marks) State three methods of reducing the system transfer reactance to improve steady-state stability.
(b) (6 marks) Derive the swing equation of a synchronous generator.
(c) (11 marks) The ABCD constants of a nominal-π network representing a three-phase transmission line are A = D = 0.94∠0.8°, B = 66∠72° Ω and C = 0.002∠91° S. If the sending-end and receiving-end voltages are both kept constant at 132 kV, determine the steady-state stability limit (i) using the constants as given; and (ii) with both series resistance and shunt admittance neglected.
Answer 1(a)
- Use series capacitive compensation to cancel part of the line’s inductive reactance.
- Use parallel transmission paths or additional circuits so the equivalent transfer reactance is lower.
- Use series-connected phase-shifting or FACTS equipment, such as a TCSC, to reduce the effective transfer reactance.
Answer 1(b)
The rotor accelerates whenever mechanical input power differs from electrical output power. Let the accelerating torque be the difference between mechanical and electromagnetic torque:
Tₐ = Tₘ − Tₑ = J d²θₘ/dt²
Multiplying by synchronous mechanical speed converts torque imbalance to accelerating power. If p is the number of poles, electrical rotor acceleration is related to mechanical acceleration by δ̈ = (p/2)θ̈ₘ. Thus, with rotor angle δ in electrical radians and damping neglected:
Pₐ = Pₘ − Pₑ = Tₐωₛ,ₘ = Jωₛ,ₘθ̈ₘ = (2Jωₛ,ₘ/p)δ̈
Mₛᵢ = 2Jωₛ,ₘ/p (SI form); Mₚᵤ = 2H/ωₛ,ₑ (per-unit form)
Here H is the inertia constant in seconds and ωₛ is synchronous electrical angular speed in rad/s. Therefore the per-unit swing equation is:
(2H/ωₛ,ₑ) d²δ/dt² = Pₘ − Pₑ (per unit)
With damping included, a common form is Mδ̈ + Dδ̇ = Pₘ − Pₑ.
Answer 1(c)
For a line with A = |A|∠α and B = |B|∠β, take Vᵣ as the phase reference and let δ = ∠Vₛ − ∠Vᵣ. From Iᵣ = (Vₛ − AVᵣ)/B, the receiving-end real power is:
Pᵣ = 3|Vᵣ||Vₛ|/|B| cos(β − δ) − 3|A||Vᵣ|²/|B| cos(β − α)
When voltages are given line-to-line, 3|Vᵣ||Vₛ| = Vᵣ,LL Vₛ,LL.
(i) Constants as stated. The variable term is greatest when δ = β = 72°. Using Vᵣ,LL = Vₛ,LL = 132 kV, α = 0.8°, β = 72°, |A| = 0.94 and |B| = 66 Ω:
Pᵣ,max = (132²/66)[1 − 0.94 cos(72° − 0.8°)] MW
Pᵣ,max = 264[1 − 0.94 cos 71.2°] = 184.0 MW
(ii) Neglect series resistance and shunt admittance. The line is treated as lossless with A = D = 1 and B = jX. Retaining the reactive component of the stated B gives X = 66 sin 72° = 62.77 Ω. Thus:
Pᵣ,max = Vᵣ,LL Vₛ,LL / X = 132² / 62.77 = 277.6 MW
This is the lossless-line limit, reached at δ = 90°.
Question 2
(a) (4 marks) Explain the operation of a synchronous phase modifier as used in transmission lines.
(b) (6 marks) With the aid of a labelled circuit and phasor diagram of a nominal-π network, show that Vₛ = (1 + YZ/2)Vᵣ + IᵣZ.
(c) (10 marks) A transmission line has a span of 280 m between level supports. The conductor has an effective diameter of 2.04 cm and weighs 0.875 kg/m. Its ultimate strength is 8,080 kg. The conductor has an ice coating of radial thickness 1.3 cm and is subjected to a wind pressure of 4.01 gm/cm² of projected area. If the factor of safety is 2, determine the vertical sag.
Answer 2(a)
A synchronous phase modifier is an unloaded synchronous motor connected in shunt to the busbars. Changing its field excitation changes the reactive power it draws or supplies. Under-excitation makes it absorb lagging reactive power; over-excitation makes it supply leading reactive power. This adjusts power factor and bus voltage and can improve transmission stability. It exchanges mainly reactive power, with a small real-power input to cover losses.
