General Aspects of Energy Management & Energy Audit Available here with full solutions — 52 questions recovered from the 2021 exam:
Objective (1 mark)
48 of 50
Short (5 marks)
2 of 8
Long (10 marks)
2 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 48
📖 §9.6 Linear Regression — E = C + M·P
1. A factory has a fixed energy consumption of 2,000 kWh/month and it consumes a total of 38,000 kWh/month for manufacturing 90,000 units of the product. The variable energy consumption in kWh/unit is ____.
0.4
2.4
2.5
none of the above
Answer: A) 0.4
Confirmed vs Book-1 §9.6 — the standard performance line is E = M·P + C, so the variable (slope) term M = (E - C)/P. M = (38,000 - 2,000)/90,000 = 36,000/90,000 = 0.4 kWh/unit. C = 2,000 kWh/month is the fixed/base-load part and is excluded before dividing by output. Answer (a).
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
2. If wet bulb and dry bulb temperatures read the same, the relative humidity is ____.
0%
50%
100%
none of the above
Answer: C) 100%
Confirmed vs Book-1 §3.4 — When WBT = DBT the air is saturated, so relative humidity is 100%. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
📖 §9.6 Plant Energy Performance (PEP) & production factor
4. Which of the following data is not used for calculating Plant Energy Performance?
Reference year energy use
Production factor
Current year energy use
Maximum electrical demand
Answer: D) Maximum electrical demand
Confirmed vs Book-1 §9.6 (M&T normalisation) — PEP% = (Reference-year equivalent energy - Current-year energy)/Reference-year equivalent × 100, where Reference-year equivalent = Reference-year energy × Production factor and Production factor = Current-year output / Reference-year output. Only reference-year energy, production factor and current-year energy are needed; maximum electrical demand (kVA) plays no part. Answer (d).
📖 §4.12 Energy audit instruments — Ultrasonic Flow Meter
5. Non-contact flow measurement can be carried out by ____.
Orifice meter
Turbine flow meter
Ultrasonic flow meter
Magnetic flow meter
Answer: C) Ultrasonic flow meter
Confirmed vs Book-1 §4.12 — Book §4.12: the ultrasonic flow meter is "one of the popular means of non-contact flow measurement" (transit-time or Doppler), clamped on the outside of the pipe. Orifice and turbine meters are intrusive/in-line devices inserted in the fluid stream, and a magnetic flow meter, though obstruction-less, is still a wetted in-line spool piece — so only the ultrasonic meter is non-contact.
📖 §4.9 Maximizing system efficiencies (continuous-improvement practice; term itself not defined in Ch4 text)
6. Which of the following means 'continuous improvement'?
Seiton
Kaizen
Seiso
Kanban
Answer: B) Kaizen
Confirmed vs Book-1 §4.9 — Kaizen is the Japanese term for continuous improvement, i.e. small ongoing improvements in operation and maintenance practice — the spirit of Book §4.9 'best operation and maintenance practices'. Seiton (set in order) and Seiso (shine/clean) are 5-S housekeeping steps, and Kanban is a pull-type production-signalling system, so none of those means continuous improvement.
📖 §5.5 Example 5.5 — concentrations (mole fraction)
7. 54 kg of water is mixed with 0.34 moles of salt to make a solution. The mole fraction of the solution is ____.
0.1
18.36
158.8
none of the above
Answer: A) 0.1
Confirmed vs Book-1 §5.5 Ex.5.5: Book computes moles as (mass)/(mol. wt), mol. wt of water = 18. Moles of water = 54/18 = 3. Mole fraction of salt = moles salt/(total moles) = 0.34/(3 + 0.34) = 0.34/3.34 = 0.102 ≈ 0.1. Hence option (a).
📖 §7.3 Financial Analysis Techniques — Time Value of Money
8. What is the future value of a cash flow at the end of the 6th year, if the Present Value is Rs. 2 Lakhs and the interest rate is 9%?
