BEE Exam Prep › Paper-1 › Chapter 11

BEE Paper-1 — Chapter 11: New & Renewable Energy

149 questions — 76 objective (1 mark), 57 short (5 marks), 16 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation.
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Objective questions (1 mark) — 76

📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

1. The typical efficiency of a solar cell in the field is

  1. 12-15%
  2. 25-30%
  3. 45-50%
  4. 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.
Source: Sep 2021
📖 §11.4 Solar Electrical Energy / §11.5 Wind Energy (Capacity factor)

2. Capacity utilization factor of a solar PV power plant is in the range of ____.

  1. 80-85%
  2. 60-65%
  3. 18-20%
  4. 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.)
Source: Sep 2021
📖 §11.1 (cross-reference: Book-1 Ch.3 — energy units, 1 toe = 10⁷ kcal)

3. What is the 'TOE' of 125 Ton of coal which has GCV of 4000 kcal/kg

  1. 40
  2. 50
  3. 400
  4. 500
Answer: B) 50
Confirmed vs Book-1 §11.1 (cross-reference: Book-1 Ch.3 — energy units, 1 toe = 10⁷ kcal) — Heat content = 125 t × 1000 kg/t × 4000 kcal/kg = 5 × 10⁸ kcal. 1 toe = 10⁷ kcal, so TOE = 5 × 10⁸ / 10⁷ = 50 toe. Answer b.
Source: Sep 2021
📖 §11.2 Fundamentals of Solar Energy (Solar Insolation / Solar Window)

4. The period when maximum sunlight is available is called?

  1. Solar constant
  2. Solar insolation
  3. Solar window
  4. 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.
Source: Sep 2021
📖 §11.5 Wind Energy (Power available from the wind turbine)

5. If wind speed increases by three times, energy output from windmill will be ____.

  1. 3 times higher
  2. 27 times higher
  3. 8 times higher
  4. 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.’
Source: Sep 2021
📖 §11.1 Concept of New and Renewable Energy (Concept of renewable energy)

6. Which of the following is a renewable energy source?

  1. bitumen
  2. bagasse
  3. Diesel oil
  4. natural gas
Answer: B) bagasse
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (Concept of renewable energy) — The book lists renewable sources as sun, wind, falling water, sea waves, geothermal heat and biomass; solid biomass explicitly includes bagasse (§11.6, Direct Combustion of Biomass). Bitumen, diesel oil and natural gas are all fossil (stock) fuels. Answer b.
Source: Apr 2010
📖 §11.1 Concept of New and Renewable Energy / §11.5 Wind Energy (Wind energy conversion)

7. A person can do the following with wind energy

  1. destroy it
  2. convert it
  3. create it
  4. burn it
Answer: B) convert it
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy / §11.5 Wind Energy (Wind energy conversion) — The book calls wind machines ‘wind energy conversion systems (WECS)’ — the turbine converts the kinetic energy of moving air into mechanical and then electrical energy. Energy can neither be created nor destroyed, and wind is not a combustible fuel. Answer b (convert it).
Source: Nov 2009
📖 §11.3 Solar Thermal Energy (Evacuated Tube Collector)

8. Which of the following statements regarding evacuated tube collectors (ETC) are true? i) ETC can reach high temperatures upto 150°C ii) Because of the vacuum between the two concentric glass tubes, a higher amount of heat is retained in the ETC iii) Heat loss due to conduction back to the atmosphere from the ETC is high iv) Performance of the evacuated tube is highly dependent upon the ambient temperature

  1. i & iii
  2. ii & iii
  3. i & iv
  4. i & ii
Answer: D) i & ii
Confirmed vs Book-1 §11.3 Solar Thermal Energy (Evacuated Tube Collector) — Book: ETC ‘can reach high temperatures upto 150°C’ (statement i true) and the vacuum between the two concentric glass tubes traps more heat than a flat plate collector (statement ii true). Statement iii is false — ‘since conduction cannot take place in vacuum, heat loss due to conduction back to atmosphere is also prevented’ (heat loss <10% vs 40% for FPC). Statement iv is false — the ETC ‘is less dependent upon ambient temperature unlike flat plate collector’. Hence i & ii, answer d.
Source: 2019
📖 §11.7 Hydro Power (Water into Watts)

9. How much power you would expect to generate from a river-based mini hydropower with flow of 40 litres/second, head of 12 metres and system efficiency of 55%.

  1. 872 kW
  2. 2.59 KW
  3. 264 kW
  4. none of the above
Answer: B) 2.59 KW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) — Book formula: P (kW) = 9.81 × Q × H × η, with Q in m³/s and H in m. Q = 40 l/s = 0.040 m³/s, H = 12 m, η = 0.55. P = 9.81 × 0.040 × 12 × 0.55 = 2.59 kW. Answer b.
Source: 2019
📖 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.2 — EC Act 2001, BEE schemes)

10. Which of the following is one of the schemes of BEE under Energy Conservation Act ?

  1. Standards and Labelling
  2. Availability based Tariff
  3. Standard of Performance of DISCOMs
  4. Renewable Energy Certificates
Answer: A) Standards and Labelling
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.2 — EC Act 2001, BEE schemes) — Standards & Labelling is one of the thrust-area schemes of BEE under the Energy Conservation Act, 2001. Availability Based Tariff and DISCOM standards of performance are CERC/regulatory instruments and RECs come under the electricity/RE regulatory framework, not BEE schemes under the EC Act. Answer a.
Source: 2019
📖 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process)

11. Bio-gas generated through anaerobic process mainly consists of

  1. only methane
  2. Methane and carbon dioxide
  3. only ethane
  4. only carbon dioxide
Answer: B) Methane and carbon dioxide
Confirmed vs Book-1 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process) — Book: bio-methane produced by anaerobic digestion ‘is composed mainly of methane and carbon dioxide’; gobar gas is ‘typically comprising of around 60% methane and 40% carbon dioxide’. It is therefore not pure methane, not ethane and not pure CO₂. Answer b.
Source: 2019
📖 §11.1 Concept of New and Renewable Energy (Concept of renewable energy)

12. Which among the following is not a renewable source of energy?

  1. Bagasse
  2. Rice husk
  3. Nuclear
  4. Wind
Answer: C) Nuclear
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (Concept of renewable energy) — The book lists renewables as wind, solar, geothermal, tidal, bio-energy and hydro, and notes that fossil AND nuclear fuels are ‘stocks of energy’, not flows. Bagasse and rice husk are solid biomass (§11.6) and wind is a flow. Nuclear is therefore the non-renewable one — answer c.
Source: 2018
📖 §11.5 Wind Energy (Yaw Control)

13. The rotor axis is aligned with the wind direction in a wind mill by ________ control

  1. yaw
  2. pitch
  3. disc break
  4. all of the above
Answer: A) yaw
Confirmed vs Book-1 §11.5 Wind Energy (Yaw Control) — Book: ‘It is necessary for rotor axis to be aligned with wind direction … sensors activate the yaw control motor, which rotates the nacelle and rotor assembly until turbine is properly aligned.’ Pitch control changes blade angle to regulate power; the disc brake only slows the rotor. Answer a.
Source: 2018
📖 §11.6 Biomass Energy (Gasification of Biomass)

14. Producer gas basically comprises of

  1. CO, H2 and CH4
  2. Only CH4
  3. CO and CH4
  4. Only CO and H2
Answer: A) CO, H2 and CH4
Confirmed vs Book-1 §11.6 Biomass Energy (Gasification of Biomass) — Book: ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’. Typical producer-gas composition given is CO = 19±3%, H₂ = 18±2%, CH₄ = 3±1%, CO₂ = 10±3%, N₂ = 50±2%. Answer a (CO, H₂ and CH₄).
Source: 2018
📖 §11.4 Solar Electrical Energy (Power Towers)

15. In a solar thermal power station Molten salt is preferred as it provides an efficient low cost medium to store ______ energy

  1. Electrical
  2. Thermal
  3. Kinetic
  4. Potential
Answer: B) Thermal
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Power Towers) — Book: ‘Molten salt is a mixture of 60% sodium nitrate and 40% potassium nitrate. It is preferred as it provides an efficient low-cost medium to store thermal energy’. Salt is heated to 566°C in the central receiver and returns at 288°C — a sensible-heat (thermal) store. Answer b.
Source: 2018
📖 §11.5 Wind Energy (Cut-out Speed / Furling Speed)

16. Furling speed of wind turbine indicates ____

  1. Cut out speed
  2. Cut in speed
  3. Rated speed
  4. None of the above
Answer: A) Cut out speed
Confirmed vs Book-1 §11.5 Wind Energy (Cut-out Speed / Furling Speed) — Book: ‘The wind speed at which shut down occurs is called the cut-out speed. Cut-out speed is also known as furling speed’ (about 20–30 m/s). Cut-in speed (~5 m/s) is where useful power starts; rated speed is where rated power is first reached. Answer a.
Source: 2018
📖 §11.8 Fuel Cell (Fuel Cell)

17. The input to a fuel cell is.

  1. Electricity
  2. Hydrogen
  3. Oxygen
  4. All of the above
Answer: B) Hydrogen
Confirmed vs Book-1 §11.8 Fuel Cell (Fuel Cell) — Book opens §11.8 with: ‘Input to a Fuel Cell is hydrogen. Hydrogen combines with oxygen to produce electricity … with water and heat as by-products.’ Oxygen is the oxidant at the cathode, not the fuel input; electricity is the output. Answer b.
Source: 2018
📖 §11.3 Solar Thermal Energy (Solar Flat Plate Collector)

18. Which of the following type of collector is used for low temperature systems?

  1. Flat plate collector
  2. Line focusing parabolic collector
  3. Parabolic trough collector
  4. None of the above
Answer: A) Flat plate collector
Confirmed vs Book-1 §11.3 Solar Thermal Energy (Solar Flat Plate Collector) — Book: ‘Flat-plate collectors heat the circulating fluid to a temperature of about 40–60°C’ — a low-temperature system. Line-focusing / parabolic trough concentrators reach about 400°C and are used for high-temperature power generation. Answer a.
Source: 2018
📖 §11.1 Concept of New and Renewable Energy (Concept of New and Renewable Energy)

19. Energy sources which are inexhaustible are known as

  1. commercial energy
  2. primary energy
  3. renewable energy
  4. secondary energy
Answer: C) renewable energy
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (Concept of New and Renewable Energy) — Book: ‘Renewable energy is energy obtained from sources that are essentially inexhaustible such as sun and wind … Renewable energy is also known as non-conventional energy.’ Commercial/primary/secondary are classifications by trade and by conversion stage, not by inexhaustibility. Answer c.
Source: 2017
📖 §11.7 Hydro Power (Water into Watts)

20. The power generation potential in mini hydro power plant for a water flow of 3 m3/sec with a head of 14 meters and with a system efficiency of 55% is

  1. 226.6 kW
  2. 76.4 kW
  3. 23.1 kW
  4. none of the above
Answer: A) 226.6 kW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) — P (kW) = 9.81 × Q × H × η = 9.81 × 3 × 14 × 0.55. 9.81 × 3 = 29.43; × 14 = 412.02; × 0.55 = 226.6 kW. Answer a.
Source: 2017
📖 §11.6 Biomass Energy (Gasification of Biomass)

21. The producer gas basically consists of

  1. Only CH4
  2. CO & CH4
  3. CO, H2 & CH4
  4. Only CO & H2
Answer: C) CO, H2 & CH4
Confirmed vs Book-1 §11.6 Biomass Energy (Gasification of Biomass) — Book: partial combustion yields ‘Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’; the methanation reaction C + 2H₂ = CH₄ supplies the CH₄. So producer gas is not only CH₄, nor only CO + H₂. Answer c.
Source: 2017
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell)

22. The energy conversion efficiency of a solar cell does not depend on

  1. solar energy insolation
  2. inverter
  3. area of the solar cell
  4. maximum power output
Answer: B) inverter
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell) — Book formula: η = (Pm / (E × A)) × 100, where Pm = maximum power output (W), E = insolation (W/m²) and A = cell area (m²). Only these three quantities appear, so cell efficiency is independent of the inverter (a downstream balance-of-system component). Answer b.
Source: 2017
📖 §11.4 Solar Electrical Energy (Building-integrated PV) — with Book-1 Ch.10/energy-efficient buildings

23. Which of the following enhances the energy efficiency in buildings?

  1. Light pipes
  2. Triple glaze windows
  3. Building integrated solar photovoltaic panels
  4. All of the above
Answer: D) All of the above
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Building-integrated PV) — with Book-1 Ch.10/energy-efficient buildings — The book describes BIPV panels integrated into the roof or façade, generating daytime electricity and also providing weather-proofing and glazing. Light pipes (daylighting) and triple-glazed windows likewise cut building energy use. All three enhance building energy efficiency — answer d.
Source: 2016
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

24. A solar _______ is connected and packaged in a solar _________, which in turn is linked with others in sequence in a solar _________.

  1. module, cell, array
  2. array, module, sequence
  3. module, array, sequence
  4. cell, module, array
Answer: D) cell, module, array
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — Book: ‘Solar cells are connected in series and parallel combinations to form modules … Modules can be connected together to form an array’ (36 cells × 0.5 V per cell = one module). The order is therefore cell → module → array. Answer d.
Source: 2016
📖 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.1 — installed generating capacity)

25. Which amongst the following sources of electricity has the highest installed capacity in India ?

  1. Gas
  2. Nuclear
  3. Oil
  4. Renewables
Answer: D) Renewables
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.1 — installed generating capacity) — Of the four options listed (gas, nuclear, oil, renewables), renewables carry by far the largest installed capacity in India — nuclear and oil-based capacity are only a few GW each. (Coal, which is the largest single source overall, is not among the options.) Answer d.
Source: 2016
📖 §11.4 Solar Electrical Energy (Power Towers)

26. In a solar thermal power station , molten salt which is a mixture of 60% sodium nitrate and 40% potassium nitrate is used. It is preferred as it provides an efficient low cost medium to store _______

  1. Electrical energy
  2. Thermal energy
  3. Kinetic energy
  4. Potential energy
Answer: B) Thermal energy
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Power Towers) — Book: molten salt = 60% sodium nitrate + 40% potassium nitrate, ‘preferred as it provides an efficient low-cost medium to store thermal energy and it is non-flamable and nontoxic’. It is charged at 566°C and returned at 288°C via the hot- and cold-salt tanks. Answer b.
Source: 2016
📖 §11.6 Biomass Energy (Gasification of Biomass)

27. The producer gas is basically

  1. CO, H2 and CH4
  2. Only CH4
  3. CO and CH4
  4. Only CO and H2
Answer: A) CO, H2 and CH4
Confirmed vs Book-1 §11.6 Biomass Energy (Gasification of Biomass) — Book: producer gas contains CO, H₂ and traces of CH₄ (typical CO 19±3%, H₂ 18±2%, CH₄ 3±1%, balance CO₂ and N₂). Its calorific value is 1000–1200 kcal/Nm³. Answer a.
Source: 2016
📖 §11.5 Wind Energy (Power available from the wind turbine)

28. What is the expected power output in watts from a wind turbine with 6m diameter rotor, a coefficient of performance 0.45, generator efficiency 0.8,a gear box efficiency 0.90 and wind speed of 11m/sec

  1. 4875 watts
  2. 1100 watts
  3. 7312 watts
  4. 73.12 kW
Answer: C) 7312 watts
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) — P = 0.5 × ρ × A × Cp × Ng × Nb × V³, with ρ = 1.2 kg/m³ and A = (π/4) × 6² = 28.27 m². P = 0.5 × 1.2 × 28.27 × 0.45 × 0.8 × 0.90 × 11³ = 7,312 W. (The book's own example with Cp = 0.30 gives 4,875 W; scaling 4875 × 0.45/0.30 = 7,312 W.) Answer c.
Source: 2016
📖 §11.7 Hydro Power (Water into Watts)

29. How much power generation potential is available in a run of river mini hydropower plant for a flow of 40 liters/second with a head of 24 metres. Assume system efficiency of 60% ?

