Skip to main content

Chapter 3

Mechanical Engineering Core

Nepal Electricity Authority past exam questions

Past questions and answers

12 questions set from this chapter. Most repeated first.

  • 2082 · 2nd · 5 marks

What is flow measurement device? Explain the working principle of Rotameter with its application.

Answer

Flow measurement device. An instrument that measures the rate of flow (volume or mass per unit time) of a fluid passing a section of a pipe or channel. Examples: orifice meter, venturi meter, pitot tube, rotameter, turbine meter, electromagnetic flow meter.

Rotameter (variable-area meter). It consists of a vertical, transparent tapered tube (narrow at the bottom, wide at the top) containing a free-moving float, with a graduated scale on the tube.

Working principle. The fluid enters at the bottom and flows upward past the float. The float rises until the upward forces balance its weight:

drag+buoyancy=weight of float\text{drag} + \text{buoyancy} = \text{weight of float}

Because this balance does not depend on the float's height, the pressure drop across the float stays constant. As the float rises, the annular area AaA_a between float and tube wall increases, so a higher flow rate is needed to hold the float at a higher position. Flow rate is thus a function of float height, which is read directly from the scale:

Q=CdAa2gVf(ρf−ρ)ρAfQ = C_d A_a \sqrt{\frac{2 g V_f(\rho_f-\rho)}{\rho A_f}}

where VfV_f is the float volume, AfA_f its cross-sectional area, and ρf\rho_f, ρ\rho the float and fluid densities.

Applications. Small liquid and gas flows in laboratories and pilot plants; purge and carrier gas measurement; chemical dosing; water treatment; monitoring of cooling water and lubricating oil.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 5 marks

Compare soldering, brazing and arc welding with respect to working temperature, filler materials and typical applications.

Answer

Comparison of Soldering, Brazing and Arc Welding

BasisSolderingBrazingArc Welding
Working temperatureBelow 450 °C (typically 180–350 °C)Above 450 °C but below the melting point of the parent metal (typically 600–900 °C)Above the melting point of the parent metal (arc temperature about 3000–6000 °C; base metal melts locally)
Filler materialSoft solder alloys — tin–lead (Sn–Pb), tin–silver, tin–copper; flux (rosin, zinc chloride)Hard filler alloys — copper–zinc (brass), copper–silver (silver solder), aluminium–silicon; borax or fluoride fluxConsumable electrode itself — mild steel, stainless steel, cast iron, nickel, aluminium rods; flux coating on electrode or shielding gas
Bonding mechanismFiller melts and wets the surfaces; no melting of parent metal (adhesion/alloying at surface only)Filler melts and diffuses into the parent metal; parent metal does not meltParent metal melts and fuses with the filler to form a homogeneous joint
Joint strengthLowest (weak, used mainly for electrical/leak-proof joints)Intermediate (stronger than soldering, weaker than welding)Highest (joint strength approaches that of the parent metal)
Typical applicationsElectronic circuits, PCB assembly, electrical terminals, copper pipe plumbing, sheet-metal seamsPipe joints, bicycle frames, air-conditioning and refrigeration tubing, carbide tool tips, heat exchangersStructural steel work, pressure vessels, pipelines, machine frames, shipbuilding, repair of broken castings

Key points for the examiner:

  • The essential difference is the temperature relative to the melting point of the parent metal: soldering < 450 °C, brazing between 450 °C and the melting point of the base metal, welding above the melting point.
  • Soldering and brazing are non-fusion processes (parent metal remains solid); arc welding is a fusion process.
  • Strength order: arc welding > brazing > soldering; distortion order is the reverse.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

What are material properties required for steel used for Radial Gate in dams in hydropower projects? What is phase transformation and heat treatment process for steel? Explain.

Answer

Material Properties Required for Steel Used in Radial Gates of Dams

A radial gate (Tainter gate) is a curved steel gate that rotates about a horizontal trunnion axis. It must withstand high hydrostatic water pressure, repeated open–close cycles, and exposure to a humid, corrosive river environment. The steel used must therefore possess the following properties:

1. High Strength (Tensile and Yield Strength) The gate skin plate, girders, radial arms and trunnion must resist large hydrostatic thrust without excessive stress. Structural steel of yield strength not less than 250 MPa (e.g., IS 2062 Grade E250) or higher grades (E350, E410) is used for main structural members. High strength permits thinner sections and reduces self-weight.

2. Adequate Ductility Ductility allows the steel to deform plastically before fracture, absorbing shock loads during operation and preventing sudden brittle failure. A minimum elongation of about 20–23% is required. Ductility also permits cold bending of the skin plate into the required radial curvature.

3. Good Weldability Radial gates are largely fabricated by welding. The steel must have low carbon equivalent (CE ≤ 0.42%) so that sound welds free from cracks can be produced without preheating or post-weld cracking problems.

4. High Toughness (Impact Resistance) The steel must resist brittle fracture at low ambient temperatures in the Himalayan region. Charpy V-notch impact energy of about 27 J at 0 °C or lower is desirable. Fine grain practice improves toughness.

5. Corrosion Resistance Since gates remain in contact with flowing water and alternate between wet and dry conditions, the steel must resist corrosion. This is achieved by:

  • Using weathering steel (e.g., Corten) or copper-bearing steel,
  • Applying protective coatings (epoxy, zinc-rich primer, coal-tar epoxy),
  • Providing a corrosion allowance of 1–2 mm on plate thickness.

6. Fatigue Resistance Gates undergo repeated loading and unloading cycles. The steel must have good fatigue strength, especially at welded joints and at the trunnion hub, to avoid fatigue cracking.

7. Homogeneity and Freedom from Defects The steel must be free from internal defects such as laminations, inclusions, porosity and segregation. Ultrasonic testing is done to ensure soundness.

8. Resistance to Abrasion and Cavitation In high-velocity flow zones (e.g., bottom seal and lip), abrasion and cavitation erosion occur. Hardness and surface finish are therefore important; sometimes stainless steel cladding is used at seals.

9. Dimensional Stability and Machinability Trunnion pins, bushings and seal seats require accurate machining; hence the steel must have uniform properties and good machinability.