Answer 2(b)
Nominal-π equivalent circuit: each end has shunt admittance Y/2 to neutral and the line’s total series impedance Z lies between the ends.
Sending end Receiving end
Vₛ Y/2 Z Y/2 Vᵣ
o-------┬───────────────[ series ]─────────────┬---------o
│ │
⏚ ⏚
neutral neutral
The shunt current at the receiving end is YVᵣ/2, so the current through the series impedance is Iᵣ + YVᵣ/2. The sending voltage is therefore:
Vₛ = Vᵣ + Z(Iᵣ + YVᵣ/2)
Vₛ = (1 + YZ/2)Vᵣ + ZIᵣ
Phasor construction: take Vᵣ as the reference. Add the phasor drop ZIᵣ to (1 + YZ/2)Vᵣ; their vector sum is Vₛ. The term YVᵣ/2 is the receiving-end charging current that first adds to Iᵣ in the line-current vector.
A ●
↗ ↑
Vₛ ↗ │ ZIᵣ
↗ ↑
O ●──────────────● B
(1 + YZ/2)Vᵣ
Answer 2(c)
Assume the standard ice density 0.915 g/cm³, level supports, and the parabolic-sag approximation. Use kgf for the load and tension values, consistent with the paper’s kilogram-force convention.
Conductor radius = 2.04/2 = 1.02 cm; outer radius with ice = 1.02 + 1.3 = 2.32 cm.
Ice area = π(2.32² − 1.02²) = 13.641 cm²
Ice load = 13.641 × 100 × 0.915 / 1000 = 1.248 kgf/m
Vertical load wᵥ = 0.875 + 1.248 = 2.123 kgf/m
The wind acts on the iced projected diameter, 4.64 cm:
Wind load wₕ = 4.01 × 4.64 × 100 / 1000 = 1.861 kgf/m
Resultant load w = √(wᵥ² + wₕ²) = 2.823 kgf/m
Allowable tension = 8,080/2 = 4,040 kgf. The vertical component of sag is calculated from the vertical load:
sᵥ = wᵥL²/(8T) = 2.123 × 280² / (8 × 4,040) = 5.15 m
Vertical sag = 5.15 m. The corresponding resultant sag is about 6.85 m.
Question 3
(a) (4 marks) Distinguish between symmetrical and unsymmetrical faults.
(b) (8 marks) The line currents of a three-phase system supplying an unbalanced load are Iᵣ = (120 + j60) A, Iᵧ = (120 − j120) A and Iᵦ = (−150 + j100) A. Determine (i) Iᵣ₀, (ii) Iᵣ₁ and (iii) Iᵣ₂.
(c) (8 marks) Figure 1 shows an unloaded three-phase system with a single-line-to-ground fault at point F. The system neutral is solidly grounded. Show that the fault current is Iᵣ = 3Eᵣ/(Z₀ + Z₁ + Z₂ + 3Zf), where Zf is fault impedance, Z₀ is zero-sequence impedance, Z₁ is positive-sequence impedance and Z₂ is negative-sequence impedance.
Answer 3(a)
A symmetrical fault affects all three phases equally, so the phase currents remain balanced; a three-phase short circuit is the usual example. An unsymmetrical fault affects phases unequally and produces unbalanced currents; examples are a single-line-to-ground fault, a line-to-line fault and a double-line-to-ground fault.
Answer 3(b)
For R-Y-B phase sequence, let a = 1∠120° and a² = 1∠240°. The symmetrical components of phase R are:
Iᵣ₀ = (Iᵣ + Iᵧ + Iᵦ)/3
Iᵣ₁ = (Iᵣ + aIᵧ + a²Iᵦ)/3
Iᵣ₂ = (Iᵣ + a²Iᵧ + aIᵦ)/3
Substituting the given currents:
Iᵣ₀ = 30 + j13.33 A = 32.83∠23.96° A
Iᵣ₁ = 108.51 + j101.28 A = 148.43∠43.03° A
Iᵣ₂ = −18.51 − j54.61 A = 57.66∠−108.72° A
Answer 3(c)
For an R-phase-to-ground fault, the positive-, negative- and zero-sequence networks are connected in series. Their sequence currents are equal. The fault-phase current is the sum of the three sequence currents, so Iᵣ = 3I₀. The fault drop is ZfIᵣ = 3ZfI₀:
Eᵣ ── Z₁ ── Z₂ ── Z₀ ── 3Zf ── neutral
I₁ = I₂ = I₀ = Iᵣ/3
Eᵣ = I₀Z₁ + I₀Z₂ + I₀Z₀ + 3I₀Zf
Iᵣ = 3I₀ = 3Eᵣ/(Z₀ + Z₁ + Z₂ + 3Zf)
Question 4
(a) (6 marks) Explain three functions of the spark gaps of a valve-type surge arrester.
