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Chapter 2

Electrical Engineering Core

Nepal Electricity Authority past exam questions

Past questions and answers

24 questions set from this chapter. Most repeated first.

  • 2083 · 1st · 5 marks

Define transients in electrical circuits. Discuss step and natural response of RL and RC circuits.

Answer

Transients in electrical circuits

A transient is the temporary, time-varying condition that exists in a circuit immediately after a sudden change — switching on or off, a fault, or a change in source or load. During this interval the circuit passes from one steady state to another, and the voltages and currents contain both a decaying part (the natural or transient response) and a forced part (the steady-state response). The transient dies out after about 5 time constants, leaving only the steady-state value.

Step and natural response of RL and RC circuits

Any first-order circuit response can be written as:

x(t)=x(∞)+[x(0+)−x(∞)] e−t/τx(t) = x(\infty) + [x(0^+) - x(\infty)]\,e^{-t/\tau}

where x(∞)x(\infty) is the steady-state (forced) value, x(0+)x(0^+) the initial value, and τ\tau the time constant. The exponential term is the natural response; the constant term is the step (forced) response.

RL circuit (series R–L with DC source VV)

Differential equation: Ldidt+Ri=VL\dfrac{di}{dt} + Ri = V

  • Time constant: τ=LR\tau = \dfrac{L}{R}
  • Current: i(t)=VR(1−e−t/τ)i(t) = \dfrac{V}{R}\left(1 - e^{-t/\tau}\right) — step response rising from 0 to V/RV/R
  • Natural response (source removed, V=0V = 0): i(t)=I0e−t/τi(t) = I_0 e^{-t/\tau}, decaying to zero
  • Inductor voltage: vL(t)=Ve−t/τv_L(t) = V e^{-t/\tau}

RC circuit (series R–C with DC source VV)

Differential equation: RCdvCdt+vC=VRC\dfrac{dv_C}{dt} + v_C = V

  • Time constant: τ=RC\tau = RC
  • Capacitor voltage: vC(t)=V(1−e−t/τ)v_C(t) = V\left(1 - e^{-t/\tau}\right) — step response rising from 0 to VV
  • Natural response (source removed): vC(t)=V0e−t/τv_C(t) = V_0 e^{-t/\tau}
  • Capacitor current: i(t)=VRe−t/τi(t) = \dfrac{V}{R}e^{-t/\tau}

Key points

  • The natural response is the solution of the source-free (homogeneous) equation and always has the form Ae−t/τAe^{-t/\tau}.
  • The step response is the complete solution: natural response + particular (steady-state) solution.
  • τ\tau is the time to reach 63.2% of the final value; steady state is practically reached in 5τ5\tau.
  • Initial conditions: inductor current and capacitor voltage cannot change instantaneously, so iL(0+)=iL(0−)i_L(0^+)=i_L(0^-) and vC(0+)=vC(0−)v_C(0^+)=v_C(0^-).

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 5 marks

What are the different types of auxiliaries required in a hydropower station? Explain any one type that you are much familiar with.

Answer

Auxiliary equipment in a hydropower station are all those supporting systems, machines and devices — other than the main generating units (turbine, generator, transformer) — that are needed for safe, reliable and continuous operation of the plant. They may be classified as follows.

1. Electrical auxiliaries

  • Station service (unit auxiliary) transformers and their switchgear
  • DC system: batteries, battery chargers, DC distribution boards
  • AC/DC emergency (standby) diesel generator set
  • Instrument transformers (CTs, PTs), protective relays, control and metering panels
  • Excitation system, AVR, synchronising equipment
  • Earthing and lightning protection, station illumination

2. Mechanical auxiliaries

  • Governing system (oil pressure unit, servomotors, speed governor)
  • Lubricating oil system for turbine and generator bearings
  • Cooling water system for generator air coolers, thrust bearing and transformer oil coolers
  • Compressed air system (HP air for governor/breaker, LP air for station services)
  • Dewatering and drainage (sump) pumps, leakage pumps
  • Fire-fighting system (hydrant, CO₂, foam, sprinkler)
  • Ventilation and air-conditioning of the powerhouse
  • Overhead travelling crane, hoists and gates

3. Hydraulic auxiliaries

  • Penstock/intake gates, trash-rack cleaning machines
  • Head race and tail race control gates, surge tank equipment
  • Cooling water pumps, water supply and treatment

4. Miscellaneous

  • SCADA/telemetry, communication system
  • Workshop, stores, staff facilities

Explanation of one familiar auxiliary — the Governing (Oil Pressure) System

The governing system is the most critical mechanical auxiliary because it continuously matches turbine output to the load demand while holding speed (frequency) within limits. It consists of:

  • Oil pressure unit (OPU): a sump with turbine oil, two motor-driven screw/gear pumps (one duty, one standby), a relief valve, non-return valves, accumulators (air-loaded pressure vessels) and an oil cooler. It maintains oil at about 40–63 bar.
  • Electro-hydraulic governor: an electronic controller senses speed (from a permanent magnet generator or speed signal) and compares it with the set frequency. The error signal is amplified and sent to a servo valve.
  • Servomotor and distributor: the servo valve admits high-pressure oil to either side of the servomotor piston, which through a linkage and the distributor (or guide vane operating ring) opens or closes the guide vanes of the turbine.
  • Feedback (follow-up) mechanism: as the servomotor moves, a feedback linkage returns the servo valve to neutral, so the guide vane position is proportional to the control signal.
  • Safety features: over-speed trip (usually 110 % of rated speed) closes the guide vanes and the main inlet valve; low oil pressure or pump failure starts the standby pump and raises an alarm.

Function: on load increase the speed falls, the governor opens the guide vanes to admit more water; on load rejection the speed rises and the governor closes the guide vanes rapidly (with a closing time set to limit water hammer in the penstock). Thus the governing system keeps frequency within about ±1 % and protects the machine against runaway speed.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 10 marks

Starting from the operating principle of a shunt DC motor, explain how its speed is controlled both above and below of its rated speed.

Answer

Speed control of a shunt DC motor is achieved by varying the voltage applied to the armature or the flux per pole.


1. Operating Principle of a Shunt DC Motor

In a shunt DC motor, the field winding is connected in parallel with the armature across the same DC supply. Therefore:

  • Field current: ( I_{sh} = \dfrac{V}{R_{sh}} ) (independent of load)
  • Armature current: ( I_a = I_L - I_{sh} )

When the armature is energized, current flows through armature conductors placed in the magnetic field produced by the shunt field poles. A force ( F = B I L ) acts on each conductor, producing torque that rotates the armature. Rotation induces a back EMF:

[ E_b = \frac{\Phi Z N P}{60 A} = K_e \Phi N ]

where ( K_e = \dfrac{ZP}{60A} ) is the machine constant.

Applying KVL to the armature circuit:

[ V = E_b + I_a R_a ]

[ \Rightarrow \quad N = \frac{V - I_a R_a}{K_e \Phi} ]

This is the speed equation of the shunt DC motor. It shows that speed depends on:

  1. Supply voltage ( V )
  2. Armature resistance drop ( I_a R_a )
  3. Flux per pole ( \Phi )

Since ( \Phi \propto I_{sh} \propto V ) (for constant ( R_{sh} )), the shunt motor is called a constant-speed motor — its speed remains nearly constant from no-load to full-load.


2. Speed Control Below Rated Speed

(a) Armature Resistance Control (Rheostatic Control)

A variable resistance ( R_{ext} ) is inserted in series with the armature.

[ N = \frac{V - I_a (R_a + R_{ext})}{K_e \Phi} ]

  • Increasing ( R_{ext} ) reduces the numerator, hence speed decreases.
  • Advantages: Simple, cheap, easy to implement.
  • Disadvantages: Large power loss ( I_a^2 R_{ext} ), poor speed regulation, speed varies with load. Suitable only for small motors or intermittent duty.

(b) Armature Voltage Control (Ward-Leonard Method)

The armature is fed from a separate variable DC source (motor-generator set or controlled rectifier) while the field is excited at rated value.

[ N \propto V_a \quad (\Phi \text{ constant}) ]

  • Reducing ( V_a ) below rated value reduces speed below rated speed.
  • Advantages: Smooth, stepless control; high efficiency; good speed regulation; constant torque operation.
  • Disadvantages: Costly (extra MG set), bulky, higher initial cost.

(c) Tapped Field / Diverter Method

A diverter resistance is connected in parallel with the field winding to reduce ( I_{sh} ), hence reduce ( \Phi ). But reducing flux increases speed — so this method is used for above rated speed, not below. For below rated speed, field flux must be increased by adding series turns (field strengthening), which is rarely used because the field is already near saturation.


2. Speed Control Above Rated Speed

Field Flux Control (Field Weakening Method)

A variable resistance ( R_c ) is inserted in series with the shunt field winding.

[ I_{sh} = \frac{V}{R_{sh} + R_c} ]

Increasing ( R_c ) decreases ( I_{sh} ), hence decreases ( \Phi ). From the speed equation:

[ N \propto \frac{1}{\Phi} ]

So speed increases above rated value.

  • Advantages: Simple, low power loss (field current is small), smooth control, economical.
  • Disadvantages: Speed can be increased only up to about 1.5–2 times rated speed due to:
    • Commutation problems (sparking at brushes)
    • Mechanical stress on armature
    • Loss of stability (speed may become unstable at very low flux)
  • Limit: ( N_{max} \approx 2 N_{rated} )

Summary Table

MethodSpeed RangePrinciplePower Loss
Armature resistance controlBelow rated( R_{ext} ) in series with ( R_a )High
Armature voltage controlBelow ratedVariable ( V_a )Low
Field flux controlAbove rated( R_c ) in series with ( R_{sh} )Low

Key Points for Full Marks

  1. Speed equation ( N = \dfrac{V - I_a R_a}{K_e \Phi} ) must be derived or stated.
  2. Below rated speed → increase denominator (flux) or decrease numerator (voltage/resistance).
  3. Above rated speed → decrease flux (field weakening).
  4. Mention practical limits: ( N_{max} \approx 2 N_{rated} ) due to commutation and mechanical constraints.
  5. Ward-Leonard method is the standard for below-rated speed control with good regulation.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 10 marks

A power utility is planning a transmission system to supply electricity to a growing urban area. Discuss the factors affecting the selection of transmission voltage and conductor size, including the role of sag and tension and the need for maintaining proper clearances and right of way. Do you expect the corona phenomenon and its impact on the transmission line's performance? State your views.

Answer

Selection of Transmission Voltage and Conductor Size

Factors Affecting the Selection of Transmission Voltage

  1. Amount of power to be transmitted: The voltage level rises with the power to be carried. For a given power, raising the voltage reduces the current, and hence the conductor cross-section and losses. As a rough guide, the line voltage in kV is taken as about 1 kV per km of line length plus 1 kV per 100 kW of power (or 5.5√L + P/100 in some texts).

  2. Distance of transmission: For long lines a higher voltage is economical, since the line drop and the I²R loss fall as the voltage is raised. Standard practice in Nepal: 11 kV for short distribution feeders, 33 kV for sub-transmission, 66 kV, 132 kV and 220 kV for transmission, and 400 kV for bulk long-distance transmission.

  3. Economy (cost comparison): The total annual cost = annual fixed cost (interest, depreciation, maintenance of conductors, insulators, towers, transformers, switchgear) + annual operating cost (I²R loss energy cost). As voltage increases, the fixed cost rises (more insulation, taller towers, larger clearances) but the loss cost falls. The economical voltage is the one giving minimum total annual cost. Kelvin's law is applied to the conductor size, and a similar cost comparison is made for voltage.

