Advanced Energy Engineering interview questions

125 questions with answers (ENE-301 to ENE-425). Try each one aloud before reading the answer.

Energy Fundamentals & Thermodynamics

ENE-301 · What is the exergy of 1000 kJ of heat available at 500 K with surroundings at 300 K?

Ex = Q (1 − T₀ ÷ T) = 1000 × (1 − 300 ÷ 500) = 400 kJ. Only 40% of the heat can, in principle, be converted to work. The rest is unavoidable rejection to the environment.

ENE-302 · A plant needs 10 MW of electricity and 20 MW of process heat. Compare separate supply with CHP.

Separate: grid at 35% needs 28.6 MW fuel, and a boiler at 85% needs 23.5 MW, totalling 52.1 MW. A CHP at 30% electrical and 60% thermal efficiency needs 33.3 MW of fuel for both outputs, saving about 36% primary energy.

ENE-303 · How does a heat pump compare with a gas boiler on primary energy?

A heat pump with seasonal COP 3.5 on a grid with 40% generation efficiency delivers 3.5 × 0.40 = 1.4 units of heat per unit of primary fuel. A 90% boiler delivers 0.9. The heat pump uses about 36% less primary energy, more as the grid decarbonises.

ENE-304 · What is pinch analysis and how is it used?

A heat integration method that plots hot and cold composite curves. The point of closest approach, set by ΔT_min, is the pinch. It gives minimum hot and cold utility targets before network design. Rules: no heat transfer across the pinch, no cold utility above it and no hot utility below it.

ENE-305 · Where is the largest exergy destruction in a fossil steam plant?

In the boiler: combustion is highly irreversible, and heat transfer occurs across large temperature differences between flame (above 1500 °C) and steam (about 550 °C). The condenser rejects much energy, but that energy has little exergy.

ENE-306 · How much heat can be recovered from 10 kg/s of flue gas cooled from 300 °C to 150 °C?

Q = ṁ c_p ΔT = 10 × 1.1 kJ/kg·K × 150 K = 1650 kW. The lower limit is often set by acid dew point, about 120–150 °C for sulphur-bearing fuels, to avoid corrosion.

ENE-307 · When is an organic Rankine cycle (ORC) preferred for waste heat?

For low-to-medium grade heat (about 80–350 °C) and small capacities. Organic fluids with low boiling points and dry expansion characteristics give compact turbines and avoid wet-steam erosion. Cycle efficiencies are typically about 10–20%.

ENE-308 · What is the exergy efficiency of an electric heater keeping a room at 20 °C when outdoors is 0 °C?

The exergy of the delivered heat is Q(1 − 273 ÷ 293) ≈ 0.068 Q. Since the electricity input equals Q, exergy efficiency is about 6.8%, showing why heat pumps are thermodynamically better.

ENE-309 · What is energy return on investment (EROI)?

The ratio of energy delivered over a system's life to the energy invested to build, operate and decommission it. Low EROI sources leave less net energy for society. Values depend strongly on system boundary and method.

ENE-310 · Estimate the CO₂ intensity of a gas-fired CCGT at 55% efficiency.

Natural gas emits about 56 kg CO₂/GJ of fuel. Per kWh of electricity: 56 × 0.0036 GJ ÷ 0.55 ≈ 0.37 kg CO₂/kWh, compared with roughly 0.9–1.0 kg/kWh for subcritical coal.

Thermal Power Plants

ENE-311 · Find Rankine efficiency given h_turbine-in = 3400, h_turbine-out = 2200, h_condensate = 190 and pump work = 10 kJ/kg.

Turbine work = 1200 kJ/kg. Net work = 1200 − 10 = 1190 kJ/kg. Boiler heat = 3400 − (190 + 10) = 3200 kJ/kg. η = 1190 ÷ 3200 ≈ 37.2%.

ENE-312 · A turbine has an isentropic enthalpy drop of 1300 kJ/kg and an actual drop of 1144 kJ/kg. What is its isentropic efficiency, and why does it matter?

η_is = 1144 ÷ 1300 = 0.88. The lost 156 kJ/kg appears as higher exhaust enthalpy that the condenser rejects. Each point of turbine efficiency translates almost directly into plant heat rate.

ENE-313 · Why can't steam pressure be raised indefinitely without reheat?

At fixed superheat temperature, higher pressure moves expansion towards the wet region, lowering exhaust dryness. Below about 0.88–0.90 dryness, droplet erosion damages last-stage blades. Reheat or higher superheat restores dryness.

ENE-314 · A coal unit's heat rate has risen 3%. How would you diagnose it?

Check condenser vacuum and cleanliness, cooling water temperature, feedwater heater outages and terminal temperature differences, main and reheat steam temperatures, excess air and flue gas exit temperature, unburnt carbon, soot deposits, turbine stage efficiencies, and auxiliary power.