3,28,540
3,35,420
2,84,980
none of the above
Answer: B) 3,35,420
Confirmed vs Book-1 §7.3 — FV = PV(1+i)^n = 2,00,000 x (1.09)^6. (1.09)^6 = 1.6771, so FV = 2,00,000 x 1.6771 = Rs.3,35,420.
The book's compounding relation is FV = NPV(1+i)^n; options (a) and (c) do not satisfy it at 9% for 6 years.
📖 §3.4 Temperature — Celsius, Fahrenheit and Kelvin scales
9. A temperature of -40 deg F will be ____ deg C?
0
-10
-40
none of the above
Answer: C) -40
Confirmed vs Book-1 §3.4 — -40 deg F equals -40 deg C; the two scales coincide at -40. Book-1 Ch.3, Temperature — Celsius, Fahrenheit and Kelvin scales.
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
10. Unit of maximum demand is ____.
kVAh
kVA
kVAr
kWh
Answer: B) kVA
Confirmed vs Book-1 §3.3 — Maximum demand is billed in kVA (apparent power). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
11. The ISO standard for energy management system is ____.
ISO 9001
ISO 50001
ISO 14000
ISO 14001
Answer: B) ISO 50001
Confirmed vs Book-1 §2 (EnMS — general) — ISO 50001 is the international standard for an Energy Management System (EnMS). ISO 9001 is quality management and ISO 14001/14000 is environmental management, which is why they are the tempting distractors. (The EnMS standard number is not printed in Book-1 Ch2; it is the standard BEE answer.)
13. For the purpose of calculating TOE for a designated consumer the calorific value of oil is taken as
10500 kcal/kg
10000 kcal/kg
5000 kcal/kg
8700 kcal/kg
Answer: B) 10000 kcal/kg
Confirmed vs Book-1 §3.5 — 1 tonne of oil equivalent (toe) is based on a calorific value of 10,000 kcal/kg (10^7 kcal/tonne). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
16. The power indicated in the name plate of a motor denotes ____.
minimum kW drawn by the motor
maximum kW drawn by the motor
maximum kVA drawn by the motor
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.3 — The motor nameplate kW indicates the rated mechanical (shaft) output power, not input/maximum kW or kVA; hence none of the above. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
18. The producer gas is basically ____.
only CH4
only CO and CH4
CO, H2 and CH4
only CO and H2
Answer: C) CO, H2 and CH4
Confirmed vs Book-1 §3.4 — Producer gas is a mixture of CO, H2 and CH4 (plus N2 and CO2). Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
19. Return on investment (ROI) is ____.
initial investment/annual return
annual cost/capital cost
annual net cash flow/capital cost
none of the above
Answer: C) annual net cash flow/capital cost
Confirmed vs Book-1 §7.3 — Book: ROI = (Annual net cash flow / Capital cost) x 100; as a plain fraction it is annual net cash flow / capital cost.
ROI is the inverse of the simple payback period (Example 7.3: 25,000/1,00,000 = 25%, payback = 4 yr).
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
20. The typical efficiency of a solar cell in the field is
12-15%
25-30%
45-50%
80-85%
Answer: A) 12-15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
The book's worked example rates a 175 W panel of 0.75 × 1.50 m = 1.125 m² at 1,000 W/m²: η = (175 / (1.125 × 1000)) × 100 = 15.6%.
The chapter-end key also puts the typical solar-cell efficiency at 10–15%.
Hence 12–15% (option a) is the only field range consistent with the book; 25–30%, 45–50% and 80–85% are far above any commercial PV cell.
📖 §11.4 Solar Electrical Energy / §11.5 Wind Energy (Capacity factor)
21. Capacity utilization factor of a solar PV power plant is in the range of ____.
80-85%
60-65%
18-20%
less than 10%
Answer: C) 18-20%
Confirmed vs Book-1 §11.4 Solar Electrical Energy / §11.5 Wind Energy (Capacity factor) —
The book defines capacity factor CF = kWh produced / (8760 × nameplate kW).