  4. Corona and radio interference: Very high voltage causes corona loss, audible noise and radio interference; this limits how high the voltage may be pushed for a given conductor size and spacing.

  5. Voltage regulation and stability: Higher voltage improves regulation and the steady-state and transient stability limits of the system. For a line of given length, the permissible voltage drop (say 5–10%) fixes the minimum voltage.

  6. Standardisation: Only standard voltages (11, 33, 66, 132, 220, 400 kV) are chosen so that transformers, insulators and switchgear are interchangeable and spare parts are available.

  7. Right of way and clearances: Higher voltage needs wider right of way and greater ground clearance, which increases cost and land acquisition problems, especially in hilly terrain and forest areas of Nepal.

  8. Future expansion: Provision is kept for future load growth so that the line need not be rebuilt soon.

Factors Affecting the Selection of Conductor Size

  1. Current-carrying capacity (ampacity): The conductor must carry the maximum continuous current without exceeding the permissible temperature rise (usually 75 °C for ACSR), otherwise annealing and loss of tensile strength occur.

  2. Voltage drop (regulation): The cross-section must be large enough that the voltage drop at the far end stays within the permissible limit.

  3. Corona loss: For EHV lines the conductor diameter must be large enough (or bundled conductors used) to keep the surface voltage gradient below the disruptive value, so that corona loss and radio interference are within limits. This often fixes the minimum size, not the current.

  4. Mechanical strength: The conductor must withstand wind and ice loading and its own weight without excessive sag or tension. A minimum size (e.g. 35 mm² or 50 mm²) is often specified for mechanical strength alone.

  5. Economy — Kelvin's law: The most economical size is that for which the annual cost of conductor material (interest + depreciation) equals the annual cost of energy lost (I²R loss). This gives the economical current density of about 1–2 A/mm² for copper and 0.5–1 A/mm² for aluminium.

  6. Type of conductor: ACSR is preferred in Nepal because of its high tensile strength, low cost and suitability for long spans in hilly terrain; copper is used where conductivity and corrosion resistance matter more.

Role of Sag and Tension

Sag is the vertical distance between the lowest point of the conductor and the line joining the two supports. For a level span,

S=wL28TS = \frac{wL^2}{8T}

where ww = weight per unit length, LL = span, TT = tension. Also, T=wL28ST = \dfrac{wL^2}{8S}.

  • Sag must be kept within limits so that the specified ground clearance is maintained under the worst conditions (maximum temperature, wind, ice).
  • Tension must not exceed the ultimate tensile strength divided by the factor of safety (usually 2 for transmission lines). Too high a tension causes fatigue and failure; too low a tension causes excessive sag.
  • The stringing chart (sag–tension–temperature) is used at the time of erection so that the conductor is strung at the correct tension for the design temperature.
  • Sag affects tower height: greater sag needs taller towers, hence more cost. A balance between tension (conductor cost) and sag (tower cost) is therefore sought.

Clearances and Right of Way

  • Ground clearance: Minimum clearance from the lowest conductor to ground, roads, railways, rivers and buildings must be maintained as per the Electricity Rules (e.g. about 6.1 m over ground for 11 kV, increasing with voltage; about 8–10 m for 132 kV and above).
  • Clearance to other lines and structures must be adequate to avoid flashover and to allow safe maintenance.
  • Right of way (RoW): A strip of land along the line must be kept free of tall trees and buildings. Typical RoW width is about 15–30 m for 132 kV and 30–50 m for 220 kV and above. In Nepal, forest clearance and land compensation for RoW are major cost and social issues, so the route is chosen to minimise them.
  • Safety: Proper clearance prevents accidents to people, animals and vehicles, and reduces the risk of flashover due to wind swinging or conductor galloping.

Corona Phenomenon and Its Impact

Yes, corona is expected on high-voltage transmission lines, and its effect must be considered in design.

Cause: When the surface voltage gradient of the conductor exceeds the disruptive value (about 30 kV/cm peak in air at NTP), the air around the conductor is ionised, producing a faint violet glow, hissing noise and ozone. It is more pronounced in humid, rainy and foggy weather, and at high altitude where air density is low — conditions common in Nepal.

Effects on performance:

  1. Power loss: Corona loss increases with voltage and frequency and reduces transmission efficiency. It is significant above 132 kV and must be estimated (Peek's formula).
  2. Radio and television interference: The high-frequency pulses disturb communication and radio reception near the line.
  3. Audible noise: Hissing and humming cause annoyance, especially in residential areas.
  4. Ozone and corrosion: Ozone produced attacks insulators and fittings and may cause corrosion.
  5. Reduction of effective capacitance and increased leakage: Corona acts as a shunt path, altering the line parameters slightly.
  6. Flashover risk: If corona becomes excessive, it may lead to flashover of insulators and eventual failure.

Methods to reduce corona:

  • Increase conductor diameter (use ACSR, hollow or expanded conductors).
  • Use bundled conductors (2, 3 or 4 sub-conductors per phase) — this reduces the surface gradient and is standard for 220 kV and above.
  • Increase conductor spacing (but this increases tower cost and RoW).
  • Use corona rings and smooth fittings at insulators and conductor ends.
  • Operate at lower voltage gradient by proper design of the line.

Conclusion: Corona cannot be eliminated but can be kept within acceptable limits by proper choice of conductor size, bundling and spacing. In the planning of the transmission system for the urban area, the voltage and conductor size should be selected by an economical comparison of fixed and operating costs, with due regard to sag, tension, clearances, right of way and corona performance, so that the supply is reliable, safe and economical.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 10 marks

Write short note on economic load dispatch in interconnected power system. Present necessary figure and mathematical equation to justify your points.

Answer

Economic Load Dispatch in Interconnected Power Systems

Definition. Economic Load Dispatch (ELD) is the process of allocating the total system load among the committed generating units of an interconnected power system so that the total operating (fuel) cost is minimised while satisfying the load demand and all unit and network constraints. It is an on-line operational problem solved at the load dispatch centre, usually every few minutes, after the unit commitment decision has already decided which units are running.

Objective function. For NN thermal units over a scheduling interval,

min⁡CT=∑i=1NCi(Pi)=∑i=1N(ai+biPi+ciPi2)Rs/h\min C_T = \sum_{i=1}^{N} C_i(P_i) = \sum_{i=1}^{N}\left(a_i + b_i P_i + c_i P_i^{2}\right) \quad \text{Rs/h}

where PiP_i is the real power output of unit ii (MW) and ai,bi,cia_i, b_i, c_i are the cost coefficients of that unit.

Constraints.

  1. Power balance (equality): ∑i=1NPi=PD+PL\sum_{i=1}^{N} P_i = P_D + P_L where PDP_D is total load demand and PLP_L is total transmission loss.

  2. Generation limits (inequality): Pimin⁡≤Pi≤Pimax⁡,i=1,2,…,NP_i^{\min} \le P_i \le P_i^{\max}, \qquad i = 1,2,\dots,N

Solution by the Lagrange multiplier method. Form the Lagrangian

L=∑i=1NCi(Pi)+λ(PD+PL−∑i=1NPi)\mathcal{L} = \sum_{i=1}^{N} C_i(P_i) + \lambda\left(P_D + P_L - \sum_{i=1}^{N} P_i\right)

Setting ∂L/∂Pi=0\partial \mathcal{L}/\partial P_i = 0 gives the coordination equation

dCidPi+λ ∂PL∂Pi=λ\frac{dC_i}{dP_i} + \lambda\,\frac{\partial P_L}{\partial P_i} = \lambda

Defining the penalty factor of unit ii as

Li=11−∂PL∂PiL_i = \frac{1}{1 - \dfrac{\partial P_L}{\partial P_i}}

the condition becomes the classical equal-incremental-cost rule:

 LidCidPi=λfor all i \boxed{\,L_i \frac{dC_i}{dP_i} = \lambda \quad \text{for all } i\,}

i.e. the penalised incremental cost of every unit must be equal to the system incremental cost λ\lambda (the Lagrange multiplier, in Rs/MWh). For the quadratic cost curve, dCidPi=bi+2ciPi\dfrac{dC_i}{dP_i} = b_i + 2c_iP_i, so

Pi=λ/Li−bi2ciP_i = \frac{\lambda/L_i - b_i}{2c_i}

Neglecting losses. If transmission losses are small compared with the load (short lines, load near generation), ∂PL/∂Pi≈0\partial P_L/\partial P_i \approx 0, so Li≈1L_i \approx 1 and the rule reduces to the simple, much-used form

dCidPi=dCjdPj=λ∀ i,j\frac{dC_i}{dP_i} = \frac{dC_j}{dP_j} = \lambda \qquad \forall\, i,j

Iterative procedure (losses included).

  1. Choose a trial value of λ\lambda.
  2. Compute PiP_i for each unit from the coordination equation, respecting Pimin⁡P_i^{\min} and Pimax⁡P_i^{\max}; units at a limit are "clamped" and dropped from the equal-λ\lambda set.
  3. Compute PLP_L using the B-coefficient (Kron's) loss formula PL=∑i∑jPiBijPj+∑iBi0Pi+B00P_L = \sum_i \sum_j P_i B_{ij} P_j + \sum_i B_{i0} P_i + B_{00}
  4. Check the power balance ∑Pi−PD−PL=0\sum P_i - P_D - P_L = 0. If the mismatch ΔP\Delta P is not within tolerance, update λ(k+1)=λ(k)−ΔP[∑i12ciLi2]−1\lambda^{(k+1)} = \lambda^{(k)} - \Delta P \left[\sum_i \frac{1}{2c_i L_i^{2}}\right]^{-1} and repeat until convergence. (Excess generation, ΔP>0\Delta P>0, lowers λ\lambda; a deficit raises it. With Li≈1L_i\approx1 the denominator is simply ∑1/2ci\sum 1/2c_i.)

Figure.

        Rs/MWh
          |            dC1/dP1        dC2/dP2
          |              \              /
          |               \            /
     λ  -|- - - - - - - - -\- - - - - /- - - - -  equal incremental cost line
          |                 \        /
          |                  \      /
          |___________________\____/________________ MW
              P1  P2  P3  ...  PN
        ┌───────────────────────────────────────────┐
        │  Load Dispatch Centre (SCADA / EMS)       │
        │  P_D ──► ELD algorithm ──► P_i set points │
        └───────────────────────────────────────────┘
             ▲                        │
             │  telemetered P_i, f     ▼
        ┌───────────────────────────────────────────┐
        │  Interconnected system: Area 1 | Area 2   │
        │  tie-line flows, losses P_L, freq. bias   │
        └───────────────────────────────────────────┘

The upper part shows the classical graphical interpretation: the system λ\lambda is the horizontal line at which the sum of the unit outputs equals the demand plus losses. The lower part shows the practical setting — the dispatch centre computes set points from telemetered data and sends them to the plants, while tie-line flows and frequency are fed back.

Why interconnection matters.

  • Loss reduction / diversity. Interconnected areas share reserves and can exploit the diversity of load patterns; the loss formula PLP_L couples all units, so a unit far from the load is penalised by a penalty factor Li>1L_i > 1 and is backed down, reducing total PLP_L.
  • Economic interchange. Area AA imports from area BB as long as the incremental cost of the importing area exceeds that of the exporting area, i.e. until λA=λB\lambda_A = \lambda_B at the tie-line limit. This is the basis of the "tie-line bias" and multi-area ELD.
  • Constraints. In practice ELD is solved subject to tie-line flow limits, ramp-rate limits, prohibited operating zones and security (contingency) constraints, which is then termed Security-Constrained Economic Dispatch (SCED).