ENE-315 · How much coal per hour does a 500 MW unit at heat rate 2400 kcal/kWh need, using coal of 3800 kcal/kg GCV?

Heat input = 500,000 kWh × 2400 = 1.2 × 10⁹ kcal/h. Coal = 1.2 × 10⁹ ÷ 3800 ≈ 316,000 kg/h, or about 316 t/h (about 7,600 t/day).

ENE-316 · What cooling water flow is needed to reject 600 MW in a condenser with a 10 °C temperature rise?

ṁ = Q ÷ (c_p ΔT) = 600,000 kW ÷ (4.186 × 10) ≈ 14,300 kg/s, or about 14.3 m³/s. This shows why thermal plants are sited near large water sources.

ENE-317 · What is the efficiency of a combined cycle with GT efficiency 38% and a bottoming steam cycle efficiency of 33%?

η_CC = η_GT + η_ST (1 − η_GT) = 0.38 + 0.33 × 0.62 ≈ 0.585, or 58.5%, assuming all GT exhaust heat reaches the steam cycle.

ENE-318 · What challenges arise when coal plants operate flexibly to balance renewables?

Low minimum load affects flame stability and needs combustion tuning. Faster ramping and frequent starts cause thermal cycling, creep-fatigue damage in thick-walled components, higher heat rate at part load, and more emissions control difficulties.

ENE-319 · What is the main trade-off of post-combustion carbon capture with amines?

It captures most flue gas CO₂, but regenerating the solvent needs large amounts of low-pressure steam extracted from the turbine, plus compression power. This causes a significant efficiency penalty, raises fuel per kWh and adds capital cost.

ENE-320 · When would you choose an air-cooled condenser over a water-cooled one?

In water-scarce sites, because air cooling nearly eliminates make-up water loss. The trade-off is higher condenser pressure, especially on hot days, which raises heat rate and reduces output, plus higher capital cost and fan power.

ENE-321 · What are pinch point and approach temperature in an HRSG?

Pinch point is the smallest gas-to-water temperature difference, at the evaporator exit. Approach temperature is the difference between saturation and economiser outlet water temperature. Smaller pinch recovers more heat but needs much more surface area.

Solar Photovoltaics

ENE-322 · Size a rooftop PV system for 6000 kWh/year at a specific yield of 1450 kWh/kWp.

Required capacity = 6000 ÷ 1450 ≈ 4.14 kWp. Using 400 Wp modules, 4140 ÷ 400 ≈ 10.4, so 11 modules (4.4 kWp). Check roof area, shading and inverter rating before finalising.

ENE-323 · How many modules (Voc = 49.5 V, β_Voc = −0.28%/°C) can be in series for a 1000 V inverter if the minimum cell temperature is −5 °C?

ΔT = −30 °C, so V_oc,max = 49.5 × (1 + 0.0028 × 30) ≈ 53.7 V. 1000 ÷ 53.7 ≈ 18.6, so a maximum of 18 modules. Also check V_mp at high temperature stays within the MPPT window.

ENE-324 · How is row spacing in a PV plant determined?

Spacing is set to avoid inter-row shading at a design sun position, often around solar noon at the winter solstice. Shadow length is h ÷ tan(solar altitude). Ground coverage ratio trades land use against shading loss.

ENE-325 · What does partial shading do to the P-V curve, and how is it handled?

Bypass diodes conduct for shaded substrings, creating multiple local maxima. Simple P&O MPPT can lock onto a local peak. Global MPPT scans, module-level electronics (optimisers, microinverters) and careful string layout reduce losses.

ENE-326 · Compare perturb-and-observe and incremental conductance MPPT.

P&O perturbs voltage and checks the power change. It is simple but oscillates around the MPP and can track wrongly under fast irradiance changes. Incremental conductance compares dI/dV with −I/V, tracking more accurately under changing conditions at higher computational cost.

ENE-327 · What is potential-induced degradation (PID)?

Power loss caused by high voltage between cells and the grounded frame, driving ion migration (often sodium from glass) and leakage currents, especially in hot, humid conditions. It is mitigated by PID-resistant modules, negative-pole grounding or night-time recovery.

ENE-328 · If a module degrades at 0.5%/year after a 2% first-year loss, what output remains after 25 years?

After year 1, 98% remains. Then 24 more years at about 0.5%: 98% × (1 − 0.005)²⁴ ≈ 98% × 0.887 ≈ 87%. This aligns with typical linear performance warranties.

ENE-329 · Estimate the output of a 400 Wp module at 1000 W/m² and 35 °C ambient, with NOCT 45 °C and γ = −0.4%/°C.

T_c = 35 + (45 − 20) × 1000 ÷ 800 ≈ 66 °C. Loss = 41 × 0.4% ≈ 16.5%. Output ≈ 400 × 0.835 ≈ 334 W.

