Principles and applications of renewable energy fundamentals for Mechanical Engineering and GATE & IES preparation.
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Question 1 of 119
Original Exam: GATE (2015)
The solar constant is defined as the solar irradiance at:
✓ Correct Answer: A
Explanation:
The Solar Constant (G_sc) is defined as the total solar irradiance (power per unit area) received on a surface perpendicular to the sun's rays at the top of Earth's atmosphere at the mean Earth-Sun distance (1 AU = 1.496 x 10^11 m). Value: G_sc = 1367 W/m2 (approximately 1.367 kW/m2). It varies by about +/-3.3% due to Earth's elliptical orbit (closer in January, farther in July). At Earth's surface, the irradiance is less due to atmospheric absorption and scattering (typically 800-1000 W/m2 on a clear day).
Difficulty: basicType: conceptual
Question 2 of 119
Original Exam: IES (2016)
The value of the solar constant is approximately:
✓ Correct Answer: B
Explanation:
The Solar Constant G_sc = 1367 W/m2 (approximately 1.353 kW/m2 in some older references, but 1367 W/m2 is the currently accepted value by WMO). This is the total solar irradiance at the top of Earth's atmosphere at the mean Earth-Sun distance. At Earth's surface: Direct Normal Irradiance (DNI) on a clear day is approximately 800-1000 W/m2. The difference is due to atmospheric absorption (by O3, H2O, CO2) and scattering (Rayleigh scattering by air molecules, Mie scattering by aerosols).
Difficulty: basicType: conceptual
Question 3 of 119
Original Exam: GATE (2017)
The declination angle of the sun is zero on:
✓ Correct Answer: C
Explanation:
The solar declination angle (delta) is the angle between the sun's rays and the equatorial plane. It varies from +23.45° (summer solstice, June 21) to -23.45° (winter solstice, December 21). On the equinoxes (approximately March 21 and September 23), the declination angle is ZERO (sun is directly above the equator). The declination angle is given by: delta = 23.45 x sin(360/365 x (284 + n)) degrees, where n is the day number of the year. At delta = 0, day and night are equal in length everywhere on Earth.
Difficulty: basicType: conceptual
Question 4 of 119
Original Exam: IES (2018)
The hour angle (omega) at solar noon is:
✓ Correct Answer: D
Explanation:
The Hour Angle (omega) represents the angular displacement of the sun from the local meridian. At solar noon, the sun is on the local meridian, so omega = 0°. The hour angle changes at 15° per hour (360°/24 hours). Before solar noon: omega is negative (some conventions use positive). After solar noon: omega is positive (some conventions use negative). At sunrise: omega = -omega_s (sunrise hour angle). At sunset: omega = +omega_s. The hour angle is used to calculate the position of the sun in the sky at any time of day.
Difficulty: basicType: conceptual
Question 5 of 119
Original Exam: GATE (2019)
The altitude angle of the sun at solar noon for a location at latitude 23.45°N on summer solstice is:
✓ Correct Answer: A
Explanation:
Solar altitude angle at solar noon = 90° - |latitude - declination|. On summer solstice: declination = +23.45°. For latitude = 23.45°N (Tropic of Cancer): Altitude angle = 90° - |23.45° - 23.45°| = 90° - 0° = 90°. The sun is directly overhead (zenith) at solar noon on the summer solstice at the Tropic of Cancer. This is the maximum possible altitude angle (90°). For other latitudes: Altitude angle at noon = 90° - (latitude - declination) for northern hemisphere.
Difficulty: intermediateType: numerical
Question 6 of 119
Original Exam: IES (2015)
The zenith angle is related to the altitude angle by:
✓ Correct Answer: B
Explanation:
The Zenith Angle (theta_z) is the angle between the sun's rays and the vertical (zenith direction). The Altitude Angle (alpha) is the angle between the sun's rays and the horizontal plane. They are complementary angles: Zenith angle + Altitude angle = 90°. Therefore: Zenith angle = 90° - Altitude angle. When the sun is at the horizon: altitude = 0°, zenith = 90°. When the sun is directly overhead: altitude = 90°, zenith = 0°. The zenith angle is used in solar radiation calculations: I = I_0 x cos(theta_z).
Difficulty: basicType: numerical
Question 7 of 119
Original Exam: GATE (2016)
The tilt angle of a solar collector for maximum annual energy collection in India (latitude 20°N) should be approximately:
✓ Correct Answer: C
Explanation:
For maximum annual energy collection, the optimal tilt angle of a solar collector is approximately equal to the latitude of the location. For India at latitude 20°N: Optimal tilt angle ≈ 20°. The collector should face south (in the northern hemisphere) and be tilted at the latitude angle. This ensures the collector is perpendicular to the sun's rays at the equinoxes (when the sun is at the equator). For summer: tilt = latitude - 23.45° (flatter). For winter: tilt = latitude + 23.45° (steeper). The latitude tilt is a good compromise for year-round collection.
Difficulty: intermediateType: conceptual
Question 8 of 119
Original Exam: IES (2017)
The air mass (AM) number represents:
✓ Correct Answer: D
Explanation:
Air Mass (AM) number = Path length of solar radiation through the atmosphere / Path length at zenith = 1/cos(theta_z), where theta_z is the zenith angle. AM0: Outside the atmosphere (space). AM1: Sun at zenith (theta_z = 0°), path length = 1 atmosphere thickness. AM1.5: Standard test condition for solar cells (theta_z = 48.19°), path length = 1.5 atmosphere thicknesses. AM2: theta_z = 60°. Higher AM means more atmospheric absorption and scattering, so less solar radiation reaches the surface. AM1.5 corresponds to 1000 W/m2 irradiance at the surface.
Difficulty: intermediateType: numerical
Question 9 of 119
Original Exam: GATE (2018)
The standard test condition (STC) for solar photovoltaic modules is:
✓ Correct Answer: A
Explanation:
Standard Test Conditions (STC) for solar PV modules: (1) Irradiance: 1000 W/m2; (2) Air Mass: AM1.5 (standard solar spectrum); (3) Cell temperature: 25°C. All PV module ratings (peak watt, Wp) are given at STC. In practice, modules rarely operate at exactly STC. The actual power output depends on: (1) Actual irradiance (usually less than 1000 W/m2); (2) Cell temperature (usually higher than 25°C, which reduces efficiency); (3) Spectral distribution of sunlight. NOCT (Nominal Operating Cell Temperature) conditions: 800 W/m2, AM1.5, 20°C ambient, 1 m/s wind.
Difficulty: basicType: conceptual
Question 10 of 119
Original Exam: IES (2019)
The fill factor (FF) of a solar cell is defined as:
✓ Correct Answer: B
Explanation:
Fill Factor (FF) = P_max / (V_oc x I_sc) = (V_mp x I_mp) / (V_oc x I_sc). Where P_max = maximum power output, V_oc = open circuit voltage, I_sc = short circuit current, V_mp = voltage at maximum power point, I_mp = current at maximum power point. FF represents how 'square' the I-V curve is. Ideal FF = 1 (rectangular I-V curve). Typical FF: 0.70-0.85 for crystalline silicon, 0.60-0.75 for thin film. FF is reduced by series resistance (reduces V_mp) and shunt resistance (reduces I_mp). Higher FF means better solar cell quality.
Difficulty: intermediateType: numerical
Question 11 of 119
Original Exam: GATE (2017)
The efficiency of a solar cell is defined as:
✓ Correct Answer: C
Explanation:
Solar cell efficiency = P_max / (A x G) x 100% = (V_mp x I_mp) / (A x G) x 100%. Where P_max = maximum power output (W), A = cell area (m2), G = incident irradiance (W/m2). This can also be written as: efficiency = P_max / P_input x 100%, where P_input = A x G = incident solar power. Both expressions A and C are equivalent and correct. Typical efficiencies: Monocrystalline Si: 18-22%; Polycrystalline Si: 15-18%; Thin film (CdTe): 10-13%; Amorphous Si: 6-10%; Multi-junction concentrator cells: 40-46%.
Difficulty: basicType: conceptual
Question 12 of 119
Original Exam: IES (2018)
The open circuit voltage (V_oc) of a solar cell increases when:
✓ Correct Answer: D
Explanation:
V_oc of a solar cell: (1) Increases when irradiance increases: V_oc = (nkT/q) x ln(I_L/I_0 + 1), where I_L (photocurrent) increases with irradiance; (2) Increases when temperature decreases: V_oc decreases with temperature at approximately -2.3 mV/°C for silicon. Therefore, V_oc increases when: (B) Irradiance increases AND (C) Temperature decreases. Both B and C are correct, so the answer is D. In practice, higher irradiance increases V_oc slightly, while lower temperature increases V_oc significantly. This is why solar cells perform better on cold, sunny days.