Conclusion. ELD minimises fuel cost by equalising penalised incremental costs, subject to demand, loss and limit constraints. In an interconnected system the loss penalty factors and tie-line interchange make the problem coupled across areas, so the dispatch centre must solve it centrally (or by hierarchical multi-area coordination) to obtain the true economic optimum.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · 1st · 10 marks

Discuss the operating principle of 3ϕ3\phi controlled rectifier. Hence, explain its application in HVDC power transmission.

Answer

Operating Principle of a 3-Phase Controlled Rectifier

A three-phase controlled rectifier converts three-phase AC into variable DC using thyristors (SCRs) whose firing angle α\alpha controls the average output voltage. The most common configuration is the 3-phase fully controlled bridge (6-pulse), using six thyristors — two per phase, one in the top group (cathodes common) and one in the bottom group (anodes common).

Key principles:

  1. Natural commutation and conduction sequence. The thyristor whose anode is at the most positive potential among the top group conducts when fired; the one whose cathode is at the most negative potential among the bottom group conducts when fired. The six thyristors fire in the sequence T1–T2–T3–T4–T5–T6 at intervals of 60°60°, each conducting for 120°120°. At any instant two thyristors conduct — one from the top group and one from the bottom group — completing the path from the most positive phase to the most negative phase.

  2. Firing angle control. If thyristors are fired at α=0°\alpha = 0° (i.e., at the natural commutation instants), the circuit behaves as a diode bridge. Delaying the firing by angle α\alpha reduces the average DC output. The output voltage waveform is the envelope of the line-to-line voltages, with the conduction interval shifted by α\alpha.

  3. Average output voltage. For a 3-phase fully controlled bridge with peak phase voltage VmV_m (RMS phase voltage Vs=Vm/2V_s = V_m/\sqrt{2}):

Vdc=33 Vmπcos⁡α=33 VLLπcos⁡α=1.35 VLLcos⁡αV_{dc} = \frac{3\sqrt{3}\,V_m}{\pi}\cos\alpha = \frac{3\sqrt{3}\,V_{LL}}{\pi}\cos\alpha = 1.35\,V_{LL}\cos\alpha

where VLLV_{LL} is the RMS line-to-line voltage. Thus:

  • α=0°\alpha = 0°: Vdc=1.35 VLLV_{dc} = 1.35\,V_{LL} (maximum)
  • α=90°\alpha = 90°: Vdc=0V_{dc} = 0
  • α>90°\alpha > 90°: VdcV_{dc} becomes negative → inverter mode (power flows from DC to AC), provided a DC source (e.g., a back-emf) exists.
  1. Ripple and harmonics. Output ripple frequency is 6f6f (six pulses per cycle), giving smaller filter requirements than single-phase rectifiers. Input current contains harmonics of order 6k±16k \pm 1 (5th, 7th, 11th, …).

  2. Two-quadrant / four-quadrant operation. With α\alpha control, the bridge operates as a rectifier (0°<α<90°0° < \alpha < 90°) or inverter (90°<α<180°90° < \alpha < 180°), enabling bidirectional power flow when two bridges are connected back-to-back.

Application in HVDC Power Transmission

HVDC transmission requires conversion of AC to DC at the sending end (rectifier) and DC back to AC at the receiving end (inverter). The 3-phase controlled rectifier is the heart of this conversion:

  1. Rectification at the sending end. A 3-phase fully controlled thyristor bridge (usually 12-pulse, formed by two 6-pulse bridges fed from star and delta transformer secondaries) converts generated AC into high-voltage DC. The firing angle α\alpha directly sets the DC link voltage, so power flow is controlled electronically without mechanical switching.

  2. Inversion at the receiving end. The same bridge, operated with α>90°\alpha > 90°, acts as a line-commutated inverter, converting DC back to AC for the receiving grid. The two ends are thus identical converter stations, each capable of rectifying or inverting.

  3. Power flow control. DC power is P=Vdc×IdcP = V_{dc} \times I_{dc} By adjusting α\alpha (and hence VdcV_{dc}) and the current order, the operator controls both magnitude and direction of transmitted power rapidly and precisely — a major advantage over AC ties.

  4. Reactive power and harmonics. Thyristor bridges consume reactive power and inject harmonics; these are managed by AC filters and shunt capacitor banks at each terminal. The 12-pulse configuration cancels the 5th and 7th harmonics, reducing filter cost.

  5. Advantages realised in HVDC. Because the controlled rectifier allows independent control of voltage and power:

    • Bulk power can be transmitted over long distances with lower losses (no skin effect, no charging current).
    • Asynchronous grids can be interconnected.
    • Submarine cable transmission becomes feasible (no capacitive charging limitation).
    • Fast, precise control aids stability and emergency power support.

Conclusion. The 3-phase controlled rectifier works on phase-controlled firing of thyristors to give a controllable DC output (Vdc∝cos⁡αV_{dc} \propto \cos\alpha), and by extending α\alpha beyond 90°90° it also performs inversion. This dual rectifier–inverter capability makes it the fundamental building block of HVDC transmission, enabling efficient, controllable, and asynchronous bulk power transfer.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 5 marks

Explain how a transmission line is protected against lightening? What is the use and importance of optical ground wire running along the apex of the transmission towers? Discuss.

Answer

Protection of transmission lines against lightning

A transmission line is protected against lightning by a combination of shielding, low tower footing resistance, and co-ordination of insulation:

  1. Overhead earth wire (shield wire): One or two earth wires are strung at the apex of the towers above the phase conductors. They intercept the lightning stroke before it reaches the phase conductors, so the stroke current flows to earth through the tower and its footing instead of flashing over the line insulation. The earth wire is placed so that the phase conductors lie within its zone of protection (shielding angle generally about 20°–30° for vertical configuration).

  2. Low tower footing resistance: The surge impedance of the tower and the footing resistance determine the tower-top potential. A low footing resistance (commonly aimed below 10 Ω) limits the rise of tower potential, so the voltage across the insulator string stays below its impulse flashover value and back-flashover is avoided.

  3. Co-ordination of insulation (basic impulse insulation level): The insulator strings and air gaps are chosen with a BIL higher than the expected surge voltage, so that any flashover occurs at the protective device (arcing horn or surge arrester) rather than across the insulator.

  4. Arcing horns / protective gaps: Horn gaps across insulator strings divert the flashover arc away from the insulator surface, preventing damage to the insulator and allowing the arc to extinguish when the fault current is cleared.

  5. Overhead ground wires combined with surge arresters: At substation approaches and on lines in high-isokeraunic areas, surge arresters and counterpoise (buried counterpoise wires) are used to further reduce surge voltage and footing resistance.

  6. Shielding failure: If a stroke bypasses the earth wire (shielding failure), the resulting flashover is cleared by the protective relays and circuit breakers, which isolate the faulty section and allow reclosing.

Use and importance of optical ground wire (OPGW) at the apex of towers

The optical ground wire is a composite cable in which optical fibres are enclosed in a metallic (usually aluminium-clad steel) tube, and the outer strands serve both as the earth wire and as the mechanical strength member. It is run along the apex of the towers, i.e. in the position of the overhead earth wire. Its importance:

  1. Dual function: It simultaneously performs the function of the overhead earth wire (lightning shielding and fault current return) and of a communication link, so no separate earth wire or separate optical cable is needed.

  2. Communication for protection and control: The fibres carry teleprotection signals, SCADA data, telemetry, and voice/data traffic between substations. This enables fast and reliable unit protection, remote monitoring, and load dispatch, which is essential for a stable grid.

  3. High bandwidth and immunity: Optical fibres have very large bandwidth, low attenuation, and are immune to electromagnetic interference and induced voltages from the power line itself, unlike pilot wires or power line carrier.

  4. Reliability and security: Being at the apex, the OGW is protected from mechanical damage and vandalism, and it is not affected by short circuits or lightning-induced surges in the same way as a separate communication cable. It also withstands the mechanical tension and wind loading of the line.

  5. Economy: It avoids the cost of a separate communication cable, its towers, and its right-of-way, and it uses the existing transmission corridor, giving a very economical communication system for the utility.

Thus the earth wire protects the line against lightning, while the OPGW at the apex provides the same protection and additionally serves as the backbone communication medium of the power system.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 5 marks

Describe three part tariff and power factor tariff.

Answer

Three Part Tariff

A three part tariff consists of three components:

  1. Fixed charge (consumer charge) — a lump sum charged per consumer per billing period, independent of the energy consumed. It covers the cost of connecting the consumer to the supply (meter, service line, billing cost).
  2. Semi-fixed charge / demand charge (capacity charge) — charged on the maximum demand (kW or kVA) recorded during the billing period. It recovers the fixed cost of generation, transmission and distribution capacity installed to meet the consumer's peak demand.
  3. Running charge (energy charge) — charged on the actual energy consumed (kWh), usually at a flat or block rate. It recovers the variable cost of fuel and operation.

Total bill = Fixed charge + (Maximum demand in kW × demand rate) + (Energy consumed in kWh × energy rate)

Example: If fixed charge = Rs 200, demand rate = Rs 300/kW, energy rate = Rs 8/kWh, and a consumer has maximum demand 10 kW and consumes 1,500 kWh:

Bill = 200 + (10 × 300) + (1,500 × 8) = 200 + 3,000 + 12,000 = Rs 15,200

Advantages: It recovers the fixed cost reliably, encourages the consumer to reduce peak demand (improving load factor), and is fair because each cost element is charged separately. It is widely used for large industrial and commercial consumers.

Power Factor Tariff

A power factor tariff is a tariff in which the electricity bill depends on the power factor of the consumer's load, in addition to the energy and demand charges. It is applied to consumers whose loads are inductive (motors, transformers, induction furnaces), which draw reactive power and reduce the power factor.

The purpose is to penalise low power factor and reward good power factor, because a low power factor means:

  • larger current for the same useful power, hence larger conductor and transformer ratings;
  • higher I²R losses and greater voltage drop;
  • poor utilisation of installed plant capacity.

Common forms:

  1. kW and kVA maximum demand tariff — the demand charge is levied on the kVA maximum demand instead of kW. A consumer with low power factor records a higher kVA demand and hence pays more.
  2. Sliding scale (power factor bonus/penalty) — a reference power factor (say 0·85 lagging) is fixed. For every 0·01 fall below it, a penalty is added to the bill; for every 0·01 rise above it, a discount is allowed.
  3. kVAh tariff — energy is charged in kVAh instead of kWh, so that the consumer pays for both active and reactive energy.

Example: A consumer has a demand charge of Rs 300/kVA and a maximum demand of 100 kW at 0·8 pf lagging.

kVA demand = 100 / 0·8 = 125 kVA Demand charge = 125 × 300 = Rs 37,500

If the power factor is improved to 0·95 lagging:

kVA demand = 100 / 0·95 = 105·26 kVA Demand charge = 105·26 × 300 = Rs 31,578

Saving = 37,500 − 31,578 = Rs 5,922 per month, which shows how a power factor tariff encourages the consumer to install shunt capacitors and improve the power factor.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 10 marks

Discuss with a neat single line diagram (SLD) a 132kV substation bus-bar with circuit breaker and isolator for single bus, main and transfer bus and double bus arrangements.

Answer

132 kV Substation Bus-Bar Arrangements

A bus-bar is the common conducting node to which all incoming lines (from generating stations or grid substations) and outgoing feeders (to load centres) are connected through switching and protective equipment. The choice of bus-bar scheme decides the reliability, flexibility of operation, ease of maintenance and cost of a 132 kV substation. The three standard arrangements used by NEA at 132 kV level are discussed below. In all cases the switching sequence rule is strictly followed: close isolator first, then circuit breaker; while opening, open circuit breaker first, then isolator — because an isolator (off-load switch) cannot interrupt or make fault current, whereas a circuit breaker (SF₆ type at 132 kV) can.