A solar PV plant generates only in daylight: with about 5 peak-sun-hours per day, CF ≈ 5/24 ≈ 0.20, i.e. 18–20%.
So option c. (20–40% in the book is the figure for wind turbines, not solar PV.)
📖 §4.12 Energy audit instruments — Speed Measurements
22. Contact type speed measurement can be carried out by ____.
Tachometer
Stroboscope
Oscilloscope
Odometer
Answer: A) Tachometer
Confirmed vs Book-1 §4.12 — Book §4.12: "a simple tachometer is a contact type instrument, which can be used where direct access is possible." The stroboscope is expressly listed as the more sophisticated and safer NON-contact alternative, so it is the tempting wrong answer here; an oscilloscope displays waveforms and an odometer measures distance travelled.
📖 §3.4 Steam properties — superheat and dryness fraction (x)
23. The 'superheat' of steam is expressed as ____.
degrees centigrade above saturation temperature
degrees centigrade above critical temperature of the steam
degrees centigrade below the boiling point of water
all of the above
Answer: A) degrees centigrade above saturation temperature
Confirmed vs Book-1 §3.4 — Superheat is the temperature of steam above its saturation temperature at a given pressure. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
📖 §3.4 Fuel properties — density, specific gravity, viscosity
25. Which of the following is not true of liquid fuels?
the viscosity of a liquid fuel is a measure of its internal resistance to flow
the viscosity of all liquid fuels decreases with increase in its temperature
higher the viscosity of liquid fuels, higher will be its heating value
viscous fuels need heat tracing
Answer: C) higher the viscosity of liquid fuels, higher will be its heating value
Confirmed vs Book-1 §3.4 — Viscosity has no direct relation to heating value; the statement is false. Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
📖 § Sec 14(h)/(i)/(l) — Energy Manager vs Accredited Energy Auditor
26. Which one of the following is not the duty of an energy manager under EC Act?
Report to BEE and state level designated agency once a year
Prepare an annual activity plan
Conduct energy audit
Prepare a scheme for efficient use of energy
Answer: C) Conduct energy audit
Confirmed vs Book-1 §2.3.6 — Sec 14(h)/(i) requires the energy audit to be got conducted by an ACCREDITED ENERGY AUDITOR — it is not the energy manager's job, so (c) is the odd one out. The energy manager designated under Sec 14(l) is in charge of activities for efficient use of energy: he plans the year's activities, prepares the scheme for efficient use of energy under Sec 14(o) and submits the annual status report on energy consumption to the designated agency.
27. Which one is not an energy consumption benchmark parameter?
kcal/kWh of electricity generated
kg/deg C
kWh/kg of fertilizer
kWh/kg of yarn
Answer: B) kg/deg C
Confirmed vs Book-1 §4.6 — Book §4.6 benchmarks always relate energy to output: kcal/kWh (power-plant heat rate), Million kcal or kWh per MT of fertilizer, kWh/kg of yarn. 'kg/deg C' relates mass to temperature and carries no energy term at all, so it cannot be a specific-energy benchmark.
28. 300 litres of water in a tank is heated from 30 deg C to 70 deg C by using a direct steam with an enthalpy of 600 kcal/kg. The mass in kg of steam used is ____.
10
200
40
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.4 — Heat to water = 300 x 1 x (70-30) = 12,000 kcal; steam mass = 12,000/600 = 20 kg, which is not among a/b/c, so none of the above. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
29. Which of the following is not a unit of energy?
Joule
Calorie
Watt
BTU
Answer: C) Watt
Confirmed vs Book-1 §3.2 — Watt is a unit of power (energy per unit time), not energy. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
30. What is the heat content of the 200 litres of water at 50 deg C in terms of the basic unit of energy in kilo Joules (kJ)?