  1. 5.6 kW
  2. 2.4 kW
  3. 4.0 kW
  4. 2.8 kW
Answer: A) 5.6 kW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) — P (kW) = 9.81 × Q × H × η with Q = 40 l/s = 0.040 m³/s, H = 24 m, η = 0.60. P = 9.81 × 0.040 × 24 × 0.60 = 5.65 kW ≈ 5.6 kW. Answer a.
Source: 2012
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

30. What percentage of the sun's energy can silicon solar panels convert into electricity?

  1. 30%
  2. 15%
  3. 75%
  4. 50%
Answer: B) 15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — The book's worked PV example gives η = (175 / (1.125 × 1000)) × 100 = 15.6%, and the chapter-end key gives typical cell efficiency as 10–15%. So a silicon panel converts roughly 15% of incident solar energy into electricity. Answer b.
Source: Jul 2022
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

31. What is the average conversion efficiency of a solar photo voltaic cell?

  1. 22%
  2. 15%
  3. 98%
  4. 50%
Answer: B) 15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — Book worked example: a 175 W panel of area 1.125 m² at 1000 W/m² → η = (175/(1.125 × 1000)) × 100 = 15.6%. The chapter-end key states typical solar cell efficiency 10–15%. Answer b (15%).
Source: 2013
📖 §11.7 Hydro Power (Water into Watts)

32. The power generation potential in mini hydro power plant for a water flow of 3 m3/sec with a head of 14 meters with system efficiency of 55% is

  1. 226.6 kW
  2. 76.4 kW
  3. 23.1 kW
  4. none of the above
Answer: A) 226.6 kW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) — P (kW) = 9.81 × Q × H × η = 9.81 × 3 m³/s × 14 m × 0.55. = 29.43 × 14 × 0.55 = 226.6 kW. Answer a.
Source: 2013
📖 §11.5 Wind Energy (Power available from the wind turbine)

33. If the wind speed doubles, energy output from a wind turbine will be:

  1. 2 times higher
  2. 4 times higher
  3. 6 times higher
  4. 8 times higher
Answer: D) 8 times higher
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) — Book: ‘Doubling the wind speed increases the power by eight times, but doubling the turbine area only doubles the power.’ This follows from P ∝ V³: 2³ = 8. Answer d.
Source: 2013
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

34. What percentage of the sun’s energy falling on a silicon solar panel gets converted into electricity?

  1. 25%
  2. 15%
  3. 75%
  4. 50%
Answer: B) 15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — Book worked example: η = (Pm/(E × A)) × 100 = (175/(1.125 × 1000)) × 100 = 15.6%; chapter-end key gives 10–15% typical. About 15% of the sun's energy falling on a silicon panel becomes electricity. Answer b.
Source: 2012
📖 §11.8 Fuel Cell (Fuel Cell)

35. A fuel cell is

  1. an electromagnetic cell
  2. a magnetic cell
  3. an electrochemical device
  4. none of the above
Answer: C) an electrochemical device
Confirmed vs Book-1 §11.8 Fuel Cell (Fuel Cell) — Book: ‘Hydrogen combines with oxygen to produce electricity through an electrochemical process … (not combustion process)’. A fuel cell has two catalyst-coated electrodes (anode, cathode) separated by an electrolyte — an electrochemical device. Answer c.
Source: 2012
📖 §11.6 Biomass Energy (Biofuels from Biomass) — chapter-end objective Q.1

36. Which among the following raw material is used for biodiesel production?

  1. leaves
  2. corn
  3. animal and vegetable fat
  4. coal
Answer: C) animal and vegetable fat
Confirmed vs Book-1 §11.6 Biomass Energy (Biofuels from Biomass) — chapter-end objective Q.1 — Book: ‘All oils extracted from plant origin, waste cooking oil and animal fat can be used as raw materials for biodiesel production’, the route being transesterification with methanol. Corn/molasses feed ethanol fermentation, not biodiesel; leaves and coal are not oil sources. Answer c.
Source: Guidebook
📖 §11.1 Concept of New and Renewable Energy (Concept of New and Renewable Energy) — chapter-end objective Q.2

37. A non-renewable energy resource:

  1. can be plugged in and recharged
  2. will eventually run out
  3. will not work for new appliances
  4. can be used over again
Answer: B) will eventually run out
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (Concept of New and Renewable Energy) — chapter-end objective Q.2 — Book: renewable sources are ‘essentially inexhaustible’ and are ‘flows of energy unlike the fossil and nuclear fuels which are considered stocks of energy’. A stock is finite, so a non-renewable resource will eventually run out. Answer b.
Source: Guidebook
📖 §11.5 Wind Energy (Cut-in Speed) — chapter-end objective Q.3

38. The rotor blades start to rotate

  1. at cut-in speed
  2. cut-out speed
  3. rated speed
  4. furling speed
Answer: A) at cut-in speed
Confirmed vs Book-1 §11.5 Wind Energy (Cut-in Speed) — chapter-end objective Q.3 — Book: ‘There is a minimum speed at which a wind turbine can reliably produce useable power. This is known as the cut-in speed which is generally around 5 m/s.’ Rated speed is where rated power is reached; cut-out (furling) speed is where the machine shuts down. Answer a. (The printed option ‘e) furling speed’ is a typographical error for d.)
Source: Guidebook
📖 §11.5 Wind Energy (Power available from the wind turbine) — chapter-end objective Q.4

39. If wind speed doubles, energy output from wind turbine will be

  1. 2 times higher
  2. 4 times higher
  3. 6 times higher
  4. 8 times higher
Answer: D) 8 times higher
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) — chapter-end objective Q.4 — Book: P = 0.5 × ρ × A × Cp × Ng × Nb × V³ and ‘Doubling the wind speed increases the power by eight times’. 2³ = 8. Answer d.
Source: Guidebook
📖 §11.2 Fundamentals of Solar Energy (Solar Constant) — chapter-end objective Q.5

40. Solar constant is

  1. 342 W/m^2
  2. 1368 W/m^2
  3. 1342 W/m^2
  4. 1380 W/m^2
Answer: B) 1368 W/m^2
Confirmed vs Book-1 §11.2 Fundamentals of Solar Energy (Solar Constant) — chapter-end objective Q.5 — Book: ‘The solar constant actually varies by about 0.3% over the 11-year solar cycle but averages about 1,368 W/m².’ 342 W/m² is the average solar insolation, i.e. one-fourth of the solar constant (1368/4 = 342). Answer b — use the guidebook value 1368 W/m² in the exam.
Source: Guidebook
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — chapter-end objective Q.6

41. A solar cell is made up of

  1. silicon
  2. titanium
  3. magnesium
  4. teflon
Answer: A) silicon
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — chapter-end objective Q.6 — Book: photons ‘knock the electrons in the silicon material out of their normal energy state’ and ‘this phenomenon is exploited to form the individual solar cells on wafers of silicon’. Other PV materials listed are GaAs, CIS and CdTe, but silicon is the common one; titanium, magnesium and teflon are not PV materials. Answer a.
Source: Guidebook
📖 §11.12 Geothermal Energy (Geothermal Energy) — chapter-end objective Q.7

42. The molten material mixed with gases in the mantle of the earth is called

  1. core
  2. lava
  3. geyser
  4. magma
Answer: D) magma
Confirmed vs Book-1 §11.12 Geothermal Energy (Geothermal Energy) — chapter-end objective Q.7 — Book: ‘The top layer of the mantle is a hot liquid rock called magma … When magma breaks through the surface of the earth in a volcano, it is called lava.’ Core is the innermost earth layer and a geyser is a surface hot-water jet. Answer d.
Source: Guidebook
📖 §11.5 Wind Energy (How Wind is Created?) — chapter-end objective Q.8

43. What is wind? Which statement is correct?

  1. wind is the product of temperature differences between atmosphere and stratosphere
  2. wind is the product of air pressure differences in the atmosphere, due mainly to solar radiation disparities
  3. wind is the movement of air caused by beating wings of birds
  4. wind is air movement caused only by earth rotation
Answer: B) wind is the product of air pressure differences in the atmosphere, due mainly to solar radiation disparities
Confirmed vs Book-1 §11.5 Wind Energy (How Wind is Created?) — chapter-end objective Q.8 — Book: the sun heats the equator more than the poles; ‘In warmer regions … the air is hot and is therefore at a high pressure, compared to colder regions, where the air is at a low pressure. Wind is the movement of air from area's of high pressure to low pressure.’ The Coriolis force from the earth's rotation only deflects this flow — it does not create it. Answer b.
Source: Guidebook
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — chapter-end objective Q.9

44. Typical efficiency of solar cell is

  1. 10-15%
  2. 25 %
  3. 50%
  4. 75%
Answer: A) 10-15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — chapter-end objective Q.9 — The book's worked PV example gives η = (175/(1.125 × 1000)) × 100 = 15.6% for a good commercial panel; typical cell efficiency is quoted as about 10–15%. 25%, 50% and 75% are far beyond commercial silicon cells. Answer a.
Source: Guidebook
📖 §11.12 Geothermal Energy (Binary Cycle Power Plant)

45. In ____, hot water from a geo-thermal well flows to a heat exchanger where the hot water is used to heat a working fluid with low boiling temperature.

  1. Flash steam power
  2. Binary cycle power plant
  3. Dry steam power plants
  4. None of the above
Answer: B) Binary cycle power plant
Confirmed vs Book-1 §11.12 Geothermal Energy (Binary Cycle Power Plant) — Book: ‘Binary cycle pumps hot water from well to a heat exchanger where hot water is used to heat a working fluid, usually organic compound with low boiling point.’ It operates on 107–182°C waters. Dry steam takes steam directly to the turbine; flash steam flashes hot water (182°C) to steam — neither uses a secondary fluid. Answer b.
Source: Mar 2023
📖 §11.5 Wind Energy (Power available from the wind turbine)

46. Upon doubling the length of a wind turbine blade, its power generation:

  1. No Change in power generation
  2. Gets increased by two times
  3. Gets increased by four times
  4. Gets increased by Eight times
Answer: C) Gets increased by four times
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) — P = 0.5 × ρ × A × Cp × Ng × Nb × V³, and the swept area A = πD²/4 = πL² for blade length (radius) L. Doubling the blade length quadruples A, so power rises 2² = 4 times. (The book's rule: ‘doubling the turbine area only doubles the power’ — here the area itself becomes 4×.) Answer c.
Source: Mar 2023
📖 §11.5 Wind Energy (Cut-out Speed / Furling Speed)

47. Speed of wind at which a wind turbine shuts down automatically so as to avoid damage is known as ____.

  1. Betz Constant
  2. Cut-in wind speed
  3. Cut-off wind speed
  4. Rated wind speed
Answer: C) Cut-off wind speed
Confirmed vs Book-1 §11.5 Wind Energy (Cut-out Speed / Furling Speed) — Book: ‘Above a certain speed beyond the rated speed, the wind turbine will need to shut down and stop operation to prevent damage to the unit … called the cut-out speed’ (about 20–30 m/s). Betz limit (59%) is an efficiency ceiling, cut-in (~5 m/s) is start-up and rated speed is where rated power is first met. Answer c.
Source: Mar 2023
📖 §11.7 Hydro Power (Water into Watts)

48. Power derived from the flowing water is ____.

  1. Directly proportional to flow rate & inversely proportional to its head
  2. Directly proportional to both its flow rate as well as its head
  3. Inversely proportional to flow rate & directly proportional to its head
  4. Inversely proportional to both its flow rate as well as its head
Answer: B) Directly proportional to both its flow rate as well as its head
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) — Book: Theoretical power P = Flow rate (Q) × Head (H) × Gravity (g), i.e. P = 9.81 × Q × H kW. Both Q and H appear in the numerator, so power is directly proportional to each. Answer b.
Source: Mar 2023
📖 §11.8 Fuel Cell (Table 11.3 Types of Fuel Cells)

49. For a Proton-exchange membrane fuel cell, choose the correct match:

  1. Anode- Methanol; Cathode - Oxygen
  2. Anode- Hydrogen; Cathode - Oxygen
  3. Anode- Synthetic Gas; Cathode - Oxygen
  4. None of the above
Answer: B) Anode- Hydrogen; Cathode - Oxygen
Confirmed vs Book-1 §11.8 Fuel Cell (Table 11.3 Types of Fuel Cells) — Table 11.3 row 1 (PEMFC): anode = Hydrogen, cathode = Oxygen, electrolyte = water-based acidic polymer membrane. Methanol at the anode is the DMFC (row 2); synthesis gas at the anode is SOFC/MCFC (rows 5–6). Answer b.
Source: Mar 2023
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

50. Monocrystalline and polycrystalline are types of ____.

  1. Geothermal heat pumps
  2. Electrical vehicle battery cell
  3. Solar PV panels
  4. None of above
Answer: C) Solar PV panels
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — Monocrystalline and polycrystalline describe how the silicon wafer is grown for photovoltaic cells — the book notes cells are formed ‘on wafers of silicon’. They are not geothermal or battery terms. Answer c.
Source: Mar 2023
📖 §11.9 Energy from Wastes / §11.10 Wave Energy / §11.12 Geothermal Energy / §11.5 Wind Energy

51. Which of the following statements are true for renewable energy? I) Methane gas produced in landfill sites, escapes into air and is a source of greenhouse gas emission. II) Magma is a solid core in earth layer and is used to produce hot water. III) Energy production from ocean waves is steady and predictable compared to wind and solar energy. IV) Wattage output of wind turbine is varies with cube of wind velocity (Wv).