1. Single Bus-Bar Arrangement

Single Line Diagram

   Line-1        Line-2        Line-3        Line-4
     |             |             |             |
    ISO           ISO           ISO           ISO
     |             |             |             |
    CB            CB            CB            CB
     |             |             |             |
     +-------------+-------------+-------------+
     |=========== 132 kV BUS-BAR  =============|
     +-------------+-------------+-------------+
     |             |             |
    CB            CB            CB
     |             |             |
    ISO           ISO           ISO
     |             |             |
   Tr-1          Tr-2          Tr-3

Description and Features

  • Only one bus-bar is used; every bay has one circuit breaker and two isolators (one bus-side, one line-side).
  • Advantages: simplest, cheapest, least land and least switchgear; simple protection and interlocking.
  • Disadvantages: a fault on the bus-bar trips the entire substation; maintenance of the bus-bar or of any bus-side isolator requires a complete shutdown; no alternative path for load transfer.
  • NEA use: 132/33 kV or 132/11 kV distribution substations of small capacity, radial load points, and as the low-priority end of a transmission line.

2. Main and Transfer (M&T) Bus-Bar Arrangement

Single Line Diagram

   Line-1     Line-2     Line-3     Line-4
     |          |          |          |
    ISO        ISO        ISO        ISO
     |          |          |          |
    CB         CB         CB         CB
     |          |          |          |
    ISO        ISO        ISO        ISO
     |          |          |          |
     +----------+----------+----------+------ MAIN BUS
     |          |          |          |
     |         BUS-COUPLER CB         |
     |          |          |          |
     +----------+----------+----------+------ TRANSFER BUS
     |          |          |          |
    ISO        ISO        ISO        ISO
     |          |          |          |
   Tr-1       Tr-2       Tr-3       Tr-4

Description and Features

  • Two bus-bars: the Main bus (normal supply) and the Transfer bus (standby). A bus-coupler circuit breaker connects them.
  • Each bay has a circuit breaker with isolators on both sides, plus an additional isolator to the transfer bus.
  • Operation: If the main bus or a bay breaker needs maintenance, the bus-coupler CB is closed, the transfer isolator of that bay is closed, the bay CB is opened and isolated, and supply continues through the bus-coupler and transfer bus.
  • Advantages: any bay breaker or the main bus can be maintained without interrupting supply; cheaper than double bus (only one extra CB — the coupler).
  • Disadvantages: a fault on the main bus still causes loss of the whole bus unless the coupler is opened; during transfer, the bus-coupler CB must be rated for the largest bay; protection becomes complicated during transfer.
  • NEA use: 132 kV substations feeding important load centres where supply continuity is required but full double-bus cost is not justified.

3. Double Bus-Bar Arrangement

Single Line Diagram

   Line-1     Line-2     Line-3     Line-4
     |          |          |          |
    ISO        ISO        ISO        ISO
     |          |          |          |
    +----------+----------+----------+------ BUS-1
    |          |          |          |
    |         BUS-COUPLER CB         |
    |          |          |          |
    +----------+----------+----------+------ BUS-2
     |          |          |          |
    ISO        ISO        ISO        ISO
     |          |          |          |
    CB         CB         CB         CB
     |          |          |          |
   Tr-1       Tr-2       Tr-3       Tr-4

Description and Features

  • Two identical bus-bars (Bus-1 and Bus-2), each capable of carrying the full station load, connected by a bus-coupler CB.
  • Each bay has two bus-side isolators (one to each bus) and one line-side isolator, so the bay can be connected to either bus.
  • Operation: Normally all bays on Bus-1 and Bus-2 kept as a reserve; on a fault on Bus-1, the bus-coupler is opened and all healthy bays are transferred to Bus-2 without loss of supply. Any bus can be taken out for maintenance while the other carries the load.
  • Advantages: highest reliability and flexibility; bus fault does not cause total shutdown; maintenance of either bus possible without interruption; load can be redistributed between buses.
  • Disadvantages: most expensive — double bus-bar, double isolators per bay, larger land and civil cost; more complex interlocking and protection; risk of parallel operation if both bus isolators of a bay are closed.
  • NEA use: 132 kV grid substations, interconnecting substations and major load centres (e.g. Kathmandu valley ring, industrial corridors) where reliability is critical.

Comparison

FeatureSingle BusMain & TransferDouble Bus
No. of bus-bars12 (1 main + 1 transfer)2 (both main)
Extra CB—1 bus-coupler1 bus-coupler
Isolators per bay233
Bus fault effectTotal shutdownPartial (with coupler)No shutdown
Maintenance without shutdownNot possiblePossiblePossible
CostLowestModerateHighest
ReliabilityLowHighHighest

Conclusion: For a 132 kV NEA substation, the single bus scheme is adopted only for small radial substations; the main-and-transfer scheme is a compromise for important load centres; and the double bus scheme is adopted for grid/interconnecting substations where continuity of supply and operational flexibility are paramount. In every case the circuit breaker performs the making and breaking of current, while the isolator only provides visible isolation after the breaker is opened.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 10 marks

What is electric shock? Write down in the proper order for the physiological effects of electric current (as current magnitude increase) in human body.

Answer

Electric shock is the sudden passage of electric current through the human body, caused when a person comes into contact with a live conductor (or a conductor that has become live due to insulation failure) while the body simultaneously provides a path to earth or to another conductor at a different potential. The severity of the shock depends mainly on the magnitude of current, its path through the body, the duration of contact, the frequency, and the body's resistance (chiefly skin resistance, which falls sharply when the skin is wet).

Physiological effects of electric current in order of increasing current magnitude (for a 50/60 Hz AC, hand-to-hand or hand-to-foot path, contact duration of about 1 second):

S.N.Current (AC, 50/60 Hz)Physiological effect
11–2 mA (threshold of perception)Faint tingling sensation; current just perceptible.
22–10 mAPainful tingling; muscular contraction begins but the person can usually still let go.
310–20 mA"Let-go" threshold — involuntary muscular contraction of the forearm/hand; the person cannot release the conductor on his own.
420–50 mASevere muscular contraction, paralysis of respiratory muscles; breathing becomes difficult or stops (suffocation).
550–100 mAVentricular fibrillation of the heart begins (fatal if not defibrillated within minutes); respiratory paralysis.
6100 mA – 1 A (≈ 0.1–1 A)Definite ventricular fibrillation; death is almost certain unless immediate resuscitation is given.
7Above 1 A (≈ 1–5 A)Sustained myocardial contraction (heart may clamp and later resume), severe burns, tissue charring; death from burns and asphyxia.

Key points to remember:

  • The let-go threshold (≈ 10–20 mA) is the most critical value — above it the victim cannot free himself, so contact duration increases and the shock becomes dangerous.
  • Ventricular fibrillation (the usual cause of death) occurs from about 50 mA upward.
  • Very high currents (several amperes) may cause the heart to clamp rather than fibrillate, but cause massive internal and external burns.
  • For the same current, longer duration and a path through the heart (hand-to-hand or left-hand-to-foot) are more dangerous.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 10 marks

Explain how Analog to Digital conversion can be achieved by using successive approximation ADC.

Answer

Successive Approximation ADC

The successive approximation register (SAR) ADC is the most widely used method for medium-speed, medium-resolution conversion. It works on a binary search (trial-and-error) principle: it guesses the output digital code bit by bit, starting from the MSB, and keeps a bit only if the trial guess does not exceed the input.

Main blocks

  1. Sample-and-Hold (S/H) circuit — samples the analog input and holds it constant at VinV_{in} during the whole conversion, since the input must not change while bits are being decided.
  2. Comparator — compares the held input VinV_{in} with the DAC output VDACV_{DAC} and gives logic 1 if Vin>VDACV_{in} > V_{DAC}, else logic 0.
  3. Successive Approximation Register (SAR) — an nn-bit register that supplies the trial digital word to the DAC and stores the accepted bits. It is driven by a clock.
  4. D/A converter (DAC) — converts the trial digital word back to an analog voltage VDACV_{DAC} for comparison.
  5. Control logic and clock — sequences the SAR from MSB to LSB and issues the end-of-conversion (EOC) signal.

Working principle (for an nn-bit ADC)

Let the full-scale voltage be VFSV_{FS}, so the resolution (step size) is VLSB=VFS2nV_{LSB} = \frac{V_{FS}}{2^n}

The conversion proceeds in nn clock cycles:

  • Cycle 1: The SAR sets the MSB = 1 and all other bits = 0, giving a trial code of 100…0100\ldots0, i.e. VDAC=VFS/2V_{DAC} = V_{FS}/2. The comparator checks:
    • If Vin≥VDACV_{in} \ge V_{DAC}, the MSB is retained (kept as 1).
    • If Vin<VDACV_{in} < V_{DAC}, the MSB is cleared back to 0.
  • Cycle 2: The next lower bit is set to 1 (keeping the decided MSB), and the same comparison is repeated. This bit is kept or cleared accordingly.
  • Cycle 3 onward: The process continues down to the LSB, each cycle fixing one bit.
  • After nn cycles the register holds the final digital word, and the EOC signal indicates that the output is ready. The conversion time is Tc=n TclkT_c = n\,T_{clk}

Illustration (3-bit ADC, VFS=8 VV_{FS} = 8\ \text{V}, Vin=5.2 VV_{in} = 5.2\ \text{V})

Step size =8/23=1 V= 8/2^3 = 1\ \text{V}.

CycleTrial codeVDACV_{DAC} (V)ComparisonDecision
110045.2>45.2 > 4MSB kept → 1
211065.2<65.2 < 6bit cleared → 0
310155.2>55.2 > 5LSB kept → 1

Final output: 101 (i.e. 5 V5\ \text{V}, the nearest level to 5.2 V5.2\ \text{V}).

Advantages

  • Only one comparator is needed (unlike flash ADC, which needs 2n−12^n - 1).
  • Good balance of speed, resolution and cost; conversion time is fixed and independent of the input value.
  • Widely used in microcontrollers and data-acquisition ICs (e.g. ADC0804, ADC0809).

Limitations

  • Slower than flash and pipeline ADCs.
  • If the input changes during conversion, the result is in error — hence the S/H circuit is essential.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2083 · 1st · 10 marks

What is importance of lab testing of an electric machine? Discuss the process of routine test of electric transformer via short circuit and open circuit testing at lab.

Answer

Importance of Lab Testing of an Electric Machine

Laboratory testing of an electric machine is carried out to verify its design, performance, and safety before it is put into service. Its main importance is as follows:

  1. Verification of design and rated performance — Tests confirm whether the machine delivers its rated voltage, current, power, efficiency, and speed as guaranteed by the manufacturer.
  2. Determination of parameters and characteristics — Equivalent-circuit parameters, losses, efficiency, regulation, and temperature rise are obtained, which are needed for system studies, protection setting, and loading decisions.
  3. Quality control and comparison — Routine tests on each unit ensure that production machines conform to the type-tested design and to standards such as IS/IEC.
  4. Safety assurance — Insulation resistance, dielectric (high-voltage) tests, and temperature-rise tests reveal weaknesses that could cause failure, flashover, or fire in service.
  5. Detection of defects — Manufacturing defects such as loose connections, shorted turns, poor insulation, or wrong core material are detected before dispatch.
  6. Basis for maintenance and condition monitoring — Baseline test data allow comparison in later years to judge deterioration of insulation and loading capability.
  7. Economy and reliability — A machine proved by test gives reliable service, reduces outage and repair cost, and protects the investment of the utility.