3000
4187
1000
41870
Answer: D) 41870
Confirmed vs Book-1 §3.4 — Heat = m c dT = 200 x 4.187 x 50 = 41,870 kJ (taking rise from 0 deg C reference). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
📖 §4.12 Energy audit instruments — Non Contact Infrared Thermometer
33. Infrared thermometer is commonly used to measure:
Surface temperature
Flue gas temperature
Steam Temperature
Hot water temperature
Answer: A) Surface temperature
Confirmed vs Book-1 §4.12 — Book §4.12: the IR thermometer computes temperature from the thermal radiation emitted by an object's SURFACE, and is used for objects in hazardous or hard-to-reach places. Flue gas, steam and hot-water temperatures are stream temperatures taken by inserting a contact thermometer (thermocouple) probe into the stream, per the Contact Thermometer entry.
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
34. Power in a 3 phase AC system is
3 x Voltage x Current
Voltage x Current
1.73 x Voltage x Current
None of the above
Answer: D) None of the above
Confirmed vs Book-1 §3.3 — Three-phase active power = sqrt3 x V x I x cos(phi); the listed forms omit power factor, so none of the above. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
35. Which industry among the following is not a designated consumer as per EC Act-2001?
fertilizers
chlor alkali
cement
nuclear power stations
Answer: D) nuclear power stations
Confirmed vs Book-1 §2.3.6 — The Schedule notifies nine energy-intensive industries as designated consumers: Thermal Power Stations, Fertilizer, Cement, Iron & Steel, Chlor-Alkali, Aluminium, Railways, Textile and Pulp & Paper. Nuclear power stations are NOT on that list — 'thermal power stations' (30,000 MTOE/yr) is the look-alike that makes (d) tempting.
📖 § 2.3.2 Standards and Labeling (S&L) — Star Ratings
36. Star rating is a ____ program of BEE
Demand Side Management
Integrated Energy Policy
Standards & Labelling
National Mission for enhanced energy efficiency
Answer: C) Standards & Labelling
Confirmed vs Book-1 §2.3.2 — Star rating is a ranking system (Star 1 = least efficient to Star 5 = most efficient) declared by the manufacturer and is part of BEE's Standards & Labelling programme, which puts energy labels on appliances. DSM manages the demand for power at the utility end and NMEEE is a NAPCC mission — neither issues star labels.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
37. Energy consumption per unit of GDP is called as:
energy elasticity
energy intensity
energy per capita
none of above
Answer: B) energy intensity
Confirmed vs Book-1 §1.11 — Energy intensity is defined as the ratio of gross inland energy consumption to GDP, i.e. energy consumed per unit of GDP (toe per million US$). 'Energy elasticity' is the ratio of % growth in energy demand to % growth in GDP, not consumption per unit GDP, and 'energy per capita' divides energy by population, not GDP.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
38. To maximize the combustion efficiency, it is required to ____ in the flue gas?
maximize O2
maximize CO2
minimize CO2
maximize NOx
Answer: B) maximize CO2
Confirmed vs Book-1 §3.4 — High combustion efficiency corresponds to maximum CO2 (minimum excess air) in flue gas. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
📖 §3.3 Example 3.6 — resistive load power varies as V²
39. An electric heater of 230 V, 10 kW rating is installed for hot water generation in a hospital. The consumption per hour at 200 V is
10 kWh
8.7 kWh
13.23 kWh
7.56 kWh
Answer: D) 7.56 kWh
Confirmed vs Book-1 §3.3 — P proportional to V^2: P = 10 x (200/230)^2 = 10 x 0.756 = 7.56 kW, so 7.56 kWh in one hour. Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
40. A sensitivity analysis is carried out for an energy saving project to make an assessment of
cash flows
risks due to assumptions
capital investment
best financing source
Answer: B) risks due to assumptions
Confirmed vs Book-1 §7.5 — Book, Section 7.5: 'Sensitivity analysis is an assessment of risk.' Cash flows rest on assumptions (capital cost, savings, escalation, project life) that carry uncertainty.