  1. I & IV
  2. II & IV
  3. III & IV
  4. I & III
Answer: A) I & IV
Confirmed vs Book-1 §11.9 Energy from Wastes / §11.10 Wave Energy / §11.12 Geothermal Energy / §11.5 Wind Energy — I is true — §11.9: ‘The methane gas produced in landfill sites normally escapes into the atmosphere and contributes to greenhouse gas emissions.’ II is false — §11.12: magma is ‘a hot liquid rock’ in the mantle, not a solid core. IV is true — §11.5: P ∝ V³, ‘doubling the wind speed increases the power by eight times’. Hence I & IV, answer a. (Caution: §11.10 also calls wave power ‘much steadier and more predictable’, so statement III is arguably true as well; the official key nevertheless marks a.)
Source: Mar 2023
📖 §11.5 Wind Energy (Betz Limit)

52. Which among the following statement is correct about wind energy?

  1. We can convert 100% of wind energy to electricity
  2. Wind turbine extracts energy by increasing wind speed
  3. Theoretically wind turbine can convert 59% of wind energy to electricity.
  4. If wind speed doubles power output of wind turbine increases by 100%
Answer: C) Theoretically wind turbine can convert 59% of wind energy to electricity.
Confirmed vs Book-1 §11.5 Wind Energy (Betz Limit) — Book: ‘The theoretical maximum amount of energy in the wind that can be collected by a wind turbines rotor is approximately 59%. This value is known as the Betz limit’ (59.3%). A turbine cannot be 100% efficient, it extracts energy by SLOWING the wind, and doubling wind speed raises power 8 times (800%), not 100%. Answer c.
Source: Mar 2023
📖 §11.6 Biomass Energy (Biofuels from Biomass)

53. Which of the following is used for Bio-Diesel production?

  1. Jatropha
  2. Light Diesel Oil
  3. High Speed Diesel
  4. Shale Oil
Answer: A) Jatropha
Confirmed vs Book-1 §11.6 Biomass Energy (Biofuels from Biomass) — Book: ‘The most economical way of producing biodiesel is by transesterification of extracted oil (e.g. Jatropha seeds oil) with alcohol such as methanol. Jatropha is a non edible tree-borne oilseed’. LDO, HSD and shale oil are petroleum products, not biodiesel feedstocks. Answer a.
Source: Mar 2023
📖 §11.6 Biomass Energy (Gasification of Biomass)

54. The producer gas consists of ____.

  1. CO
  2. H2
  3. CH4
  4. All of the Above
Answer: D) All of the Above
Confirmed vs Book-1 §11.6 Biomass Energy (Gasification of Biomass) — Book: partial combustion products are ‘combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’; typical composition CO 19±3%, H₂ 18±2%, CH₄ 3±1%. All three species are therefore present. Answer d (All of the above).
Source: Jul 2022
📖 §11.7 Hydro Power (Table 11.2 Classification of Hydropower by Size)

55. Micro hydro will generate ____.

  1. less than 10 kW
  2. 11kW up to 100 kW
  3. 101 kW to 2 MW
  4. None of the above
Answer: B) 11kW up to 100 kW
Confirmed vs Book-1 §11.7 Hydro Power (Table 11.2 Classification of Hydropower by Size) — Table 11.2: Pico-hydro up to 10 kW; Micro-hydro ‘From 11 kW up to 100 kW’; Mini-hydro 101 kW to 2 MW; Small-hydro 2001 kW–25 MW; Large-hydro >25 MW. So micro-hydro = 11 kW to 100 kW. Answer b.
Source: Jul 2022
📖 §11.6 Biomass Energy (Gasification of Biomass)

56. Producer gas consists of:

  1. CO, H₂, CH₄
  2. CO, CH₄
  3. CO, H₂
  4. Only CH₄
Answer: A) CO, H₂, CH₄
Corrected (was c) — Book-1 §11.6 Biomass Energy (Gasification of Biomass): Book: ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’, and the chapter-end key to objective Q.10 is ‘CO, H₂ and CH₄’. The methanation reaction C + 2H₂ = CH₄ in the reduction zone supplies the methane, and Typical Producer Gas Composition lists CH₄ = 3 ± 1%. So producer gas is CO + H₂ + CH₄ — option a, not ‘CO, H₂’ only.
Source: Sep 2025
📖 §11.2 Fundamentals of Solar Energy (Fundamentals of Solar Energy)

57. Solar radiation consists of:

  1. X-rays, Gamma rays, and Microwaves
  2. Ultra-violet, Visible, and Infra-red radiation
  3. Visible, Infra-red, and Radio waves
  4. Ultra-violet, X-rays, and Cosmic rays
Answer: B) Ultra-violet, Visible, and Infra-red radiation
Confirmed vs Book-1 §11.2 Fundamentals of Solar Energy (Fundamentals of Solar Energy) — Book opens §11.2: ‘Solar radiation is radiant energy emitted by the sun comprising of ultra-violet, visible and infra-red radiation.’ X-rays, gamma rays, microwaves and radio waves are not the constituents named by the book. Answer b.
Source: Sep 2025
📖 §11.6 Biomass Energy (Average conversion efficiency of a gasifier)

58. Biomass gasifier using 1 kg wood (4,000 kCal/kg) producing 2 m³ gas (1,000 kCal/m³). What would be the efficiency?

  1. 25%
  2. 50%
  3. 75%
  4. 100%
Answer: B) 50%
Confirmed vs Book-1 §11.6 Biomass Energy (Average conversion efficiency of a gasifier) — Book formula: ηgas = (calorific value of gas per kg of fuel) / (avg. calorific value of 1 kg of fuel). Gas energy = 2 m³/kg × 1000 kcal/m³ = 2000 kcal; fuel energy = 4000 kcal/kg. η = 2000/4000 = 50%. Answer b. (Compare the book's solved example: 46,000/64,000 = 71.88%.)
Source: Sep 2025
📖 §11.5 Wind Energy (Yaw Control)

59. The roto axis is aligned with wind direction in windmill by ____________ control?

  1. Yaw
  2. Pitch
  3. Disc Break
  4. Both A and B
Answer: A) Yaw
Confirmed vs Book-1 §11.5 Wind Energy (Yaw Control) — Book: yaw control aligns the rotor axis with the wind direction — ‘sensors activate the yaw control motor, which rotates the nacelle and rotor assembly until turbine is properly aligned’. Pitch adjusts blade angle for power regulation; the disc brake only slows the rotor. Answer a.
Source: Sep 2024
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

60. One Silicon cell in PV modules typically produces

  1. 0.5 V
  2. 1.0 V
  3. 1.5 V
  4. 2.0 V
Answer: A) 0.5 V
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — Book: ‘One silicon cell generally produces 0.5 Volts. 36 such cells connected together are called a PV module and it has enough voltage to charge 12 V battery’. Check: 36 × 0.5 = 18 V, adequate to charge a 12 V battery. Answer a.
Source: Sep 2024
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell)

61. The energy conversion efficiency of solar cell does not depend on

  1. Solar Energy Insolation
  2. Inverter
  3. Area of the Solar Cell
  4. Maximum Power Output
Answer: B) Inverter
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell) — Book formula: η = (Pm / (E × A)) × 100 — only maximum power output Pm, insolation E and cell area A appear. The inverter is a downstream balance-of-system component and does not enter the cell efficiency. Answer b.
Source: Sep 2024
📖 §11.5 Wind Energy (Power available from the wind turbine)

62. If wind speed triples, the energy output from wind turbine will be

  1. 3 Times
  2. 6 Times
  3. 9 Times
  4. None of the above
Answer: D) None of the above
Corrected (was c) — Book-1 §11.5 Wind Energy (Power available from the wind turbine): Book: P = 0.5 × ρ × A × Cp × Ng × Nb × V³ — power varies as the CUBE of wind speed, and ‘doubling the wind speed increases the power by eight times’ (2³). Tripling the speed therefore gives 3³ = 27 times, NOT 9 times (9 would be a square law, which is wrong). Since 27 times does not appear among options a–c, the book-consistent answer is d) None of the above.
Source: Sep 2024
📖 §11.8 Fuel Cell (Operation of Fuel Cell)

63. In a fuel cell, ___ combines with ____ to generate electricity and ____ comes out as a by-product.

  1. Hydrogen, Oxygen, water
  2. Hydrogen, Nitrogen, nitrous oxide
  3. Carbon, hydrogen, methane
  4. Carbon, oxygen, carbon dioxide
Answer: A) Hydrogen, Oxygen, water
Confirmed vs Book-1 §11.8 Fuel Cell (Operation of Fuel Cell) — Book: ‘Hydrogen combines with oxygen to produce electricity through an electrochemical process with water and heat as by-products.’ Hydrogen enters the anode, oxygen the cathode; protons cross the electrolyte and recombine with electrons and oxygen to form water. Answer a.
Source: Sep 2024
📖 §11.4 Solar Electrical Energy — Solar Photovoltaic Technology

64. A single silicon solar PV cell generally produces a voltage of approximately:

  1. 0.5 V
  2. 1.5 V
  3. 5 V
  4. 12 V
Answer: A) 0.5 V
Confirmed vs Book-1 §11.4 — ‘One silicon cell generally produces 0.5 Volts.’ 36 such cells in series form a PV module with enough voltage to charge a 12 V battery and run a pump/motor; modules combine into arrays. Question reworded to the book’s own phrasing (it says the cell ‘produces 0.5 V’; it never uses the term open-circuit voltage). Answer letter unchanged.
Source: AI practice
📖 §11.4 Solar Electrical Energy — Solar Photovoltaic Technology (PV module)

65. Approximately how many silicon solar cells are connected in series in a standard module to charge a 12 V battery?

  1. 12 cells
  2. 24 cells
  3. 36 cells
  4. 100 cells
Answer: C) 36 cells
Confirmed vs Book-1 §11.4 — ‘36 such cells connected together are called a PV module and it has enough voltage to charge 12 V battery and run pump and motor.’ Check: 36 × 0.5 V = 18 V, comfortably above 12 V. Module output is rated in peak Watt (Wp), the maximum power under standard test conditions; a single module is made from 5 Wp up to 120 Wp.
Source: AI practice
📖 §11.4 Solar PV — energy conversion efficiency of a PV cell

66. A 200 Wp solar PV panel of area 1.4 m^2 receives an insolation of 1000 W/m^2. Its conversion efficiency is approximately:

  1. 7.0%
  2. 14.3%
  3. 20.0%
  4. 28.6%
Answer: B) 14.3%
Confirmed vs Book-1 §11.4 — η = [Pₘ / (E × A)] × 100, where Pₘ = maximum power output (W), E = insolation (W/m²), A = cell area (m²). Working: η = 200 / (1000 × 1.4) × 100 = 200/1400 × 100 = 14.3%. Same method as the book’s worked example (175 W panel, 0.75 × 1.50 m, 1000 W/m² → 15.6%). Chapter-end Q.9: typical solar cell efficiency is 10–15%.
Source: AI practice
📖 §11.2 Fundamentals of Solar Energy — solar insolation / India’s solar potential

67. India receives solar energy of about 5 to 7 kWh/m² for 300 to 330 days in a year. As per the book, this energy is sufficient to set up a solar power plant of what capacity per square kilometre of land area?

  1. 2 MW
  2. 10 MW
  3. 20 MW
  4. 200 MW
Answer: C) 20 MW
Rewritten (the original asked a ‘peak-sun-hours’ calculation, a term the 2014 guidebook never uses) — Book-1 §11.2: ‘India receives solar energy in the region of 5 to 7 kWh/m² for 300 to 330 days in a year. This energy is sufficient to set up 20 MW solar power plant per square kilometre land area.’ Learn the companion figures from the same section: solar constant = 1368 W/m² (the book’s value — write 1368, not 1367) and average insolation = one-fourth of the solar constant = 342 W/m². Insolation is quoted in kWh/m²/day.
Source: AI practice
📖 §11.5 Wind Energy — Power available from a wind turbine

68. The power available in the wind is proportional to which power of the wind speed? Consequently, if wind speed doubles, the available power increases by:

  1. Proportional to v; power doubles (×2)
  2. Proportional to v^2; power increases ×4
  3. Proportional to v^3; power increases ×8
  4. Proportional to v^3; power increases ×6
Answer: C) Proportional to v^3; power increases ×8
Confirmed vs Book-1 §11.5 — ‘Power extracted by wind turbine is proportional to the cross sectional area… and cube of the wind speed’, P = 0.5 × ρ × A × Cp × Ng × Nb × V³. Working: P ∝ V³, so doubling V gives 2³ = 8 times the power (the book states ‘Doubling the wind speed increases the power by eight times’); tripling gives 3³ = 27 times. Doubling the turbine AREA only doubles the power. This is chapter-end objective Q.4, and it is why siting turbines in the highest-wind areas matters most.
Source: AI practice
📖 §11.5 Wind Energy — Power available from a wind turbine (P = ½·ρ·A·V³)

69. For an ideal wind turbine with rotor diameter 10 m, in air of density 1.2 kg/m^3 and wind speed 12 m/s, the ideal power available (P = ½·ρ·A·v^3) is approximately:

  1. 8.1 kW
  2. 40.7 kW
  3. 81.4 kW
  4. 163 kW
Answer: C) 81.4 kW
Confirmed vs Book-1 §11.5 — from KE = ½·mass·V² with mass = ρ·A·V, the power in the wind is P = ½·ρ·A·V³. Working: A = πD²/4 = π × 10²/4 = 78.54 m²; P = 0.5 × 1.2 × 78.54 × 12³ = 0.5 × 1.2 × 78.54 × 1728 ≈ 81,433 W ≈ 81.4 kW. This is the IDEAL (wind) power. Real output multiplies by Cp (0.33–0.59, Betz ceiling ~59%), generator efficiency Ng and gearbox efficiency Nb — as in the book’s worked example (6 m rotor, Cp 0.30, Ng 0.8, Nb 0.90, 11 m/s → 4875 W).
Source: AI practice
📖 §11.5 Wind Energy — Betz limit

70. The Betz limit states that the maximum fraction of wind kinetic energy a turbine can theoretically convert to mechanical energy is about:

  1. 25%
  2. 33%
  3. 59%
  4. 100%
Answer: C) 59%
Confirmed vs Book-1 §11.5 — ‘The theoretical maximum amount of energy in the wind that can be collected by a wind turbines rotor is approximately 59%. This value is known as the Betz limit.’ A rotor can never be 100% efficient because some air must keep moving to pass through the blades. Related figures: coefficient of performance Cp varies between 0.33 and 0.59; after gearbox and generator losses only about 15–25% of the wind energy becomes useful power.
Source: AI practice
📖 §11.5 Wind Energy — Components of a wind turbine system (Yaw control)

71. In a horizontal-axis wind turbine, which control system rotates the entire nacelle so the rotor axis aligns with the wind direction?

  1. Pitch control
  2. Yaw control
  3. Disc brake
  4. Anemometer
Answer: B) Yaw control
Confirmed vs Book-1 §11.5 — ‘Large wind turbines with upwind rotors require yaw control to align the rotor with the wind… sensors activate the yaw control motor, which rotates the nacelle and rotor assembly until turbine is properly aligned.’ Do not confuse with pitch (blade angle), the disc brake (on the main or high-speed shaft, to slow the rotor) or the anemometer/wind vane (they only sense wind speed and direction).
Source: AI practice
📖 §11.5 Wind Energy — Operating characteristics of a wind turbine

72. The wind speed at which a turbine first starts to produce usable power is called the:

  1. Rated speed
  2. Cut-out (furling) speed
  3. Cut-in speed
  4. Betz speed
Answer: C) Cut-in speed
Confirmed vs Book-1 §11.5 — ‘There is a minimum speed at which a wind turbine can reliably produce useable power. This is known as the cut-in speed which is generally around 5 m/s.’ Chapter-end Q.3 asks the same thing (‘the rotor blades start to rotate… at cut-in speed’). Rated speed = the speed giving rated (maximum) power, often 1.5 × site mean wind speed; cut-out / furling speed = 20–30 m/s, where the turbine shuts down to avoid damage. ‘Betz speed’ does not exist. (Option repaired: the ‘≈ 5 m/s’ hint printed inside the correct option gave the answer away; the value is now in the explanation.)
Source: AI practice
📖 §11.6 Biomass Energy — Gasification of biomass (producer gas)

73. Producer gas obtained from biomass gasification consists mainly of which combustible gases?

  1. CO2 + N2 + O2
  2. CO + H2 + traces of CH4
  3. SO2 + NOx + H2O
  4. Pure methane (CH4) only
Answer: B) CO + H2 + traces of CH4
Confirmed vs Book-1 §11.6 — ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄) and non useful products like tar and dust.’ Chapter-end Q.10 answers ‘CO, H₂ and CH₄’. Typical composition: CO 19±3%, H₂ 18±2%, CH₄ 3±1%, CO₂ 10±3%, N₂ 50±2%. Calorific value 1000–1200 kcal/Nm³; gasifier conversion efficiency 60–70%; gasification is partial combustion with sub-stoichiometric air at about 1000 °C.
Source: AI practice
📖 §11.6 Biomass Energy — Biofuels from biomass (biodiesel)

74. Biodiesel is produced by the transesterification of plant oils. Which non-edible crop, suitable for dry/arid wasteland, is a major raw material in India?