Routine Tests of a Transformer by Open-Circuit and Short-Circuit Tests

The two tests together give the equivalent circuit of the transformer and allow the losses (and hence efficiency and regulation) to be computed without actually loading the transformer.

(a) Open-Circuit (No-Load) Test

Purpose: To determine the no-load (iron) loss, the no-load current, and the shunt-branch parameters (R_0) and (X_0).

Circuit: LV winding supplied at rated voltage, HV winding kept open. A wattmeter, voltmeter, and ammeter are connected on the LV side.

Procedure:

  1. Connect the LV winding to a variable AC supply through the wattmeter, ammeter, and voltmeter.
  2. Keep the HV winding open-circuited.
  3. Raise the voltage gradually to the rated value of the LV winding.
  4. Record the voltmeter reading (V_0), ammeter reading (I_0), and wattmeter reading (W_0).
  5. Since no load is connected, the copper loss in the LV winding is negligible and the wattmeter reading gives practically the iron loss (P_i = W_0).

Calculations:

Pi=W0,Iw=I0cos⁡ϕ0=W0V0,Iμ=I02−Iw2P_i = W_0, \qquad I_w = I_0\cos\phi_0 = \frac{W_0}{V_0}, \qquad I_\mu = \sqrt{I_0^2 - I_w^2}

R0=V0Iw,X0=V0IμR_0 = \frac{V_0}{I_w}, \qquad X_0 = \frac{V_0}{I_\mu}

The no-load power factor is (\cos\phi_0 = \dfrac{W_0}{V_0 I_0}).

(b) Short-Circuit Test

Purpose: To determine the full-load copper loss and the series-branch parameters (R_{eq}) and (X_{eq}).

Circuit: LV winding short-circuited (ammeter in the shorting link), reduced voltage applied to the HV winding through wattmeter and voltmeter.

Procedure:

  1. Short-circuit the LV terminals with a thick copper bar carrying an ammeter.
  2. Apply a small variable voltage (about 5–10 % of rated) to the HV winding.
  3. Increase the voltage slowly until the ammeter in the shorted winding reads the rated current.
  4. Record (V_{sc}), (I_{sc}) (rated current), and wattmeter reading (W_{sc}).
  5. Since the applied voltage is very small, the iron loss is negligible and (W_{sc}) gives the full-load copper loss (P_{cu}).

Calculations:

Zeq=VscIsc,Req=WscIsc2,Xeq=Zeq2−Req2Z_{eq} = \frac{V_{sc}}{I_{sc}}, \qquad R_{eq} = \frac{W_{sc}}{I_{sc}^2}, \qquad X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2}

The short-circuit power factor is (\cos\phi_{sc} = \dfrac{W_{sc}}{V_{sc} I_{sc}}).

Use of the Test Results

  • Iron loss (P_i) is constant at all loads; copper loss varies as the square of the load current.
  • Efficiency at any load fraction (x):

η=x (kVA)cos⁡ϕx (kVA)cos⁡ϕ+Pi+x2Pcu\eta = \frac{x\,(\text{kVA})\cos\phi}{x\,(\text{kVA})\cos\phi + P_i + x^2 P_{cu}}

  • Maximum efficiency occurs when (x^2 P_{cu} = P_i), i.e. variable loss equals constant loss.
  • Voltage regulation can be found from (R_{eq}) and (X_{eq}):

Regulation=x(IReqcos⁡ϕ±IXeqsin⁡ϕV)\text{Regulation} = x\left(\frac{I R_{eq}\cos\phi \pm I X_{eq}\sin\phi}{V}\right)

Thus, the open-circuit test gives the shunt branch (iron loss) and the short-circuit test gives the series branch (copper loss), together fully defining the transformer's performance without loading it.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

What are the problems associated with low power factor of the system and how can it be resolved? Discuss. 2+3+5=10

Answer

1. Concept of power factor (2 marks)

Power factor (PF) is the ratio of active power to apparent power in an AC system:

PF=cos⁡ϕ=PS=PP2+Q2\text{PF} = \cos\phi = \frac{P}{S} = \frac{P}{\sqrt{P^2+Q^2}}

where (P) = active power (kW), (Q) = reactive power (kVAr), and (S) = apparent power (kVA). It shows how effectively the current drawn from the supply is converted into useful work. A low power factor means a large amount of reactive power is being drawn by inductive loads such as induction motors, transformers, arc furnaces, discharge lamps, and air-conditioning equipment. NEA generally requires industrial consumers to maintain PF above 0.9, and its tariff includes power factor penalty/incentive provisions.

2. Problems associated with low power factor (3 marks)

  • Increased current and losses: For the same active power, current is (I = P/(V\cos\phi)). If PF falls, current rises, causing higher (I^2R) losses in transmission lines, transformers, and generators. This increases generation requirement and operating cost.
  • Poor voltage regulation: Low PF increases reactive current, causing larger voltage drop along feeders and transformers. Consumers receive low voltage, and the system needs extra voltage regulation or tap changing.
  • Reduced equipment capacity: Generators, transformers, cables, and switchgear are rated in kVA. At low PF, their useful kW capacity decreases. For example, a 100 kVA transformer delivers only 70 kW at PF 0.7, but 90 kW at PF 0.9. Thus equipment gets overloaded and must be oversized.
  • Overheating and reduced life: Higher current causes overheating of windings and insulation, reducing equipment life and increasing maintenance cost.
  • Higher cost and penalty: NEA and consumers face power factor penalties, higher energy charges, larger capital investment in equipment, and increased system losses.
  • System instability and inefficiency: Large reactive power flow reduces stability margin, may cause voltage collapse, and lowers overall system efficiency.

3. Remedial measures to improve power factor (5 marks)

  • Shunt capacitors: This is the most common and economical method. Capacitors supply leading reactive power locally, reducing the reactive power drawn from the supply. They can be fixed or automatically switched using a PF controller and contactors. They reduce line losses, improve voltage, and release system capacity. However, their output varies with voltage and they may cause resonance in harmonic-rich systems unless detuned reactors are used.
  • Synchronous condensers: A synchronous motor running at no load and overexcited acts as a synchronous condenser. It provides continuously variable leading reactive power and good voltage control. It is suitable for large substations and transmission systems, but has higher cost, losses, and maintenance.
  • Synchronous motors and phase advancers: Industrial synchronous motors can be operated in overexcited mode to supply reactive power to the plant. For slip-ring induction motors, phase advancers inject an emf into the rotor circuit to improve power factor at the motor terminals.
  • Static VAR compensators and FACTS devices: SVC, STATCOM, DSTATCOM, and other FACTS devices provide fast, dynamic reactive power compensation. They are used in transmission and distribution networks to control voltage, improve power factor, and enhance stability.
  • Proper load and equipment management: Avoid oversized or idling motors, use energy-efficient motors, replace old rewound motors, use variable frequency drives with proper filters, and use high-PF lighting such as LEDs and electronic ballasts. Balance single-phase loads and schedule loads to avoid peak reactive demand.
  • Tariff and regulatory measures: NEA’s power factor penalty and incentive scheme encourages consumers to install capacitors and maintain PF above the prescribed limit. This is a cost-effective demand-side management tool.
  • Design and maintenance: Use short feeders, adequate conductor sizes, and proper transformer tap settings. Regularly maintain capacitors, motors, and switchgear to ensure they operate at high power factor.

In conclusion, low power factor increases losses, reduces voltage quality and equipment capacity, and raises cost. A combination of shunt capacitors, synchronous machines, FACTS devices, proper load management, and NEA’s tariff incentives can effectively improve the power factor and make the power system efficient and reliable.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

What is load duration curve? Explain its significance in generation planning and system operation. 2+3+5=10

Answer

Load Duration Curve

1. Definition (2 marks)

A load duration curve (LDC) is a curve obtained by rearranging the chronological load curve of a power system in descending order of magnitude. The loads are plotted against the percentage of time (or total duration in hours) for which each load level is equalled or exceeded during a given period (day, month, or year).

If the load curve gives load as a function of time, the LDC gives load as a function of duration:

P=f(t),where t is the time for which load P is equalled or exceededP = f(t), \quad \text{where } t \text{ is the time for which load } P \text{ is equalled or exceeded}

The area under the LDC equals the total energy demand of the system over that period, exactly as the area under the chronological load curve does — only the order of occurrence is lost.

2. Construction and Key Features (3 marks)

Construction: The load values from the chronological load curve are sorted from highest to lowest. The highest load (peak) is plotted at the left (duration = 0), and the lowest load at the right (duration = total hours). The resulting monotonically decreasing curve is the LDC.

Key features:

  • Peak load — the maximum value, at the extreme left of the curve.
  • Base load — the minimum load level, which persists for the whole period (24 h or 8760 h).
  • Load factor — the ratio of average load to peak load; numerically equal to the ratio of the area under the LDC to the area of the rectangle formed by the peak load and the total duration:

Load Factor=Average LoadPeak Load=Area under LDCPmax×T\text{Load Factor} = \frac{\text{Average Load}}{\text{Peak Load}} = \frac{\text{Area under LDC}}{P_{max} \times T}

  • Duration axis — the horizontal axis is graduated in hours or as a percentage of the total period.

3. Significance in Generation Planning and System Operation (5 marks)

(a) Load forecasting and demand assessment The LDC, when projected for future years, gives the planner the expected peak demand and energy requirement. The area under the curve fixes the total energy to be generated, while the peak fixes the installed capacity needed. This forms the basis of capacity addition programmes.

(b) Economic selection of generating units (base load vs peak load) The LDC clearly separates base, intermediate and peak portions:

  • Base load (flat, long-duration portion) is supplied by plants with low running cost but high fixed cost — large hydro, nuclear, or large thermal plants — since they run for most hours and their high capital cost is recovered over high energy output.
  • Peak load (short-duration, high-load portion) is supplied by plants with low fixed cost but high running cost — gas turbines, diesel sets, pumped storage, or small hydro — since they operate for few hours only.

This matching of plant type to position on the LDC is the essence of optimal generation expansion planning.

(c) Reserve and reliability assessment The area to the right of the LDC (unserved duration) and the peak value indicate the reserve margin required. Planners use the LDC together with the load–duration statistics to decide spinning reserve and to carry out probabilistic reliability studies (LOLP, LOLE).

(d) Scheduling of maintenance and outages The low-load (right-hand) portion of the LDC identifies periods when surplus capacity exists. Preventive maintenance of generating units and transmission lines is scheduled in these valley periods, so that no load shedding occurs and plant availability is maximised during peak hours.

(e) Economic dispatch and unit commitment For daily operation, the LDC tells the system operator the order in which units should be brought on line (merit order) and the duration for which each must run. It also fixes the spinning reserve to be kept on part load. This minimises the total operating cost while meeting the load at every instant.

(f) Tariff and load management studies The LDC, through the load factor, is used in tariff formulation (two-part tariff, TOD tariff) and in evaluating demand-side management measures such as load shifting, which flattens the curve and improves the load factor.

(g) Interconnection and energy exchange The LDC of two interconnected systems, when superposed, shows the diversity benefit — a reduced combined peak and hence reduced total installed capacity. This justifies interconnection and power trading (e.g., Nepal–India cross-border exchange).

Conclusion: The load duration curve is thus a compact, single-line representation of the system's load behaviour. It converts a time-dependent problem into a duration-dependent one, and is indispensable for deciding what capacity to build, which plant to run, when to maintain it, and how to price electricity — making it a fundamental tool in both long-term generation planning and day-to-day system operation.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Explain the torque-speed characteristics of an induction motor. A three-phase induction motor has slip of 4%. If the supply frequency is 50Hz, calculate the rotor frequency and synchronous speed. 10

Answer

Torque–Speed Characteristics of a Three-Phase Induction Motor

The slip of an induction motor is

[ s=\frac{N_s-N_r}{N_s} ]

where (N_s) is synchronous speed and (N_r) is rotor speed.