It answers 'what if one or more factors are not as favourable as predicted', i.e. it quantifies the risk in the assumptions.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
41. The specific gravity of water is expressed as ____.
1
1 kg/m3
1 g/cc
1000 kg/m3
Answer: A) 1
Confirmed vs Book-1 §3.4 — Specific gravity is a dimensionless ratio; for water it is 1. Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
42. An activity in a project is having an optimistic time of 8 days, a most likely time of 15 days and a pessimistic time of 16 days. Its expected time of completion is
14 days
13 days
12 days
none of the above
Answer: A) 14 days
Confirmed vs Book-1 §8.3 — PERT expected time T_E = (T_O + 4T_M + T_P)/6. Here (8 + 4×15 + 16)/6 = (8 + 60 + 16)/6 = 84/6 = 14 days.
The most-likely time carries a weight of 4, so the answer is not the plain average (13 days). Option (a).
43. From an activity in a project, latest start time is 8 weeks; latest finish time is 12 weeks. The slack time for the activity is ____.
1 week
5 weeks
4 weeks
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §8.3 — Book-1 defines float only as LS − ES or LF − EF. Here only LS = 8 and LF = 12 are given.
LF − LS = 12 − 8 = 4 weeks is the activity DURATION t (since LS = LF − t), not the slack.
With no ES or EF supplied the float cannot be computed, so (d) none of the above.
44. Which of the following is not a greenhouse gas?
Water Vapour
SO2
CO2
CH4
Answer: B) SO2
Confirmed vs Book-1 §10.5 — The greenhouse gases named in the book are water vapour, CO2, methane, nitrous oxide, ozone, CFCs/HFCs, PFCs and SF6. SO2 is an acid-rain / air-pollution gas (§10.3), not a greenhouse gas.
📖 §11.2 Fundamentals of Solar Energy (Solar Insolation / Solar Window)
45. The period when maximum sunlight is available is called?
Solar constant
Solar insolation
Solar window
Solar irradiance
Answer: C) Solar window
Confirmed vs Book-1 §11.2 Fundamentals of Solar Energy (Solar Insolation / Solar Window) —
The book's margin note reads: ‘Solar Window is the period, typically 9 AM – 3 PM, when maximum sunlight is available.’
Solar constant (1368 W/m²) is a radiation rate at the top of the atmosphere and insolation is the daily energy per m² — neither is a time period.
Answer c.
📖 §11.5 Wind Energy (Power available from the wind turbine)
46. If wind speed increases by three times, energy output from windmill will be ____.
3 times higher
27 times higher
8 times higher
none of the above
Answer: B) 27 times higher
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) —
Book formula: P = 0.5 × ρ × A × Cp × Ng × Nb × V³ — power varies with the CUBE of wind speed.
Tripling the speed gives a factor 3³ = 27.
Answer b (27 times higher). The book states the parallel case: ‘Doubling the wind speed increases the power by eight times.’
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)
48. The number of moles in 90 kg of water is ____.
5
18
2
none of the above
Answer: A) 5
Confirmed vs Book-1 §3.5 — Moles = mass/molar mass = 90,000 g/18 g/mol... in kmol: 90/18 = 5 kmol. Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
1. A furnace shell (4 tonnes) is to be cooled from 95 C to 45 C. The maximum permissible rise in water temperature is 5 C. Compute the quantity of water required. (Cp shell = 0.122 kcal/kg.C, Cp water = 1 kcal/kg.C)
Model answer: Heat to be removed Q = m x Cp x dT = 4000 x 0.122 x (95-45) = 24,400 kcal. For water: Q = m x Cp x dT, so 24,400 = m x 1 x 5, giving m = 24,400/5 = 4,880 kg of water.
Same formula on both sides: Q = m × Cp × ΔT for the shell gives the heat to be removed, and the same Q for water gives the mass required. Convert first: 4 tonnes = 4000 kg. Cp of water = 1 kcal/kg·°C is what makes the water side easy. Common mistake: using the shell's 50 °C drop for the water — the water is only allowed a 5 °C rise, and that is the ΔT you divide by.