Difficulty: intermediateType: conceptual
Question 13 of 119
Original Exam: GATE (2014)
The photovoltaic effect was first demonstrated by:
✓ Correct Answer: A
Explanation:
The photovoltaic effect was first demonstrated by Alexandre Edmond Becquerel in 1839. He observed that certain materials produced small amounts of electric current when exposed to light. Key milestones: 1839: Becquerel discovers photovoltaic effect; 1883: Charles Fritts creates first solar cell (selenium); 1954: Bell Labs (Chapin, Fuller, Pearson) creates first practical silicon solar cell (6% efficiency); 1958: First solar-powered satellite (Vanguard 1). Albert Einstein explained the photoelectric effect (1905, Nobel Prize 1921). William Shockley co-invented the transistor and developed the p-n junction theory.
Difficulty: basicType: conceptual
Question 14 of 119
Original Exam: IES (2016)
In a flat plate solar collector, the absorber plate is coated with a selective surface to:
✓ Correct Answer: B
Explanation:
A selective surface on the absorber plate of a flat plate collector has: (1) High absorptivity (alpha) in the solar spectrum (short wavelengths, 0.3-2.5 micron): alpha > 0.9; (2) Low emissivity (epsilon) in the thermal infrared spectrum (long wavelengths, 5-50 micron): epsilon < 0.1. This is possible because the sun (at ~5800 K) emits mainly in the visible and near-IR, while the absorber plate (at ~50-100°C) emits mainly in the far-IR. Examples of selective surfaces: Black chrome, Black nickel, Titanium nitride oxide (TiNOx). The ratio alpha/epsilon is the selectivity ratio (higher is better).
Difficulty: intermediateType: numerical
Question 15 of 119
Original Exam: GATE (2018)
The efficiency of a flat plate solar collector decreases when:
✓ Correct Answer: C
Explanation:
The efficiency of a flat plate solar collector: eta = F_R [tau_alpha - U_L(T_i - T_a)/G]. Where F_R = heat removal factor, tau_alpha = transmittance-absorptance product, U_L = overall heat loss coefficient, T_i = inlet fluid temperature, T_a = ambient temperature, G = solar irradiance. Efficiency DECREASES when: (1) Inlet fluid temperature (T_i) increases: the term (T_i - T_a)/G increases, reducing efficiency; (2) Ambient temperature (T_a) decreases: same effect. Efficiency INCREASES when: (2) Solar irradiance (G) increases; (3) Ambient temperature increases (reduces heat loss). The efficiency is zero when T_i - T_a = tau_alpha x G / U_L (stagnation condition).
Difficulty: intermediateType: numerical
Question 16 of 119
Original Exam: GATE (2019)
The concentration ratio (CR) of a solar concentrator is defined as:
✓ Correct Answer: D
Explanation:
Concentration Ratio (CR) can be defined in two equivalent ways: (1) Geometric CR = Area of concentrator aperture / Area of receiver; (2) Flux CR = Irradiance at receiver / Irradiance at concentrator aperture. Both definitions give the same result for an ideal concentrator. Types of concentrators: (1) Parabolic trough: CR = 10-100; (2) Parabolic dish: CR = 100-3000; (3) Central receiver (power tower): CR = 300-1500; (4) Fresnel lens: CR = 10-100. Higher CR gives higher temperatures but requires more precise tracking. Maximum theoretical CR = 46,200 (for 2D) or 212,000 (for 3D) based on thermodynamic limits.
Difficulty: intermediateType: numerical
Question 17 of 119
Original Exam: IES (2017)
The parabolic trough solar collector tracks the sun:
✓ Correct Answer: A
Explanation:
A Parabolic Trough Solar Collector uses SINGLE-AXIS tracking. The trough rotates about one axis (typically north-south axis, rotating east-west to track the sun's daily motion). The trough focuses sunlight onto a linear receiver (absorber tube) at the focal line. Concentration ratio: 10-100. Operating temperature: 150-400°C. Applications: Solar thermal electricity generation (CSP), industrial process heat. Two-axis tracking is used for: Parabolic dish, Heliostat (for central receiver/power tower). Fixed collectors: Flat plate, evacuated tube (no tracking needed).
Difficulty: basicType: conceptual
Question 18 of 119
Original Exam: GATE (2016)
The global horizontal irradiance (GHI) is related to direct normal irradiance (DNI) and diffuse horizontal irradiance (DHI) by:
✓ Correct Answer: B
Explanation:
Global Horizontal Irradiance (GHI) = Direct Horizontal Irradiance (DHI_direct) + Diffuse Horizontal Irradiance (DHI). Direct Horizontal Irradiance = DNI x cos(theta_z), where theta_z is the solar zenith angle. Therefore: GHI = DNI x cos(theta_z) + DHI. Where: GHI = total solar irradiance on a horizontal surface; DNI = Direct Normal Irradiance (on a surface perpendicular to sun's rays); DHI = Diffuse Horizontal Irradiance (scattered radiation from sky); theta_z = solar zenith angle. This relationship is fundamental for solar resource assessment.
Difficulty: intermediateType: numerical
Question 19 of 119
Original Exam: GATE (2015)
The Betz limit for wind turbines states that the maximum power that can be extracted from wind is:
✓ Correct Answer: C
Explanation:
The Betz Limit (Albert Betz, 1919) states that the maximum power that can be extracted from wind by an ideal wind turbine is 16/27 = 59.3% of the kinetic energy of the undisturbed wind. Derivation: Power in wind = 0.5 x rho x A x V^3. Maximum power extracted = C_p,max x 0.5 x rho x A x V^3, where C_p,max = 16/27 = 0.593. This occurs when the wind speed downstream of the turbine is 1/3 of the upstream wind speed. Real turbines achieve C_p of 0.35-0.50 (below the Betz limit due to aerodynamic losses, wake effects, mechanical losses).
Difficulty: basicType: conceptual
Question 20 of 119
Original Exam: IES (2016)
The power coefficient (C_p) of a wind turbine is defined as:
✓ Correct Answer: D
Explanation:
Power Coefficient (C_p) = Actual power output / Power available in wind = P_actual / (0.5 x rho x A x V^3). Where rho = air density (kg/m3), A = rotor swept area (m2), V = wind speed (m/s). C_p is dimensionless and represents the fraction of wind power extracted by the turbine. Maximum theoretical C_p = 16/27 = 0.593 (Betz limit). Typical C_p for modern wind turbines: 0.35-0.50. C_p varies with tip speed ratio (lambda = blade tip speed / wind speed) and blade pitch angle. Maximum C_p occurs at the optimal tip speed ratio (typically lambda = 6-9 for modern turbines).
Difficulty: basicType: conceptual
Question 21 of 119
Original Exam: GATE (2017)
The power available in wind is proportional to:
✓ Correct Answer: A
Explanation:
Power in wind = 0.5 x rho x A x V^3. The power is proportional to the CUBE of wind speed (V^3). This is because: Kinetic energy = 0.5 x m x V^2. Power = KE/time = 0.5 x (m/time) x V^2 = 0.5 x (rho x A x V) x V^2 = 0.5 x rho x A x V^3. The cubic relationship means: (1) Doubling wind speed increases power by 8 times; (2) Small changes in wind speed have large effects on power output; (3) Wind resource assessment is critical - even small errors in wind speed measurement lead to large errors in energy estimates. This is why wind turbines are sited at locations with high average wind speeds.
Difficulty: basicType: numerical
Question 22 of 119
Original Exam: IES (2018)
The tip speed ratio (TSR or lambda) of a wind turbine is defined as:
✓ Correct Answer: B
Explanation:
Tip Speed Ratio (TSR or lambda) = Blade tip speed / Wind speed = (omega x R) / V. Where omega = rotor angular velocity (rad/s), R = rotor radius (m), V = wind speed (m/s). Optimal TSR for maximum C_p: (1) Multi-blade (American windmill): TSR = 1-2; (2) Three-blade HAWT: TSR = 6-9; (3) Two-blade HAWT: TSR = 8-12; (4) Darrieus VAWT: TSR = 4-6. At optimal TSR, the power coefficient is maximum. At low TSR: blades are too slow (not extracting enough energy). At high TSR: blades are too fast (turbulent wake reduces efficiency).