ENE-330 · When is a single-axis tracker worth it over fixed tilt?

Trackers typically add roughly 15–25% energy at sunny, low-to-mid latitude sites with high direct irradiance. The gain must outweigh higher capital and O&M cost, wider spacing and wind-loading risk. Bifacial modules on trackers add further gain.

ENE-331 · Why are perovskite–silicon tandem cells attractive?

A wide-band-gap perovskite top cell absorbs high-energy photons while silicon absorbs the rest, reducing thermalisation loss and allowing efficiencies beyond the single-junction limit. Long-term stability and scale-up remain the main challenges.

ENE-332 · Design an off-grid system for a 5 kWh/day load with 5 peak sun hours and 2 days of autonomy.

PV = 5 ÷ (5 × 0.75 system derate) ≈ 1.33 kWp. Battery = 5 × 2 ÷ (0.8 DoD × 0.9 efficiency) ≈ 13.9 kWh. Size the charge controller for array current and the inverter for peak load.

Solar Thermal

ENE-333 · What is solar multiple in CSP design?

The ratio of solar field thermal output at design point to the thermal power the turbine needs. A value of 1 matches the turbine only at peak sun. Plants with storage use about 2–3 to charge storage while running the turbine.

ENE-334 · Size molten-salt storage for a 100 MWe CSP turbine to run 7.5 hours at 40% cycle efficiency, with ΔT = 275 K and c_p = 1.5 kJ/kg·K.

Thermal energy = 100 × 7.5 ÷ 0.40 = 1875 MWh_th, or 6.75 × 10⁹ kJ. Salt mass = 6.75 × 10⁹ ÷ (1.5 × 275) ≈ 1.64 × 10⁷ kg, about 16,400 t.

ENE-335 · What are the main optical losses in a parabolic trough field?

Cosine loss from incidence angle, mirror reflectivity and soiling, glass envelope transmittance, absorber absorptance, intercept factor from tracking and mirror errors, end losses and row shading.

ENE-336 · Why is there an optimum receiver temperature in CSP?

Raising receiver temperature improves power cycle (Carnot) efficiency, but radiative losses rise roughly with T⁴, reducing receiver efficiency. Overall efficiency peaks at an intermediate temperature that rises with concentration ratio.

ENE-337 · Find efficiency of a flat-plate collector with F_R(τα) = 0.75, F_R U_L = 5 W/m²·K, T_i = 50 °C, T_a = 25 °C and G = 800 W/m².

η = 0.75 − 5 × (50 − 25) ÷ 800 = 0.75 − 0.156 ≈ 0.59, or 59%.

ENE-338 · How would you select a solar collector for industrial process heat?

By required temperature: flat-plate for below about 80 °C, evacuated tube for about 80–120 °C, and concentrating collectors (compact troughs, linear Fresnel) for about 120–400 °C steam. Consider DNI, roof or land area, load profile and integration point.

ENE-339 · What are the trade-offs of dry cooling in CSP plants?

CSP sites are often arid. Dry cooling cuts water use dramatically but raises condenser temperature, lowering cycle efficiency and annual output by a few percent and raising capital cost. Hybrid wet-dry cooling is a compromise.

ENE-340 · How is molten salt freezing prevented in CSP plants?

By keeping salt above its freezing point (about 220–240 °C for solar salt) using heat tracing on pipes and valves, keeping the cold tank hot, draining receivers at night, and recirculation. Freezing can damage piping and cause long outages.

ENE-341 · Why combine CSP with PV in a hybrid plant?

PV supplies cheap daytime electricity, while CSP stores heat for evening and night dispatch. The hybrid gives a firm, flexible profile at lower cost than CSP alone and uses shared grid connection and land.

ENE-342 · How is overheating managed in solar water heating systems?

By drainback designs that empty collectors when the pump stops, heat dump radiators, night cooling, pressure relief valves, expansion vessels sized for steam formation, and glycol mixtures rated for stagnation temperatures.

Wind Energy

ENE-343 · What power does a 120 m diameter rotor produce at 8 m/s with Cp = 0.45 and ρ = 1.225 kg/m³?

A = π × 60² ≈ 11,310 m². P = 0.5 × 1.225 × 11,310 × 512 × 0.45 ≈ 1.6 MW.

ENE-344 · How is the Betz limit derived?

From actuator disc theory with axial induction factor a: rotor velocity is v(1 − a) and power coefficient is Cp = 4a(1 − a)². Setting dCp/da = 0 gives a = 1/3, so Cp,max = 16/27 ≈ 0.593.

ENE-345 · What rotor speed gives optimum tip speed ratio of 7 at 8 m/s for a 60 m blade?

ω = λ v ÷ R = 7 × 8 ÷ 60 ≈ 0.933 rad/s, or N = 0.933 × 60 ÷ 2π ≈ 8.9 rpm.