  1. Sugarcane
  2. Jatropha
  3. Wheat
  4. Rice
Answer: B) Jatropha
Confirmed vs Book-1 §11.6 — ‘The most economical way of producing biodiesel is by transesterification of extracted oil (e.g. Jatropha seeds oil) with alcohol such as methanol. Jatropha is a non edible tree-borne oilseed which grows in dry and arid land.’ Contrast ethanol, which is made by FERMENTATION of molasses (a sugar by-product), sugar beet, sweet corn or ligno-cellulosic material. Chapter-end Q.1: animal and vegetable fat are also biodiesel raw materials.
Source: AI practice
📖 §11.7 Hydro Power — Water into Watts

75. A micro-hydro scheme has a flow of 0.5 m^3/s and a net head of 10 m. Taking overall efficiency as 60% (P = 9.81 × Q × H × η), the power output is approximately:

  1. 2.94 kW
  2. 29.4 kW
  3. 49 kW
  4. 58.9 kW
Answer: B) 29.4 kW
Confirmed vs Book-1 §11.7 — theoretical power P = 9.81 × Q × H (kW) with Q in m³/s and H in m; multiply by the overall efficiency for the delivered power. Working: P = 9.81 × 0.5 × 10 × 0.6 = 29.43 ≈ 29.4 kW. (Without η the theoretical figure would be 49 kW — distractor c.) Same method as the book’s examples: 9.81 × 0.3 × 10 × 0.5 = 14.7 kW, and 9.81 × (20/1000) × 12 × 0.6 = 1.4 kW. Book note: small turbines rarely exceed 80% efficiency, and ~50% is a rough guide for a few-kW system.
Source: AI practice
📖 §11.6 Biomass Energy — Gasification of biomass (chapter-end objective Q.10)

76. The producer gas is basically

  1. only CH₄
  2. CO and CH₄
  3. CO, H₂ and CH₄
  4. only CO and H₂
Answer: C) CO, H₂ and CH₄
Confirmed vs Book-1 §11.6 Biomass Energy (Gasification of Biomass) — Book: ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄) and non useful products like tar and dust.’ Typical Producer Gas Composition: CO = 19 ± 3%, H₂ = 18 ± 2%, CH₄ = 3 ± 1%, CO₂ = 10 ± 3%, N₂ = 50 ± 2%; CH₄ comes from the methanation reaction C + 2H₂ = CH₄. So producer gas is CO + H₂ + CH₄ — answer c. Its CV is 1000–1200 kcal/Nm³ and gasifier conversion efficiency is 60–70%.
Source: Book EOC

Short questions (5 marks) — 57

📖 §11.1 Concept of New and Renewable Energy

1. Define renewable energy and state how it differs from conventional (fossil/nuclear) energy.

Model answer: Renewable energy is energy obtained from sources that are essentially inexhaustible such as the sun and wind. Examples include solar, wind, geothermal, tidal, bio-energy and hydropower. A renewable energy system converts the energy in sunlight, wind, falling water, sea waves, geothermal heat or biomass into useful heat or electricity, generally without releasing harmful pollutants. It is also called non-conventional energy. The key difference: renewable sources are flows of energy, whereas fossil and nuclear fuels are considered stocks of energy.
Flows vs stocks is the textbook distinction; renewable = non-conventional.
Source: 2014
📖 §11.2 Fundamentals of Solar Energy (Solar Constant)

2. What is the solar constant? Give its value.

Model answer: The solar constant is the rate at which solar energy, at all wavelengths, is received per unit area at the top of the Earth's atmosphere (on a surface normal to the sun). Its value averages about 1,368 W/m². It actually varies by about 0.3% over the 11-year solar cycle. Each planet has its own planetary solar constant.
Objective Q5 answer = 1368 W/m². Do not confuse with average insolation 342 W/m².
Source: 2014
📖 §11.2 Fundamentals of Solar Energy (Solar Insolation)

3. What is solar insolation? In what units is it expressed and what is its average value?

Model answer: Solar insolation is the amount of solar energy that strikes a square metre of the Earth's surface in a single day. It is greatest when the surface is normal to the sun; as the angle increases it is reduced in proportion to the cosine of the angle. The average incoming radiation (solar insolation) is one-fourth of the solar constant, i.e. about 342 W/m². For site assessment it is expressed in kWh/m²/day. India receives 5 to 7 kWh/m² for 300 to 330 days a year.
Avg insolation = ¼ × solar constant = 342 W/m²; field units = kWh/m²/day.
Source: 2014
📖 §11.2 Fundamentals of Solar Energy (Solar Window / India's solar potential)

4. What is meant by the 'solar window', and what is India's solar power potential per square kilometre?

Model answer: The solar window is the period, typically 9 AM to 3 PM, when the maximum sunlight is available. India receives solar energy in the region of 5 to 7 kWh/m² for 300 to 330 days a year. This level of insolation is sufficient to set up a 20 MW solar power plant per square kilometre of land area.
Solar window 9 AM–3 PM; land-scale rule ≈ 20 MW per km².
Source: 2014
📖 §11.2 Fundamentals of Solar Energy (Fundamentals of Solar Energy)

5. Name the components of solar radiation and the two routes by which solar energy can be used.

Model answer: Solar radiation is the radiant energy emitted by the sun, comprising ultra-violet, visible and infra-red radiation. The amount reaching a location depends on geographic location, time of day, season, landscape and local weather. Solar energy can be used through two routes: (1) Solar Thermal Energy, where the sun's heat is collected and converted into heat energy, and (2) Solar Electric (Solar Photovoltaic) Energy, where sunlight is converted directly into electricity.
Radiation = UV + Visible + Infra-red; routes = thermal & PV.
Source: 2014
📖 §11.3 Solar Thermal Energy (Solar Water Heating System)

6. Describe a solar water heating system and its main components.

Model answer: A solar water heating system consists of a flat-plate or evacuated-tube solar collector, a storage tank and connecting pipes. It is generally installed on a roof or open ground with the collector facing the sun and connected to a continuous water supply. The collector absorbs the sun's energy and transfers it to the water. Because the storage tank is insulated and heat losses are small, the water stored remains hot overnight.
Components: collector + insulated storage tank + connecting pipes.
Source: 2014
📖 §11.3 Solar Thermal Energy (Solar Flat Plate Collector)

7. Describe the construction and operating temperature of a solar flat-plate collector (FPC).

Model answer: The flat-plate collector (FPC) is the most common solar collector. It heats the circulating fluid to about 40-60°C. It usually comprises copper tubes welded to copper sheets (both coated with a highly absorbing black coating), with a toughened glass sheet on top as cover and insulating material at the bottom, the entire assembly placed in a flat box. Its performance is highly dependent on ambient temperature, giving good efficiency only when ambient temperature is high; hence heat output is higher in summer than in winter.
FPC: 40-60°C, copper-on-copper black absorber, glass cover, ambient-dependent.
Source: 2014
📖 §11.3 Solar Thermal Energy (Evacuated Tube Collector) — chapter-end short question S-5

8. Why is an evacuated tube collector (ETC) more efficient than a flat-plate collector?

Model answer: An evacuated tube collector uses two concentric glass tubes fused at the ends, with the air evacuated from the gap, providing thermal insulation like a Thermos bottle. The vacuum stops conductive heat loss back to the atmosphere, and the selective absorbing coating on the inner tube converts short-wave radiation to long-wave radiation, preventing re-radiation. Because of this, far more heat is trapped: heat loss is less than 10% compared with about 40% for a flat-plate collector. It is also less dependent on ambient temperature and can reach high temperatures up to 150°C.
Directly answers OCR short question S-5. ETC loss <10% vs FPC ~40%; reaches 150°C.
Source: 2014
📖 §11.4 Solar Electrical Energy (Power Towers)

9. Explain the working of a solar power tower (central receiver) plant.

Model answer: In a power tower, sunlight is concentrated and directed from a large field of heliostats (mirrors) onto a central receiver on a tall tower. Molten salt from the cold salt tank is pumped through the receiver where it is heated to 566°C, then stored in the hot salt thermal storage tank. The hot molten salt is pumped through a steam generator that creates steam, which drives a steam turbine to generate electricity. The cooled salt at 288°C flows back to the cold salt tank and is reused. The molten salt is a mixture of 60% sodium nitrate and 40% potassium nitrate, preferred because it is an efficient, low-cost, non-flammable and non-toxic heat-storage medium.
Heliostats → central receiver → molten salt (566°C) → steam → turbine. Salt = 60% NaNO3 + 40% KNO3.
Source: 2014
📖 §11.4 Solar Electrical Energy (Parabolic Trough Collector)

10. Describe the parabolic trough collector for solar thermal power generation.

Model answer: The parabolic trough collector is currently the most proven solar thermal electric technology. It uses a series of parabolic, trough-shaped reflectors that focus the sun's energy onto a receiver tube running along the focus of the reflector. Because of their parabolic shape, the troughs can focus the sun at 30-60 times its normal intensity on the receiver pipe, heating the heat-transfer fluid in the receiver to about 400°C. Large arrays provide high-temperature fluid to drive a steam turbine. The collectors are aligned on an east-west axis and the troughs rotate to follow the sun, maximising energy input.
Most proven; 30-60× concentration; ~400°C; east-west axis, sun-tracking.
Source: 2014
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

11. Explain the photovoltaic effect and the working of a solar PV cell.

Model answer: The photoelectric or photovoltaic effect is the process in which two dissimilar materials in close contact produce an electrical voltage when struck by light or radiant energy. Discrete packets of light energy called photons strike the PV cell and knock electrons in the silicon material out of their normal energy state, putting them in a position to be conducted as electricity. The effect occurs only when a photon of the correct energy strikes an atom in the cell, so cells are tuned to absorb the most intense part of the spectrum. Since silicon is naturally reflective, each cell is covered with an anti-reflective coating to minimise reflection loss.
Photons knock electrons free in silicon → DC current; anti-reflective coating needed.
Source: 2014
📖 §11.4 Solar Electrical Energy (PV cell, module and array)

12. Distinguish between a solar PV (SPV) cell, module and array, and state the voltage of a silicon cell.

Model answer: A solar PV cell is the basic unit; one silicon cell generally produces about 0.5 Volts. Cells are connected in series and parallel to form a module (solar panel): 36 such cells connected together form a module with enough voltage to charge a 12 V battery and run a pump and motor. Modules connected together form an array to generate more power. A complete PV system comprises PV panels (modules), a battery system, a charge controller and an inverter. PV cells are made of silicon (Si), gallium arsenide (GaAs), copper indium diselenide (CIS), cadmium telluride (CdTe), etc.
Cell ≈ 0.5 V; 36 cells = module → 12 V battery; modules → array. Material = silicon (Objective Q6).
Source: 2014
📖 §11.4 Solar Electrical Energy (Peak Watt, Wp rating)

13. What is the 'peak Watt' (Wp) rating of a PV module?

Model answer: The wattage output of a PV module is rated in terms of peak Watt (Wp). The peak-Watt output power of a module is defined as the maximum power output that the module could deliver under standard test conditions (STC). A single PV module can be manufactured with capacity ranging from 5 Wp to 120 Wp.
Wp = max output at Standard Test Conditions; single module 5-120 Wp.
Source: 2014
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — chapter-end short question S-1

14. Why is solar cell efficiency very low?

Model answer: Solar cell efficiency is low because the cell cannot convert all the different wavelengths of light hitting it. The photovoltaic effect occurs only when a photon has the exact amount of energy needed to knock an electron loose; that energy requirement depends on the cell material. Photons with too little energy are not absorbed, and the excess energy of higher-energy photons is wasted as heat. Reflection losses also occur (silicon is naturally reflective), which is why an anti-reflective coating is used. As a result, typical solar cell efficiency is only about 10-15%.
Answers OCR short question S-1; ties to Objective Q9 (efficiency 10-15%).
Source: 2014
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell)

15. Give the formula for the energy conversion efficiency of a solar PV cell, defining each term with units.

Model answer: Energy conversion efficiency η (%) = [ Pm / (E × A) ] × 100, where Pm = maximum power output (watts), E = solar insolation (watts per square metre), and A = area of the solar cell (square metres). For example, a 175 W panel measuring 0.75 × 1.50 m (area = 1.125 m²) at an insolation of 1000 W/m² gives η = (175 / (1.125 × 1000)) × 100 = 15.6%, i.e. it converts 15.6% of the available solar energy into electricity.
η = Pm/(E·A)·100. Worked example from guidebook = 15.6%.
Source: 2014
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency — worked numerical)

16. A 375 W solar panel of size 1.20 m × 1.50 m is installed on a rooftop of 10 m × 15 m. Find the panel conversion efficiency if solar insolation is 1000 W/m².

Model answer: Maximum power output Pm = 375 W; insolation E = 1000 W/m²; panel (cell) area A = 1.20 × 1.50 = 1.8 m². Conversion efficiency η = (Pm / (E × A)) × 100 = (375 / (1000 × 1.8)) × 100 = 20.83%. (The rooftop dimension 10 × 15 m is extra data not needed for panel efficiency.)
Verified exam numerical; uses panel area not roof area.
Source: Jul 2022 Exam
📖 §11.4 Solar Electrical Energy (Rooftop SPV sizing numerical)

17. 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.

Model answer: (a) Usable area = 1200 × (1 − 0.20) = 960 m²; capacity = 960 / 10 = 96 kWp. (b) Daily generation = 5 peak-sun-hours × 1 kWp = 5 kWh/day per kWp. (c) Annual generation = 96 × 5 × 250 = 1,20,000 kWh; CO2 avoided = 1,20,000 × 0.82 = 98,400 kg CO2/year.
Verified past-exam numerical. Rule: 1 kWp ≈ 10 m² shadow-free area.
Source: Sep 2021 Paper-1
📖 §11.4 Solar Electrical Energy (Stand-alone SPV vs Grid-connected Solar System)

18. Differentiate between a stand-alone SPV power plant and a grid-connected solar system.

Model answer: A stand-alone SPV power plant is used where conventional grid supply is unavailable or irregular; electricity is centrally generated and supplied through a local grid in stand-alone mode, commonly for electrifying remote villages, hospitals, hotels, communication equipment, railway stations and border outposts (it requires batteries). A grid-connected solar system uses an inverter that synchronises with the utility power; it does not generally require batteries (though batteries can give backup), and is easier to install and maintain than a stand-alone system.
Stand-alone = no grid + battery; grid-connected = inverter synchronised, usually no battery.
Source: 2014
📖 §11.4 Solar Electrical Energy (Building-integrated PV Systems)

19. What is a Building-Integrated Photovoltaic (BIPV) system and what are its advantages?

Model answer: In a building-integrated photovoltaic (BIPV) system, PV panels are integrated into the roof or façade of a building. BIPV provides photovoltaic power as well as weatherproofing and glazing for the building. The SPV panels generate electricity during the daytime to meet part of the building's electrical needs. Because the PV cells are integrated into the building structure, no separate costly mountings are required.
BIPV = PV built into roof/façade; gives power + weatherproofing, no separate mountings.
Source: 2014
📖 §11.5 Wind Energy (How Wind is Created?)