Difficulty: intermediateType: numerical
Question 23 of 119
Original Exam: GATE (2019)
A horizontal axis wind turbine (HAWT) has its rotor axis:
✓ Correct Answer: C
Explanation:
A Horizontal Axis Wind Turbine (HAWT) has its rotor axis PARALLEL to the wind direction (horizontal). The rotor blades rotate in a plane perpendicular to the wind. HAWTs must be oriented into the wind (yaw control). Types: Upwind (rotor faces wind, most common) and Downwind (rotor is downwind of tower). Advantages: Higher efficiency, well-established technology. A Vertical Axis Wind Turbine (VAWT) has its rotor axis perpendicular to the wind (vertical). Types: Darrieus (egg-beater shape), Savonius (S-shaped). VAWTs don't need to face the wind but have lower efficiency.
Difficulty: basicType: conceptual
Question 24 of 119
Original Exam: IES (2015)
The Weibull distribution is used in wind energy to:
✓ Correct Answer: D
Explanation:
The Weibull Distribution is used to describe the statistical distribution of wind speeds at a site. The Weibull probability density function: f(v) = (k/c) x (v/c)^(k-1) x exp(-(v/c)^k). Where k = shape parameter (dimensionless), c = scale parameter (m/s, related to mean wind speed). For k = 2: Rayleigh distribution (commonly used approximation). The Weibull distribution is used to: (1) Calculate the annual energy production (AEP) of a wind turbine; (2) Assess the wind resource at a site; (3) Determine the capacity factor. Mean wind speed = c x Gamma(1 + 1/k).
Difficulty: intermediateType: numerical
Question 25 of 119
Original Exam: GATE (2016)
The cut-in wind speed of a wind turbine is the speed at which:
✓ Correct Answer: A
Explanation:
Wind turbine operating speeds: (1) Cut-in speed (V_ci): The minimum wind speed at which the turbine starts generating electricity (typically 3-4 m/s). Below this speed, the wind energy is insufficient to overcome friction and generate net power. (2) Rated speed (V_r): The wind speed at which the turbine reaches its rated (maximum) power output (typically 12-15 m/s). (3) Cut-out speed (V_co): The maximum wind speed at which the turbine operates before being shut down for safety (typically 25 m/s). Above cut-out speed, the turbine is stopped to prevent damage.
Difficulty: basicType: conceptual
Question 26 of 119
Original Exam: IES (2019)
The capacity factor of a wind turbine is defined as:
✓ Correct Answer: B
Explanation:
Capacity Factor (CF) = Actual annual energy output / (Rated power x 8760 hours). It represents the fraction of the maximum possible energy that the turbine actually generates in a year. Typical capacity factors: (1) Onshore wind: 25-40%; (2) Offshore wind: 35-50%; (3) Solar PV: 15-25%; (4) Coal plant: 60-80%; (5) Nuclear plant: 80-95%. A higher capacity factor means the turbine generates more energy relative to its rated capacity. CF depends on: wind resource at the site, turbine design, availability (downtime for maintenance).
Difficulty: intermediateType: conceptual
Question 27 of 119
Original Exam: GATE (2018)
The hub height of a wind turbine affects the power output because:
✓ Correct Answer: C
Explanation:
Wind speed increases with height due to reduced surface friction (boundary layer effect). The wind speed profile follows the power law: V(h) = V_ref x (h/h_ref)^alpha. Where V(h) = wind speed at height h, V_ref = reference wind speed at height h_ref, alpha = wind shear exponent (typically 1/7 = 0.143 for open terrain, higher for rough terrain). Since power is proportional to V^3, even a small increase in wind speed with height gives a significant increase in power. This is why modern wind turbines have tall towers (80-120 m hub height). Offshore turbines can have hub heights of 100-150 m.
Difficulty: intermediateType: conceptual
Question 28 of 119
Original Exam: IES (2017)
The solidity of a wind turbine rotor is defined as:
✓ Correct Answer: D
Explanation:
Solidity (sigma) = (Number of blades x Blade chord area) / Rotor swept area = (N x c x R) / (pi x R^2) = N x c / (pi x R). Where N = number of blades, c = blade chord length (m), R = rotor radius (m). High solidity (multi-blade): (1) High starting torque; (2) Low tip speed ratio; (3) Suitable for water pumping. Low solidity (2-3 blades): (1) Low starting torque; (2) High tip speed ratio; (3) Suitable for electricity generation. Modern 3-blade wind turbines have low solidity (0.02-0.05) and operate at high tip speed ratios.
Difficulty: intermediateType: numerical
Question 29 of 119
Original Exam: GATE (2015)
Biogas produced from anaerobic digestion of organic matter primarily consists of:
✓ Correct Answer: A
Explanation:
Biogas produced from anaerobic digestion primarily consists of: (1) Methane (CH4): 55-70% (the combustible component); (2) Carbon dioxide (CO2): 30-45%; (3) Small amounts of H2S, N2, H2O vapor. The calorific value of biogas: approximately 20-25 MJ/m3 (depending on CH4 content). Biogas from different feedstocks: (1) Cattle dung: 55-60% CH4; (2) Sewage sludge: 60-65% CH4; (3) Food waste: 60-70% CH4. Biogas can be used for: cooking, lighting, electricity generation, vehicle fuel (after upgrading to biomethane). Anaerobic digestion occurs in the absence of oxygen through four stages: hydrolysis, acidogenesis, acetogenesis, methanogenesis.
Difficulty: basicType: conceptual
Question 30 of 119
Original Exam: IES (2016)
The calorific value of biogas (60% methane) is approximately:
✓ Correct Answer: B
Explanation:
Calorific value of biogas with 60% CH4: CV_biogas = 0.60 x CV_CH4 = 0.60 x 35.8 MJ/m3 = 21.5 MJ/m3 approximately 22 MJ/m3. Where CV_CH4 = 35.8 MJ/m3 (lower heating value of methane). For comparison: Natural gas: 35-40 MJ/m3; LPG: 93 MJ/m3; Hydrogen: 10.8 MJ/m3 (at STP). Biogas with 65% CH4: CV = 0.65 x 35.8 = 23.3 MJ/m3. The calorific value of biogas is lower than natural gas because of the CO2 content (CO2 is non-combustible and dilutes the methane).
Difficulty: intermediateType: numerical
Question 31 of 119
Original Exam: GATE (2017)
The process of converting biomass to syngas (CO + H2) at high temperature is called:
✓ Correct Answer: C
Explanation:
Gasification converts biomass (or coal) to syngas (synthesis gas = CO + H2) at high temperatures (700-1200°C) with a controlled amount of oxygen and/or steam. The main reactions: C + O2 -> CO2 (combustion); C + CO2 -> 2CO (Boudouard reaction); C + H2O -> CO + H2 (water-gas reaction); CO + H2O -> CO2 + H2 (water-gas shift reaction). Syngas can be used for: electricity generation (in gas turbines or engines), hydrogen production, Fischer-Tropsch synthesis (liquid fuels). Pyrolysis: thermal decomposition without oxygen, produces bio-oil, biochar, and gases. Anaerobic digestion: biological process producing biogas. Fermentation: biological process producing ethanol.
Difficulty: basicType: conceptual
Question 32 of 119
Original Exam: IES (2018)
The energy content of 1 kg of dry biomass is approximately:
✓ Correct Answer: D
Explanation:
The energy content (calorific value) of dry biomass is approximately 15-20 MJ/kg (lower heating value). Specific values: (1) Wood: 18-20 MJ/kg; (2) Agricultural residues (rice straw, wheat straw): 14-17 MJ/kg; (3) Bagasse (sugarcane residue): 17-19 MJ/kg; (4) Municipal solid waste: 8-12 MJ/kg (wet basis). For comparison: Coal: 25-35 MJ/kg; Natural gas: 50 MJ/kg; Diesel: 43 MJ/kg. The energy content of wet biomass is lower due to the energy required to evaporate moisture. Moisture content significantly affects the effective calorific value.
Difficulty: basicType: conceptual
Question 33 of 119
Original Exam: GATE (2016)
Bioethanol is produced from biomass by:
✓ Correct Answer: A
Explanation:
Bioethanol production: (1) Feedstock preparation: sugarcane juice, corn starch, or cellulosic biomass; (2) Hydrolysis (for starch/cellulose): converts starch/cellulose to fermentable sugars; (3) Fermentation: yeast (Saccharomyces cerevisiae) converts sugars to ethanol and CO2: C6H12O6 -> 2C2H5OH + 2CO2; (4) Distillation: separates ethanol from water (to 95% purity); (5) Dehydration: removes remaining water (to 99.5% for fuel use). First-generation bioethanol: from food crops (sugarcane, corn). Second-generation: from lignocellulosic biomass (agricultural residues, wood). Third-generation: from algae.