(b) (4 marks) With the aid of a labelled typical voltage-surge waveform, describe the 1/60 voltage surge.
(c) (10 marks) A 100 km long, three-phase, 50 Hz transmission line has resistance per phase per kilometre 0.2 Ω, inductance per phase per kilometre 2 mH and line-to-neutral capacitance per kilometre 0.015 μF. If it supplies a star-connected load of 50 MW at 132 kV and 0.8 power factor lagging, use the nominal-T method to determine the sending-end (i) voltage and (ii) current.
Answer 4(a)
- At normal system voltage, the gaps remain non-conducting and isolate the non-linear valve resistor from the line.
- When a surge raises the voltage above the sparkover level, the gaps break down and connect the valve resistor, providing a path for surge current to earth.
- The series gaps divide the voltage and help extinguish the power-frequency follow current after the surge has passed, restoring insulation between line and earth.
Answer 4(b)
The paper’s 1/60 notation describes a steep impulse: a short front to the crest followed by a longer tail. The front time is approximately 1 μs; the tail time is measured to the point where the voltage has fallen to half its crest value, approximately 60 μs.
Voltage
│ crest Vₚ
│ ●
│ ／ ＼
│ ／ ＼
│ ／ ＼_____ 0.5 Vₚ
│___________／____________________________ Time
↑ ↑
virtual front ≈ 1 μs half-value tail ≈ 60 μs
Answer 4(c)
Per phase, the total series impedance and shunt admittance are:
Z = 100(0.2 + j2π50 × 0.002) = 20 + j62.832 Ω
Y = j2π50(100 × 0.015 μF) = j0.00047124 S
Receiving phase voltage is Vᵣ = 132/√3 = 76.210 kV∠0°. The load current is:
Iᵣ = 50 MW/(√3 × 132 kV × 0.8) = 273.4∠−36.87° A
For the nominal-T model, first find the midpoint voltage, then the shunt current, and then the sending-end values:
Vₘ = Vᵣ + (Z/2)Iᵣ = 83.71∠3.58° kV per phase
Iₛ = Iᵣ + YVₘ = 249.6∠−29.96° A
Vₛ = Vₘ + (Z/2)Iₛ = 90.27∠6.85° kV per phase
|Vₛ,LL| = √3|Vₛ| = 156.4 kV
Sending end: 156.4 kV line-to-line and 249.6 A line current.
Question 5
(a) (6 marks) With reference to protective relays, explain (i) pick-up current, (ii) plug-setting multiplier and (iii) current setting.
(b) (8 marks) With the aid of a labelled diagram, describe the operation of an induction-type directional power relay.
(c) (6 marks) A three-phase transformer having a line-voltage ratio of 415 V/11 kV is connected star-delta. Protective transformers on the 415 V side have a current ratio of 600/6 A. Determine the current ratio of the protective transformers on the 11 kV side.
Answer 5(a)
- Pick-up current: the minimum current at which a relay just begins to operate.
- Plug-setting multiplier (PSM): actual relay-coil current divided by the relay’s pick-up current.
- Current setting: the selected pick-up current, usually expressed as a percentage of the relay’s rated current.
Answer 5(b)
A current transformer feeds the relay’s current coil, while a potential transformer supplies its voltage (polarizing) coil. A phase-shifting circuit sets the reference angle between their magnetic fluxes. The two alternating fluxes induce eddy currents in an aluminium disc; their interaction produces operating torque. Reversing the direction of real power reverses the torque, so the disc moves toward the trip contact only for the selected direction. When it reaches the contact, the relay completes the circuit to the circuit-breaker trip coil.
Line current ── CT ── current coil ──┐
│ induction disc ── trip contact
Bus voltage ─── PT ─ phase shifter ─ voltage / polarising coil
│
circuit-breaker trip
The directional torque is proportional to the in-phase power component:
T ∝ VI cos(θ − τ)
Here θ is the voltage-current phase angle and τ is the relay’s maximum-torque angle.