20. How is wind created? (What is wind?)

Model answer: The sun heats the earth unevenly, with the majority of heat received at the equator and gradually less towards the poles. In warmer regions the air is hot and at high pressure compared with colder regions where air is at low pressure. Wind is the movement of air from areas of high pressure to areas of low pressure. The rotation of the earth adds the Coriolis force, a swirling action on the winds, creating a series of wind circulations in both hemispheres. Thus wind is produced mainly by air-pressure differences in the atmosphere arising from solar radiation disparities.
Objective Q8 answer = wind from air-pressure differences due to solar radiation disparities; Coriolis = swirl from earth's rotation.
Source: 2014
📖 §11.5 Wind Energy (Description of Wind Energy Technology)

21. What is a wind turbine (WECS)? Describe the common rotor configurations.

Model answer: Modern windmills are called wind turbines as their function is similar to gas and steam turbines; they are also called wind energy conversion systems (WECS), and those generating electricity are called wind generators. The most common type is the horizontal-axis machine, with the main rotor shaft and generator at the top of a tower, which must be pointed into the wind. Rotors can be single-, two-, three- or multi-bladed; two- and three-bladed rotors are common for power generation (three-bladed rotors run more smoothly and quietly). Multi-bladed rotors have large starting torque in light winds and are used for water pumping and low-frequency mechanical power.
WECS = Wind Energy Conversion System; HAWT most common; multi-blade = water pumping.
Source: 2014
📖 §11.5 Wind Energy (Components of wind turbine system)

22. List and briefly describe the main components of a wind turbine system.

Model answer: (1) Rotor and blades — capture the wind. (2) Low-speed shaft — main shaft connected to the rotor hub, turning at 30-60 rpm. (3) Gear box — steps the speed up from 30-60 rpm to 1000-1800 rpm. (4) High-speed shaft — driven via the gearbox at 1000-1800 rpm, drives the generator. (5) Generator — usually an induction generator producing 50-cycle AC. (6) Nacelle — housing at the top of the tower containing the gearbox, generator and controls. (7) Disc brake — slows the rotor. (8) Yaw control — rotates the nacelle so the rotor axis aligns with wind direction. (9) Anemometer and wind vane — measure wind speed and direction.
Directly answers OCR long question L-2 in short form. Yaw = nacelle alignment.
Source: 2014
📖 §11.5 Wind Energy (Yaw Control)

23. What is the purpose of yaw control in a wind turbine?

Model answer: Yaw control aligns the rotor axis with the wind direction so as to extract as much of the wind's kinetic energy as possible. Large wind turbines with upwind rotors require yaw control. When the wind direction changes, sensors activate the yaw control motor, which rotates the nacelle and rotor assembly until the turbine is properly aligned with the wind.
Yaw = whole nacelle rotation to face wind (distinct from pitch = blade angle).
Source: 2014
📖 §11.5 Wind Energy (Operating Characteristics of Wind Turbine)

24. Define cut-in speed, rated speed and cut-out (furling) speed of a wind turbine.

Model answer: Cut-in speed is the minimum wind speed at which a turbine can reliably produce usable power, generally around 5 m/s. Rated speed is the minimum wind speed at which the turbine generates its maximum (rated) power; it is often about 1.5 times the site mean wind speed. Cut-out speed (also called furling speed) is the speed above the rated speed at which the turbine must shut down to prevent damage; it varies by manufacturer from about 20 to 30 m/s.
Cut-in ≈5 m/s; rated ≈1.5× site mean; cut-out/furling 20-30 m/s. Objective Q3: rotor starts at cut-in speed.
Source: 2014
📖 §11.5 Wind Energy (Betz Limit) — chapter-end short question S-4

25. Explain the term Betz limit.

Model answer: It is impossible for the blades of a wind turbine to be 100% efficient, because some of the wind energy must pass through the blades to make the turbine turn (the turbine extracts energy by slowing the wind). The theoretical maximum amount of energy in the wind that can be collected by a wind turbine's rotor is approximately 59%. This value is known as the Betz limit. Considering the Betz limit together with efficiency losses through the generator, gearbox, etc., only about 15-25% of the wind energy is converted into useful power.
Answers OCR short question S-4. Betz limit ≈59%; usable after losses 15-25%.
Source: 2014
📖 §11.5 Wind Energy (Rotor Efficiency, Cp)

26. What is the coefficient of performance (Cp) of wind turbine blades and its typical range?

Model answer: The ability of a turbine rotor to extract the wind's power depends on its efficiency, expressed by a non-dimensional coefficient of performance of the blades, Cp. Cp is included in the power equation to express the turbine's power output. It varies with speed and generally lies between 0.33 and 0.59 (the upper bound corresponding to the Betz limit).
Cp = 0.33–0.59; appears in the wind power formula.
Source: 2014
📖 §11.5 Wind Energy (Power available from the wind turbine)

27. Give the formula for the power available from a wind turbine, defining each term with units.

Model answer: The power produced by a wind turbine is P = 0.5 × ρ × A × Cp × Ng × Nb × V³, where P = power produced by the generator (watts); ρ = air density (kg/m³, about 1.2); A = cross-sectional/swept area intercepted by the turbine (m², = πD²/4); Cp = coefficient of performance of the blades; Ng = generator efficiency; Nb = gearbox efficiency; and V = wind speed (m/s). The ideal form (without losses) is P = ½ρAV³. Power is proportional to the swept area and to the cube of wind speed, so doubling the wind speed increases power eight-fold, while doubling the area only doubles the power.
P ∝ V³ → doubling speed = ×8 (Objective Q4). Includes term definitions with units.
Source: 2014
📖 §11.5 Wind Energy (Power available from the wind turbine — worked example)

28. A wind turbine has a 6 m diameter rotor, Cp = 0.30, generator efficiency 0.8, gearbox efficiency 0.90 and wind speed 11 m/s (ρ = 1.2 kg/m³). Find the expected power output.

Model answer: Swept area A = πD²/4 = (3.14/4) × 6² = 28.27 m². P = 0.5 × ρ × A × Cp × Ng × Nb × V³ = 0.5 × 1.2 × 28.27 × 0.30 × 0.8 × 0.90 × 11³. This gives P ≈ 4875 watts, i.e. about 4.875 kW.
Worked guidebook example: P ≈ 4.875 kW.
Source: 2014
📖 §11.5 Wind Energy (Capacity Factor) — chapter-end short question S-2

29. What is the capacity factor of a wind turbine? Give its formula and typical range.

Model answer: The capacity factor of a wind turbine is the actual energy output of the turbine over a given period (usually one year) compared with its theoretical maximum energy output for the same period. CF = kWh produced / (8760 × nameplate rating of the wind turbine in kW), where 8760 is the number of hours in a year. Typical capacity factors are 20-40%, with values at the upper end at particularly favourable sites. Example: a 2.5 MW turbine producing 5,000,000 kWh/yr gives CF = 5,000,000 / (2500 × 8760) = 22.8%.
Answers OCR short question S-2. CF = actual / (8760 × kW rated); typical 20-40%.
Source: 2014
📖 §11.5 Wind Energy (Table 11.1 Wind Speed vs Power Generation Suitability)

30. Give the guideline relating average wind speed to suitability for power generation.

Model answer: As per the guidebook table: up to 4 m/s (15 km/h) is no good; 5 m/s (18 km/h) is poor; 6 m/s (22 km/h) is moderate; 7 m/s (25 km/h) is good; and 8 m/s (29 km/h) is excellent for power generation. Sites with higher wind speed generate more power, so siting turbines in the highest-wind-speed areas gives significant economic benefit.
≤4 no good · 5 poor · 6 moderate · 7 good · 8 excellent.
Source: 2014
📖 §11.6 Biomass Energy (Biomass Energy)

31. What is biomass energy, and why is biomass considered carbon neutral?

Model answer: Biomass is basically organic matter such as wood, straw, crops, algae, sewage sludge, animal waste and other biological waste. Bioenergy is the energy derived from biomass. In energy terms biomass is a form of stored solar energy, since the sun's energy is captured and stored via photosynthesis in the biomass material. It is considered carbon neutral because the carbon dioxide released during burning of the biomass is largely balanced by the carbon dioxide absorbed/captured during its growth.
Carbon neutral = CO2 released on burning ≈ CO2 absorbed during growth.
Source: 2014
📖 §11.6 Biomass Energy (Methods of generating energy from biomass)

32. List the different methods used to generate energy from biomass.

Model answer: The four main methods used to generate energy from biomass are: (1) Direct combustion of biomass (burning in a grate, stoker or fluidised bed for heat/steam/electricity); (2) Gasification of biomass (partial combustion to produce combustible producer gas); (3) Biomethanation / anaerobic digestion (production of biogas, mainly methane and carbon dioxide); and (4) Biofuels (conversion into liquid fuels such as ethanol and biodiesel).
Four routes: combustion, gasification, biomethanation, biofuels.
Source: 2014
📖 §11.6 Biomass Energy (Direct Combustion of Biomass)

33. Explain direct combustion of biomass and the need for pelleting/briquetting.

Model answer: Direct combustion is the burning of biomass in a grate, stoker or fluidised bed with excess air, capturing the released energy to provide steam or hot water for process heating and/or electricity, in devices ranging from small domestic boilers to multi-megawatt power plants. Solid biomasses include coconut shells, rice husks, bagasse, wood waste and oil-seed cakes such as de-oiled bran (DOB). Biomasses of low bulk density are processed into pellets or briquettes (briquetting) to make them easier and more efficient to store, handle and burn.
Low-density biomass → pellets/briquettes for handling/combustion.
Source: 2014
📖 §11.6 Biomass Energy (Gasification of Biomass)

34. What is producer gas? How is it produced and what is its calorific value?

Model answer: Producer gas is the combustible gas produced by gasification of biomass. Gasification is partial (incomplete) combustion carried out by supplying air less than the stoichiometric requirement, at a temperature of about 1000°C. The products are combustible gases — carbon monoxide (CO), hydrogen (H2) and traces of methane (CH4) — plus non-useful tar and dust. Producer gas has a relatively low calorific value of 1000 to 1200 kcal/Nm³, and the gasification conversion efficiency is about 60-70%. Used in a dual-fuel DG set it can give 65-85% diesel savings.
Producer gas = CO + H2 + traces CH4 (Objective Q10 = c). CV 1000-1200 kcal/Nm³; efficiency 60-70%.
Source: 2014
📖 §11.6 Biomass Energy (Biomass gasifier zones and gasification stages)

35. Describe the four zones of a biomass gasifier and the four stages of a gasification system.

Model answer: A biomass gasifier is a thermo-chemical reactor in which the biomass passes through four zones (top to bottom): Drying/Distillation Zone, Pyrolysis Zone, Combustion Zone and Reduction Zone, after which it is converted into high-quality combustible producer gas. A gasification system consists of four main stages: (1) feeding of feedstock, (2) gasifier reactions where gasification takes place, (3) cleaning of the resultant gas, and (4) utilisation of the cleaned gas.
Zones: Drying → Pyrolysis → Combustion → Reduction. Stages: feed, react, clean, use.
Source: 2014
📖 §11.6 Biomass Energy (Typical Producer Gas Composition)

36. Give the typical composition of producer gas.

Model answer: The typical composition of producer gas is: carbon monoxide (CO) = 19 ± 3%, hydrogen (H2) = 18 ± 2%, methane (CH4) = 3 ± 1%, carbon dioxide (CO2) = 10 ± 3%, and nitrogen (N2) = 50 ± 2%. The combustible constituents are CO, H2 and CH4, while the large nitrogen fraction (from the air supplied) gives the gas its relatively low calorific value.
CO ~19% · H2 ~18% · CH4 ~3% · CO2 ~10% · N2 ~50%.
Source: 2014
📖 §11.6 Biomass Energy (Average conversion efficiency of a gasifier — solved example)

37. Give the gasifier conversion efficiency formula and find the efficiency if 20 kg of wood (CV 3200 kcal/kg) produces 46 m³ of producer gas (CV 1000 kcal/Nm³).

Model answer: Gasifier conversion efficiency = (calorific value of gas produced per kg of fuel) / (average calorific value of 1 kg of fuel) × 100, i.e. Heat output as producer gas / Heat input as fuel × 100. Here heat input = 20 × 3200 = 64,000 kcal; heat output = 46 × 1000 = 46,000 kcal. Therefore conversion efficiency = (46,000 / 64,000) × 100 = 71.88%.
Solved example from guidebook = 71.88%. η = gas-out kcal / fuel-in kcal × 100.
Source: 2014
📖 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process)

38. Explain biomethanation (anaerobic digestion) and the composition of biogas (Gobar gas).

Model answer: Biomethanation is the production of bio-methane gas (biogas) by the biological/anaerobic digestion of biomass in the absence of air, by specific bacteria, best at temperatures of 35-40°C. Raw materials include manure, sewage sludge, municipal solid waste and other biodegradable wastes; a valuable by-product is high-grade manure. Biogas produced from cow dung is called Gobar gas and is a mixture typically comprising about 60% methane (CH4) and 40% carbon dioxide (CO2). Biogas offers higher energy efficiency than direct burning of dung and can be used for electricity generation or, when purified, as a vehicle fuel.
Gobar gas ≈ 60% CH4 + 40% CO2; anaerobic, 35-40°C; by-product = manure.
Source: 2014
📖 §11.6 Biomass Energy (Four-stage anaerobic digestion)

39. Describe the four-stage process of anaerobic digestion of biomass.

Model answer: Anaerobic digestion is a four-stage bacterial process. Stage 1: acidic bacteria dismantle complex organic molecules into smaller molecules. Stage 2: these molecules break down further into organic acids, carbon dioxide and ammonia. Stage 3: a second type of bacteria (methanogenic bacteria) begins converting these molecules into acetates and hydrogen. Stage 4: these are converted into methane. The methane-producing bacteria are strongly influenced by ambient conditions, which can slow or halt the process if conditions are unfavourable.
4 stages: hydrolysis → acidogenesis → acetogenesis → methanogenesis (textbook phrasing).
Source: 2014
📖 §11.6 Biomass Energy (Figure 11.17 Floating-drum Biogas Plant)

40. Describe the working of a floating-drum biogas plant prevalent in India.

Model answer: A floating-drum biogas plant consists of an underground digester tank or well with a partition wall to prevent mixing of incoming fresh dung slurry with the outgoing spent slurry. Dung-and-water slurry is fed through an inlet; the gas produced is trapped under a floating plastic or metallic drum (gas-holder). As more gas is produced and trapped, the drum rises, acting as a storage unit, and when the tap is opened the gas is discharged at more or less constant pressure. A non-return valve in the outlet prevents air being drawn into the digester, which would destroy the bacteria and create an explosive mixture.
Floating drum = gas-holder that rises; gives constant-pressure gas; partition wall separates slurries.
Source: 2014
📖 §11.6 Biomass Energy (Biofuels from Biomass)

41. Describe ethanol and biodiesel as biofuels from biomass, including their feedstocks and production methods.

Model answer: Biomass can be converted into liquid biofuels to partially replace petroleum fuels. Ethanol is commonly produced by fermentation of molasses (a by-product of sugar manufacture) or other carbohydrate-rich feedstock (starch, sugar or cellulose); it is used as a fuel additive to cut carbon monoxide and smog emissions, and flexible-fuel vehicles use up to 85% ethanol. Biodiesel is a good diesel substitute produced most economically by transesterification of extracted plant oil (e.g. Jatropha seed oil — a non-edible oilseed grown on dry/arid land) with an alcohol such as methanol; raw materials include plant oils, waste cooking oil and animal fat. Biodiesel reduces vehicle emissions by about 20%.
Ethanol from molasses (fermentation); biodiesel from Jatropha by transesterification. Objective Q1 = animal & vegetable fat.
Source: 2014
📖 §11.7 Hydro Power (Table 11.2 Classification of Hydropower by Size)

42. Give the classification of hydropower by size (large, small, mini, micro, pico).

Model answer: Hydropower is classified by size as: Large-hydro — more than 25 MW, feeding into a utility grid; Small-hydro — 2001 kW to 25 MW, usually feeding a grid; Mini-hydro — 101 kW to 2 MW, either stand-alone or feeding a grid; Micro-hydro — from 11 kW up to 100 kW, usually providing power to remote areas away from the grid; and Pico-hydro — from a few hundred watts up to 10 kW.
Large >25 MW · Small 2-25 MW · Mini 101 kW-2 MW · Micro 11-100 kW · Pico <10 kW.
Source: 2014
📖 §11.7 Hydro Power (Run-of-the-river micro-hydro scheme)

43. What is a run-of-the-river micro-hydro scheme and its main components?

Model answer: Micro-hydro power is the small-scale harnessing of energy from falling water, for example powering a small factory or village from a local river. A run-of-the-river micro-hydro scheme requires no water storage; instead it diverts some water from the river, which is channelled along the side of a valley before being dropped into the turbine via a penstock. Its main components are an intake weir and settling basin, a channel, a forebay tank, the penstock, and a power house containing the turbine and generator.
No storage; intake weir → channel → forebay → penstock → turbine.
Source: 2014
📖 §11.7 Hydro Power (Water into Watts)

44. Give the formula for hydropower potential and calculate the power for a flow of 20 litres/second at a head of 12 m with 60% system efficiency.