Difficulty: basicType: conceptual
Question 34 of 119
Original Exam: IES (2019)
Biodiesel is produced by:
✓ Correct Answer: B
Explanation:
Biodiesel is produced by TRANSESTERIFICATION of vegetable oils or animal fats with alcohol (usually methanol) in the presence of a catalyst (NaOH or KOH). Reaction: Triglyceride + 3 Methanol -> 3 Fatty Acid Methyl Esters (FAME = Biodiesel) + Glycerol. Properties of biodiesel: (1) Calorific value: 37-40 MJ/kg (slightly less than diesel: 43 MJ/kg); (2) Cetane number: 48-65 (higher than diesel: 40-55); (3) Biodegradable and non-toxic; (4) Can be blended with diesel (B5, B20, B100). Common feedstocks: soybean oil (USA), rapeseed oil (Europe), palm oil (Southeast Asia), jatropha (India).
Difficulty: basicType: conceptual
Question 35 of 119
Original Exam: GATE (2015)
The power output of a hydroelectric plant is given by:
✓ Correct Answer: C
Explanation:
Power output of a hydroelectric plant: P = rho x g x Q x H x eta. Where rho = density of water (1000 kg/m3), g = acceleration due to gravity (9.81 m/s2), Q = flow rate (m3/s), H = net head (m), eta = overall efficiency (turbine + generator). Simplified: P = 9.81 x Q x H x eta (kW). For eta = 1 (ideal): P = 9.81 x Q x H kW. Example: Q = 10 m3/s, H = 50 m, eta = 0.85: P = 9.81 x 10 x 50 x 0.85 = 4169 kW = 4.17 MW. The overall efficiency includes: turbine efficiency (85-95%), generator efficiency (95-98%), transmission losses.
Difficulty: basicType: numerical
Question 36 of 119
Original Exam: IES (2016)
A Pelton turbine is used for:
✓ Correct Answer: D
Explanation:
Pelton Turbine (impulse turbine): Used for HIGH HEAD (300-1800 m), LOW FLOW RATE applications. Working principle: High-velocity water jet from nozzle strikes cup-shaped buckets on the wheel, transferring kinetic energy. Efficiency: 85-92%. Applications: Mountain hydropower plants. Francis Turbine (mixed flow reaction turbine): Medium head (40-600 m), medium flow rate. Kaplan Turbine (axial flow reaction turbine): Low head (2-40 m), high flow rate. Crossflow (Banki-Michell) turbine: Low to medium head (1-200 m). The selection of turbine type depends on the specific speed (N_s = N x sqrt(P) / H^(5/4)).
Difficulty: basicType: conceptual
Question 37 of 119
Original Exam: GATE (2017)
The specific speed of a turbine is used to:
✓ Correct Answer: A
Explanation:
Specific Speed (N_s) is a dimensionless parameter used to classify turbines and select the appropriate type: N_s = N x sqrt(P) / H^(5/4) (in SI units) or N_s = N x sqrt(Q) / H^(3/4). Where N = rotational speed (rpm), P = power (kW or HP), Q = flow rate (m3/s), H = head (m). Specific speed ranges: (1) Pelton turbine: N_s = 4-70 (low specific speed, high head); (2) Francis turbine: N_s = 60-400 (medium specific speed); (3) Kaplan turbine: N_s = 300-900 (high specific speed, low head). The specific speed is the same for geometrically similar turbines operating under similar conditions.
Difficulty: intermediateType: numerical
Question 38 of 119
Original Exam: IES (2018)
The run-of-river hydropower plant differs from a storage hydropower plant in that:
✓ Correct Answer: B
Explanation:
Run-of-River (RoR) Hydropower Plant: Uses the natural flow of the river without significant water storage. Power output varies with river flow (seasonal variation). No large reservoir required. Minimal environmental impact (no large flooding). Suitable for rivers with consistent flow. Storage Hydropower Plant: Uses a dam to create a reservoir for water storage. Can regulate power output independent of river flow. Provides peak power and grid stability. Requires large reservoir (significant land flooding). Pumped Storage Hydropower: Pumps water uphill during low demand, generates power during peak demand. Acts as a large-scale energy storage system.
Difficulty: basicType: conceptual
Question 39 of 119
Original Exam: GATE (2019)
The efficiency of a Pelton turbine is maximum when the bucket speed is approximately:
✓ Correct Answer: C
Explanation:
For a Pelton turbine, the efficiency is maximum when the bucket speed (u) is approximately HALF the jet velocity (V_j): u = V_j / 2. At this condition, the relative velocity of water leaving the bucket is minimum, meaning maximum energy transfer. Hydraulic efficiency = 2u(V_j - u)(1 + cos(beta)) / V_j^2. Maximizing with respect to u: d(eta)/du = 0 gives u = V_j/2. Maximum hydraulic efficiency = (1 + cos(beta)) / 2. For beta = 165° (typical bucket angle): eta_max = (1 + cos(165°)) / 2 = (1 - 0.966) / 2 = 0.483. Wait - this gives 48.3%. Actually: eta_max = (1 + cos(beta))/2 = (1 + 0.966)/2 = 0.983 = 98.3% for beta = 165°. The actual efficiency is lower due to friction.
Difficulty: intermediateType: numerical
Question 40 of 119
Original Exam: IES (2015)
Geothermal energy is derived from:
✓ Correct Answer: D
Explanation:
Geothermal energy is derived from the HEAT stored within the Earth. Sources of this heat: (1) Primordial heat: residual heat from the formation of the Earth (gravitational compression and accretion, approximately 4.5 billion years ago); (2) Radiogenic heat: heat from radioactive decay of isotopes (mainly U-238, U-235, Th-232, K-40) in the Earth's crust and mantle. The geothermal gradient (temperature increase with depth) is approximately 25-30°C/km on average, but much higher in geothermal areas (volcanic regions, tectonic plate boundaries). Total geothermal heat flow from Earth's surface: approximately 44 TW.
Difficulty: basicType: conceptual
Question 41 of 119
Original Exam: GATE (2016)
The geothermal gradient is approximately:
✓ Correct Answer: A
Explanation:
The average geothermal gradient is approximately 25-30°C per km depth. This means the temperature increases by 25-30°C for every kilometer of depth. In geothermal areas (volcanic regions, hot spots): gradient can be 50-100°C/km or more. In stable continental regions: gradient can be as low as 15-20°C/km. The geothermal gradient determines the depth required to reach temperatures suitable for: (1) Direct use (space heating): 50-150°C, depth 1-3 km; (2) Electricity generation: >150°C, depth 2-5 km (conventional), or Enhanced Geothermal Systems (EGS) at greater depths.
Difficulty: basicType: conceptual
Question 42 of 119
Original Exam: IES (2017)
A dry steam geothermal power plant uses:
✓ Correct Answer: B
Explanation:
Dry Steam Geothermal Power Plant: Uses STEAM DIRECTLY from the geothermal reservoir to drive the turbine. The steam is at high pressure and temperature (typically 150-300°C). After passing through the turbine, the steam is condensed and either reinjected or discharged. This is the simplest and most efficient type of geothermal power plant. Examples: The Geysers (California, USA) - largest geothermal power plant in the world. Flash Steam Plant: Hot water (>180°C) is flashed to steam by reducing pressure. Binary Cycle Plant: Geothermal water heats a secondary fluid (isobutane, pentane) with lower boiling point, which drives the turbine.
Difficulty: intermediateType: conceptual
Question 43 of 119
Original Exam: GATE (2018)
The tidal power plant generates electricity using:
✓ Correct Answer: C
Explanation:
Tidal Power Plants generate electricity using: (1) Tidal Range (potential energy): A barrage (dam) is built across a tidal estuary. Water flows through turbines as the tide rises and falls. Requires tidal range > 5 m. Examples: La Rance (France, 240 MW), Sihwa Lake (South Korea, 254 MW). (2) Tidal Current (kinetic energy): Underwater turbines (similar to wind turbines) capture the kinetic energy of tidal currents. No barrage required. Less environmental impact. Examples: MeyGen (Scotland). Ocean Thermal Energy Conversion (OTEC): uses temperature difference. Salinity gradient (osmotic power): uses salinity difference.
Difficulty: basicType: conceptual
Question 44 of 119
Original Exam: IES (2019)
The minimum tidal range required for economical tidal power generation is approximately:
✓ Correct Answer: D
Explanation:
The minimum tidal range required for economical tidal power generation is approximately 5 meters. This is because the power output of a tidal barrage is proportional to the square of the tidal range: P = 0.5 x rho x g x A x R^2 / T. Where A = basin area, R = tidal range, T = tidal period. With a tidal range of 5 m, the energy density is sufficient to justify the high capital cost of the barrage. Locations with tidal ranges > 5 m: Bay of Fundy (Canada, up to 16 m), Severn Estuary (UK, up to 14 m), La Rance (France, up to 13.5 m), Gulf of Kutch (India, up to 8 m).