ENE-346 · If wind speed is 6 m/s at 10 m, estimate the speed and power gain at 100 m with α = 0.14.

v₁₀₀ = 6 × (100 ÷ 10)^0.14 ≈ 6 × 1.38 ≈ 8.3 m/s. Power ratio = (8.3 ÷ 6)³ ≈ 2.6, so taller towers greatly increase output.

ENE-347 · How should turbines be spaced in a wind farm?

Commonly about 3–5 rotor diameters apart crosswind and 5–10 diameters downwind along the prevailing direction. Layout optimisation balances wake losses, turbulence loading, land or seabed constraints and cable cost.

ENE-348 · How are offshore wind foundations chosen by water depth?

Monopiles suit shallow water, up to roughly 30–40 m. Jackets and tripods suit transitional depths. Floating platforms (spar, semi-submersible, tension-leg) suit deep water where fixed foundations become uneconomic.

ENE-349 · What grid code requirements apply to wind farms?

Fault ride-through (LVRT/HVRT), reactive power and voltage control capability, active power and ramp rate control, frequency response, power quality limits (flicker, harmonics), and real-time communication with the system operator.

ENE-350 · Estimate wind power density at a site with mean speed 7 m/s, assuming a Rayleigh distribution.

For Rayleigh, mean of v³ ≈ (6 ÷ π) v̄³ ≈ 1.91 v̄³. Power density = 0.5 × 1.225 × 1.91 × 343 ≈ 401 W/m², almost twice the value calculated from the mean speed alone.

ENE-351 · A turbine underperforms its warranted power curve. What would you check?

Yaw misalignment, pitch angle calibration, blade soiling, icing or leading-edge erosion, nacelle anemometer calibration and transfer function, controller derating settings, air density correction, and site turbulence and wake effects.

ENE-352 · What are the trade-offs between direct-drive and geared wind turbines?

Direct-drive avoids the gearbox, a major failure item, improving reliability, especially offshore. It needs large, heavy, low-speed permanent-magnet generators and rare-earth materials. Geared designs use smaller high-speed generators but need more gearbox maintenance.

ENE-353 · What causes leading-edge erosion of blades, and why does it matter?

Rain, hail, sand and particles striking blade tips at high speed (around 80–100 m/s) roughen the leading edge. This reduces lift and increases drag, lowering annual energy, and requires protective coatings, tapes or repairs.

Hydropower

ENE-354 · What upper reservoir volume is needed for 1000 MWh of pumped storage at 300 m head and 90% generating efficiency?

V = E ÷ (ρ g H η) = 3.6 × 10¹² J ÷ (1000 × 9.81 × 300 × 0.9) ≈ 1.36 × 10⁶ m³.

ENE-355 · Select a turbine for a site with 400 m net head and 5 m³/s flow.

Power ≈ 9.81 × 5 × 400 × 0.9 ≈ 17.7 MW. High head and low flow give low specific speed, so a multi-jet Pelton is suitable. A high-head Francis is possible but more prone to cavitation and silt wear.

ENE-356 · How is the economic penstock diameter determined?

By minimising total cost: a larger diameter raises capital (steel) cost but reduces friction head loss, h_f ∝ L v² ÷ D, and the value of lost energy. The optimum often gives velocities of a few m/s.

ENE-357 · Estimate the water hammer pressure rise for a sudden 3 m/s velocity change with a 1000 m/s wave speed.

Δp = ρ c Δv = 1000 × 1000 × 3 = 3 MPa, equivalent to about 306 m of water head. Closure slower than the pipe period 2L/c greatly reduces this.

ENE-358 · What is the jet velocity and optimum bucket speed for a Pelton wheel under 400 m head?

v_jet = C_v √(2gH) ≈ 0.98 × √(2 × 9.81 × 400) ≈ 86.8 m/s. Optimum bucket speed is about 0.46–0.50 × v_jet, so about 40–43 m/s.

ENE-359 · How is design flow chosen for a run-of-river project?

From the flow duration curve, by optimising annual energy against cost. Design flow is often set near a flow exceeded about 30–40% of the time, after subtracting environmental flow, and checked with economic analysis.

ENE-360 · What are the benefits of variable-speed pumped storage?

It allows power control during pumping mode for frequency regulation, improves efficiency across varying head, and gives faster response. The cost is larger converters, using DFIM or full-converter designs.

ENE-361 · Why is hydropower valuable for grid balancing?

Fast ramping (from standstill to full load in minutes), wide operating range, black-start capability, spinning reserve, inertia from synchronous machines, and storage-based dispatch flexibility to complement variable renewables.

ENE-362 · How can silt erosion of turbines be mitigated?