Phase Transformation in Steel

Phase transformation is the change of one crystal structure (phase) of steel into another due to change in temperature, pressure or composition. In steel, the important phases are:

PhaseCrystal StructureDescription
Austenite (γ)FCCStable at high temperature (above 723 °C for eutectoid)
Ferrite (α)BCCSoft, ductile, stable at room temperature
Cementite (Fe₃C)OrthorhombicHard, brittle compound
PearliteLamellar (α + Fe₃C)Eutectoid mixture at 0.8% C
MartensiteBCTFormed by rapid quenching; hard and brittle
BainiteAcicularFormed at intermediate cooling rates

Transformation on heating (for eutectoid steel, 0.8% C):

  • At 723 °C (A₁), pearlite transforms to austenite: Pearlite (α + Fe3C)→723 ∘CAustenite (γ)\text{Pearlite (α + Fe}_3\text{C)} \xrightarrow{723\,^\circ\text{C}} \text{Austenite (γ)}
  • Above A₃ (910 °C for pure iron), ferrite completely dissolves into austenite.

Transformation on cooling:

  • Slow cooling (furnace): Austenite → Ferrite + Pearlite (equilibrium structure).
  • Moderate cooling (air): Austenite → Fine pearlite (the normalized structure). Bainite forms only at intermediate rates, e.g. isothermal holding at 250–400 °C (austempering).
  • Rapid cooling (quenching in water/oil): Austenite → Martensite (diffusionless shear transformation). Martensite is hard but brittle and requires tempering.

The TTT (Time–Temperature–Transformation) diagram and CCT (Continuous Cooling Transformation) diagram describe these transformations and are the basis of heat treatment.


Heat Treatment Processes for Steel

Heat treatment is the controlled heating and cooling of steel to obtain desired mechanical properties (hardness, toughness, ductility, strength) without changing its chemical composition.

1. Annealing

  • Process: Heat steel to 30–50 °C above A₃ (about 850–900 °C), hold for sufficient time (soaking), then cool very slowly in the furnace.
  • Purpose: Softening, relieving internal stresses, improving machinability, refining grain structure, removing segregation.
  • Result: Coarse pearlite; low hardness, high ductility.

2. Normalizing

  • Process: Heat to 30–50 °C above A₃, soak, then cool in still air.
  • Purpose: Refine grain size, homogenize structure, improve machinability and toughness.
  • Result: Fine pearlite; slightly harder and stronger than annealed steel.

3. Hardening (Quenching)

  • Process: Heat to above A₃ (for hypoeutectoid) or above A₁ (for hypereutectoid), soak, then quench rapidly in water, oil or brine.
  • Purpose: Produce martensite for maximum hardness and wear resistance.
  • Result: Very hard, brittle martensite; high internal stresses.

4. Tempering

  • Process: Reheat hardened steel to a temperature below A₁ (150–650 °C), hold, then cool (usually in air).
  • Purpose: Relieve internal stresses, reduce brittleness, and adjust hardness–toughness balance.
  • Types:
    • Low-temperature tempering (150–250 °C): For tools, cutting edges — retains high hardness.
    • Medium-temperature tempering (350–500 °C): For springs — good elastic limit.
    • High-temperature tempering (500–650 °C): For structural parts — tough, strong (tempered martensite).
  • Result: Tempered martensite — the best combination of strength and toughness.

5. Case Hardening (Surface Hardening)

Used when a hard wear-resistant surface is needed with a tough core.

  • Carburizing: Add carbon to surface at 900–950 °C, then quench and temper.
  • Nitriding: Diffuse nitrogen at 500–550 °C; no quenching needed.
  • Induction/Flame hardening: Rapid surface heating followed by quenching.

6. Stress Relieving

  • Process: Heat to 550–650 °C, soak, then cool slowly.
  • Purpose: Relieve residual stresses from welding, casting or machining — very important for large radial gate fabrications to prevent distortion and cracking.

7. Austempering and Martempering

  • Austempering: Quench into a salt bath at 250–400 °C, hold to form bainite, then cool in air. Gives high toughness with good strength.
  • Martempering: Quench into a salt bath just above Mₛ, hold until temperature equalizes, then cool in air. Reduces distortion and cracking.

Application to Radial Gates

For radial gate steel (e.g., IS 2062 E250/E350):

  • Normalizing is often specified for plates to refine grains and ensure uniform properties.
  • Stress relieving after welding is essential to remove residual stresses in girders and trunnion assemblies.
  • Quenching and tempering is applied to trunnion pins and seal bolts to obtain high strength with toughness.
  • Surface hardening may be applied to seal contact surfaces to resist wear.

Thus, proper selection of material properties and correct heat treatment ensure that radial gates remain safe, durable and serviceable throughout the design life of the hydropower project.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

Explain working principle of Rankine cycle. How can you increase the thermal efficiency of Rankine cycle? Discuss.

Answer

Working Principle of the Rankine Cycle

The Rankine cycle is the ideal thermodynamic cycle on which steam power plants (including NEA's thermal plants) operate. It converts heat energy from fuel into mechanical work using water as the working fluid, which alternately changes phase between liquid and vapour. It consists of four reversible processes:

1. Isentropic expansion in turbine (1→2): High-pressure, high-temperature steam from the boiler enters the turbine and expands isentropically (ideally) to a lower pressure. Enthalpy drops from h1h_1 to h2h_2, producing shaft work: Wt=h1−h2W_t = h_1 - h_2

2. Constant-pressure heat rejection in condenser (2→3): The exhaust steam at low pressure is condensed at constant pressure by rejecting heat to cooling water, becoming saturated liquid. Heat rejected: qout=h2−h3q_{out} = h_2 - h_3

3. Isentropic compression in feed pump (3→4): The condensate is pumped back to boiler pressure. Pump work is small because water is incompressible: Wp=h4−h3≈vf(p1−p2)W_p = h_4 - h_3 \approx v_f(p_1 - p_2)

4. Constant-pressure heat addition in boiler (4→1): The compressed water is heated in the economiser, evaporated in the evaporator, and superheated in the superheater at constant pressure, receiving heat: qin=h1−h4q_{in} = h_1 - h_4

Thermal efficiency: ηth=Wnetqin=(h1−h2)−(h4−h3)h1−h4\eta_{th} = \frac{W_{net}}{q_{in}} = \frac{(h_1 - h_2) - (h_4 - h_3)}{h_1 - h_4}

The cycle is closed and continuously repeated. Its advantage over the Carnot cycle is that the pump handles liquid (not a two-phase mixture), so the compression work is small and practically achievable.