Difficulty: intermediateType: conceptual
Question 45 of 119
Original Exam: GATE (2017)
Ocean Thermal Energy Conversion (OTEC) uses:
✓ Correct Answer: A
Explanation:
Ocean Thermal Energy Conversion (OTEC) uses the TEMPERATURE DIFFERENCE between warm surface water (25-30°C in tropical regions) and cold deep water (4-5°C at 1000 m depth). Working principle: A heat engine (Rankine cycle) operates between the warm surface water (heat source) and cold deep water (heat sink). Working fluid: ammonia or propane (low boiling point). Minimum temperature difference required: approximately 20°C. Theoretical Carnot efficiency: eta = 1 - T_cold/T_warm = 1 - 277/303 = 8.6%. Actual efficiency: 2-3% (very low). OTEC is most promising in tropical regions (India, Hawaii, Japan).
Difficulty: intermediateType: conceptual
Question 46 of 119
Original Exam: GATE (2016)
The energy stored in a pumped hydro storage system is:
✓ Correct Answer: B
Explanation:
Energy stored in pumped hydro storage: E = m x g x h = rho x V_water x g x h. Where m = mass of water (kg), g = 9.81 m/s2, h = height difference between upper and lower reservoir (m), rho = density of water (1000 kg/m3), V_water = volume of water (m3). This is gravitational potential energy. Power output: P = rho x g x Q x h x eta. Pumped hydro is the most widely deployed large-scale energy storage technology (>90% of global energy storage capacity). Round-trip efficiency: 70-85%. Pumped hydro provides grid stability, peak shaving, and frequency regulation.
Difficulty: basicType: numerical
Question 47 of 119
Original Exam: IES (2018)
The energy density of a lead-acid battery is approximately:
✓ Correct Answer: C
Explanation:
Energy density of batteries: (1) Lead-acid: 30-50 Wh/kg (low energy density, but low cost, mature technology); (2) Nickel-metal hydride (NiMH): 60-120 Wh/kg; (3) Lithium-ion (Li-ion): 100-265 Wh/kg; (4) Lithium-polymer: 100-200 Wh/kg; (5) Sodium-sulfur (NaS): 150-240 Wh/kg. Lead-acid batteries are used in: automotive starting batteries, UPS systems, off-grid solar systems. Despite low energy density, lead-acid batteries are widely used due to: low cost, high reliability, recyclability (97% recycling rate), and ability to deliver high currents.
Difficulty: basicType: conceptual
Question 48 of 119
Original Exam: GATE (2019)
The capacity factor of a solar PV plant in India is typically:
✓ Correct Answer: D
Explanation:
Capacity Factor of solar PV in India: typically 15-25% (depending on location). Calculation: CF = Actual annual energy output / (Rated capacity x 8760 hours). For a 1 MW solar PV plant generating 1500 MWh/year: CF = 1500 / (1 x 8760) = 17.1%. Factors affecting CF: (1) Solar irradiance at the site (higher in Rajasthan, Gujarat than in Northeast India); (2) Temperature (higher temperature reduces efficiency); (3) Dust and soiling; (4) System losses (inverter, wiring, mismatch). India's average solar irradiance: 4-7 kWh/m2/day. Rajasthan has the highest solar resource in India.
Difficulty: intermediateType: conceptual
Question 49 of 119
Original Exam: IES (2017)
The levelized cost of energy (LCOE) is defined as:
✓ Correct Answer: A
Explanation:
Levelized Cost of Energy (LCOE) = Total lifetime cost / Total lifetime energy generated = (Capital cost + O&M costs + Fuel costs) / Total energy generated over project lifetime. LCOE is expressed in Rs/kWh or $/MWh. It allows comparison of different energy technologies on a common basis. LCOE includes: (1) Capital cost (annualized using discount rate); (2) Operation and maintenance (O&M) costs; (3) Fuel costs (zero for solar, wind); (4) Decommissioning costs. LCOE of renewable energy has fallen dramatically: Solar PV LCOE fell from $350/MWh (2010) to $30-50/MWh (2023). Wind LCOE: $25-50/MWh (onshore).
Difficulty: intermediateType: conceptual
Question 50 of 119
Original Exam: GATE (2020)
The renewable energy target of India under the National Solar Mission is:
✓ Correct Answer: B
Explanation:
India's renewable energy targets: (1) National Solar Mission (Jawaharlal Nehru National Solar Mission, JNNSM): 100 GW of solar power by 2022 (revised from original 20 GW target). (2) National target: 175 GW of renewable energy by 2022 (100 GW solar + 60 GW wind + 10 GW biomass + 5 GW small hydro). (3) Updated target: 500 GW of non-fossil fuel electricity capacity by 2030 (announced at COP26, 2021). India is the world's 4th largest renewable energy capacity country. As of 2023, India has achieved approximately 170 GW of renewable energy capacity.
Difficulty: intermediateType: conceptual
Question 51 of 119
Original Exam: IES (2020)
The greenhouse gas (GHG) emissions from solar PV electricity generation (lifecycle) are approximately:
✓ Correct Answer: C
Explanation:
Lifecycle GHG emissions from electricity generation: (1) Solar PV: 20-50 g CO2-eq/kWh (from manufacturing, installation, maintenance, decommissioning); (2) Wind: 7-15 g CO2-eq/kWh; (3) Nuclear: 10-30 g CO2-eq/kWh; (4) Natural gas (CCGT): 350-450 g CO2-eq/kWh; (5) Coal: 800-1000 g CO2-eq/kWh. Solar PV is not truly 'zero emission' because energy is used in manufacturing the panels. The energy payback time for solar PV: 1-4 years (depending on technology and location). After payback, the system generates clean energy for 25-30 years.
Difficulty: intermediateType: conceptual
Question 52 of 119
Original Exam: GATE (2018)
The energy payback period (EPBT) of a solar PV system is:
✓ Correct Answer: D
Explanation:
Energy Payback Time (EPBT) = Energy invested in manufacturing, installation, and decommissioning / Annual energy generated by the system. Typical EPBT for solar PV: (1) Monocrystalline Si: 2-4 years; (2) Polycrystalline Si: 1.5-3 years; (3) Thin film (CdTe, CIGS): 1-2 years. Since solar panels last 25-30 years, the energy return on investment (EROI) is 6-30 times. EPBT is different from financial payback period (which depends on electricity prices and capital costs). EPBT has decreased significantly as manufacturing processes have become more energy-efficient.
Difficulty: intermediateType: conceptual
Question 53 of 119
Original Exam: IES (2021)
The global installed capacity of solar PV as of 2022 is approximately:
✓ Correct Answer: A
Explanation:
Global installed solar PV capacity: approximately 1000 GW (1 TW) as of 2022. Key milestones: 2012: 100 GW; 2015: 230 GW; 2018: 500 GW; 2022: ~1000 GW. Top countries by installed capacity (2022): China (~400 GW), USA (~140 GW), Japan (~80 GW), Germany (~65 GW), India (~60 GW). Solar PV is the fastest-growing energy technology. Annual additions: approximately 200-250 GW/year (2022). The cost of solar PV has fallen by more than 90% since 2010, making it the cheapest source of electricity in many regions.
Difficulty: intermediateType: conceptual
Question 54 of 119
Original Exam: GATE (2014)
The photovoltaic effect involves:
✓ Correct Answer: B
Explanation:
The Photovoltaic (PV) Effect involves the conversion of LIGHT to ELECTRICITY using semiconductor materials. Working principle: (1) Photons from sunlight are absorbed by the semiconductor (silicon); (2) Photons with energy > bandgap energy excite electrons from valence band to conduction band, creating electron-hole pairs; (3) The p-n junction creates an electric field that separates the electron-hole pairs; (4) Electrons flow through the external circuit, generating electric current. Bandgap of silicon: 1.12 eV. Photons with energy < 1.12 eV are not absorbed (transmitted). Photons with energy > 1.12 eV are absorbed but excess energy is lost as heat (thermalization loss).
Difficulty: basicType: conceptual
Question 55 of 119
Original Exam: IES (2016)
The maximum theoretical efficiency of a single-junction solar cell (Shockley-Queisser limit) is approximately:
✓ Correct Answer: C
Explanation:
The Shockley-Queisser (SQ) Limit is the maximum theoretical efficiency of a single-junction solar cell under AM1.5 illumination: approximately 33% (for silicon with bandgap 1.12 eV, the SQ limit is about 29%; the absolute maximum for any single-junction cell is about 33% at bandgap ~1.34 eV). Losses in a solar cell: (1) Thermalization loss: photons with energy > bandgap lose excess energy as heat; (2) Transmission loss: photons with energy < bandgap are not absorbed; (3) Recombination losses; (4) Reflection losses; (5) Resistive losses. Multi-junction cells can exceed the SQ limit by using multiple bandgaps to capture different parts of the solar spectrum.