With efficient desilting chambers, hard coatings such as HVOF tungsten carbide on runners and guide vanes, erosion-resistant designs, operating restrictions during high-silt monsoon flows, and planned overhauls.

ENE-363 · What are the main steps in a small hydro feasibility study?

Hydrological analysis and the flow duration curve, head survey, layout options, turbine and generator selection, energy estimate, environmental and social assessment, grid connection study, cost estimate, and financial analysis (LCOE, IRR).

Biomass & Bioenergy

ENE-364 · Estimate biogas and electricity from 1000 kg/day of fresh cattle dung.

At about 0.036 m³ biogas per kg of fresh dung, output is about 36 m³/day. At about 6 kWh/m³, that is roughly 216 kWh of fuel energy. A gas engine at 30% gives about 65 kWh of electricity per day.

ENE-365 · How much biomass does a 10 MWe plant at 25% net efficiency need, using fuel with LHV 15 MJ/kg?

Fuel input = 10 ÷ 0.25 = 40 MW_th = 40 MJ/s. Mass flow = 40 ÷ 15 ≈ 2.67 kg/s, or about 9.6 t/h and about 230 t/day at full load.

ENE-366 · What are the key supply-chain risks for a biomass power plant?

Seasonal availability, competing uses that raise fuel prices, variable moisture and quality, low bulk density raising transport cost (limiting economic collection radius), storage losses and fire risk, and dependence on many small suppliers.

ENE-367 · Why do some biomass fuels cause slagging, fouling and corrosion?

Agricultural residues are rich in potassium, sodium and chlorine. Alkali compounds lower ash melting points, forming deposits on superheaters, while chlorides drive high-temperature corrosion. Mitigation includes lower steam temperatures, additives (kaolin, sulphur), fuel blending and sootblowing.

ENE-368 · What is BECCS?

Bioenergy with carbon capture and storage. Biomass absorbs CO₂ as it grows, and capturing and storing the CO₂ released at the plant can yield net negative emissions. Its credibility depends on sustainable sourcing and permanent storage.

ENE-369 · A region burns rice straw in fields. What energy options would you evaluate?

Ex-situ collection for compressed biogas plants, pellets or briquettes for co-firing in coal plants, dedicated boilers for power, and 2G ethanol. Compare collection logistics, short harvest windows, silica-rich ash, offtake contracts and farmer incentives.

ENE-370 · Why does a gas engine derate on producer gas?

Producer gas from air gasification has a low LHV, about 4–6 MJ/Nm³, because of nitrogen dilution, giving a lower energy content per cylinder charge than natural gas. Engines typically derate substantially and need gas cleaning to protect against tar and particulates.

ENE-371 · Why can biofuels have higher lifecycle emissions than expected?

Fertiliser N₂O, fossil energy for cultivation and processing, and especially direct or indirect land-use change, such as clearing forest or displacing food crops, can release large carbon stocks that take decades to repay.

ENE-372 · What are the main pathways for sustainable aviation fuel (SAF)?

HEFA (hydroprocessed esters and fatty acids from waste oils), alcohol-to-jet, Fischer–Tropsch synthesis from gasified biomass or waste, and power-to-liquid e-fuels from green hydrogen and captured CO₂.

ENE-373 · What digester volume is needed for 10 m³/day of slurry at a 30-day retention time?

V = Q × HRT = 10 × 30 = 300 m³ of active volume. Add gas storage and headspace allowance, typically about 10–20% more.

Nuclear Energy

ENE-374 · What is reactor period?

The time for reactor power to change by a factor of e, from P(t) = P₀ e^(t/τ). Short periods mean rapid power rise, so reactors trip on short-period signals, especially during start-up.

ENE-375 · Estimate annual uranium fuel needed for a 1000 MWe LWR at 33% efficiency, 90% capacity factor and 45 MWd/kg burnup.

Thermal power ≈ 3030 MW. Energy = 3030 × 365 × 0.9 ≈ 995,000 MWd. Fuel = 995,000 ÷ 45 ≈ 22,100 kg, about 22 tonnes of enriched uranium per year.

ENE-376 · What were the main engineering lessons from Fukushima Daiichi?

The tsunami caused prolonged station blackout and loss of decay heat removal, leading to core melt and hydrogen explosions. Lessons: protection against beyond-design-basis floods, diverse and mobile backup power, passive cooling, hydrogen recombiners, filtered venting and spent fuel pool monitoring.

ENE-377 · What are the advantages and challenges of small modular reactors (SMRs)?

Advantages include factory fabrication, smaller capital per unit, passive safety, flexible siting and possible use for process heat. Challenges include first-of-a-kind costs, loss of economies of scale, licensing, supply chain and fuel availability for some designs.

ENE-378 · What happens during a large-break loss-of-coolant accident (LOCA) in a PWR?