Methods to Increase Thermal Efficiency

1. Increasing boiler pressure (raising p1p_1): Higher pressure raises the saturation temperature at which heat is added, increasing the mean temperature of heat addition and hence ηth\eta_{th}. However, excessive pressure reduces the moisture content at turbine exhaust (should remain ≥ 88–90%), causing blade erosion. This is countered by reheating.

2. Superheating the steam: Raising the temperature of steam entering the turbine (T1T_1) increases the mean temperature of heat addition and also lowers the exhaust moisture content. Efficiency improves and blade life increases. Limited by metallurgical limits of superheater and turbine materials (about 540–600 °C).

3. Reducing condenser pressure: Lowering the back pressure lowers the saturation temperature at which heat is rejected, increasing the temperature difference across the cycle and thus ηth\eta_{th}. Limited by the temperature of the available cooling water and by the practical vacuum achievable (about 0.07–0.1 bar).

4. Reheat cycle: Steam is partially expanded in a high-pressure turbine, returned to the boiler and reheated at constant pressure, then expanded in a low-pressure turbine. This raises the average temperature of heat addition and reduces exhaust moisture, improving efficiency and protecting blades.

5. Regenerative feedwater heating: A portion of steam is bled from intermediate turbine stages to heat the feedwater in closed or open feedwater heaters. This raises the average temperature of heat addition in the boiler (heat is added only above the feedwater temperature), improving ηth\eta_{th} and reducing boiler fuel consumption.

6. Combined reheating and regeneration: Practically all modern steam plants use both together, giving efficiencies of 40–45%.

7. Improving component performance: Reducing irreversibilities — using efficient turbines (high isentropic efficiency), minimising pressure drops in piping and condenser, and reducing pump work — brings the actual cycle closer to the ideal, raising overall efficiency.

8. Binary (combined) cycle: The steam cycle's heat rejection is used to drive a low-temperature cycle (e.g., mercury–steam or gas–steam combined cycle), raising the overall plant efficiency beyond that of the Rankine cycle alone.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

How do the project evaluation techniques IRR, Method, Benefit and Cost analysis, Breakeven Analysis differ from each other? What are their roles in engineering economic decision making? Elucidate.

Answer

Project Evaluation Techniques: IRR, Benefit–Cost Analysis and Break-even Analysis

1. Internal Rate of Return (IRR)

Concept. The IRR is the discount rate at which the Net Present Value (NPV) of a project becomes zero, i.e. the rate at which the present worth of benefits equals the present worth of costs.

NPV=∑t=0nBt−Ct(1+r)t=0⇒r=IRR\text{NPV} = \sum_{t=0}^{n}\frac{B_t - C_t}{(1+r)^t} = 0 \quad \Rightarrow \quad r = \text{IRR}

Decision rule. Accept the project if IRR ≥ MARR (Minimum Attractive Rate of Return); among mutually exclusive projects, do not simply pick the highest IRR (it can rank projects of different scale wrongly); use NPV, or the IRR of the incremental cash flow (larger minus smaller project) against MARR.

Nature. It is a rate-based (percentage) criterion — it measures the efficiency of capital, not its absolute size. It is computed from the project's own cash-flow stream and does not require an externally fixed discount rate to be assumed (though MARR is needed for the accept/reject test).

2. Benefit–Cost (B/C) Analysis

Concept. A ratio of the present worth of benefits to the present worth of costs:

BC=∑t=0nBt/(1+i)t∑t=0nCt/(1+i)t\frac{B}{C} = \frac{\sum_{t=0}^{n} B_t/(1+i)^t}{\sum_{t=0}^{n} C_t/(1+i)^t}

Decision rule. Accept if B/C ≥ 1; among alternatives, choose the highest B/C.

Nature. It is a relative (ratio) criterion expressed in dimensionless terms. It is widely used for public/social projects (roads, irrigation, rural electrification) where benefits are not directly priced in the market and must be valued socially. It is sensitive to how benefits and costs are classified (e.g. whether a cost is treated as a disbenefit).

3. Break-even Analysis

Concept. Determines the level of output (or sales, or time) at which total revenue equals total cost, so that profit is zero:

QBE=FP−VQ_{BE} = \frac{F}{P - V}

where F = fixed cost, P = selling price per unit, V = variable cost per unit.

Decision rule. Operate/proceed only if expected demand exceeds the break-even quantity; the margin of safety = (actual output − Q_BE)/actual output.

Nature. It is a volume/quantity-based technique. It ignores the time value of money and is essentially a short-run, cost–volume–profit (CVP) tool rather than a full investment appraisal method.

4. How They Differ

BasisIRRB/C AnalysisBreak-even Analysis
Result formRate (%)Ratio (dimensionless)Quantity/volume (units)
Time value of moneyConsideredConsideredNot considered
Discount rate neededOnly MARR for decisionYes, discount rate iNo
ScopeWhole project lifeWhole project lifeShort-run output level
Main usePrivate/commercial investmentPublic/social projectsCapacity, pricing, output planning
Key limitationMultiple IRRs; ignores scaleSensitive to classification of B and CStatic; ignores time value and risk

5. Role in Engineering Economic Decision Making

  1. Screening and ranking. IRR and B/C screen projects against a hurdle rate and rank competing alternatives; IRR is preferred for commercial ventures (hydropower IPPs, transmission upgrades), B/C for public-funded schemes.
  2. Social justification. B/C analysis is indispensable where benefits are external (flood control, rural supply) and cannot be captured by IRR alone.
  3. Operational planning. Break-even analysis fixes the minimum load factor, tariff or sales volume at which a plant or feeder becomes viable, and quantifies the margin of safety against demand shortfall.
  4. Risk and sensitivity. Together they allow sensitivity testing — how IRR, B/C or Q_BE shift with changes in cost, tariff and demand — supporting robust decisions under uncertainty.
  5. Complementary use. A project may show a high IRR but a low B/C (scale effect) or a comfortable break-even but a sub-MARR IRR. Sound practice is to use all three together: IRR for profitability, B/C for social worth, break-even for operational feasibility.

Conclusion. The three techniques answer different questions — at what rate? (IRR), is it worth it to society? (B/C), and at what volume? (break-even). Together they give a complete engineering-economic picture for project selection.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

What is CPM in Project Management? Take suitable example to explain why we use Critical Path Analysis. Discuss the procedure and steps to find critical path in project.