Answer 5(c)
At 600 A on the 415 V side, the corresponding 11 kV line current is found from three-phase power balance. For star-delta transformer protection, CTs are delta-connected on the star side and star-connected on the delta side to compensate the phase shift.
I₁₁ₖV = 600 × 415/11,000 = 22.64 A
The 415 V CT secondary phase current is 6 A; its delta-connected bank gives a line current of √3 × 6 = 10.39 A. The 11 kV star-connected CTs must match that current:
CT ratio₁₁ₖV = 22.64/10.39 = 2.18 : 1 ≈ 13.1/6 A
This is the exact matching ratio for the stated 600/6 A CTs, before selecting a standard CT rating or relay tap.
Section B: Electromagnetic Field Theory
Question 6
(a) (4 marks) (i) State two properties of electromagnetic waves. (ii) Explain why a three-dimensional coordinate system is used to analyse electromagnetic fields.
(b) (10 marks total) (i) State Stokes’ theorem. (ii) Figure 2 shows an elemental electric charge located in space. Using cylindrical coordinates, write the expressions for the surface areas of the element. (iii) Figure 3 shows a line-charge distribution in a cylindrical charge tube oriented along the Z-axis. Its line charge density is ρL = 2z, where z is distance from the bottom end of the 10 cm tube. Determine the total charge contained.
(c) (6 marks) A vector field is G = [2x/(1 + y²)]aₓ + (y + z + 11)aᵧ + (5x − z²)a_z. Determine the unit vector in the direction of G at (1, 2, −3).
Answer 6(a)
(i) Electromagnetic waves are transverse: the electric field, magnetic field and direction of propagation are mutually perpendicular. They carry energy and can propagate through a vacuum without a material medium.
(ii) Fields vary in three-dimensional space and have direction as well as magnitude. Three coordinates are needed to locate a point, resolve vector components and describe spatial derivatives, surfaces and volumes.
Answer 6(b)(i)–(ii)
Stokes’ theorem equates the circulation of a vector field around a closed contour to the flux of its curl through any surface bounded by that contour:
∮C A·dℓ = ∬S (∇ × A)·dS
For an elemental cylindrical-coordinate cell (r, φ, z), the three independent differential face areas, with their positive unit normals, are:
dSᵣ = r dφ dz aᵣ (r-constant face)
dSφ = dr dz aφ (φ-constant face)
dS_z = r dr dφ a_z (z-constant face)
The opposite faces have the same areas with negative normals; the volume element is dV = r dr dφ dz.
Answer 6(b)(iii)
Take z in metres and ρL in C/m, as required for a consistent SI integral. The tube length is 10 cm = 0.10 m:
Q = ∫₀⁰·¹ ρL dz = ∫₀⁰·¹ 2z dz
Q = [z²]₀⁰·¹ = 0.010 C = 10 mC
Answer 6(c)
Substitute x = 1, y = 2 and z = −3 into each component:
G = (2/5)aₓ + 10aᵧ − 4a_z
|G| = √(0.4² + 10² + (−4)²) = √116.16 = 10.778
âG = G/|G| = 0.0371aₓ + 0.9278aᵧ − 0.3711a_z
Question 7
(a) (2 marks) State two items of equipment that use electrostatic fields in their operation.
(b) (i) (2 marks) State Coulomb’s law of electrostatics. (ii) (10 marks) A point charge Q₁ = 2 μC is at P₁(3, 7, −4) in free space and a second charge Q₂ = −5 μC is at P₂(2, 4, −1). Determine the total electric field strength at (12, 15, 18) due to both charges.
(c) (6 marks total) (i) Distinguish between convection and conduction currents in electromagnetic field study. (ii) State Maxwell’s equations for time-varying fields in integral form.
(d) (2 marks) Figure 4 shows a parallel-plate capacitor connected to an alternating generator of voltage V. Redraw the circuit and indicate displacement current Id and capacitor current Ic.
Answer 7(a)
Examples include an electrostatic precipitator and a photocopier. Electrostatic spray-painting equipment is another valid example.