📖 §11.4 Solar Electrical Energy (Rooftop SPV sizing numerical)
2. A rooftop of 1200 m² has 20% shading. If 1 kWp SPV needs 10 m² and peak output is 5 hours/day: (a) suggested kWp, (b) daily generation per kWp, (c) kg CO2/year avoided for 250 days at 0.82 kg/kWh.
📖 §11.4 Solar Electrical Energy (Power Towers & Parabolic Trough Collector)
1. Explain the two main types of solar thermal (concentrating) power stations - the power tower and the parabolic trough collector.
Model answer: Solar thermal power stations concentrate sunlight to raise steam and drive a steam turbine. There are two basic types: the power tower and the parabolic trough collector.
Power Tower (central receiver): A large field of sun-tracking mirrors called heliostats concentrates and directs sunlight onto a central receiver mounted on a tall tower. Molten salt from a cold-salt tank is pumped through the receiver, where it is heated to about 566 C. The hot salt is stored in a hot-salt thermal storage tank, then pumped through a steam generator that raises steam; the steam drives a turbine-generator to produce electricity. The cooled salt (about 288 C) returns to the cold-salt tank and is reused. The molten salt is a mixture of 60% sodium nitrate and 40% potassium nitrate - an efficient, low-cost, non-flammable and non-toxic medium for storing thermal energy (which allows generation even when the sun is not shining).
Parabolic Trough Collector: This is currently the most proven solar thermal electric technology. It uses long, parabolic (curved) trough-shaped reflectors that focus sunlight onto a receiver tube running along the focal line of the trough. Because of the parabolic shape the troughs can focus the sun at 30 to 60 times its normal intensity on the receiver pipe. A heat-transfer fluid (such as water) in the receiver is heated to about 400 C. The collectors are aligned on an east-west axis and the troughs rotate to follow the sun so as to maximise the energy captured. Large arrays are coupled together to provide high-temperature fluid that drives a steam turbine, producing many megawatts of electricity - but only where solar insolation is sufficient.
Book-verified (OCR Sec 11.4). Exam favourite. Marks: heliostats -> central receiver -> molten salt 566 C -> storage -> steam turbine; salt = 60% NaNO3 + 40% KNO3; trough focuses 30-60x, HTF ~400 C, east-west axis, most proven technology.
📖 §11.4 Solar Electrical Energy (Stand-alone SPV, Grid-connected Solar and BIPV systems)
2. Explain the difference between stand-alone (off-grid) and grid-connected (on-grid) solar PV systems. What is a building-integrated PV (BIPV) system?
Model answer: Stand-alone (off-grid) SPV power plant: Used where a conventional grid supply is not available or is irregular. Electricity is centrally generated and supplied to users through a local grid in stand-alone mode. Because power must be available when there is no sunlight, these systems use a battery bank (with a charge controller) to store energy. Common uses are electrification of remote villages, hospitals, hotels, communication equipment, railway stations and border outposts.
Grid-connected (on-grid) solar system: Uses an inverter that synchronises with the utility power. These systems do not generally require batteries (though batteries may be added for backup if the utility fails). Grid-connected solar is easier to install and maintain than a stand-alone system because storage is not essential and excess power can be fed to the grid.
Key differences: (i) Storage - stand-alone needs batteries, grid-connected usually does not; (ii) Grid link - stand-alone supplies an isolated local load, grid-connected feeds/draws from the utility through a synchronising inverter; (iii) Application - stand-alone for remote/no-grid areas, grid-connected for locations with a reliable utility; (iv) Cost/maintenance - grid-connected is easier and cheaper to install and maintain.
Building-Integrated PV (BIPV): PV panels are integrated into the roof or facade of a building instead of being separately mounted. BIPV provides photovoltaic power as well as weather-proofing and glazing for the building, generating electricity during the day to meet part of the building's needs. Since the cells are built into the structure, no separate costly mountings are required.
Book-verified (OCR Sec 11.4). Actual Sep-2021 exam question. Marks: batteries yes/no; synchronising inverter for grid-tie; application context; BIPV = PV integrated in roof/facade giving power + weatherproofing, no separate mounts.