Answer

CPM in Project Management

CPM (Critical Path Method) is a deterministic network scheduling technique used in project management to plan, schedule, and control projects. It was developed in the 1950s by DuPont and Remington Rand. CPM uses a single time estimate for each activity and determines:

  • the minimum time required to complete the project,
  • the sequence of activities that must be completed on time,
  • the amount of float (slack) available in non-critical activities.

The sequence of activities with zero float is called the critical path. Any delay in a critical activity directly delays the whole project.


Example to Explain Why We Use Critical Path Analysis

Consider a small construction project with the following activities:

ActivityDescriptionImmediate predecessorDuration (days)
ASite clearance—2
BFoundationA4
CEquipment procurementA3
DSuperstructureB5
EInstallationC, D3
FTesting and commissioningE2

The network logic is:

  • Start → A
  • A → B and A → C
  • B → D
  • C → E and D → E
  • E → F → End

Possible paths

  1. A → B → D → E → F
    Duration = (2 + 4 + 5 + 3 + 2 = 16) days

  2. A → C → E → F
    Duration = (2 + 3 + 3 + 2 = 10) days

The longest path is A → B → D → E → F = 16 days. Therefore, the project cannot be completed in less than 16 days. This is the critical path.

Forward and backward pass

ActivityESEFLSLFTotal floatCritical?
A02020Yes
B26260Yes
C258116No
D6116110Yes
E111411140Yes
F141614160Yes

Why this analysis is useful

  • Activity C has 6 days of total float. It can be delayed by up to 6 days without delaying the project. So the manager need not panic if equipment procurement is slightly late.
  • Activities A, B, D, E, F have zero float. If foundation B is delayed by 1 day, the whole project is delayed by 1 day.
  • Therefore, the project manager focuses attention, manpower, and monitoring on the critical activities.
  • If the project must be finished earlier, extra resources can be added to critical activities such as B or D (crashing). Adding resources to C alone will not reduce project duration.
  • Non-critical activities can be rescheduled within their float to level resources and reduce cost.

Thus, Critical Path Analysis tells us where delay matters and where it does not, and helps in effective time and resource control.


Procedure and Steps to Find the Critical Path

  1. List all activities
    Prepare a work breakdown structure. For each activity, identify its immediate predecessor and duration.

  2. Draw the network diagram
    Use either Activity-on-Arrow (AOA) or Activity-on-Node (AON). AON is more common today. Arrows show dependencies. Avoid loops and dangling activities.

  3. Number the events/nodes
    In AOA, number events systematically so that each activity is identified by its start and end event, e.g. (i-j). Use dummy activities where necessary to show correct logic.

  4. Perform the forward pass
    Calculate Earliest Start (ES) and Earliest Finish (EF) for each activity: [ ES_j = \max(EF_i) \quad \text{for all predecessors } i ] [ EF = ES + \text{Duration} ] The project duration is the maximum EF at the end event.

  5. Perform the backward pass
    Calculate Latest Finish (LF) and Latest Start (LS): [ LF_i = \min(LS_j) \quad \text{for all successors } j ] [ LS = LF - \text{Duration} ]

  6. Compute float
    Total float: [ TF = LS - ES = LF - EF ] Free float: [ FF = ES_j - EF_i ] where (j) is the immediate successor with the smallest ES.

  7. Identify the critical path
    The critical path consists of activities with total float = 0 (or minimum total float). It is a continuous path from start to end. There may be more than one critical path.

  8. Verify and use
    Check the network logic. Use the critical path for scheduling, resource allocation, progress monitoring, and crashing if the project duration must be reduced.


Conclusion

CPM is a powerful deterministic scheduling tool. It identifies the critical path, which is the longest path through the network and determines the minimum project duration. Critical Path Analysis helps managers concentrate on activities that directly affect completion time, use float in non-critical activities for flexibility, and take corrective action such as resource shifting or crashing when delays occur.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 5 marks

Draw a typical daily load curve for a hydropower plant and explain how it helps in planning the plant operation and sizing.

Answer

Typical daily load curve of a hydropower plant

A daily load curve is a plot of the load (power demand, in MW) on the vertical axis against time (hours of the day, 0–24 h) on the horizontal axis. For a hydropower-dominated system such as Nepal's, a typical curve has the following shape:

Load (MW)
  |
  |            ****  (evening peak, ~18:00–21:00)
  |           *    *
  |          *      *
  |         *        *
  |    *****          ****
  |  **                    **
  | *                        *   (night base, ~23:00–05:00)
  |*                          *________________
  +----|----|----|----|----|----|----|----|----> Time (h)
  0    4    8   12   16   20   24
      base     day      evening
      (night)  plateau   peak

Key features:

  • Base load (00:00–05:00 and 22:00–24:00): minimum demand, supplied by run-of-river plants and the must-run portion of storage plants.
  • Day plateau (08:00–17:00): moderate, fairly steady demand from industry, irrigation pumping and commercial loads.
  • Evening peak (18:00–21:00): sharp maximum when domestic lighting, cooking and entertainment loads coincide; often 1.5–2 times the base load.
  • A smaller morning rise around 06:00–08:00 is also common.

How the load curve helps in planning plant operation and sizing

  1. Load factor and capacity factor. Load factor = average load / peak load. A low load factor (typical for Nepal, ~0.5–0.6) means the plant is lightly loaded most of the day, so the designer must decide how much capacity to install to meet the peak without excessive idle capacity.

  2. Sizing of installed capacity. The peak of the curve fixes the minimum installed capacity required (plus spinning reserve and transmission losses). The area under the curve gives the energy demand (MWh/day), which fixes the energy (firm and secondary) the plant must generate. Comparing the two decides the number and size of units.

  3. Unit commitment and scheduling. The curve shows when to start/stop units. Base-load units run continuously; peaking units are brought in only during the evening peak. This minimises start-up/shut-down losses and fuel (or water) consumption.

  4. Reservoir operation and pondage sizing. For a storage or pondage plant, the curve indicates the daily drawdown: water stored during off-peak hours is released during the peak. The pondage volume required is obtained from the area between the load curve and the available inflow curve.

  5. Selection of turbine type and rated discharge. A flat base plus a sharp peak favours a Francis or Pelton unit with good part-load efficiency; the rated discharge is set so that the plant operates near best efficiency at the average load, not at the peak.

  6. Spinning reserve and reliability. The difference between peak demand and available capacity determines the reserve margin needed to cover forced outages and load forecasting error.