Answer 7(b)
(i) Coulomb’s law states that the electrostatic force between two point charges is along the line joining them, proportional to the product of their charges, and inversely proportional to the square of their separation:
F = (1/(4πε₀))(Q₁Q₂/R²) aR
(ii) For a charge Q at r′, the field at r is E = (1/(4πε₀))Q(r − r′)/|r − r′|³. The displacement vectors from each charge to the observation point are R₁ = (9, 8, 22) m and R₂ = (10, 11, 19) m.
|R₁| = √629 m |R₂| = √582 m
E₁ = (10.255aₓ + 9.116aᵧ + 25.068a_z) V/m
E₂ = (−32.006aₓ − 35.206aᵧ − 60.811a_z) V/m
E = E₁ + E₂ = (−21.751aₓ − 26.091aᵧ − 35.743a_z) V/m
|E| = 49.31 V/m âE = −0.4411aₓ − 0.5291aᵧ − 0.7249a_z
Answer 7(c)
(i) Conduction current is charge motion through a conducting material under an electric field; its current density is Jc = σE. Convection current is charge transported by the bulk motion of charged particles through a non-conducting medium or free space; its current density is J = ρᵥu.
(ii) Maxwell’s equations for time-varying fields in integral form are:
∯S D·dS = Qfree,enclosed
∯S B·dS = 0
∮C E·dℓ = −d/dt ∬S B·dS
∮C H·dℓ = ∬S J·dS + d/dt ∬S D·dS
Answer 7(d)
Conduction current Ic flows in the connecting wires. Across the dielectric gap between the capacitor plates, the changing electric field produces displacement current Id. For an ideal capacitor their instantaneous magnitudes are equal:
Ic = C dV/dt = Id
Id = d/dt ∬S D·dS
Ic →
┌──────────────────────────────┐
│ │
───┤ upper plate │ ~ V(t)
│ ↓ Id across dielectric │ AC source
───┤ lower plate │
│ │
└──────────────────────────────┘
← Ic
Question 8
(a) (6 marks total) (i) State Biot–Savart’s law. (ii) Figure 5 shows a cylindrical Gaussian surface at the boundary between two media. Using Gauss’s law, show that the normal component of magnetic flux density B is continuous across the boundary.
(b) (6 marks total) (i) Define a uniform plane wave as used in electromagnetic fields. (ii) A wave in a lossless medium is described by Ey(x,t) = C₁ cos(ωt − βx) + C₂ cos(ωt + βx). (I) Obtain the expression for the wave travelling in the positive direction. (II) Determine its wave velocity. (III) Describe the wave when C₁ = C₂.
(c) (8 marks) A 9.4 × 10⁹ Hz uniform plane wave propagates in a medium where μr = 2 and εr = 3. Determine (i) velocity of propagation, (ii) phase constant and (iii) intrinsic impedance.
Answer 8(a)
(i) Biot–Savart’s law gives the magnetic field contribution from a current element I dℓ:
dB = (μ₀/4π) I dℓ × aR / R²
(ii) Apply Gauss’s law for magnetism to a thin pillbox crossing the boundary. As its height tends to zero, the side flux tends to zero. The outward flux through the top and bottom faces must sum to zero:
B₁ₙ ΔS − B₂ₙ ΔS = 0
B₁ₙ = B₂ₙ
Thus the normal component of magnetic flux density is continuous. The paper calls the two materials “dielectrics” and labels their parameters μ₁ and μ₂; μ denotes magnetic permeability here, but the continuity result holds regardless of the media.
Answer 8(b)
(i) A uniform plane wave has constant field magnitude and direction over any plane normal to its direction of travel; its fields vary only with distance along that direction and time.
(ii)(I) The positive-x travelling component is:
Eᵧ⁺(x,t) = C₁ cos(ωt − βx)
(II) The wave travels at phase velocity:
vₚ = ω/β = 1/√(με)
(III) If C₁ = C₂, use cos A + cos B = 2cos[(A+B)/2]cos[(A−B)/2]:
Eᵧ(x,t) = 2C₁ cos(ωt) cos(βx)
This is a standing wave formed by equal-amplitude waves travelling in opposite directions. Nodes and antinodes remain fixed in space.
Answer 8(c)
Use μ = μ₀μr and ε = ε₀εr. The velocity, phase constant and intrinsic impedance are:
v = c/√(μrεr) = 2.998 × 10⁸/√6 = 1.224 × 10⁸ m/s
β = 2πf/v = 482.6 rad/m
η = η₀√(μr/εr) = 376.73√(2/3) = 307.6 Ω