Model answer: Theoretical power P = Flow rate (Q) × Head (H) × Gravity (g), i.e. P = 9.81 × Q × H (kW), with Q in m³/s, H in metres and g = 9.81 m/s². For small systems the overall efficiency is roughly 50% (turbines rarely exceed 80%). For Q = 20 L/s = 0.020 m³/s, H = 12 m and η = 60%: P = 9.81 × 0.020 × 12 × 0.6 ≈ 1.4 kW.
Worked guidebook example = 1.4 kW. P = ρgQH; multiply by efficiency.
Source: 2014
📖 §11.8 Fuel Cell (Operation of Fuel Cell)

45. Explain the principle and operation of a fuel cell.

Model answer: The input to a fuel cell is hydrogen, which combines with oxygen to produce electricity through an electrochemical process (not combustion), giving water and heat as by-products — so it is clean, quiet and highly efficient. A fuel cell has two catalyst-coated electrodes (an anode and a cathode) surrounding an electrolyte. Hydrogen molecules enter the anode, where the catalyst separates them into positively charged protons and negatively charged electrons. The electrolyte allows protons to pass through to the cathode but blocks the electrons, so the electrons are directed through an external circuit, creating electric current. At the cathode, oxygen combines with the protons and the returning electrons to produce water and heat.
Answers OCR long question L-1 in short form. H2 at anode, O2 at cathode, electrons via external circuit; by-product = water + heat.
Source: 2014
📖 §11.8 Fuel Cell (Fuel Cell — hydrogen as a secondary energy resource)

46. Why is hydrogen called a secondary energy source, and what is the biggest hurdle to large-scale fuel cell use?

Model answer: All fuel cells require hydrogen as fuel. Hydrogen is a secondary energy resource, meaning it does not occur freely and must be made from another fuel. It can be produced in several ways, such as steam reforming of natural gas, electrolysis of water, or gasification of biomass. The biggest hurdle to large-scale commercial exploitation of fuel cells is the high cost of producing this hydrogen. Individual fuel cells are placed in series to form a fuel cell stack to power a vehicle or provide stationary power to a building.
Hydrogen must be manufactured; high H2 production cost is the main barrier. Cells stacked for higher power.
Source: 2014
📖 §11.8 Fuel Cell (Table 11.3 Types of Fuel Cells)

47. Name the main types of fuel cells and one key feature of each.

Model answer: (1) PEMFC (Proton Exchange / Polymer Electrolyte Membrane Fuel Cell) — polymer membrane electrolyte, platinum catalyst, operates up to ~200°C, ideal for vehicles. (2) DMFC (Direct Methanol Fuel Cell) — runs on methanol, 60-130°C, convenient for portable power below 250 W. (3) PAFC (Phosphoric Acid Fuel Cell) — liquid phosphoric acid electrolyte, ~180°C, used in stationary generators 100-400 kW. (4) AFC (Alkaline Fuel Cell) — KOH electrolyte, ~70°C, needs pure H2/O2, used on NASA shuttles. (5) SOFC (Solid Oxide Fuel Cell) — solid ceramic electrolyte, 800-1000°C. (6) MCFC (Molten Carbonate Fuel Cell) — molten carbonate electrolyte, ~650°C, used in MW-scale plants. Fuel cells produce power in the 1 W to 10 MW range.
PEMFC, DMFC, PAFC, AFC, SOFC, MCFC with electrolyte/temperature/use.
Source: 2014
📖 §11.9 Energy from Wastes (Energy from Wastes)

48. How is energy recovered from wastes in a Waste-to-Energy (WTE) plant?

Model answer: Energy can be recovered from wastes (trash) by combustion in incinerators to generate power. In a typical Waste-to-Energy (WTE) plant: trash is dumped on a tipping floor, then picked up by a crane and dropped into an incinerator; the burning trash heats water, creating steam; the steam turns a turbine, generating electricity; and emissions are filtered as they leave the smokestacks. The remaining ash occupies only about one-tenth of the original volume.
WTE: incinerate MSW → heat water → steam → turbine; ash ≈ 1/10 volume.
Source: 2014
📖 §11.9 Energy from Wastes (Power generation from landfill gas)

49. What is landfill gas, and how is power generated from it?

Model answer: Landfill gas is the biogas produced from landfill sites by anaerobic digestion, as bacteria decompose organic matter naturally in the absence of oxygen over time. It is composed mainly of methane and carbon dioxide. Normally the methane escapes into the atmosphere and contributes to greenhouse gas emissions, but if perforated pipes are inserted into the landfill, the gas travels through the pipes under natural pressure and can be collected and used as an energy source to drive a generator and produce electricity.
Landfill gas = mainly CH4 + CO2 (anaerobic); collected via perforated pipes → generator.
Source: 2014
📖 §11.10 Wave Energy (Wave Energy)

50. Explain wave energy and how a wave power device (oscillating water column) works.

Model answer: Sea waves result from the concentration of energy from natural sources such as the sun, wind, tides, ocean currents, the moon and the earth's rotation. Waves originate from wind and storms and travel long distances with little energy loss, so power output is steadier and more predictable and continues round the clock. Wave energy contains roughly 1000 times the kinetic energy of wind, and wave power varies as the square of the wave height (whereas wind power varies as the cube of speed); water is about 850 times denser than air, giving higher power. In an oscillating water column device, the rising wave enters a chamber and forces air out through a turbine that spins a generator; when the wave recedes, air flows back through the turbine, generating power in both directions. Roughly 40 MW/km of coast is available for 1 m waves and up to 1000 MW/km for 5 m waves.
Wave power ∝ (height)²; round-the-clock; oscillating water column drives air turbine.
Source: 2014
📖 §11.11 Tidal Energy (Tidal Energy) — chapter-end short question S-3

51. What causes tides, and what is the basic requirement for tapping tidal energy?

Model answer: Tides are generated by the combination of the moon's and sun's gravitational forces; the greatest effects occur at spring tides when the sun and moon combine forces, and cycles of low and high tides occur twice a day. Bays and inlets amplify the tide. When the tide comes in, water is trapped in reservoirs behind dams (barrages), and when the tide drops the trapped water is let out through turbines just like a hydroelectric plant; the turbines are driven in both directions. The basic requirement is that the height difference (tidal range) needs to be at least 5 metres for tidal energy to be practicable, together with a suitable bay/estuary site for the barrage.
Answers OCR short question S-3. Tidal range ≥5 m required; moon+sun gravity, spring tides.
Source: 2014
📖 §11.12 Geothermal Energy (Geothermal Energy)

52. What is geothermal energy and how does the earth's temperature vary with depth?

Model answer: Geothermal energy is heat energy from inside the earth. The earth's structure is crust, then mantle, whose top layer is hot liquid rock called magma; when magma breaks through the surface in a volcano it is called lava. For every 100 metres below ground, the temperature of the rock increases about 3°C, so at about 3000 m depth the rock is hot enough to boil water. Where water reaches hot rock it becomes boiling hot water (over 148°C) or steam; emerging through a crack it forms a hot spring. Where there is enough steam/hot water, holes are drilled and the steam/water raised to generate electricity, with no fuel burned.
Magma = molten rock (Objective Q7); +3°C per 100 m; ~3000 m to boil water.
Source: 2014
📖 §11.12 Geothermal Energy (Types of geothermal power plants)

53. Describe the three types of geothermal power plants.

Model answer: (1) Dry steam power plant — draws steam directly from a hydrothermal production well and sends it to a turbine/generator; the steam is then condensed and returned to the reservoir via an injection well. (2) Flash steam power plant — draws hot water (about 182°C) from the well into a flash tank where a drop in pressure 'flashes' it to steam, which drives the turbine/generator; the condensed steam and remaining hot water are returned via an injection well. (3) Binary cycle power plant — operates on lower-temperature water (107-182°C); the hot water heats a separate low-boiling-point working fluid (e.g. iso-butane) in a heat exchanger, and this fluid vaporises to drive the turbine. The two fluids stay in separate closed loops, so there are no emissions to the air.
Dry steam · Flash steam (~182°C) · Binary cycle (107-182°C, second working fluid, no emissions).
Source: 2014
📖 §11.1 Concept of New and Renewable Energy (Concept of New and Renewable Energy)

54. State the main advantages and limitations of renewable energy sources.

Model answer: Advantages: renewable sources are essentially inexhaustible (flows of energy, not stocks); they can be used without releasing harmful pollutants; biomass is carbon neutral; and they can serve remote/off-grid areas (e.g. stand-alone SPV, micro-hydro). Limitations: many are intermittent and variable (solar depends on sunshine and time of day; wind speed fluctuates, giving capacity factors of only 20-40%); conversion efficiencies are limited (solar cells 10-15%, Betz limit 59% for wind); they are site-specific (good insolation, high wind speed, ≥5 m tidal range, or suitable geothermal/hydro sites); and some have high costs (e.g. the high cost of producing hydrogen for fuel cells).
Synthesised from the chapter's recurring themes; verified facts but no single advantages/limitations list in OCR.
Source: 2014
📖 §11.1–§11.9 — abbreviations used across the chapter (SPV, WECS, BIPV, WTE, MSW)

55. Give the full form of the following renewable-energy acronyms: SPV, FPC, ETC, WECS, OTEC, WTE, MSW, BIPV, MNRE.

Model answer: SPV = Solar Photovoltaic; FPC = Flat-Plate Collector; ETC = Evacuated Tube Collector; WECS = Wind Energy Conversion System; OTEC = Ocean Thermal Energy Conversion; WTE = Waste-to-Energy; MSW = Municipal Solid Waste; BIPV = Building-Integrated Photovoltaic; MNRE = Ministry of New and Renewable Energy (the nodal Indian government ministry for renewable energy). Book check: SPV, WECS and BIPV are spelled out in the 2014 Ch-11 text; ‘flat plate collector’, ‘evacuated tube collector’, ‘waste-to-energy’ and ‘municipal solid waste’ appear in full (FPC/ETC/WTE/MSW are the standard short forms). OTEC and MNRE are standard sector acronyms not printed in this chapter — hence verified=false.
SPV, FPC, ETC, WECS, WTE, MSW, BIPV appear in OCR. OTEC and MNRE are standard sector acronyms not defined in the 2014 chapter text, hence verified=false.
Source: AI practice
📖 Cogeneration — NOT in Book-1 Ch-11 (2014); covered in BEE Book-2/Book-3. Listed here only because it appears in the Paper-1 coverage spec

56. Define cogeneration and distinguish between a topping cycle and a bottoming cycle.

Model answer: Cogeneration (combined heat and power) is the simultaneous generation of both electrical (or mechanical) power and useful thermal energy (heat) from a single fuel input, giving much higher overall fuel-use efficiency. In a topping cycle, fuel is first used to generate electricity (e.g. in a turbine/engine), and the heat rejected from the prime mover is then recovered for process heating. In a bottoming cycle, fuel is first used to produce high-temperature process heat (e.g. in a furnace), and the waste heat leaving the process is then used to generate electricity, typically through a waste-heat boiler and steam turbine. Supporting figures: cogeneration raises overall fuel-utilisation efficiency to about 75–90% against roughly 35–40% for power generation alone. Topping cycle (power first, then process heat) is by far the more common; a bottoming cycle needs a high-temperature process such as a furnace, cement kiln or glass/steel plant. NOTE: this topic is not in the 2014 Book-1 Ch-11 text — verified=false.
Cogeneration is referenced in the user's coverage spec; this topic is treated in detail in BEE Book-2/Book-3, not in this 2014 Ch11 OCR, hence verified=false.
Source: AI practice
📖 §11.8 Fuel Cell (hydrogen as a secondary energy resource) + §11.4/§11.11 storage examples

57. Briefly describe hydrogen as an energy carrier and the need for energy storage with renewables.

Model answer: Hydrogen is a clean secondary energy carrier: it is not freely available and must be produced from another source (steam reforming of natural gas, electrolysis of water, or gasification of biomass), and on use it combines with oxygen in a fuel cell to give electricity with only water and heat as by-products. Because renewable sources such as solar and wind are intermittent, energy storage is needed to match generation with demand. Common storage methods include batteries (in stand-alone SPV systems), thermal storage (e.g. molten salt in solar power towers, hot water tanks in solar heating), pumped/stored water behind dams (hydro and tidal), and chemical storage as hydrogen produced from surplus renewable electricity.
Hydrogen and storage are listed in the user's coverage spec but are not given as a standalone section in the OCR; built from verified chapter facts, hence verified=false.
Source: AI practice

Long questions (10 marks) — 16

📖 §11.8 Fuel Cell (Fuel Cell, Figure 11.20 & Table 11.3) — chapter-end long question L-1