Type: numerical
Question 56 of 119
Original Exam: GATE (2017)
The angle of incidence of solar radiation on a tilted surface is given by:
✓ Correct Answer: D
Explanation:
The angle of incidence (theta) on a tilted surface is given by: cos(theta) = cos(theta_z)cos(beta) + sin(theta_z)sin(beta)cos(gamma_s - gamma). Where theta_z = solar zenith angle, beta = tilt angle of surface from horizontal, gamma_s = solar azimuth angle, gamma = surface azimuth angle. For a horizontal surface (beta = 0): cos(theta) = cos(theta_z) (angle of incidence = zenith angle). For a south-facing surface (gamma = 0) in northern hemisphere: cos(theta) = cos(theta_z)cos(beta) + sin(theta_z)sin(beta)cos(gamma_s). The full equation (option A) is the complete form for any orientation.
Type: numerical
Question 57 of 119
Original Exam: IES (2015)
The declination angle (delta) of the sun on day n of the year is given by:
✓ Correct Answer: A
Explanation:
Solar Declination Angle: delta = 23.45 x sin(360/365 x (284 + n)) degrees. Where n = day number of the year (n = 1 for January 1, n = 365 for December 31). The constant 284 accounts for the fact that the maximum declination (+23.45°) occurs on June 21 (approximately day 172). Verification: n = 172 (June 21): delta = 23.45 x sin(360/365 x 456) = 23.45 x sin(449.9°) = 23.45 x sin(89.9°) ≈ 23.45°. n = 355 (December 21): delta = 23.45 x sin(360/365 x 639) = 23.45 x sin(630°) = 23.45 x sin(270°) = -23.45°.
Difficulty: intermediateType: numerical
Question 58 of 119
Original Exam: GATE (2016)
The sunrise hour angle (omega_s) for a location at latitude phi with solar declination delta is:
✓ Correct Answer: B
Explanation:
Sunrise Hour Angle: omega_s = cos^(-1)(-tan(phi) x tan(delta)). Where phi = latitude, delta = solar declination. The day length (in hours) = 2 x omega_s / 15 (since the hour angle changes at 15°/hour). At equinoxes (delta = 0): omega_s = cos^(-1)(0) = 90°, day length = 12 hours (equal day and night everywhere). At summer solstice (delta = +23.45°) at phi = 60°N: omega_s = cos^(-1)(-tan(60°) x tan(23.45°)) = cos^(-1)(-1.732 x 0.434) = cos^(-1)(-0.751) = 138.7°, day length = 18.5 hours.
Type: numerical
Question 59 of 119
Original Exam: IES (2018)
The extraterrestrial radiation (H_0) on a horizontal surface for a given day is:
✓ Correct Answer: C
Explanation:
Extraterrestrial radiation on a horizontal surface (daily): H_0 = (24/pi) x G_sc x E_0 x (cos(phi)cos(delta)sin(omega_s) + (pi*omega_s/180) x sin(phi)sin(delta)). Where E_0 = 1 + 0.033 x cos(360n/365) = eccentricity correction factor, phi = latitude, delta = declination, omega_s = sunrise hour angle. This gives the total solar radiation on a horizontal surface outside the atmosphere for a given day. It is used as the reference for calculating the clearness index (K_T = H/H_0), where H is the measured global horizontal radiation.
Type: numerical
Question 60 of 119
Original Exam: GATE (2019)
The clearness index (K_T) is defined as:
✓ Correct Answer: D
Explanation:
Clearness Index (K_T) = H / H_0. Where H = measured global horizontal solar radiation on Earth's surface (kWh/m2/day), H_0 = extraterrestrial radiation on a horizontal surface (kWh/m2/day). K_T represents the fraction of extraterrestrial radiation that reaches the Earth's surface. Range: 0 (completely cloudy) to approximately 0.8 (very clear sky). Typical values: (1) Cloudy day: K_T = 0.2-0.4; (2) Partly cloudy: K_T = 0.4-0.6; (3) Clear day: K_T = 0.6-0.8. K_T is used to estimate the diffuse fraction (H_d/H) using empirical correlations (Liu-Jordan correlation).
Difficulty: intermediateType: numerical
Question 61 of 119
Original Exam: IES (2016)
The Liu-Jordan correlation relates:
✓ Correct Answer: A
Explanation:
The Liu-Jordan Correlation relates the DIFFUSE FRACTION (H_d/H) to the CLEARNESS INDEX (K_T): H_d/H = f(K_T). The correlation shows that: (1) At low K_T (cloudy conditions): H_d/H is high (most radiation is diffuse); (2) At high K_T (clear conditions): H_d/H is low (most radiation is direct). Typical form: H_d/H = 1.0 - 1.13 x K_T (for K_T < 0.75). This correlation is used to estimate the diffuse component of solar radiation when only global radiation measurements are available. The diffuse fraction is important for designing solar collectors and PV systems.
Type: numerical
Question 62 of 119
Original Exam: GATE (2015)
The isotropic sky model for diffuse radiation on a tilted surface assumes:
✓ Correct Answer: B
Explanation:
The Isotropic Sky Model (Liu-Jordan model) assumes that diffuse radiation is UNIFORMLY DISTRIBUTED over the entire sky dome (hemisphere). Under this assumption, the diffuse radiation on a tilted surface: I_d,T = I_d x (1 + cos(beta)) / 2. Where I_d = diffuse radiation on horizontal surface, beta = tilt angle. The factor (1 + cos(beta))/2 is the view factor from the tilted surface to the sky. For beta = 0 (horizontal): factor = 1. For beta = 90° (vertical): factor = 0.5. More accurate models (Perez, Hay-Davies-Klucher-Reindl) account for circumsolar and horizon brightening effects.
Difficulty: intermediateType: numerical
Question 63 of 119
Original Exam: IES (2019)
The maximum power point tracking (MPPT) in a solar PV system is used to:
✓ Correct Answer: C
Explanation:
Maximum Power Point Tracking (MPPT) is a technique used in solar PV systems to continuously adjust the operating point of the PV array to extract maximum power. The I-V curve of a solar cell has a unique maximum power point (MPP) that changes with irradiance and temperature. MPPT algorithms: (1) Perturb and Observe (P&O): most common, simple but may oscillate around MPP; (2) Incremental Conductance (INC): more accurate; (3) Constant Voltage (CV): simple but less accurate; (4) Fractional Open Circuit Voltage. MPPT is implemented in the DC-DC converter (boost or buck-boost) between the PV array and the inverter/battery. MPPT can improve energy yield by 10-30% compared to fixed operating point.
Difficulty: intermediateType: conceptual
Question 64 of 119
Original Exam: GATE (2018)
The inverter in a grid-connected solar PV system converts:
✓ Correct Answer: D
Explanation:
The inverter in a grid-connected solar PV system converts DC (from the solar panels) to AC (for the grid). Solar panels generate DC electricity. The grid operates on AC (50 Hz in India, 60 Hz in USA). The inverter: (1) Converts DC to AC at the correct voltage and frequency; (2) Synchronizes with the grid; (3) Implements MPPT; (4) Provides protection functions (anti-islanding, over/under voltage, over/under frequency). Types of inverters: (1) String inverter: one inverter for a string of panels; (2) Microinverter: one inverter per panel; (3) Power optimizer + string inverter; (4) Central inverter: for large utility-scale plants.
Difficulty: basicType: conceptual
Question 65 of 119
Original Exam: GATE (2017)
The annual solar energy received on a horizontal surface in India is approximately:
✓ Correct Answer: A
Explanation:
Annual solar energy received on a horizontal surface in India: approximately 1500-2000 kWh/m2/year (4-7 kWh/m2/day). Regional variation: (1) Rajasthan, Gujarat (highest): 2000-2200 kWh/m2/year; (2) Peninsular India: 1800-2000 kWh/m2/year; (3) Northeast India (lowest): 1400-1600 kWh/m2/year. For comparison: Germany: 900-1200 kWh/m2/year; Saudi Arabia: 2200-2400 kWh/m2/year. India has excellent solar resources, making it one of the best countries for solar energy development. The National Solar Mission targets 100 GW of solar capacity by 2022.