Rapid depressurisation and coolant loss trip the reactor. Accumulators then inject water, followed by low-pressure injection and long-term sump recirculation. Analysis must show peak cladding temperature and oxidation stay within regulatory limits.

ENE-379 · Estimate decay heat one hour after shutdown of a 3000 MWth reactor.

Using the Way–Wigner approximation, P/P₀ ≈ 0.066 t^(−0.2), with t in seconds. For t = 3600 s, P/P₀ ≈ 0.066 × 0.194 ≈ 1.3%, so about 38 MW must still be removed.

ENE-380 · Why is controlled fusion so difficult?

Plasma must reach about 100–150 million K and satisfy the Lawson triple product (n T τ) for net gain, while being confined magnetically or inertially. Other challenges are plasma instabilities, materials under 14 MeV neutron flux, tritium breeding and heat exhaust.

ENE-381 · Can nuclear plants load-follow?

Yes, within limits. Many PWRs can ramp at a few percent per minute using control rods ("grey" rods) and boron adjustment. Xenon transients, fuel pellet-cladding interaction and economics, since fuel cost is low and capital cost high, often favour baseload operation.

ENE-382 · What is India's three-stage nuclear programme?

Stage 1: PHWRs on natural uranium produce plutonium. Stage 2: fast breeder reactors use plutonium and breed more fissile material, including U-233 from thorium blankets. Stage 3: thorium–U-233 reactors exploit India's large thorium reserves.

ENE-383 · How does deep geological disposal isolate high-level waste?

Through a multi-barrier system: a stable waste form (vitrified glass or spent fuel), corrosion-resistant canisters (copper or steel), a buffer such as bentonite clay, and a stable host rock hundreds of metres deep that limits groundwater flow.

Energy Storage & Batteries

ENE-384 · What storage options exist and how do they map to duration?

Seconds to minutes: supercapacitors, flywheels, SMES. Hours: Li-ion and other batteries, flow batteries. Hours to days: pumped hydro and compressed air. Days to seasons: hydrogen and synthetic fuels, and large thermal storage.

ENE-385 · Size a BESS to shave a 2 MW peak for 3 hours with 90% usable DoD.

Usable energy = 2 × 3 = 6 MWh. Nameplate = 6 ÷ 0.9 ≈ 6.7 MWh at a 2 MW power rating (about 0.3C). Add allowance for round-trip losses and degradation over the project life, or plan augmentation.

ENE-386 · What is levelized cost of storage (LCOS)?

Discounted lifetime costs (capex, O&M, charging energy, augmentation, end of life) divided by discounted lifetime energy discharged. It depends heavily on cycles per year, round-trip efficiency, degradation and duration.

ENE-387 · How is battery degradation handled over a 15–20 year project?

By oversizing at installation, or by augmentation, adding modules over time to maintain contracted capacity. Other measures are operating within SoC and temperature windows, limiting C-rate, and warranty-based throughput limits.

ENE-388 · What are key fire safety design measures for a Li-ion BESS?

Cell and module selection with propagation testing (such as UL 9540A), BMS protection, off-gas and smoke detection, deflagration venting or explosion control, suppression, thermal barriers, enclosure spacing, and emergency response planning with fire services.

ENE-389 · Which technology would you choose for 8-hour grid storage?

Compare cost per kWh for energy (not power), cycle life, round-trip efficiency, footprint and siting. Li-ion LFP is competitive but energy cost scales linearly. Flow batteries, pumped hydro and CAES scale energy more cheaply with tanks or reservoirs, but need site suitability.

ENE-390 · What is revenue stacking in storage projects?

Combining several value streams from one asset, such as energy arbitrage, frequency regulation, capacity payments, renewable firming and network deferral, while respecting conflicts in SoC and availability commitments.

ENE-391 · Configure a ~10 kWh 48 V pack using 3.2 V, 100 Ah LFP cells.

16 cells in series give 51.2 V nominal. One string stores 51.2 × 100 = 5.12 kWh, so two parallel strings (16s2p) give about 10.24 kWh with 32 cells.

ENE-392 · Why is thermal management critical in large battery systems?

Li-ion performs and ages best at about 15–35 °C. Heat accelerates degradation and raises thermal runaway risk, while cold reduces power and risks lithium plating during charging. Air or liquid cooling must also keep temperature uniform across cells.

ENE-393 · What are the challenges of second-life EV batteries for stationary storage?

Heterogeneous state of health and chemistry, testing and grading cost, uncertain remaining life, mismatched module formats, BMS integration, warranty and liability questions, and competition from falling new-cell prices.

ENE-394 · How much energy does a flywheel with I = 50 kg·m² deliver when slowing from 6000 to 3000 rpm?

ω₁ = 628 rad/s, ω₂ = 314 rad/s. ΔE = ½ × 50 × (628² − 314²) ≈ 25 × 296,000 ≈ 7.4 MJ, about 2.1 kWh. That is 75% of the energy stored at 6000 rpm.