At slip (s), the rotor induced emf and rotor reactance are

[ E_{2s}=sE_2,\qquad X_{2s}=sX_2 ]

Rotor current:

[ I_2=\frac{sE_2}{\sqrt{R_2^2+(sX_2)^2}} ]

Torque developed:

[ T=\frac{K s E_2^2 R_2}{R_2^2+(sX_2)^2} ]

or

[ T\propto \frac{sR_2}{R_2^2+(sX_2)^2} ]

Important regions of the torque–slip curve

  1. At synchronous speed, (s=0):
    Rotor emf is zero, rotor current is zero, and torque is zero. The motor cannot run exactly at synchronous speed.

  2. Low slip region, (0<s<s_m):
    Here (sX_2 \ll R_2), so

    [ T\propto s ]

    Torque increases almost linearly with slip. This is the stable operating region. Normal induction motors operate at small slip, usually (2%) to (5%).

  3. At maximum torque, (s=s_m):
    Maximum torque occurs at

    [ s_m=\frac{R_2}{X_2} ]

    and its value is

    [ T_{\max}=\frac{K E_2^2}{2X_2} ]

    This is also called breakdown torque or pull-out torque.

  4. Beyond maximum torque, (s>s_m):
    Torque decreases as slip increases. This is the unstable region. If the load torque exceeds (T_{\max}), the motor stalls.

  5. At standstill, (s=1):
    Starting torque is

    [ T_{st}=\frac{K E_2^2 R_2}{R_2^2+X_2^2} ]

Speed–torque curve

As speed decreases from synchronous speed, torque first increases up to (T_{\max}) at speed

[ N_m=N_s(1-s_m) ]

and then decreases as speed falls further. The stable operating point is the intersection of the motor torque curve and the load torque curve on the rising part of the curve.

If rotor resistance (R_2) is increased, (s_m) shifts to a higher slip value and starting torque increases, but (T_{\max}) remains unchanged. This principle is used in slip-ring induction motors with external rotor resistance to obtain high starting torque.


Numerical

Given:

[ s=4%=0.04 ]

[ f=50\text{ Hz} ]

1. Rotor frequency

[ f_r=sf ]

[ f_r=0.04\times 50 ]

[ \boxed{f_r=2\text{ Hz}} ]

2. Synchronous speed

[ N_s=\frac{120f}{P} ]

where (P) is the number of poles. Since (P) is not given in the question, synchronous speed cannot be uniquely determined. For a usual 4-pole, 50 Hz induction motor:

[ N_s=\frac{120\times 50}{4} ]

[ N_s=\frac{6000}{4} ]

[ \boxed{N_s=1500\text{ rpm}} ]

Rotor speed:

[ N_r=N_s(1-s) ]

[ N_r=1500(1-0.04) ]

[ N_r=1500\times 0.96 ]

[ \boxed{N_r=1440\text{ rpm}} ]

Slip speed:

[ N_s-N_r=1500-1440=60\text{ rpm} ]

Note: If the number of poles is not specified, synchronous speed is (N_s=\frac{6000}{P}) rpm. For example, for (P=2), (N_s=3000) rpm; for (P=6), (N_s=1000) rpm; for (P=8), (N_s=750) rpm. The rotor frequency remains (2\text{ Hz}) in all cases.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

When does transient occur in an alternator? List out the various transient periods and explain any one transient period in detail. 10

Answer

Transient in an alternator occurs when the machine suddenly changes its operating condition — a sudden short circuit, sudden load throw-off (load rejection), sudden application of load, or sudden change in excitation — so that the internal fluxes, currents and voltages pass through a time-varying state before settling to a new steady-state value. The transient condition arises because the flux linkages of the field winding and damper (amortisseur) windings cannot change instantaneously; induced currents in these closed circuits oppose the change and decay with their own time constants.

Various transient periods

  1. Sub-transient period — the first few cycles (about 1 to 2 cycles, ~0.02–0.05 s) immediately after the disturbance. Both the damper winding and the field winding carry induced currents; the machine reactance is minimum, the sub-transient reactance (X_d'').
  2. Transient period — the next period (about 0.05 s to 1 s). The damper-winding currents have decayed, but the field-winding induced currents still persist; the reactance is the transient reactance (X_d').
  3. Steady-state period — after the transient currents in the field winding have also decayed (after about 1 s or more). The machine reactance is the synchronous reactance (X_d) and the armature current is the sustained steady-state value.

Sub-transient period (explained in detail)

When a sudden short circuit occurs at the armature terminals, the armature current tends to rise suddenly. By Lenz's law, the damper (amortisseur) windings, which are short-circuited bars embedded in the pole faces, immediately develop induced currents that oppose the change in armature flux. These induced currents strongly demagnetise the air gap and reduce the effective reactance of the machine to its lowest value, the sub-transient reactance (X_d'') (typically 0.1–0.25 p.u.). Consequently the initial symmetrical short-circuit current is very large:

I′′=EXd′′I'' = \frac{E}{X_d''}

The damper-winding currents decay rapidly with the sub-transient time constant (T_d'') (about 0.02–0.05 s), so this period lasts only a few cycles. During this short interval the current envelope falls steeply, and the machine behaves almost like a transformer with a short-circuited secondary. The sub-transient period is important for selecting circuit breakers and protective relays, because the peak and initial rms fault currents — and hence the mechanical and thermal stresses on the machine and switchgear — are determined by Xd′′X_d'' and Td′′T_d''.

After the damper currents die out, the field-winding induced currents govern the slower decay of the transient period (Xd′X_d', Td′T_d'), and finally the machine reaches the steady-state period with current E/XdE/X_d.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

What is corona effect? How can it be minimized? Explain the factors affecting the corona loss. 2+3+5=10

Answer

Corona Effect

Corona effect (corona discharge) is the phenomenon of partial ionization of the air surrounding a conductor when the potential gradient at the conductor surface exceeds the dielectric strength of air (about 30 kV/cm peak, i.e. 21.1 kV/cm rms at NTP). The air envelope breaks down locally, producing a faint violet glow, a hissing sound, ozone smell, and power loss. It occurs only around the conductor (not between conductors), so it is a local breakdown, not a complete flashover.

When the surface gradient exceeds the breakdown value, air ions are formed; ions of opposite polarity are attracted and give up their charge to the conductor, while similar ions are repelled. This continuous back-and-forth movement of ions causes energy loss called corona loss.

Methods of Minimizing Corona

  1. Increase conductor diameter — use larger conductors (e.g. ACSR, expanded ACSR) so that the surface gradient falls below the disruptive value. Since corona loss ∝ (V − V_d)², raising the disruptive voltage V_d reduces loss sharply.
  2. Use bundled conductors — two, three or four sub-conductors per phase increase the effective diameter and reduce the surface gradient; this is the standard practice on EHV lines (400 kV and above).
  3. Increase conductor spacing — larger spacing reduces the voltage gradient at the conductor surface (though it increases inductance and cost).
  4. Use corona rings / grading rings — these distribute the field uniformly at insulators and conductor ends, avoiding high local gradients.
  5. Avoid sharp points and edges — smooth, large-radius surfaces prevent concentration of electric stress.
  6. Operate at lower voltage or use higher system voltage with proper design — keep the ratio V/V_d low.
  7. Keep the line dry and clean — moisture, dust and roughness increase corona; regular cleaning and smooth conductor surface help.
  8. Use higher altitude correction — at high altitude air density is low, so V_d falls; compensate by increasing diameter or spacing.

Factors Affecting Corona Loss

Corona loss depends on the following factors:

  1. Supply voltage (V): Loss increases sharply as voltage rises above the disruptive value. Empirically, P ∝ (V − V_d)². Below V_d, corona loss is negligible.

  2. Conductor surface gradient / diameter: A larger conductor diameter reduces the surface gradient, hence reduces loss. Irregular surface (stranded, rough, dirty) increases the local gradient and loss.

  3. Conductor spacing (D): Increasing spacing reduces the gradient at the conductor surface and lowers corona loss. For a single-phase line, the disruptive voltage is Vd=21.1 m δ r ln⁡ ⁣(Dr) kV (rms)V_d = 21.1 \, m \, \delta \, r \, \ln\!\left(\frac{D}{r}\right) \text{ kV (rms)} so loss falls as D increases.

  4. Air density factor (δ): Corona loss is inversely related to air density. At high altitude or high temperature, δ is low, so V_d decreases and loss increases. The air density factor is δ=3.92 b273+t\delta = \frac{3.92 \, b}{273 + t} where b = atmospheric pressure in cm of Hg and t = temperature in °C.

  5. Surface irregularity factor (m): A smooth conductor has m ≈ 1; weathered or stranded conductors have m ≈ 0.8–0.9. Lower m means lower V_d and higher loss.

  6. Atmospheric conditions: Rain, fog, snow, dust and humidity increase corona loss considerably (loss may rise 5–10 times in rain). Fair weather gives minimum loss.

  7. Frequency (f): Corona loss increases with frequency. For a given voltage, higher frequency means more rapid reversals of ionization, hence greater loss. Loss ∝ f in the empirical Peek's formula.

  8. Line length and number of conductors: Loss is proportional to the length of the line and to the number of phases/conductors.

Peek's empirical formula for fair weather corona loss per phase per km:

P=241×10−5 f+25δrD (V−Vd)2 kW/km/phaseP = 241 \times 10^{-5} \, \frac{f + 25}{\delta} \sqrt{\frac{r}{D}} \, (V - V_d)^2 \text{ kW/km/phase}

where r = conductor radius (cm), D = spacing (cm), V and V_d in kV (rms).

Thus corona loss is minimized mainly by increasing conductor diameter (bundling), increasing spacing, and improving surface and atmospheric conditions, while it is aggravated by high voltage, low air density (high altitude), high frequency, and bad weather.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Discuss the behavior of medium transmission lines in terms of voltage regulation and power transfer capability. Explain the challenges associated with modeling medium lines and their impact on system performance. 5+5=10

Answer

1. Behaviour of Medium Transmission Lines (5 marks)

Definition. A transmission line is classified as medium when its length lies roughly between 80 km and 250 km and the operating voltage between 20 kV and 100 kV. At these lengths the shunt capacitance (line-to-line and line-to-neutral) becomes significant enough that it can no longer be lumped at one end as in a short line, yet the line is not long enough to require exact distributed-parameter (hyperbolic) treatment. Standard models are the nominal-π and nominal-T models, in which the total series impedance (Z = zl) and total shunt admittance (Y = yl) are lumped at the midpoint or at the two ends respectively.

(a) Voltage regulation

Voltage regulation is defined as

[ %VR = \frac{|V_{r,NL}| - |V_{r,FL}|}{|V_{r,FL}|}\times 100 ]

where (V_{r,NL}) is the receiving-end voltage at no load and (V_{r,FL}) that at full load, the sending-end voltage (V_s) being held constant. For the nominal-π model, (V_{r,NL}=V_s/A) with (A = 1+ZY/2).