  7. Tariff and demand-side planning. The curve identifies peak hours, which are used to set time-of-day tariffs and to plan load shifting (e.g. off-peak irrigation pumping), reducing the required peaking capacity.

In short, the daily load curve is the basic planning tool: its peak sizes the plant, its area fixes the energy requirement, and its shape dictates how the plant (and its reservoir) is operated hour by hour.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 5 marks

What happens when a synchronous generator is connected to an isolated load versus when it is connected to a large power system? Discuss in terms of frequency and voltage control.

Answer

Synchronous generator on isolated load vs. on a large power system

1. Connected to an isolated load

  • The generator alone supplies the load, so the load demand directly determines the operating point.
  • Frequency control: Frequency is set by the balance between mechanical input (prime-mover torque) and the electrical load. Any change in load causes the rotor speed to change — there is no other machine to share the change. Frequency is controlled only by adjusting the prime-mover governor (speed governor) of that single unit. A load increase causes frequency to fall until the governor opens the valve/gate to restore speed.
  • Voltage control: Voltage at the terminals is controlled solely by the generator's own excitation (field current) through the AVR. A change in load current causes a change in the internal voltage drop (armature reaction and synchronous reactance drop), so terminal voltage varies unless the AVR acts. The generator must supply both active and reactive power demanded by the load.
  • The system is small and weak: frequency and voltage are highly sensitive to even small load changes, and there is no inertia support from other machines. Stability and power quality are difficult to maintain.

2. Connected to a large (infinite bus) power system

  • The generator is now one of many in parallel, and the system behaves as an infinite bus (constant frequency and constant voltage).
  • Frequency control: Frequency is fixed by the whole system, not by this generator. The generator's real-power output is controlled by its prime-mover input (governor setting): increasing steam/water input increases the power delivered to the system, but frequency stays essentially constant. Load–frequency control is shared among all units through droop characteristics and the system's automatic generation control (AGC). A single generator cannot change system frequency appreciably.
  • Voltage control: Terminal voltage is held nearly constant by the system. The generator's reactive-power output is controlled by its excitation (field current): over-excitation delivers reactive power to the system (generator acts like a capacitor), under-excitation absorbs reactive power. The AVR adjusts field current to control reactive power sharing, not to set voltage.
  • The machine operates synchronously with the grid; its speed is locked to system frequency. Load changes are absorbed by the entire system, so frequency and voltage remain stable.

Summary table

AspectIsolated loadLarge power system
FrequencySet by generator governor; varies with loadFixed by system; generator adjusts its power output
VoltageSet by generator AVR; varies with loadFixed by system; excitation controls reactive power
Active power controlGovernor (speed)Prime-mover input (governor setting)
Reactive power controlExcitation (AVR)Excitation (AVR)
SensitivityHigh (weak system)Low (strong system)

In short, on an isolated load the generator must control both frequency and voltage itself, whereas on a large system frequency and voltage are imposed by the grid, and the generator controls only its own active and reactive power output.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

What are the major electrical and mechanical components of hydroelectric power plants? Explain the hydro-turbine selection process. Also, elaborate the working principle of governor.

Answer

Major Components of a Hydroelectric Power Plant

Electrical Components

  1. Synchronous Generator (Alternator) — Converts mechanical energy of the turbine into electrical energy. Usually a 3-phase salient-pole machine directly coupled to the turbine.
  2. Excitation System / Exciter — Supplies DC field current to the rotor poles; controls terminal voltage and reactive power. Modern plants use static thyristor exciters.
  3. Power Transformer (Step-up) — Raises generator voltage (e.g., 11 kV) to transmission level (e.g., 132/220 kV).
  4. Switchgear — Circuit breakers, isolators, CTs, PTs, surge arresters, and busbars for protection and switching.
  5. Busbars and Auxiliary Transformer — Distribute power and supply station auxiliaries (cooling pumps, lighting, controls).
  6. Protection, Control and SCADA System — Relays (differential, overcurrent, earth fault), PLC/SCADA for monitoring and automation.
  7. Excitation Transformer, AVR, and DC Battery Bank — For field supply, voltage regulation, and emergency/control power.

Mechanical Components

  1. Dam / Diversion Weir — Stores or diverts water to create the required head.
  2. Intake Structure with Trash Rack and Gate — Screens debris; controls entry of water.
  3. Headrace Tunnel / Canal / Penstock — Conveys water from intake to powerhouse under pressure.
  4. Surge Tank — Absorbs pressure surges (water hammer) during load rejection or acceptance.
  5. Turbine — Converts hydraulic energy into mechanical (rotational) energy.
  6. Governor — Regulates turbine speed and power output by controlling the guide vanes / needle.
  7. Draft Tube — Guides water from turbine outlet to tailrace, recovering kinetic energy.
  8. Tailrace — Returns water to the river downstream.
  9. Main Inlet Valve (Butterfly / Spherical) — Isolates turbine for maintenance.
  10. Shaft, Bearings, Coupling — Transmit torque from turbine to generator; thrust and guide bearings support the rotating mass.
  11. Cooling System — Air coolers for generator, oil coolers for bearings and transformer.
  12. EOT Crane, Gates, Hoists — For erection and maintenance.

Hydro-Turbine Selection Process

Turbine type is selected mainly on the basis of net head (H) and discharge (Q), along with specific speed, capacity, and site conditions.

Step 1 — Determine net head and design discharge Compute gross head minus losses (friction, bends, valves) to get net head. Design discharge comes from hydrological study (Q₄₀ or firm flow).

Step 2 — Compute power P=ρgQHηP = \rho g Q H \eta

Step 3 — Compute specific speed Ns=NPH5/4N_s = \frac{N\sqrt{P}}{H^{5/4}} where N is in rpm, P in kW (per jet for Pelton), H in m. N_s decides the runner type.

Step 4 — Select turbine type by head range

TurbineHead rangeSpecific speed (metric)
Pelton (impulse)High head, H > 250 m10–50 (per jet)
Francis (reaction)Medium head, 30 m < H < 250 m60–300
Kaplan / PropellerLow head, H < 30 m300–900

Step 5 — Check cavitation (for reaction turbines) Compute Thoma's cavitation factor σ and compare with critical σ_c; set the setting level (suction head) accordingly: σ=Ha−Hv−HsH\sigma = \frac{H_a - H_v - H_s}{H}

Step 6 — Fix number of units and speed Choose synchronous speed from grid frequency (N = 120f/P poles) and match with specific speed limits. Decide number of units from part-load efficiency and maintenance requirements.