Hydrogen & Fuel Cells

ENE-395 · What is the difference between green, blue and grey hydrogen?

Grey hydrogen comes from natural gas reforming (SMR) with CO₂ vented, about 9–10 kg CO₂ per kg H₂. Blue hydrogen is SMR or ATR with carbon capture. Green hydrogen comes from electrolysis powered by renewables, with near-zero direct emissions.

ENE-396 · What electrolyser capacity is needed for 10 t/day of hydrogen at 52 kWh/kg?

Energy = 10,000 × 52 = 520 MWh/day, or an average of 21.7 MW. If powered by solar with about 25% capacity factor and no storage, the electrolyser would need roughly four times this rating, so hybrid wind-solar supply or storage is used.

ENE-397 · What drives the levelized cost of green hydrogen?

Mainly electricity price and electrolyser utilisation. At 52 kWh/kg, $30/MWh power contributes about $1.56/kg. Electrolyser capex, stack life, efficiency, water treatment, compression and storage add to this. Low capacity factor spreads capex over less hydrogen.

ENE-398 · What is the round-trip efficiency of power-to-hydrogen-to-power?

Electrolysis at about 65–70% (LHV) times fuel cell or turbine conversion at about 40–55% gives roughly 30–35% (lower once compression and storage are included). This is far below batteries, so it suits only long-duration or seasonal storage.

ENE-399 · Size a 100 kW PEM stack at 0.7 V per cell and 1 A/cm² for a 350 V bus.

Cells = 350 ÷ 0.7 = 500. Stack current = 100,000 ÷ 350 ≈ 286 A. Active area per cell = 286 ÷ 1 ≈ 286 cm².

ENE-400 · What energy is needed to liquefy hydrogen?

Current large plants need roughly 10–13 kWh/kg, about a third of hydrogen's LHV (33.3 kWh/kg). Boil-off losses during storage and transport add further penalties.

ENE-401 · How do alkaline and PEM electrolysers differ when coupled to variable renewables?

PEM responds within seconds, operates from low part load and produces high-pressure hydrogen, suiting fluctuating supply. Traditional alkaline units have narrower load range and slower response but lower capex and no platinum-group metals. Newer pressurised alkaline designs are improving flexibility.

ENE-402 · How is hydrogen used to decarbonise steelmaking?

In hydrogen direct reduction, iron ore is reduced with hydrogen: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O. The sponge iron is then melted in an electric arc furnace, replacing coke-based blast furnaces and the CO₂ they emit.

ENE-403 · What are the main PEM fuel cell degradation mechanisms?

Platinum dissolution and agglomeration, carbon support corrosion during start-stop cycles, membrane chemical and mechanical degradation from humidity and load cycling, and contamination by CO, sulphur and ammonia.

ENE-404 · When are fuel cells preferable to batteries for heavy vehicles?

For long-range, high-utilisation duty such as long-haul trucks, buses and some rail, where battery mass and long charging times limit payload and uptime. Batteries win where daily range is moderate, because of higher efficiency and lower energy cost.

Grid Integration & Smart Grids

ENE-405 · What are the main challenges of integrating high shares of renewables?

Variability and forecast uncertainty, reduced inertia and system strength, ramping needs, reverse power flow and voltage rise in distribution, protection challenges from low fault current, transmission congestion, and curtailment.

ENE-406 · What technology layers make up a smart grid?

Sensing (PMUs, smart meters, line sensors), communication networks, control and automation (SCADA, DMS, ADMS, substation automation), distributed energy resource management, demand response platforms, data analytics and cybersecurity.

ENE-407 · Estimate initial RoCoF when 1000 MW trips in a 50,000 MVA system with H = 4 s at 50 Hz.

RoCoF = ΔP × f₀ ÷ (2 H S) = 1000 × 50 ÷ (2 × 4 × 50,000) = 0.125 Hz/s. Lower inertia raises RoCoF, which can trip protection relays.

ENE-408 · What is short-circuit ratio (SCR) and why does it matter for renewables?

SCR = fault level at the connection point ÷ plant MW rating. Low SCR (a weak grid) causes voltage instability and control interactions for grid-following inverters. Remedies include synchronous condensers, STATCOMs and grid-forming inverters.

ENE-409 · How does a PV plant raise feeder voltage, and how is the rise estimated?

Injection reduces or reverses current. ΔV ≈ (P R + Q X) ÷ V. Exporting P raises voltage, so absorbing reactive power (negative Q) offsets it, which is especially effective on feeders with higher X/R ratio.

ENE-410 · What are volt-VAR and volt-watt functions in smart inverters?

Volt-VAR adjusts reactive power based on local voltage, absorbing VARs at high voltage and injecting at low voltage. Volt-watt curtails active power when voltage exceeds set limits. Together they increase feeder hosting capacity.