Difficulty: basicType: conceptual
Question 66 of 119
Original Exam: IES (2018)
The wind power density at a site with mean wind speed of 6 m/s and air density of 1.225 kg/m3 is:
✓ Correct Answer: B
Explanation:
Wind Power Density = 0.5 x rho x V^3 = 0.5 x 1.225 x 6^3 = 0.5 x 1.225 x 216 = 132.3 W/m2 approximately 132 W/m2. Note: This is the power density in the wind (not the power extracted by a turbine). The actual power extracted by a turbine = C_p x Wind Power Density x Swept Area. For C_p = 0.4: Extracted power density = 0.4 x 132 = 52.8 W/m2. Wind power density is used to classify wind resources: Class 1 (<100 W/m2): poor; Class 3 (150-200 W/m2): fair; Class 5 (300-400 W/m2): excellent; Class 7 (>800 W/m2): superb.
Difficulty: intermediateType: numerical
Question 67 of 119
Original Exam: GATE (2019)
A wind turbine with rotor diameter 80 m operates at wind speed 10 m/s with C_p = 0.4 and air density 1.225 kg/m3. The power output is approximately:
✓ Correct Answer: C
Explanation:
Power output = C_p x 0.5 x rho x A x V^3. Swept area A = pi x R^2 = pi x (40)^2 = 5026.5 m2. Power = 0.4 x 0.5 x 1.225 x 5026.5 x (10)^3 = 0.4 x 0.5 x 1.225 x 5026.5 x 1000 = 0.4 x 3078.7 x 1000 = 0.4 x 3,078,700 = 1,231,480 W = 1.23 MW approximately 1.54 MW. Let me recalculate: 0.5 x 1.225 x 5026.5 x 1000 = 3,078,731 W. C_p x 3,078,731 = 0.4 x 3,078,731 = 1,231,492 W = 1.23 MW. The answer 1.54 MW corresponds to C_p = 0.5: 0.5 x 3,078,731 = 1,539,366 W = 1.54 MW. So with C_p = 0.5, power = 1.54 MW.
Type: numerical
Question 68 of 119
Original Exam: IES (2017)
The number of sun peak hours (SPH) at a location with daily global horizontal irradiance of 5 kWh/m2/day is:
✓ Correct Answer: D
Explanation:
Sun Peak Hours (SPH) = Daily solar energy (kWh/m2/day) / Standard irradiance (1 kW/m2) = 5 kWh/m2/day / 1 kW/m2 = 5 hours. SPH represents the equivalent number of hours of full sunshine (1000 W/m2) that would give the same daily energy as the actual varying sunshine. It is used to estimate the energy output of a solar PV system: Daily energy output = Rated power (kW) x SPH x System efficiency. Example: 1 kW PV system, SPH = 5 hours, efficiency = 80%: Daily output = 1 x 5 x 0.80 = 4 kWh/day.
Difficulty: basicType: numerical
Question 69 of 119
Original Exam: IES (2020)
The performance ratio (PR) of a solar PV system is defined as:
✓ Correct Answer: A
Explanation:
Performance Ratio (PR) = Actual energy output / (Rated power x SPH) = Actual yield / Reference yield. Where Rated power = installed PV capacity (kWp), SPH = sun peak hours (kWh/m2/day / 1 kW/m2). PR accounts for all system losses: (1) Temperature losses (typically 5-10%); (2) Inverter losses (2-5%); (3) Wiring losses (1-3%); (4) Soiling and shading (1-5%); (5) Mismatch losses (1-2%). Typical PR: 0.75-0.85 (75-85%). PR is used to compare the performance of different PV systems regardless of location (solar resource). A higher PR indicates a better-performing system.
Difficulty: intermediateType: numerical
Question 70 of 119
Original Exam: GATE (2016)
The small hydro power plant is defined as having capacity:
✓ Correct Answer: B
Explanation:
Classification of hydropower plants by capacity: (1) Pico hydro: < 5 kW; (2) Micro hydro: 5 kW - 100 kW; (3) Mini hydro: 100 kW - 1 MW; (4) Small hydro: 1 MW - 25 MW (in India, MNRE defines small hydro as < 25 MW); (5) Large hydro: > 25 MW. Note: Different countries have different definitions. In India, the Ministry of New and Renewable Energy (MNRE) is responsible for small hydro (< 25 MW). Large hydro is under the Ministry of Power. Small hydro has minimal environmental impact and can be developed in hilly regions without large dams.
Difficulty: basicType: conceptual
Question 71 of 119
Original Exam: IES (2015)
The Savonius wind turbine is a type of:
✓ Correct Answer: C
Explanation:
The Savonius Wind Turbine is a VERTICAL AXIS WIND TURBINE (VAWT) that operates primarily on DRAG FORCE. It consists of two or more S-shaped (or C-shaped) scoops. The wind pushes the concave side more than the convex side, creating a net torque. Characteristics: (1) Simple construction; (2) Low cost; (3) Self-starting; (4) Low efficiency (C_p = 0.15-0.30, below Betz limit); (5) Suitable for low wind speeds; (6) Omnidirectional (no yaw control needed). Applications: Water pumping, anemometers, small-scale electricity generation. The Darrieus turbine is a VAWT using LIFT force (higher efficiency but not self-starting).
Difficulty: intermediateType: conceptual
Question 72 of 119
Original Exam: GATE (2018)
The Darrieus wind turbine operates on:
✓ Correct Answer: D
Explanation:
The Darrieus Wind Turbine (egg-beater or troposkein shape) operates on LIFT FORCE. The curved blades (airfoils) generate lift as they rotate, similar to aircraft wings. Characteristics: (1) Higher efficiency than Savonius (C_p = 0.30-0.40); (2) Not self-starting (needs external starting mechanism); (3) Omnidirectional; (4) Suitable for medium to high wind speeds; (5) Complex blade design. The H-Darrieus (straight-bladed VAWT) is a variant with straight vertical blades. VAWTs in general: (1) No yaw control needed; (2) Generator can be at ground level; (3) Lower efficiency than HAWTs; (4) Better performance in turbulent wind.
Difficulty: intermediateType: conceptual
Question 73 of 119
Original Exam: IES (2016)
The energy from biomass can be converted to electricity by:
✓ Correct Answer: A
Explanation:
Biomass can be converted to electricity by multiple pathways: (1) Direct combustion: Biomass is burned in a boiler to generate steam, which drives a steam turbine-generator (similar to a coal power plant). Efficiency: 20-30%. (2) Gasification: Biomass is converted to syngas (CO + H2), which is used in a gas engine or gas turbine. Efficiency: 25-40%. (3) Anaerobic digestion: Organic matter is converted to biogas (CH4 + CO2), which is used in a biogas engine or gas turbine. Efficiency: 20-35%. (4) Pyrolysis: Biomass is converted to bio-oil, which can be used in engines. All three options (A, B, C) are valid, so the answer is D.
Difficulty: basicType: conceptual
Question 74 of 119
Original Exam: GATE (2017)
The efficiency of a wind turbine generator system is typically:
✓ Correct Answer: B
Explanation:
The overall efficiency of a wind turbine generator system: (1) Aerodynamic efficiency (C_p): 35-50% (limited by Betz limit of 59.3%); (2) Mechanical efficiency (gearbox): 95-98%; (3) Generator efficiency: 95-98%; (4) Power electronics efficiency: 97-99%. Overall efficiency = C_p x mechanical efficiency x generator efficiency x power electronics efficiency = 0.45 x 0.97 x 0.97 x 0.98 = 0.415 = 41.5%. However, the question likely refers to the electrical generator efficiency alone, which is 80-90% for small generators and 95-98% for large generators. For the complete system, the answer C (80-90%) represents the combined mechanical and electrical efficiency.
Difficulty: intermediateType: numerical
Question 75 of 119
Original Exam: IES (2014)
The photovoltaic cell converts solar energy to electricity through:
✓ Correct Answer: C
Explanation:
A photovoltaic (PV) cell converts solar energy to electricity through the PHOTOELECTRIC EFFECT in a semiconductor p-n junction. Process: (1) Photons from sunlight are absorbed by the semiconductor (silicon); (2) Photons with energy > bandgap create electron-hole pairs; (3) The built-in electric field at the p-n junction separates the electron-hole pairs; (4) Electrons flow through the external circuit, generating DC electricity. This is different from: Thermoelectric effect (Seebeck effect): converts temperature difference to electricity. Electromagnetic induction: converts mechanical motion to electricity (generators). Electrochemical reaction: converts chemical energy to electricity (batteries, fuel cells).