1. Explain the operation of a fuel cell with a sketch. Also name the common types of fuel cells.

Model answer: A fuel cell is an electrochemical device that converts the chemical energy of a fuel (hydrogen) directly into electricity, with water and heat as the only by-products. Because conversion is by an electrochemical process and NOT by combustion, a fuel cell is clean, quiet and highly efficient. Construction: A fuel cell consists of two catalyst-coated electrodes surrounding an electrolyte. One electrode is the anode and the other is the cathode. Operation (step by step): 1. Hydrogen molecules are fed to the anode. 2. The catalyst coating on the anode separates each hydrogen atom into a positively charged proton (H+) and a negatively charged electron (e-). 3. The electrolyte allows only the protons to pass through to the cathode; it blocks the electrons. 4. The electrons are therefore forced to travel through an external circuit, and this flow of electrons is the electric current (DC) that does useful work. 5. Meanwhile oxygen (from air) is fed to the cathode. 6. At the cathode the oxygen combines with the protons (that passed through the electrolyte) and the electrons (that returned through the external circuit) to produce water and heat. Overall: 2H2 + O2 -> 2H2O + electricity + heat. As long as hydrogen and oxygen are supplied, the cell keeps producing power. Individual cells are stacked in series to form a fuel-cell stack for vehicles or stationary building power. All fuel cells need hydrogen, which is a secondary energy source produced by steam reforming of natural gas, electrolysis of water or gasification of biomass; the high cost of producing hydrogen is the biggest hurdle to large-scale use. Types: PEMFC (polymer/proton-exchange membrane, platinum catalyst, up to 200 C, ideal for vehicles), DMFC (direct methanol, portable, <250 W), PAFC (phosphoric acid, ~180 C, stationary 100-400 kW), AFC (alkaline KOH, ~70 C, used on NASA shuttles), SOFC (solid ceramic oxide, 800-1000 C) and MCFC (molten carbonate, ~650 C, MW-scale plants). Sketch: draw a rectangle split into three vertical zones - ANODE (left) fed with H2, ELECTROLYTE/membrane (centre), CATHODE (right) fed with O2/air; connect anode and cathode by an external wire through a load (lamp) showing electron flow e- ; show H+ crossing the electrolyte left-to-right; show H2O + heat leaving the cathode side.
Book-verified (OCR Sec 11.8, Fig 11.20, Table 11.3). Guaranteed 10-mark long. Key marks: electrochemical (not combustion); anode=H2, cathode=O2; electrolyte passes protons/blocks electrons; electrons via external circuit = current; by-products water+heat; overall reaction; labelled sketch; at least name PEMFC/AFC/PAFC/SOFC/MCFC/DMFC.
Source: Chapter end-question L-1
📖 §11.5 Wind Energy (Components of wind turbine system, Figures 11.12–11.13) — chapter-end long question L-2

2. List and explain the various components of a typical (horizontal-axis) wind turbine.

Model answer: A modern wind turbine (wind energy conversion system, WECS) is usually a horizontal-axis machine that must be pointed into the wind. Its main components are: 1. Rotor and Blades: Aerofoil-shaped blades mounted on a hub capture the kinetic energy of the wind and convert it into rotary motion. The blade shape and its angle to the wind decide aerodynamic performance. Two- and three-bladed rotors are common for power generation (three-bladed run more smoothly and quietly); multi-bladed rotors give high starting torque and are used for water pumping. 2. Hub: Connects the blades to the low-speed (main) shaft. 3. Low-speed shaft (main shaft): Connected directly to the rotor hub; it turns slowly at about 30 to 60 rpm. 4. Gear box: Steps up the rotational speed from about 30-60 rpm to about 1000-1800 rpm, the speed most generators need to produce electricity. 5. High-speed shaft: Driven by the gearbox at 1000-1800 rpm; it drives the generator. 6. Generator: Converts the turning motion into electricity; the most common is an induction generator producing 50-cycle (50 Hz) AC. 7. Nacelle: The housing at the top of the tower that contains the gearbox, generator, shafts and control components. 8. Disc brake: Located on the main shaft (before the gearbox) or on the high-speed shaft (after the gearbox) to slow down or stop the rotor for safety/maintenance. 9. Yaw control (yaw drive and yaw motor): Rotates the whole nacelle so the rotor axis stays aligned with the wind direction, to extract maximum kinetic energy. 10. Pitch control: Adjusts the angle (pitch) of the blades to regulate speed/power and to protect the machine in high winds. 11. Anemometer and wind vane: Measure wind speed and direction and signal the controller/yaw system. 12. Controller: Manages start-up, operation, and cut-out/shut-down. 13. Tower: Supports the nacelle and rotor at the required hub height; taller towers reach stronger, steadier wind. (Remember: Yaw = turns the whole nacelle to face the wind; Pitch = twists the individual blades.)
Book-verified (OCR Sec 11.5, Fig 11.12/11.13). One of the two guaranteed 10-mark longs. Marks for the full labelled list; do NOT confuse yaw (nacelle) with pitch (blades); mention rpm figures (30-60 -> 1000-1800) and induction generator/50 Hz for full credit.
Source: Chapter end-question L-2
📖 §11.3 Solar Thermal Energy (Solar Water Heating System, FPC & ETC)

3. Describe a solar water-heating system. Compare the flat-plate collector and the evacuated tube collector, and explain why the evacuated tube collector is more efficient.

Model answer: A solar water-heating system consists of a solar collector (flat-plate or evacuated tube), an insulated storage tank and connecting pipes. It is installed on a roof or open ground with the collector facing the sun; in the southern hemisphere collectors face a north-facing roof. Water heated in the collector is stored in the insulated tank and stays hot overnight because heat losses are small. Flat-plate collector (most common): comprises copper tubes welded to a copper sheet, both coated with a highly absorbing black coating, with a toughened glass sheet on top and insulating material at the bottom, all placed in a flat box. It heats the circulating fluid to about 40-60 C. Its performance is highly dependent on ambient temperature - efficiency is good when ambient temperature is high, so heat output is higher in summer than winter. Heat loss is about 40%. Evacuated tube collector (for higher temperatures): uses two concentric glass tubes fused at the ends, with the air evacuated from the gap between them, giving thermal insulation like a Thermos flask. The outer tube is clear; the inner tube carries a special selective absorbing coating. No separate cover sheet or insulating box is needed. It can reach temperatures up to 150 C and its efficiency does not drop with ambient temperature. Water enters through an innermost feeder tube and hot water flows out in the annulus. Why the evacuated tube is more efficient: (i) the vacuum stops conduction of heat back to the atmosphere; (ii) the selective coating converts short-wave radiation to long-wave radiation and prevents re-radiation to the atmosphere; (iii) it is nearly independent of ambient temperature. As a result its heat loss is less than 10%, compared with about 40% for a flat-plate collector, so it traps much more heat.
Book-verified (OCR Sec 11.3). Covers descriptive collector + the S-5 comparison. Numbers to lock: flat plate 40-60 C and ~40% loss; evacuated tube up to 150 C and <10% loss; vacuum stops conduction + selective coating stops re-radiation.
Source: Chapter-theme (solar thermal water heating)
📖 §11.4 Solar Electrical Energy (Power Towers & Parabolic Trough Collector)

4. 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.
Source: Sep 2021 Paper-1 (verified & expanded)
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology & conversion efficiency)

5. Explain the working of a solar photovoltaic (PV) system from cell to module to array. Then find the energy-conversion efficiency of a 175 W solar panel measuring 0.75 m x 1.50 m when the solar insolation is 1000 W/m2.

Model answer: Working principle (photovoltaic effect): Direct conversion of solar energy to electricity takes place through the photoelectric/photovoltaic effect - when two dissimilar materials in close contact are struck by light they produce a voltage. Discrete packets of light energy called photons strike the PV cell and knock electrons in the silicon out of their normal energy state at the p-n junction, so they can be conducted as electricity (DC). Only photons with the correct energy cause the effect, so cells are tuned to the most intense part of the spectrum, and because silicon is naturally reflective each cell is given an anti-reflective coating. Cell -> Module -> Array: One silicon cell produces about 0.5 V. Cells are connected in series and parallel to form a module (panel); 36 cells connected together give enough voltage to charge a 12 V battery and run a pump/motor. Several modules are connected to form an array to generate more power. Module output is rated in peak Watt (Wp) - the maximum power under Standard Test Conditions - and a single module ranges from 5 Wp to 120 Wp. A complete PV system also has a charge controller, a battery (in stand-alone systems) and an inverter to convert DC to AC. Numerical - PV conversion efficiency: Formula: n (%) = [ Pm / (E x A) ] x 100 where Pm = maximum power output (W), E = insolation (W/m2), A = cell area (m2). Given: Pm = 175 W, panel = 0.75 m x 1.50 m, E = 1000 W/m2. Area A = 0.75 x 1.50 = 1.125 m2 n = [ 175 / (1.125 x 1000) ] x 100 n = (175 / 1125) x 100 n = 15.6 % Result: the panel converts about 15.6% of the available solar energy into electrical energy (typical solar-cell efficiency is 10-15%).
Book-verified (OCR Sec 11.4 incl. the 175 W worked example = 15.6%). Marks: photovoltaic effect at p-n junction; 0.5 V/cell, 36 cells->12 V, Wp rating; correct formula and full working to 15.6%.
Source: Chapter-theme + guidebook worked numerical
📖 §11.4 Solar Electrical Energy (Stand-alone SPV, Grid-connected Solar and BIPV systems)

6. 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.
Source: Sep 2021 Paper-1 (verified & expanded)
📖 §11.5 Wind Energy (Wind Energy — Betz limit, power formula and capacity factor)

7. Explain how a wind turbine converts wind energy into electricity, state the Betz limit, and using P = 1/2 rho A Cp Ng Nb V^3 calculate the output of a turbine of 6 m rotor diameter with Cp = 0.30, generator efficiency 0.8, gearbox efficiency 0.90 and wind speed 11 m/s (air density 1.2 kg/m3).

Model answer: Energy conversion: A wind turbine extracts energy by slowing down the wind. The kinetic energy of moving air (KE = 1/2 x mass x velocity^2, with mass flow = rho x A x V) turns the aerofoil blades; the rotor drives the low-speed shaft, a gearbox steps the speed up, and the high-speed shaft drives the generator to produce AC electricity. The power extracted is proportional to the swept area and to the CUBE of the wind speed - so doubling wind speed increases power eight times, while doubling the area only doubles the power; hence siting turbines in the highest-wind areas is critical. Betz limit: The blades cannot be 100% efficient because some wind must pass through the rotor to keep it turning. The theoretical maximum fraction of wind energy that a rotor can capture is about 59% - this is the Betz limit. After gearbox and generator losses only about 15-25% of the wind energy actually becomes useful power. The rotor's coefficient of performance Cp varies between about 0.33 and 0.59. Formula: P = 0.5 x rho x A x Cp x Ng x Nb x V^3 where rho = air density, A = swept area = pi D^2 / 4, Cp = blade coefficient, Ng = generator efficiency, Nb = gearbox efficiency, V = wind speed. Numerical: Given D = 6 m, Cp = 0.30, Ng = 0.8, Nb = 0.90, V = 11 m/s, rho = 1.2 kg/m3. Swept area A = pi/4 x D^2 = (3.14/4) x 6^2 = 0.785 x 36 = 28.27 m2 V^3 = 11^3 = 1331 P = 0.5 x 1.2 x 28.27 x 0.30 x 0.8 x 0.90 x 1331 P = 0.6 x 28.27 x 0.30 x 0.8 x 0.90 x 1331 P ~= 4875 W = 4.875 kW Result: the expected power output is about 4875 W (4.875 kW).
Book-verified (OCR Sec 11.5, worked example = 4875 W). Core numerical. Marks: P proportional to v^3 (double->x8); Betz 59% and 15-25% usable; A = piD^2/4; full substitution to 4.875 kW.
Source: Chapter-theme + guidebook worked numerical
📖 §11.6 Biomass Energy (Gasification of Biomass — reactions, composition and efficiency)

8. Describe the stages of the biomass gasification process (with reactions and gas composition). Find the conversion efficiency of a gasifier if 20 kg of wood of calorific value 3200 kcal/kg produces 46 m3 of producer gas of average calorific value 1000 kcal/Nm3.

Model answer: Gasification: Biomass contains carbon, hydrogen and oxygen. Complete combustion gives CO2 and water vapour, but combustion under controlled conditions (partial combustion with air LESS than the stoichiometric requirement) at about 1000 C produces the combustible gases carbon monoxide (CO) and hydrogen (H2). This gas is called producer gas. It has a relatively low calorific value of 1000-1200 kcal/Nm3, and the conversion efficiency of gasification is about 60-70%. In a dual-fuel DG set it can give 65-85% diesel saving. Four main stages of a gasification system: 1. Feeding of the feedstock (biomass). 2. Gasifier reactions where gasification takes place. 3. Cleaning of the resultant gas (removing tar and dust). 4. Utilisation of the cleaned gas. Inside the gasifier the biomass passes through four zones: Drying/Distillation -> Pyrolysis -> Combustion -> Reduction, emerging as producer gas. Reactions: Oxidation (exothermic): C + O2 -> CO2 ; H2 + 1/2 O2 -> H2O Reduction: C + CO2 -> 2CO ; C + H2O -> CO + H2 Water-gas: CO2 + H2 -> CO + H2O Methanation: C + 2H2 -> CH4 Typical producer-gas composition: CO 19%, H2 18%, CH4 3%, CO2 10%, N2 50%. Numerical (conversion efficiency): Heat input in the gasifier = 20 kg x 3200 kcal/kg = 64,000 kcal Heat output as producer gas = 46 m3 x 1000 kcal/Nm3 = 46,000 kcal Conversion efficiency = (Heat output / Heat input) x 100 = (46,000 / 64,000) x 100 = 71.88 % Result: gasifier conversion efficiency = 71.9%.
Book-verified (OCR Sec 11.6 + solved example p.288 = 71.88%). High frequency. Marks: partial combustion below stoichiometric at ~1000 C; producer gas = CO + H2 + CH4 (low CV 1000-1200 kcal/Nm3); four stages/zones; reactions; correct efficiency 71.88%.
Source: 2012 & 2018 Paper-1 (verified & re-solved)
📖 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process)

9. Explain biomethanation (anaerobic digestion) of biomass. Describe the composition of biogas/Gobar gas, the four-stage process and a floating-drum biogas plant.

Model answer: Biomethanation is the conversion of biomass into bio-methane gas by biological/anaerobic digestion - the breakdown of organic matter by specific bacteria in the ABSENCE of air, best at 35-40 C. Unlike other routes it gives an extra advantage: high-grade manure as a by-product. Raw materials include cattle dung, sewage sludge, municipal solid waste, fruit/vegetable waste, food waste and distillery wastes. Biogas produced from cow dung is called Gobar gas - a gas mixture of about 60% methane (CH4) and 40% carbon dioxide (CO2). Bio-methane can completely replace natural gas in boilers, furnaces and IC engines, and biogas offers much higher energy efficiency than directly burning dung (device efficiency ~55% vs ~10%). Biogas can be used for electricity generation or purified for use as a vehicle fuel. Four-stage anaerobic digestion: 1. Acidic bacteria break down complex organic molecules into smaller molecules. 2. These molecules break down further into organic acids, carbon dioxide and ammonia. 3. Methanogenic bacteria convert these into acetates and hydrogen. 4. The acetates/hydrogen are converted into methane. Methane-producing bacteria are sensitive to ambient conditions, which can slow or halt the process. Floating-drum biogas plant (common in India): It has an underground digester tank with a partition wall that keeps fresh dung slurry (influent) from mixing with the spent slurry (outgoing). Dung-and-water slurry is fed from a mixing/inlet tank; the gas produced is trapped under a floating metal or plastic drum (gas-holder) that rises as more gas collects, storing the gas and delivering it at nearly constant pressure through the outlet tap. A non-return valve in the outlet prevents air being drawn into the digester (which would kill the bacteria and create an explosive mixture). Spent slurry overflows to an outlet as manure.
Book-verified (OCR Sec 11.6, Fig 11.17). Marks: anaerobic (no air) at 35-40 C; Gobar gas = 60% CH4 + 40% CO2; by-product manure; four bacterial stages; floating-drum design with partition wall and non-return valve.
Source: Chapter-theme (biomethanation / biogas)
📖 §11.6 Biomass Energy (Direct Combustion of Biomass & Biofuels from Biomass)

10. Explain how energy is obtained from biomass by direct combustion (including briquetting) and by conversion into liquid biofuels (ethanol and biodiesel).