ENE-411 · A rural feeder with high rooftop PV sees midday overvoltage. What solutions would you propose?

Inverter volt-VAR or volt-watt settings, adjusting transformer taps, line voltage regulators with reverse-flow settings, conductor reinforcement, community batteries, demand shifting (water heating at midday), and dynamic export limits.

ENE-412 · Why do inverter-based resources complicate protection?

Inverters limit fault current to about 1.1–1.5 per unit, much lower than synchronous machines, and sequence components depend on control. Overcurrent relays may fail to detect faults or lose coordination, so adaptive, differential or distance-based schemes are needed.

ENE-413 · What flexibility options support high variable renewable shares?

Flexible thermal and hydro generation, storage, demand response, larger balancing areas and interconnection, better forecasting, sector coupling (EVs, heat pumps, electrolysers), and market redesign with shorter dispatch intervals.

ENE-414 · When is HVDC preferred, and what is the difference between LCC and VSC?

For long distances, submarine cables and asynchronous links. LCC uses thyristors, suits very high power and needs a strong AC grid. VSC uses IGBTs, controls P and Q independently, can supply weak grids and black start, and is preferred for offshore wind.

ENE-415 · How is a microgrid controlled?

Through hierarchical control. Primary droop or grid-forming control shares load locally. Secondary control restores voltage and frequency. Tertiary control (EMS) optimises dispatch and grid exchange. Seamless islanding and resynchronisation are required.

Energy Audit, Conservation, Economics & Policy

ENE-416 · What is LCOE and how is it calculated?

Levelized cost of energy is the constant price that recovers lifetime costs: LCOE = Σ[(I_t + O&M_t + F_t) ÷ (1 + r)^t] ÷ Σ[E_t ÷ (1 + r)^t]. It allows comparison across technologies, but ignores system integration and timing value.

ENE-417 · Estimate LCOE for PV with capex $1000/kW, O&M $15/kW·yr, 20% CF, 8% discount rate and 25 years.

CRF = 0.08 × 1.08²⁵ ÷ (1.08²⁵ − 1) ≈ 0.0937. Annual cost = 93.7 + 15 = $108.7/kW. Energy = 0.2 × 8760 = 1752 kWh/kW. LCOE ≈ $0.062/kWh.

ENE-418 · An efficiency retrofit costs ₹10 lakh and saves ₹3 lakh/year for 5 years. Find NPV at 10%.

Annuity factor = (1 − 1.1⁻⁵) ÷ 0.1 ≈ 3.791. PV of savings = 3 × 3.791 ≈ ₹11.37 lakh. NPV ≈ +₹1.37 lakh, so it is viable, although simple payback is 3.3 years.

ENE-419 · What are the IPMVP measurement and verification options?

Option A: retrofit isolation with key parameter measurement. Option B: retrofit isolation with all parameters measured. Option C: whole-facility metering with utility bills. Option D: calibrated simulation. Choice depends on measure size, interactions and cost.

ENE-420 · How do you normalise an energy baseline?

Regress historical consumption against independent variables such as production volume, heating or cooling degree days and occupancy. Savings are calculated as adjusted baseline minus actual consumption, so weather and output changes are not mistaken for savings.

ENE-421 · What are the annual savings from replacing a 75 kW, 92% efficient motor with a 95.5% efficient motor running 8000 h/year at ₹8/kWh?

Input: 75 ÷ 0.92 = 81.5 kW versus 75 ÷ 0.955 = 78.5 kW. The saving of 3.0 kW × 8000 ≈ 23,900 kWh is worth about ₹1.9 lakh per year.

ENE-422 · What capacitor kVAr is needed to raise PF of a 500 kW load from 0.80 to 0.95?

Q_c = P (tan φ₁ − tan φ₂) = 500 × (0.750 − 0.329) ≈ 211 kVAr. kVA demand falls from 625 to about 526 kVA.

ENE-423 · How does a carbon price of $50/t CO₂ affect coal and gas generation costs?

Coal at about 0.95 kg CO₂/kWh adds about $0.048/kWh. A CCGT at about 0.37 kg/kWh adds about $0.019/kWh. Carbon pricing therefore shifts the merit order towards gas and renewables.

ENE-424 · How is an energy management system structured under ISO 50001?

Using a Plan-Do-Check-Act cycle: energy policy, energy review identifying significant energy uses, baselines and EnPIs, objectives and action plans, operational control, monitoring and measurement, internal audit, management review and continual improvement.

ENE-425 · Compare feed-in tariffs, competitive auctions and net metering as renewable policies.

Feed-in tariffs give fixed long-term prices and investor certainty but risk overpaying. Auctions discover low prices competitively but may favour large players and risk underbidding. Net metering suits rooftop systems but can shift network costs to non-solar consumers.

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