Step 7 — Verify with model tests and economic comparison Confirm efficiency, runaway speed, and cost; finalise the manufacturer's selection.


Working Principle of Governor

The governor is a closed-loop speed (frequency) control system that maintains the turbine-generator at synchronous speed (constant frequency) irrespective of load changes.

Principle:

  1. A speed sensing element (flyball / permanent magnet generator / tachometer) continuously measures the actual turbine speed.
  2. This is compared with the speed reference (corresponding to 50 Hz) to produce an error signal.
  3. The error is processed by the controller (mechanical-hydraulic or electro-hydraulic with PID action).
  4. The controller commands a servomotor / servo-valve which moves the guide vanes (Francis/Kaplan) or the needle / deflector (Pelton).
  5. Changing the vane opening changes the water flow Q, hence the torque and speed.

Response to load change:

  • Load increases → speed falls → error signal → guide vanes open → more water → speed restored.
  • Load decreases → speed rises → guide vanes close → less water → speed restored.

Additional functions:

  • Speed droop / regulation for parallel operation of multiple units.
  • Load sharing between interconnected machines.
  • Emergency shutdown on overspeed (runaway) via deflector or quick-closing valve.
  • Starting, synchronizing, and load-frequency control in modern digital governors.

Thus the governor ensures stable frequency, prevents runaway, and enables reliable parallel operation of the plant with the grid.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

Describe thermal power plants with their thermal efficiencies. Explain working principle of gas turbine cycle. How can you improve performance of gas turbine power plant? Clarify.

Answer

Thermal Power Plants

A thermal power plant converts the chemical (or nuclear) energy of a fuel into heat, uses that heat to raise the temperature and pressure of a working fluid (steam, combustion gas or air), and then expands the fluid in a prime mover to produce mechanical work, which drives a generator. In Nepal, NEA's thermal units are mainly diesel/gas-turbine based (e.g. Duhabi, Hetauda, Balaju), used for peaking and dry-season support.

Types and typical thermal efficiencies

Type of plantWorking fluid / prime moverTypical overall thermal efficiency
Coal-fired steam plant (subcritical)Steam / steam turbine30 – 35 %
Supercritical & ultra-supercritical steam plantSteam / steam turbine40 – 45 %
Nuclear steam plantSteam / steam turbine30 – 33 %
Diesel engine plantCombustion gas / reciprocating engine35 – 42 %
Simple open-cycle gas turbineCombustion gas / gas turbine25 – 35 %
Gas turbine with regenerationCombustion gas / gas turbine30 – 38 %
Combined cycle gas turbine (CCGT)Gas + steam / two turbines50 – 60 %
Cogeneration (CHP) plantGas + steam, heat also used70 – 80 % (energy utilisation)

The thermal efficiency is

ηth=net work outputheat supplied=WnetQin\eta_{th}=\frac{\text{net work output}}{\text{heat supplied}}=\frac{W_{net}}{Q_{in}}

The low efficiency of steam plants is due to the condenser rejecting a large amount of latent heat (Carnot limitation), boiler and piping losses, and irreversibilities in the turbine and compressor/pump. Gas turbine plants are further limited because a large fraction of turbine work is consumed by its own compressor.


Working Principle of the Gas Turbine (Brayton) Cycle

A gas turbine plant has three essential components on a single shaft:

  1. Compressor – draws atmospheric air and compresses it (typically pressure ratio 8–20).
  2. Combustion chamber – fuel is burnt continuously at essentially constant pressure, raising gas temperature to 1100–1500 °C.
  3. Gas turbine – hot high-pressure gas expands, producing work; part of it drives the compressor and the rest drives the generator.

The ideal cycle is the air-standard Brayton cycle:

ProcessDescription
1–2Isentropic compression in compressor
2–3Constant-pressure heat addition in combustor
3–4Isentropic expansion in turbine
4–1Constant-pressure heat rejection to atmosphere

Thermal efficiency:

ηth=1−T1T2=1−1(rp)γ−1γ,rp=P2P1\eta_{th}=1-\frac{T_1}{T_2}=1-\frac{1}{(r_p)^{\frac{\gamma-1}{\gamma}}},\qquad r_p=\frac{P_2}{P_1}

Work ratio:

rw=Wt−WcWt=1−WcWtr_w=\frac{W_t-W_c}{W_t}=1-\frac{W_c}{W_t}

For a simple cycle the back work ratio Wc/WtW_c/W_t is about 0.5–0.6, i.e. more than half the turbine work is spent on driving the compressor — this is the main reason for the low efficiency of a simple gas turbine.


Methods to Improve Gas Turbine Plant Performance

1. Regeneration (recuperation)

The turbine exhaust at 450–600 °C still contains much heat. A recuperator/regenerator transfers this heat to the air leaving the compressor before it enters the combustor, reducing fuel input for the same turbine work.

ηth=1−T1T3 (rp)γ−1γ\eta_{th}=1-\frac{T_1}{T_3}\,(r_p)^{\frac{\gamma-1}{\gamma}}

Regeneration is beneficial only when T4>T2T_4>T_2, i.e. at low pressure ratios; at high rpr_p it becomes ineffective or even harmful.

2. Intercooling

Compression in two or more stages with cooling between them reduces compressor work (because work ∝ temperature) and lowers T2T_2, so more air mass can be handled. Net work increases, but efficiency falls unless combined with regeneration.

3. Reheating

Expanding in two turbine stages with reheating to the original inlet temperature between them increases turbine work and raises the mean temperature of heat addition, improving efficiency when combined with regeneration.

4. Higher turbine inlet temperature

Raising T3T_3 increases both work output and efficiency. It is limited by blade material strength; achieved by air/film blade cooling, thermal barrier coatings and single-crystal superalloys (modern units reach 1500 °C).

5. Higher pressure ratio

Efficiency rises with rpr_p; however, specific work peaks at an optimum rpr_p (≈ 12–15 for simple cycle), so the choice is a compromise between efficiency and output.

6. Combined cycle (the most effective method)

The high-temperature exhaust of the gas turbine is passed through a heat recovery steam generator (HRSG) to raise steam for a bottoming steam turbine. Overall efficiency reaches 50–60 % (e.g. 1 gas turbine + 1 steam turbine). This is the standard choice for new base-load thermal plants.