Model answer: Biomass is organic matter (wood, straw, crops, husk, bagasse, animal waste, sewage sludge). It is stored solar energy captured through photosynthesis, and it is considered carbon neutral because the CO2 released on burning is largely balanced by the CO2 absorbed during growth. Direct combustion: Biomass is burnt in a grate, stoker or fluidized bed with excess air, and the released energy provides steam or hot water for process heating and/or electricity - from small domestic boilers up to multi-megawatt power plants. Solid biomass fuels include coconut shells, rice husk, bagasse, wood waste and oil-seed cakes such as de-oiled bran (DOB). Briquetting/pelleting: biomass of low bulk density is compacted into pellets or briquettes, which raises its density and handling/combustion quality so it can be burned efficiently as a solid fuel. Liquid biofuels: Biomass can be converted into liquid fuels to partially replace petroleum fuels. - Ethanol: commonly produced by fermentation of molasses (a by-product of sugar manufacture), or of any carbohydrate-rich feedstock (starch, sugar or cellulose) such as sugar beet, sweet corn and ligno-cellulosic straw/wood waste which are cheaper than molasses. Ethanol is used as a fuel additive to cut carbon monoxide and other smog-causing emissions; flexible-fuel vehicles can run on blends up to 85% ethanol. - Biodiesel: the most economical route is transesterification of an extracted plant oil (e.g. Jatropha seed oil) with an alcohol such as methanol. Jatropha is a non-edible tree-borne oilseed that grows on dry, arid land. All plant oils, waste cooking oil and animal fat can be raw materials. Biodiesel can be used as an additive to reduce vehicle emissions (typically by 20%) or in pure form as a renewable diesel substitute.
Book-verified (OCR Sec 11.6). Marks: carbon-neutral biomass; direct combustion in grate/stoker/fluidized bed; briquetting/pelleting of low-density biomass; ethanol by fermentation of molasses (up to 85% blends); biodiesel by transesterification of Jatropha oil with methanol.
Source: Chapter-theme (biomass combustion & biofuels)
📖 §11.7 Hydro Power (Hydro Power — classification, micro-hydro and Water into Watts)

11. Explain hydro power and its classification by size, describe a run-of-river micro-hydro scheme, and calculate the power potential of a run-of-river plant for a flow of 20 litres/second at a head of 12 m with 60% system efficiency.

Model answer: Hydro power harnesses moving water, most commonly by using a turbine turned by water falling in a controlled manner. Large dams store water for industry and grid electrification, while smaller schemes supply remote areas without dams. Classification by size: - Large-hydro: more than 25 MW, feeding a utility grid. - Small-hydro: 2001 kW to 25 MW, usually feeding a grid. - Mini-hydro: 101 kW to 2 MW, stand-alone or feeding a grid. - Micro-hydro: 11 kW up to 100 kW, usually for remote off-grid areas. - Pico-hydro: a few hundred watts up to 10 kW. Run-of-river micro-hydro scheme: requires no water storage. Some water is diverted from the river at an intake weir/settling basin, carried along a channel on the side of the valley to a forebay tank, and then dropped through a penstock into the turbine-generator in the power house; the water returns to the river. Water into Watts - formula: Theoretical power P = rho x g x Q x H, i.e. P (kW) = 9.81 x Q x H, with Q in m3/s and H in m. For small systems the overall efficiency is about 50% (turbines rarely exceed 80% and pipe friction adds losses), so multiply by the efficiency. Numerical: Given Q = 20 litres/s = 20/1000 = 0.020 m3/s, H = 12 m, efficiency = 60% (0.6). P = 9.81 x Q x H x efficiency P = 9.81 x 0.020 x 12 x 0.6 P = 9.81 x 0.020 = 0.1962 ; x 12 = 2.3544 ; x 0.6 = 1.413 kW Result: power potential ~= 1.4 kW. (Guidebook cross-check: a set at 10 m head and 0.3 m3/s at 50% eff = 9.81 x 0.3 x 10 x 0.5 = 14.715 kW.)
Book-verified (OCR Sec 11.7, both worked examples). Marks: size classification table; run-of-river components (weir->channel->forebay->penstock->turbine); P = 9.81 x Q x H x efficiency; convert 20 L/s to 0.020 m3/s; answer 1.4 kW.
Source: Chapter-theme + guidebook worked numericals
📖 §11.12 Geothermal Energy (Geothermal Energy — dry steam, flash steam and binary cycle plants)

12. Explain geothermal energy and describe the three main types of geothermal power plants: dry steam, flash steam and binary cycle.

Model answer: Geothermal energy uses the heat inside the earth. Below the crust is the mantle, whose top layer is hot liquid rock called magma (magma that breaks the surface is lava). The temperature rises about 3 C for every 100 m of depth, so at about 3000 m it is hot enough to boil water; water reaching the hot rock can turn into hot water (over 148 C) or steam and emerge as a hot spring. Where enough steam/hot water exists, wells are drilled and the steam or hot water is piped up to generate electricity. A geothermal power plant is like an ordinary steam power plant except that no fuel is burned. There are three main types: 1. Dry Steam Power Plant: Draws dry steam directly from a hydrothermal production well and sends it to a turbine-generator. The steam turns the turbine to make electricity, is then condensed, and the condensate is returned to the reservoir through an injection well. 2. Flash Steam Power Plant: Draws hot water (about 182 C) from the production well into a flash tank, where a sudden drop in pressure flashes part of the hot water into steam. The steam drives a turbine-generator; the steam is then condensed and, with the remaining hot water (waste brine) not flashed, is returned to the reservoir through an injection well. 3. Binary Cycle Power Plant: Operates on lower-temperature water (about 107-182 C). It uses two separate closed loops: hot geothermal water (brine) is pumped through a heat exchanger where it heats a secondary working fluid - an organic compound with a low boiling point such as iso-butane. The working fluid vaporises and drives the turbine, is condensed and recirculated. Because the geothermal water and the working fluid stay in separate closed loops, there are no emissions to the air. This makes the binary cycle suitable for lower-temperature resources.
Book-verified (OCR Sec 11.12). Marks: magma/lava, ~3 C per 100 m; dry steam (steam straight to turbine); flash steam (hot water ~182 C flashed by pressure drop); binary cycle (two closed loops, low-boiling working fluid like iso-butane, lower temp 107-182 C, no emissions).
Source: Chapter-theme (geothermal energy)
📖 §11.10 Wave Energy (Wave Energy) & §11.11 Tidal Energy (Tidal Energy)

13. Explain how energy is obtained from sea waves and from tides. State the special features and the basic requirement of each.

Model answer: Wave energy: Sea waves result from the concentration of energy from the sun, wind, tides, ocean currents, moon and earth's rotation. Waves originate from wind and storms far out at sea and travel long distances with little energy loss, so wave power is steadier and more predictable and continues round the clock - unlike wind and solar. Wave energy contains roughly 1000 times the kinetic energy of wind, so smaller devices can produce power; wave power varies as the SQUARE of the wave height (whereas wind power varies with the cube of air speed), and because water is about 850 times as dense as air the power produced (averaged over time) is much higher. Theoretically about 40 MW per km of coast can be extracted from gentle 1 m waves and up to 1000 MW per km where waves are 5 m. Working (oscillating water column): the wave rises into a chamber and forces the trapped air out through a turbine, which spins a generator; when the wave recedes, air flows back into the chamber through the turbine (through normally-closed doors), so power is produced in both directions. Tidal energy: Tidal energy is another form of ocean energy. Tides are generated by the combined gravitational forces of the moon and the sun; the greatest tides (spring tides) occur when the sun and moon combine their pull, and bays and inlets amplify the tide. Low and high tides occur twice a day. When the tide comes in it is trapped in a reservoir behind a dam (barrage); when the tide drops, the water behind the dam is released, like a hydroelectric plant, and drives turbines. The turbines are driven by the sea in BOTH directions (as the tide comes in and as it goes out). The basic requirement for tidal energy to be practicable is a tidal height difference of at least 5 m.
Book-verified (OCR Sec 11.10 & 11.11). Marks: wave power proportional to (height)^2, round-the-clock, ~40 MW/km (1 m) to 1000 MW/km (5 m), oscillating water column; tidal from moon+sun gravity, spring tides, twice daily, barrage/reservoir, turbines work both directions, minimum 5 m tidal range.
Source: Chapter-theme (ocean energy: wave & tidal)
📖 §11.9 Energy from Wastes (Energy from Wastes and landfill gas)

14. Explain how energy is recovered from wastes through a waste-to-energy (incineration) plant and through landfill gas.

Model answer: Energy can be recovered from waste (trash) either by directly burning it or by collecting the gas it produces as it decays. Waste-to-Energy (direct combustion) plant: The steps are (1) trash is dumped on a tipping floor; (2) a crane picks it up and drops it into an incinerator; (3) the burning trash heats water and creates steam; (4) the steam turns a turbine, which drives a generator to produce electricity; (5) the emissions/flue gases are filtered as they leave through the smokestacks; and (6) the remaining ash occupies only about one-tenth of the original volume of the waste. This both generates power and greatly reduces the volume of waste going to disposal. Power generation from landfill gas: Biogas produced from a landfill is called landfill gas. It forms by anaerobic digestion - bacteria decompose organic matter naturally in the absence of oxygen over time - and is composed mainly of methane and carbon dioxide. Normally the methane escapes into the atmosphere and adds to greenhouse-gas emissions. However, if perforated pipes (gas wells) are inserted into the landfill, the landfill gas travels through the pipes under its own natural pressure and is collected; it is then fed to an engine/generator (through a transformer) to produce electricity for the grid. This captures a harmful greenhouse gas and turns it into useful energy.
Book-verified (OCR Sec 11.9). Marks: WtE sequence tipping floor->incinerator->steam->turbine->power, filtered emissions, ash ~1/10 volume; landfill gas = anaerobic decay, mainly CH4+CO2, perforated pipes/gas wells collect it under natural pressure for engine/generator.
Source: Chapter-theme (energy from wastes)
📖 Cogeneration — NOT in Book-1 Ch-11 (2014); BEE Book-2/Book-3 topic, retained because it is in the Paper-1 coverage spec

15. Explain the concept of cogeneration and distinguish between topping cycle and bottoming cycle cogeneration.

Model answer: Cogeneration (also called combined heat and power, CHP) is the simultaneous production of two useful forms of energy - usually electrical (or mechanical) power and useful process heat - from a single fuel input. Because the heat that is normally wasted in ordinary power generation is instead recovered and used, cogeneration raises the overall fuel-utilisation efficiency to about 75-90%, compared with roughly 35-40% for power generation alone. It reduces fuel consumption, cost and emissions and is widely used in industries that need both power and process heat/steam (sugar, paper, textiles, refineries, chemicals). Depending on the sequence in which power and heat are produced, cogeneration is of two types: 1. Topping cycle: Fuel is first used to generate electrical/mechanical power (the high-grade energy comes 'on top' first), and the heat rejected from the prime mover (exhaust gas or back-pressure/extraction steam) is then recovered for process heating. This is the most common arrangement. Examples: a gas turbine or reciprocating engine whose hot exhaust raises steam in a waste-heat recovery boiler; or a back-pressure/extraction-condensing steam turbine where the exhaust steam is used for process. Here electricity is the primary output and heat is the by-product. 2. Bottoming cycle: Fuel is first used to meet a high-temperature process heat demand (for example in a furnace, cement kiln or glass/steel plant), and the hot waste gases leaving the process - which would otherwise be lost - are then used to raise steam and generate power in a 'bottoming' turbine. Here process heat is the primary output and power is generated from the residual heat at the bottom of the cycle. Bottoming cycles are less common and are applied where a suitable high-temperature waste-heat stream is available.
SYNTHESIZED - cogeneration is not part of the Chapter 11 OCR text (it belongs to another chapter). Content is standard BEE cogeneration material (topping = power first then heat; bottoming = process heat first then power; overall efficiency 75-90%). Verify wording against the guidebook's cogeneration chapter before final use.
Source: Cross-chapter theme (cogeneration) - synthesized
📖 §11.1 Concept of New and Renewable Energy (Book-1 p.263), with supporting figures from §11.2–§11.12

16. Discuss the advantages and limitations of renewable (non-conventional) energy sources.

Model answer: Renewable energy is energy obtained from essentially inexhaustible sources such as the sun, wind, falling water, sea waves, tides, geothermal heat and biomass. Renewable sources are flows of energy, unlike fossil and nuclear fuels which are finite stocks. They are also called non-conventional energy. Advantages: 1. Inexhaustible - the sources (sun, wind, water, tides, geothermal, biomass) are naturally replenished and will not run out, giving long-term energy security. 2. Clean/low pollution - they can be used without releasing harmful pollutants; solar, wind, hydro, tidal and geothermal generate power with little or no CO2 or other emissions, and biomass is broadly carbon neutral. 3. Widely distributed and decentralised - resources are available almost everywhere, so they suit stand-alone/off-grid supply to remote villages, telecom sites, hospitals and border areas, reducing transmission needs and losses. 4. Low running cost and no fuel cost - once installed, sunlight, wind and water are free, so operating costs and fuel-price risk are low. 5. Modular and scalable - systems range from a few watts (pico-hydro, small PV) to many megawatts, and can be added incrementally. 6. Reduce dependence on imported fossil fuels and help meet environmental/climate targets. Limitations: 1. Intermittent and variable - solar depends on day/season/weather and wind on continuously fluctuating wind speed, so output is not always available and often needs storage or backup. 2. Low energy density and diffuse - solar insolation and wind carry relatively little energy per unit area, so large collector areas or many turbines are needed. 3. Site specific - good sites are limited: wind needs high-wind areas, hydro needs head and flow, tidal needs at least a 5 m range, geothermal needs suitable hot fields, and solar thermal power needs high insolation. 4. High initial capital cost - equipment (PV modules, turbines, collectors) and storage batteries are costly, even though running costs are low. 5. Low conversion efficiency - e.g. solar cells typically convert only 10-15%, and wind is limited by the Betz limit (~59%, ~15-25% useful). 6. Storage and integration challenges - batteries add cost, and grid integration of variable sources needs synchronising inverters and management. Conclusion: renewable sources are clean, secure and inexhaustible and are essential for sustainable development, but their intermittency, low/diffuse energy density, site dependence and high capital cost mean they are often used alongside conventional supply and with suitable storage.
SYNTHESIZED essay - the OCR gives the renewable-energy concept (Sec 11.1) but no explicit advantages/limitations list; points are compiled from chapter facts (carbon-neutral biomass, intermittency, 10-15% PV efficiency, Betz limit, storage, site needs, 5 m tidal range). Book-consistent but drafted, so flagged ai/unverified.
Source: Cross-cutting essay (advantages & limitations) - synthesized