For the nominal-π model the two-port relations are:

[ V_s = \left(1 + \frac{ZY}{2}\right)V_r + Z,I_r ]

[ I_s = Y\left(1 + \frac{ZY}{4}\right)V_r + \left(1 + \frac{ZY}{2}\right)I_r ]

Key behavioural points:

  • Charging current effect. The shunt capacitance draws leading current, which partly cancels the lagging load current. The net line current is therefore lower than the load current, reducing the series (I^2R) and (IX) drops. Consequently the regulation of a medium line is better than that of a short line of the same impedance carrying the same load.
  • Load dependence. Regulation is worst at lagging (inductive) power factor, improves as the power factor approaches unity, and can become negative at leading power factor — the receiving-end voltage may actually rise above the sending-end voltage (Ferranti-type effect) because the shunt capacitance over-compensates the series drop. This is why medium lines supplying light loads or capacitive loads need shunt reactors or switched compensation.
  • Magnitude. For a typical 132 kV, 100 km line at full load, (VR) is of the order of 5–12 % at 0.8 lagging pf; neglecting the shunt susceptance (short-line model) overestimates it.
  • Model sensitivity. The nominal-π model gives slightly higher (more pessimistic) regulation than the nominal-T model for the same line, because lumping all capacitance at the receiving end exaggerates the voltage rise there. The π-model is therefore preferred in practice as it errs on the safe side.

(b) Power transfer capability

The steady-state power transfer is governed by the ABCD constants:

[ P_r = \frac{|V_s||V_r|}{|B|}\cos(\delta - \beta) - \frac{|A||V_r|^2}{|B|}\cos(\beta - \alpha) ]

For a lossless medium line this reduces to the familiar form

[ P_r = \frac{|V_s||V_r|}{X}\sin\delta ]

with the steady-state stability limit reached at (\delta = 90^\circ):

[ P_{max} = \frac{|V_s||V_r|}{X} ]

Behavioural points:

  • Capacitance raises the limit. The shunt admittance effectively reduces the equivalent series reactance seen by the load (the π-model gives (B = Z), but the shunt branches supply part of the reactive demand locally). The surge impedance loading (SIL) concept applies: (SIL = V^2/Z_c), with (Z_c=\sqrt{z/y}). A medium line loaded near its SIL has near-flat voltage profile and near-unity regulation.
  • Reactive power balance. Because the line generates (Q_c = \omega C V^2) and consumes (Q_l = I^2X), the net reactive requirement is small near SIL. Loading beyond SIL makes the line a net consumer of reactive power, depressing voltage and reducing transferable power; loading below SIL makes it a net generator, raising voltage.
  • Voltage dependence. Since (P \propto |V_s||V_r|), transfer capability is highly sensitive to voltage. At a fixed angle, a 5 % depression of one end voltage costs about 5 % of transferable power (about 10 % if both end voltages fall by 5 %) — hence the importance of voltage support (capacitors, SVCs, tap changers) on medium-length lines.
  • Thermal vs stability limit. For medium lines the thermal limit of the conductor often binds before the stability limit, so the practical capability is set by conductor rating, sag, and ambient temperature rather than by (\delta = 90^\circ).

2. Challenges in Modelling Medium Lines and Impact on System Performance (5 marks)

Challenge 1 — Distributed vs lumped parameters. The true line has uniformly distributed (R, L, C). Lumping (Z) and (Y) at discrete points (π or T) introduces error that grows with line length. The correction factor (ZY/4) in the π-model only partially compensates. For a 200 km line the error in computed regulation can reach 2–4 %, which matters when setting protection and voltage schedules.

Challenge 2 — Frequency dependence. Skin effect and proximity effect make (R) and (L) frequency dependent. A single 50 Hz value is inadequate for harmonic studies, switching transients, or relay transient studies. Ignoring this underestimates losses and distorts travelling-wave behaviour.

Challenge 3 — Non-transposition and asymmetry. Untransposed lines have unequal mutual coupling between phases, producing negative- and zero-sequence components even under balanced load. The symmetrical-component model then fails, and voltage unbalance appears at the receiving end, degrading motor performance and increasing losses.

Challenge 4 — Corona and shunt conductance. Corona loss and leakage are nonlinear and weather dependent. They are usually neglected in the π-model, so computed efficiency and voltage profile deviate from field measurements, particularly in hilly terrain with high altitude and low air density (relevant to Nepali conditions).

Challenge 5 — Load modelling. Constant-impedance, constant-current and constant-power (ZIP) loads behave differently as voltage varies. A medium line modelled with a single constant-power load will over-predict voltage collapse margin compared with a realistic composite load.

Impact on system performance:

AspectEffect of modelling error
Voltage regulationUnder/over-estimation of drop; wrong tap-changer settings; possible over-voltage at light load
Power transferOver-optimistic stability margin; risk of undetected proximity to collapse
LossesUnder-estimated (I^2R) and corona loss; poor energy accounting
ProtectionImpedance relay reach miscalculated; risk of maloperation on long zones
Reactive compensationWrong sizing of shunt capacitors/reactors; poor voltage control
Transient studiesInaccurate switching over-voltages; insulation coordination errors

Conclusion. The nominal-π model is adequate for steady-state load-flow and regulation studies of lines up to about 200 km, but accurate transient, harmonic, unbalanced and stability studies require distributed-parameter, frequency-dependent, and multi-phase models. In the Nepali context — long radial 132 kV lines over hilly terrain with light, unbalanced loads — the capacitance and corona effects are pronounced, so a purely lumped model can seriously misrepresent voltage regulation and the true power transfer capability.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · 2nd · 5 marks

Discuss the different types of switchgear used in power system including circuit breakers, switches, disconnectors and fuses. 5

Answer

Switchgear is the assembly of switching and protective devices used to control, protect and isolate electrical equipment in a power system. The main types are:

1. Circuit breakers — Automatic devices that make and break a circuit under both normal and fault conditions. They extinguish the arc (oil, air-blast, SF₆ or vacuum) and are the primary means of clearing faults. Types: oil (bulk/oil-minimum), air-blast, SF₆ and vacuum circuit breakers. Used at all voltage levels, from LV to EHV.

2. Switches (isolators/load switches) — Manually or motor-operated devices for making and breaking a circuit under normal load conditions only. They cannot interrupt fault currents. Examples: air-break switches, oil switches and load-break switches. Used for routine on/off operations and load transfer.

3. Disconnectors (isolators) — Off-load devices that provide a visible air gap to isolate a section for maintenance. They are operated only after the circuit breaker has cleared the circuit, since they have no arc-quenching capability. They ensure safety of personnel working on de-energised equipment.

4. Fuses — The simplest and cheapest protective device. A metal element melts and breaks the circuit when current exceeds a rated value. Types: rewirable (kit-kat), cartridge (HRC) and dropout fuses. They give one-shot, non-reclosing protection and are widely used for LV distribution, small transformers and capacitor banks.

Summary: Circuit breakers handle fault interruption; switches and disconnectors handle normal switching and isolation; fuses provide cheap backup protection — together they ensure safe, reliable and selective operation of the power system.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Explain how a substation and the equipments in the substation are protected from lightning strokes. 10

Answer

Protection of a substation and its equipment from lightning strokes — 10 marks

A substation is the most exposed and most expensive node of a power system, so it is protected by a coordinated scheme of shielding, diverting, discharging and earthing the lightning surge before it can flash over insulation. The measures are:

1. Shielding by overhead earth wires (sky wires) Overhead earth wires are strung above the busbars, transformers and line bays, and are earthed at every tower/structure. They intercept the stroke before it reaches the phase conductors. The zone of protection is designed by the protective angle method (θ ≈ 30°–45° for tall structures) or by the rolling sphere method (sphere radius equal to the striking distance, ~ 30–60 m). For the switchyard itself, tall lightning masts / spikes are erected so that all equipment lies inside the protected cone.

2. Proper earthing (grounding) system A low-resistance earthing grid is the backbone of all lightning protection. A mesh/ grid of buried conductors (rod, pipe or plate electrodes, bonded together) keeps the earth resistance low (≈ 1 Ω or less for a substation) and the step and touch voltages safe. All structures, masts, earth wires, transformer tanks, neutral points and equipment frames are bonded to this grid, so that the surge current has a low-impedance path to ground and no dangerous potential difference appears between equipment and earth.

3. Surge diverters (lightning arresters) Every incoming line and every major equipment (transformer, busbar, circuit breaker) is protected by a surge arrester connected between phase and earth. On a lightning surge the arrester becomes conducting and diverts the surge current to earth, then reseals and restores normal insulation. Types used:

  • Rod gap / horn gap — cheap, but does not reseal and causes line outage.
  • Valve-type / SiC arrester — nonlinear resistance discs in series with a spark gap.
  • Metal Oxide Varistor (MOV / ZnO) arrester — modern standard, gapless, highly nonlinear, fast response, no follow current.

The arrester must be mounted as close as possible to the equipment it protects (short leads, low inductance) so that the protective level of the arrester is below the basic insulation level (BIL) of the equipment, with a margin.

3. Overhead earth wire on the incoming transmission line The last span or two before the substation is shielded by earth wire, and the tower footing resistance is kept low, so that a stroke is intercepted and discharged at the tower itself, away from the substation.

4. Earthing of the line and tower structures Towers, earth wires and the substation grid are interconnected, so the surge is drained into the general earthing system before entering the yard.

5. Surge capacitors and protective gaps Surge capacitors are connected at the transformer terminals to flatten the steep wavefront (reduce dv/dt), and protective spark gaps are provided as a backup across critical insulation.

6. Shielding of control and communication cables Control, protection and SCADA cables are laid in earthed metallic conduits/trays or armoured cables, and surge protective devices are fitted at the cable entries, so that induced surges do not damage relays and control equipment.

7. Insulation coordination The BIL of each equipment is selected higher than the arrester protective level, and the arrester rating is chosen above the maximum system voltage, so that the whole substation is protected in a graded manner — the arrester always sparks over before the equipment insulation fails.

Conclusion: A substation is protected by a combination of overhead earth wires and lightning masts (to prevent the stroke from reaching the equipment), low-resistance mesh earthing (to carry the surge safely to ground), and surge arresters with surge capacitors (to limit the surge voltage below the equipment BIL). No single device is sufficient — protection is achieved only by this coordinated scheme.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Explain with examples about the common techniques used for demand side management. 10

Answer

Demand Side Management (DSM) — Common Techniques

Demand Side Management (DSM) refers to the planning, implementation and monitoring of those utility activities designed to influence customer use of electricity in ways that will produce desired changes in the utility's load shape — i.e. changes in the timing and magnitude of electricity demand. In the Nepali context, where the Nepal Electricity Authority (NEA) faces a winter peak deficit (supply shortage in the dry season) and surplus in the monsoon, DSM is a cost-effective alternative to building new peaking plants. The common techniques are grouped below.

1. Load Shifting (Peak Clipping and Valley Filling)

This technique shifts the use of electricity from peak hours to off-peak hours without reducing total energy consumption. The load curve is flattened — the peak is clipped and the valley is filled.

Example: NEA's Time of Day (TOD) tariff for industrial and commercial consumers charges a higher rate during peak hours (approximately 6 pm–11 pm) and a lower rate during off-peak hours (11 pm–6 am). A cement factory running its crushers and mills at night instead of during the evening peak is practising load shifting. Similarly, NEA's "Ujyalo Nepal" / off-peak promotional tariff encourages consumers to use electricity at night. The result is a reduced peak demand (avoiding load-shedding or expensive imports from India) and better utilisation of existing generation and transmission capacity.

2. Peak Clipping (Load Curtailment)

Here the utility reduces the peak demand directly, either by curtailing supply or by persuading consumers to reduce consumption during peak hours. It may be done through direct load control or through voluntary demand response.

Example: NEA's demand-side bidding / demand response programme for large industrial consumers, where industries agree to shed a contracted load (say 5 MW each) during system peak in return for a rebate. Another example is ripple control or time switches used to switch off water heaters and irrigation pumps during peak hours. In the past, NEA resorted to load-shedding as a crude form of load curtailment; modern DSM replaces this with incentive-based voluntary curtailment.