7. Inlet air cooling

Power output falls in hot weather because air density and mass flow drop. Evaporative coolers, fogging or absorption/mechanical chillers cool inlet air, increasing mass flow and output by 10–20 % in summer — very relevant to the hot Terai climate of Nepal.

8. Steam or water injection

Injecting steam into the combustor increases mass flow and power, and reduces NOₓ formation (as in the Cheng cycle).

9. Cogeneration / waste-heat utilisation

Using exhaust heat for process steam or space heating raises overall energy utilisation to 70–80 %, though electrical efficiency is unchanged.

10. Improved component efficiencies and operation

Reducing compressor and turbine losses through better blade profiles, tip-clearance control, low-loss diffusers, inlet guide vane (IGV) modulation for part-load operation, washing of compressor blades to remove fouling, and proper maintenance all raise actual efficiency towards the ideal value.

Conclusion: The simple gas turbine is inherently inefficient because of the large compressor work and high exhaust loss. Regeneration, intercooling and reheating improve the cycle, but the greatest practical gain comes from the combined cycle with inlet air cooling, which lifts efficiency from about 30 % to over 55 % and is therefore the preferred option for NEA's future thermal capacity.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

Discuss maintenance planning and concept of various maintenance technique/practice for power plants. How can we use condition monitoring in hydropower plants? What are its advantages over other maintenance techniques? What are the parameters that can be monitored in hydropower plants using condition monitoring? Explain.

Answer

Maintenance Planning and Maintenance Techniques for Power Plants

1. Maintenance Planning

Maintenance planning is the process of deciding what maintenance work is to be done, when, by whom, with what resources, and at what cost, so that plant availability, reliability and safety are maximised at minimum life-cycle cost. For a power plant it involves:

  • Asset register and criticality analysis — listing every equipment (turbine, generator, transformer, penstock, gates, governor, excitation system, switchgear) and ranking it by failure consequence (safety, generation loss, repair cost).
  • Work identification and backlog management — collecting defects from logs, inspections and condition-monitoring data, then prioritising them.
  • Scheduling — long-term (annual overhaul/shutdown plan, aligned with the dry season when water is scarce), medium-term (monthly/quarterly) and short-term (weekly/daily) schedules.
  • Resource planning — manpower (in-house vs. contractor), spare parts and inventory, special tools, cranes, and budget.
  • Outage coordination — coordinating unit shutdowns with load demand, reservoir operation and the grid dispatch centre so that generation loss is minimised.
  • Feedback and continuous improvement — recording actual work done, failure history and costs, then revising intervals and strategies (a closed PDCA loop).

2. Maintenance Techniques / Practices

TechniqueBasisTypical use in power plants
Breakdown (Corrective) MaintenanceRun to failure, then repairNon-critical, cheap, redundant items (lighting, small pumps)
Preventive (Time-based) MaintenanceFixed calendar/running-hour intervalsLubrication, filter change, bearing greasing, annual overhaul
Predictive / Condition-Based Maintenance (CBM)Measured condition vs. limitsTurbine vibration, transformer DGA, partial discharge, oil analysis
Proactive MaintenanceRoot-cause elimination, precision installation, contamination controlAlignment, balancing, oil cleanliness targets
Reliability-Centred Maintenance (RCM)Failure-mode analysis decides the task for each modeDeciding which of PM/CBM/redesign suits each component
Total Productive Maintenance (TPM)Operator involvement, 5S, autonomous maintenanceOperator daily checks, cleaning, tightening
Design-out / Improvement MaintenanceRedesign to remove the failure modeReplacing a weak seal design, upgrading a cooling system

The trend in modern hydropower stations is to move from purely time-based PM towards CBM supported by RCM, using time-based tasks only where no measurable condition parameter exists.

3. Condition Monitoring in Hydropower Plants

Condition monitoring is the continuous or periodic measurement of physical parameters that indicate the health of a machine, so that a developing fault can be detected, located and trended before it becomes functional failure.

How it is applied in a hydropower plant:

  1. Sensors are installed at selected points — accelerometers on turbine/generator bearings, proximity probes for shaft displacement, RTDs/thermocouples in bearings and stator windings, pressure and flow transmitters, and oil sampling points.
  2. Data acquisition — signals are collected online (permanently wired to a monitoring unit/SCADA) or offline (portable analyser on a route).
  3. Baseline and trending — a "healthy" signature is recorded after commissioning/overhaul; subsequent readings are compared and trended against alarm and trip limits.
  4. Diagnosis — spectral analysis, envelope analysis, orbit plots, dissolved gas analysis, etc., identify the fault type (unbalance, misalignment, looseness, rub, bearing defect, insulation degradation).
  5. Decision — maintenance is planned at the most convenient time (e.g., next dry-season shutdown) instead of an emergency stop.

4. Advantages over Other Maintenance Techniques

Compared with breakdown and time-based preventive maintenance, CBM offers:

  • Reduced unplanned downtime — faults are caught early, so failures and forced outages are avoided.
  • Maintenance only when needed — components are not opened or replaced prematurely, saving labour, spares and cost.
  • Longer asset life — unnecessary dismantling and re-assembly wear is avoided.
  • Fewer secondary damages — a failing bearing or seal is replaced before it damages the shaft or runner.
  • Better spare-part and outage planning — lead time is known in advance.
  • Improved safety — catastrophic failures (runaway, winding flashover, transformer burst) are prevented.
  • Higher availability and revenue — generation loss is minimised, which matters greatly for a run-of-river plant where water not used is water lost.
  • Root-cause data — trends give evidence for design improvements and for revising PM intervals.

Its limitations: higher initial cost of instrumentation and expertise, and it only works for failure modes that give a measurable precursor.