3. Strategic Conservation (Energy Efficiency Improvement)

This reduces total energy consumption (kWh) by using more efficient end-use devices, thereby also reducing peak demand. It is often called "negawatt" generation.

Example: Replacement of incandescent bulbs with LED bulbs (NEA/UJALA distribution), use of 5-star rated fans, refrigerators and air conditioners, promotion of induction cooktops instead of resistive heaters, and use of IE3 high-efficiency motors in industries. NEA's Energy Efficiency Programme and the Nepal Energy Efficiency Programme (NEEP) under the Ministry of Energy promote such measures. A 9 W LED replacing a 60 W bulb saves about 51 W per bulb; across a million households this saves about 51 MW of peak demand — equivalent to avoiding a medium-sized hydropower plant.

4. Strategic Growth (Load Building / Valley Filling)

This technique deliberately increases electricity consumption during off-peak periods to improve the load factor. It is useful in Nepal during the monsoon when there is surplus hydropower that would otherwise be spilled.

Example: Promotion of electric cooking (induction/electric pressure cooker) and electric vehicles (EVs) charging at night; NEA's off-peak tariff for irrigation pumps and cold storage; encouraging night-time water pumping for drinking water supply. This converts surplus monsoon energy into productive use and improves NEA's financial position.

5. Flexible Load Shape (Demand Response and Reliability Options)

The load shape is made flexible so that the utility can call on consumers to reduce or shift load during contingencies (forced outage of a plant or transmission line). Consumers receive a tariff incentive or penalty.

Example: Interruptible tariff for industries — they pay a lower energy rate but must curtail load when NEA signals a system emergency. Real-time pricing / critical peak pricing pilots where consumers are informed a day ahead of a peak event and respond by reducing load. In Nepal, the proposed Time-of-Day tariff for domestic consumers and smart-meter-based demand response are steps in this direction.

6. Peak Shaving through Distributed Generation and Storage

Consumers install rooftop solar PV, small standby diesel generators or battery storage to meet their own peak load, reducing the draw from the grid at peak times.

Example: Rooftop solar PV on a commercial building in Kathmandu reduces the building's demand from NEA during daytime; a battery bank charged at off-peak night tariff and discharged during the evening peak shaves the consumer's peak. NEA's net metering / net billing provisions for rooftop solar (up to 1 MW for some categories) encourage this.

Summary Table

TechniqueObjectiveNepali Example
Load shiftingFlatten load curveTOD tariff; night-time industrial operation
Peak clippingReduce peak demandDemand response; interruptible load
Strategic conservationReduce total kWhLED bulbs, efficient motors, induction cooking
Strategic growthFill valleyNight irrigation pumping, EV charging, electric cooking
Flexible load shapeReliability during emergenciesInterruptible tariff, critical peak pricing
Distributed generation/storageSelf-generation at peakRooftop solar with net metering, battery storage

Conclusion

DSM techniques — load shifting, peak clipping, strategic conservation, strategic growth, flexible load shape and distributed generation — together allow NEA to balance its load curve, reduce winter peak deficit, utilise monsoon surplus and defer costly capacity addition. In Nepal, the TOD tariff, energy-efficient appliance programmes, off-peak promotional tariffs and rooftop solar net metering are the principal DSM instruments currently in use.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

What is step potential and touch potential? How can they be mitigated? Discuss. 10

Answer

Step Potential and Touch Potential

Step potential is the potential difference that exists between two points on the ground surface separated by a distance of one step (usually taken as 1 m) when a fault current flows into the earth through a grounding electrode. A person standing near the point where current enters the earth has one foot nearer the electrode (at higher potential) and the other foot farther away (at lower potential). The difference of potential between the two feet drives current through the person's legs — this is the step potential. It is maximum close to the electrode and decreases as one moves away.

Touch potential is the potential difference between a grounded metallic structure that a person can touch with the hand and the ground surface on which the person stands, when fault current flows through that structure to earth. The current path is hand-to-feet (through the heart), which makes it far more dangerous than step potential. It is maximum at the base of the faulted structure and decreases with distance from it.

Mitigation measures:

  1. Ground grid / mesh design: Lay a properly designed earthing grid (mesh) with adequate conductor spacing so that the surface potential gradient is flattened. A denser mesh reduces both step and touch potentials.

  2. Surface layer of high-resistivity material: Spread crushed stone (gravel) of resistivity 2000–3000 Ω·m and thickness 75–150 mm over the switchyard. This increases contact resistance between feet and soil, raising the tolerable limits of step and touch voltage.

  3. Buried ground rods and deep-driven electrodes: Drive rods at grid corners and at equipment bases to lower grid resistance and improve current distribution.

  4. Increase conductor burial depth: Burying grid conductors at 0.5–0.8 m depth reduces surface potential gradients.

  5. Equipotential bonding: Bond all metallic structures, fences, and equipment frames to the ground grid so that no dangerous potential difference appears between them and the ground.

  6. Insulated/isolated platforms: Provide insulating mats or platforms at operating positions, and insulate hand rails or gate handles at the periphery of the substation.

  7. Restrict access and use safety procedures: Fence the switchyard, display warning signs, and enforce the use of safety shoes and gloves during maintenance.

  8. Reduce fault clearing time: Fast relay operation and circuit breakers limit the duration of fault current, reducing the energy a person may be exposed to.

  9. Maintain low grid resistance: Keep grid resistance below the permissible value (typically ≤ 1 Ω for large substations) so that the rise of earth potential is limited.

Conclusion: Step potential affects the legs (foot-to-foot path) while touch potential affects the hand-to-feet path through vital organs. Both are controlled mainly by proper ground grid design, a gravel surface layer, bonding, and fast fault clearance, so that the tolerable limits given by IEEE 80 / IS 3043 are not exceeded.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Explain switching overvoltage and how these are created in a power system. How are the system and associated installations protected against switching overvoltage? Describe. 5+5=10

Answer

Switching Overvoltage

Switching overvoltage (also called switching surge) is a transient overvoltage that appears across the insulation of system equipment when a circuit is deliberately opened or closed by a switching device (circuit breaker, isolator, or switch), or when a fault is cleared. Unlike lightning overvoltage, which originates from an external source, switching overvoltage is generated within the system itself. Its magnitude is typically 2.0 to 3.5 p.u. of the peak phase-to-neutral voltage, its waveform is oscillatory with a front of the order of hundreds of microseconds to a few milliseconds, and its duration is comparatively long (milliseconds). Because of the long duration, switching surges are more severe than lightning surges for internal insulation and for equipment such as transformers and reactors.

How Switching Overvoltages Are Created in a Power System

  1. Energisation of an unloaded line (closing surge): When a long transmission line is switched on, the line capacitance is charged suddenly. The travelling wave reflects at the open (receiving) end and doubles, producing an overvoltage of about 2 p.u. at the far end. With three-phase energisation and trapped charge, this may reach 2.5–3 p.u.

  2. Interruption of small inductive currents (current chopping): When a circuit breaker interrupts a small inductive current (e.g. magnetising current of an unloaded transformer or shunt reactor), the arc may be chopped before natural current zero. The energy stored in the inductance, ½Li², is transferred to the capacitance of the equipment, producing a high-frequency oscillation and a voltage peak given by v = i·√(L/C). This can give 3–4 p.u. across the transformer or reactor terminals.

  3. Interruption of capacitive currents (restriking and re-ignition): When switching out an unloaded line or capacitor bank, the capacitance remains charged at peak voltage. If the breaker restrikes, the capacitance is recharged to a voltage up to 2 p.u. and, on subsequent restrikes, the trapped charge can build up the voltage to 3 p.u. or more.

  4. Fault clearing and load rejection: When a fault is cleared, the sudden removal of load causes the generator to speed up and the receiving-end voltage to rise (Ferranti effect plus overspeed), giving sustained power-frequency overvoltage. The restriking of the arc during fault clearing also produces oscillatory switching surges.

  5. Switching of capacitor banks and shunt reactors: Capacitor bank energisation gives inrush transients, and reactor de-energisation gives current-chopping surges as described above.

  6. Out-of-phase switching and reclosure: Closing a breaker between two systems that are out of phase, or autoreclosing onto a line with trapped charge, produces a large potential difference across the contacts and hence a severe surge.

Protection of the System and Associated Installations Against Switching Overvoltage

  1. Surge arresters (metal-oxide / ZnO arresters): These are the primary protective devices. They are connected between line and earth at substation entrances, transformer terminals, and line ends. The non-linear ZnO elements conduct heavily above the reference voltage and clamp the surge to the protective level (residual voltage), dissipating the energy to earth. They protect transformer and switchgear insulation against both switching and lightning surges.

  2. Pre-insertion resistors in circuit breakers: A resistor of a few hundred ohms is inserted in series with the main contacts for a few milliseconds during closing. This damps the travelling-wave oscillation and reduces the trapped-charge voltage, limiting the closing surge to about 1.5–2 p.u. Similarly, pre-insertion inductors are used in some breakers.

  3. Controlled (synchronous) closing: The breaker is closed at the instant when the supply voltage passes through zero (or at the optimum point of the voltage wave), so that the prospective trapped charge and the supply voltage are in phase and the surge is minimised. This is done by controlled switching devices with point-on-wave control.

  4. Damping resistors and drain coils: Resistors connected across breaker contacts or across reactor windings damp the high-frequency oscillation and reduce the peak of the current-chopping surge. Neutral grounding reactors or drain coils are used in some installations.

  5. Neutral earthing (grounding): Solid or low-resistance neutral earthing keeps the phase-to-earth voltage low and reduces the overvoltage factor during arcing faults and switching. An effectively earthed system has a lower switching surge level than an isolated-neutral system.

  6. Shunt reactors and shunt capacitors: Shunt reactors at line ends absorb the capacitive charging current and control the Ferranti overvoltage during light load and after load rejection. Shunt capacitors are switched in steps with breakers fitted with pre-insertion resistors or controlled switching to limit surges.

  7. Reduction of trapped charge: Discharge resistors or voltage transformers connected across the line capacitance, or the use of breakers with restrike-free performance (e.g. SF₆ breakers with low restrike probability), prevent the build-up of trapped charge on the line before reclosure.

  8. Autoreclosure with a suitable dead time: A dead time long enough for the trapped charge to decay (through discharge resistors and corona losses) before reclosure avoids the severe surge of reclosing onto a charged line. Single-pole or three-pole autoreclosure schemes are chosen accordingly.

  9. Insulation coordination: The insulation levels of transformers, switchgear, bushings, and lines are selected on the basis of the withstand voltage corresponding to the protective level of the arrester, so that the arrester always sparks over before the equipment insulation fails. This coordination ensures that the protected equipment is not overstressed.

  10. Shielding and earthing of structures: Overhead earth wires, earthed substation structures, and low earthing resistance reduce the potential rise and protect against both switching and lightning surges at substations.

By combining surge arresters, pre-insertion resistors, controlled switching, neutral earthing, shunt reactors, and proper insulation coordination, the switching overvoltage in a power system is limited to a level that the insulation of the system and its associated installations can safely withstand.

Answer checked by Claude Haiku · Oct 21, 2026

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Questions from NEA Question for Engineer 7th Level Lok Sewa (Engineer (Level 7) · 1st · 2083), NEA Exam Question for Electrical (Electrical Engineer 7th · 2nd · 2082), NEA Exam Question for Mechanical (Mechanical Engineer 7th · 2nd · 2082), NEA Exam Question for Civil Engineer (Engineer 7th Level, Civil · 2nd · 2082) and NEA Lok Sewa Engineer/Survey Engineer (Engineer/ Survey Engineer (7th Level) · 1st · 2081). Answers are written for this site; check them against your class notes.