5. Parameters Monitored in Hydropower Plants

ParameterEquipment monitoredFault indicated
Vibration (velocity, acceleration, displacement)Turbine and generator bearings, shaftUnbalance, misalignment, bent shaft, bearing defects, cavitation, rubbing, looseness
Shaft displacement / eccentricity / air gapGuide and thrust bearings, generator air gapBearing wear, shaft bow, rotor–stator eccentricity
Bearing temperature (metal & oil)Thrust, guide and turbine bearingsLubrication failure, overload, oil degradation, cooling water loss
Oil quality (viscosity, acidity, water content, particle count, wear debris)Bearing and governor oil systemsOil degradation, contamination, wear of bearing surfaces
Partial discharge / insulation resistance, tan-deltaGenerator stator winding, bushings, cablesInsulation ageing, voids, moisture ingress
Dissolved gas analysis (DGA) & oil temperaturePower transformer, generator transformerOverheating, arcing, partial discharge in insulation
Vibration & temperature of transformer core/clampingTransformerCore looseness, winding deformation
Pressure, flow, head, and efficiencyPenstock, turbine, draft tubeCavitation, blockage, guide-vane/runner wear, efficiency drop
Water conductivity / pH / silt contentTurbine water pathErosion and corrosion of runner and guide vanes
Shaft speed, vibration of governor & guide-vane servomotorGovernor and hydraulic systemGovernor instability, oil leakage, servomotor sticking
Air gap and magnetic fluxGeneratorRotor eccentricity, winding short circuit
Vibration of gates, hoists and trash racksIntake, spillwayLooseness, structural fatigue, blockage
Corrosion potential / coating conditionPenstock, spiral casingInternal corrosion

Conclusion: For a hydropower plant, planned maintenance based on RCM and executed through condition monitoring gives the best balance of reliability, cost and safety. Vibration and temperature monitoring of rotating parts, oil analysis, DGA of transformers, and efficiency/head measurements together cover most critical failure modes, allowing faults to be corrected during scheduled dry-season shutdowns rather than through costly forced outages.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

  • 2082 · 2nd · 10 marks

How do you select air conditioning system? Explain components of air conditioning system with their function. Discuss the maintenance of an air conditioning system.

Answer

Selection, Components and Maintenance of an Air Conditioning System

1. Selection of an Air Conditioning System

The correct system is chosen by matching the application, load and building constraints. The main steps are:

  1. Define the application and comfort requirement. Fix the indoor design conditions (dry-bulb temperature, relative humidity, air velocity and cleanliness) as per the intended use — e.g. office, hospital, computer room, cinema hall or industrial process. Critical spaces need tighter control than ordinary comfort spaces.

  2. Estimate the cooling load. Compute the total heat gain from:

    • External sources: solar radiation through glass, conduction through walls, roof and floor.
    • Internal sources: occupants (sensible + latent), lighting, motors, appliances and equipment.
    • Ventilation/infiltration: outside air brought in for fresh-air requirement and leakage.
    • Add a safety factor (usually 5–10%) to get the design cooling load in kW or TR (1 TR = 3.517 kW).
  3. Decide the system type based on load, space and budget:

    • Window / split units — small rooms, low first cost, easy installation.
    • Package units — single large space, factory-assembled, easy maintenance.
    • Central ducted system (AHU + chiller/condenser) — large buildings, many rooms, good control and aesthetics.
    • VRF/VRV system — part-load efficiency, individual room control, no ductwork.
    • Evaporative (desert) cooler — hot dry climate, low running cost, but no humidity control.
  4. Check technical suitability: availability of power and water, ceiling/floor space for ducts and plant, noise level, zoning requirement, part-load performance (COP/ISEER), refrigerant type (environment-friendly, e.g. R-410A, R-32) and statutory requirements.

  5. Compare economics: first cost, running cost (energy consumption), maintenance cost and life-cycle cost; then select the system giving the lowest life-cycle cost while meeting comfort and reliability needs.

2. Components and Their Functions

ComponentFunction
CompressorHeart of the system. Raises the pressure and temperature of the low-pressure refrigerant vapour from the evaporator so that it can reject heat in the condenser. Types: reciprocating, scroll, screw, centrifugal.
CondenserRejects heat absorbed in the evaporator plus compressor work to the cooling medium (air or water), condensing high-pressure vapour into liquid. Air-cooled or water-cooled.
Expansion device (capillary tube, thermostatic expansion valve, electronic expansion valve)Throttles the high-pressure liquid to low pressure, lowering its temperature, and regulates refrigerant flow to the evaporator to maintain the required superheat.
EvaporatorAbsorbs heat from the conditioned space (or chilled water) and evaporates the low-pressure liquid refrigerant, thereby cooling and dehumidifying the air.
RefrigerantWorking fluid that carries heat from the evaporator to the condenser by alternately evaporating and condensing.
Air handling unit (AHU)Houses cooling/dehumidifying coil, heating coil, filters and fan; conditions and circulates air through ducts.
FiltersRemove dust, dirt and airborne particles from the air to protect coils and improve indoor air quality.
Ducts and diffusersDistribute conditioned air to the space and return it back to the AHU.
Fan / blowerCreates the pressure difference to circulate air over the coils and through the ductwork.
Chiller / cooling tower (central system)Chiller produces chilled water; cooling tower rejects condenser heat to the atmosphere.
Controls (thermostat, humidistat, sensors, BMS)Maintain the set temperature, humidity and air quality by switching the compressor, fan and dampers automatically.
Refrigerant piping, valves and driersCarry refrigerant, isolate sections for service and remove moisture/acid from the circuit.

3. Maintenance of an Air Conditioning System

Maintenance is carried out in three tiers:

A. Routine (daily/weekly)

  • Check and record suction and discharge pressures, temperatures and running current.
  • Clean or replace air filters; clean return-air grilles and diffusers.
  • Check for unusual noise, vibration, oil leakage and refrigerant leakage.
  • Verify thermostat/humidistat settings and control operation.

B. Preventive (monthly/seasonal)

  • Clean the evaporator and condenser coils; clean the cooling tower and check water quality.
  • Lubricate fan and compressor bearings; check belt tension and alignment.
  • Tighten electrical terminals, check contactors, overload relays and capacitor condition.
  • Check refrigerant charge; top up if required; check for moisture with a sight glass.
  • Clean condensate drain pan and drain line to prevent overflow and bacterial growth.
  • Before summer, service the unit fully; before winter, cover/protect idle units.

C. Breakdown (corrective)

  • Attend to complaints such as insufficient cooling, icing of the coil, compressor tripping or water leakage.
  • Diagnose faults (low charge, choked capillary, faulty compressor, dirty coil), repair or replace defective parts, and re-commission the system.

Safety and good practice: switch off and isolate the supply before servicing, use proper PPE, recover refrigerant instead of venting it, maintain a maintenance log/register for each unit, and follow the manufacturer's schedule and the Nepal Electricity Authority/statutory safety rules. A planned maintenance programme reduces energy consumption, extends equipment life and ensures reliable, healthy indoor conditions.

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

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.