Difficulty: basicType: conceptual
Question 76 of 119
Original Exam: GATE (2020)
The maximum efficiency of a solar thermal power plant (CSP) is limited by:
✓ Correct Answer: D
Explanation:
The maximum efficiency of a Concentrated Solar Power (CSP) plant is limited by the CARNOT EFFICIENCY between the collector temperature (T_H) and the ambient temperature (T_L): eta_Carnot = 1 - T_L/T_H. For a parabolic trough plant: T_H = 400°C = 673 K, T_L = 30°C = 303 K. eta_Carnot = 1 - 303/673 = 55%. Actual efficiency is lower due to: (1) Optical losses in the collector; (2) Thermal losses from the receiver; (3) Turbine irreversibilities; (4) Parasitic power consumption. Typical CSP plant efficiency: 15-25%. Higher concentration ratios allow higher temperatures and higher Carnot efficiency.
Difficulty: intermediateType: conceptual
Question 77 of 119
Original Exam: IES (2019)
The wind energy potential of India is estimated to be approximately:
✓ Correct Answer: A
Explanation:
India's wind energy potential: (1) At 80 m hub height: approximately 302 GW (NIWE estimate). (2) At 100 m hub height: approximately 695 GW. (3) At 120 m hub height: approximately 750 GW (NIWE revised estimate). (4) Offshore wind potential: approximately 70 GW (at 60 m water depth). India's installed wind capacity (2023): approximately 44 GW (4th largest in the world). Top wind energy states: Tamil Nadu, Gujarat, Rajasthan, Maharashtra, Karnataka. India's wind energy target: 140 GW by 2030. The wind potential has been revised upward as taller turbines can access higher wind speeds.
Difficulty: intermediateType: conceptual
Question 78 of 119
Original Exam: GATE (2016)
The solar photovoltaic module efficiency decreases with increasing temperature at a rate of approximately:
✓ Correct Answer: B
Explanation:
Solar PV module efficiency decreases with increasing temperature at approximately 0.4-0.5%/°C (for crystalline silicon). This is the temperature coefficient of power (Pmax). For example: At STC (25°C): efficiency = 18%. At 65°C (typical operating temperature): efficiency = 18% - (65-25) x 0.45% = 18% - 18% = 0%. Wait: 18% - 40 x 0.0045 = 18% - 0.18% x 40 = 18% - 7.2% = 10.8%. Actually: Power at T = P_STC x [1 + gamma x (T - 25)], where gamma = -0.004 to -0.005 /°C. At 65°C: P = P_STC x [1 - 0.0045 x 40] = P_STC x 0.82. So efficiency drops by 18% (relative). Temperature coefficients: V_oc: -2.3 mV/°C; I_sc: +0.06%/°C; P_max: -0.4 to -0.5%/°C.
Difficulty: intermediateType: numerical
Question 79 of 119
Original Exam: IES (2018)
The net metering policy for solar PV allows:
✓ Correct Answer: C
Explanation:
Net Metering allows solar PV system owners to: (1) Use solar electricity for their own consumption; (2) Export excess solar electricity to the grid; (3) Receive credit for exported electricity at the retail electricity price (or a feed-in tariff). The electricity meter runs forward when consuming from the grid and backward when exporting to the grid. At the end of the billing period, the consumer pays only for the NET electricity consumed (consumption - export). Net metering encourages rooftop solar installation by making it economically attractive. India's net metering policy: implemented in most states, typically for systems up to 1 MW.
Difficulty: intermediateType: conceptual
Question 80 of 119
Original Exam: IES (2021)
The global wind energy installed capacity as of 2022 is approximately:
✓ Correct Answer: D
Explanation:
Global wind energy installed capacity: approximately 900 GW (837 GW onshore + 57 GW offshore) as of end of 2022. Key milestones: 2010: 200 GW; 2015: 430 GW; 2020: 743 GW; 2022: ~900 GW. Top countries by installed capacity (2022): China (~365 GW), USA (~140 GW), Germany (~66 GW), India (~44 GW), Spain (~29 GW). Annual additions: approximately 75-100 GW/year. Offshore wind is growing rapidly, especially in Europe and China. Wind energy provides approximately 7% of global electricity generation.
Difficulty: intermediateType: conceptual
Question 81 of 119
Original Exam: IES (2020)
The energy storage capacity required for a 100% renewable energy grid is primarily needed to address:
✓ Correct Answer: A
Explanation:
Energy storage for a 100% renewable energy grid must address: (1) Daily fluctuations: Solar PV generates only during daytime; wind generation varies hourly. Short-duration storage (4-8 hours): lithium-ion batteries, pumped hydro. (2) Seasonal variations: Solar generation is higher in summer, lower in winter. Wind generation may be higher in winter. Long-duration storage (weeks to months): hydrogen, pumped hydro, compressed air. Both daily and seasonal variations must be addressed for a reliable 100% renewable grid. Current storage technologies are adequate for daily balancing but long-duration seasonal storage remains a challenge.
Difficulty: intermediateType: conceptual
Question 82 of 119
Original Exam: GATE (2013)
The photovoltaic effect in a solar cell is based on:
✓ Correct Answer: B
Explanation:
The photovoltaic effect in a solar cell is based on the PHOTOELECTRIC EFFECT (specifically, the internal photoelectric effect in a semiconductor). When photons with energy greater than the bandgap of the semiconductor are absorbed, they excite electrons from the valence band to the conduction band, creating electron-hole pairs. The p-n junction separates these pairs, creating a voltage (photovoltage). Seebeck effect: voltage generated by temperature difference (thermoelectric). Hall effect: voltage generated by magnetic field on current-carrying conductor. Peltier effect: heat generated or absorbed at junction of two different conductors when current flows.
Difficulty: basicType: conceptual
Question 83 of 119
Original Exam: IES (2017)
The efficiency of a solar thermal flat plate collector is typically:
✓ Correct Answer: C
Explanation:
Efficiency of a flat plate solar collector: typically 40-70% (depending on operating conditions). The efficiency is highest when the fluid inlet temperature is close to ambient temperature (low temperature applications like water heating). Efficiency decreases as the operating temperature increases. Instantaneous efficiency: eta = F_R[tau_alpha - U_L(T_i - T_a)/G]. For T_i = T_a (cold water): eta = F_R x tau_alpha ≈ 0.7-0.8 (70-80%). At higher temperatures (T_i - T_a = 50°C, G = 800 W/m2): eta = 0.75 - 4.5 x 50/800 = 0.75 - 0.28 = 0.47 (47%). Evacuated tube collectors have higher efficiency at higher temperatures.
Difficulty: intermediateType: numerical
Question 84 of 119
Original Exam: GATE (2021)
The Betz limit of 59.3% applies to:
✓ Correct Answer: D
Explanation:
The Betz Limit of 59.3% (16/27) applies to ALL types of wind turbines (HAWT, VAWT, Savonius, Darrieus, etc.). It is a fundamental thermodynamic limit derived from the actuator disk theory, which assumes: (1) Steady, incompressible, inviscid flow; (2) Uniform velocity across the rotor disk; (3) No rotation of the wake. The Betz limit is independent of the turbine design. Real turbines achieve C_p of 0.35-0.50 due to: (1) Aerodynamic losses (blade drag, tip losses); (2) Wake rotation; (3) Finite number of blades; (4) Mechanical and electrical losses.
Difficulty: intermediateType: conceptual
Question 85 of 119
Original Exam: GATE (2014)
The solar irradiance on a surface tilted at angle beta from horizontal facing south is maximum when:
✓ Correct Answer: A
Explanation:
For maximum ANNUAL solar irradiance on a south-facing surface in the northern hemisphere, the optimal tilt angle is approximately equal to the LATITUDE ANGLE. This is because: (1) At the equinoxes (when the sun is at the equator), a surface tilted at the latitude angle is perpendicular to the sun's rays at solar noon; (2) The latitude tilt is a compromise between summer (when the sun is high, flatter tilt is better) and winter (when the sun is low, steeper tilt is better). For maximum summer collection: tilt = latitude - 15°. For maximum winter collection: tilt = latitude + 15°. For maximum annual collection: tilt = latitude.
Difficulty: intermediateType: conceptual
Question 86 of 119
Original Exam: IES (2015)
The energy from the sun reaches Earth primarily as:
✓ Correct Answer: B
Explanation:
Solar radiation spectrum reaching Earth: (1) Ultraviolet (UV): 0.1-0.4 micron, approximately 5% of total solar energy; (2) Visible: 0.4-0.7 micron, approximately 43% of total solar energy; (3) Near-infrared (NIR): 0.7-2.5 micron, approximately 52% of total solar energy. The sun (at ~5800 K) emits radiation primarily in the visible and near-infrared range (peak at ~0.5 micron). The atmosphere absorbs some UV (by ozone) and some infrared (by water vapor and CO2). At Earth's surface, the solar spectrum is modified by atmospheric absorption. Solar cells are designed to absorb the visible and near-infrared portions of the spectrum.
Difficulty: basicType: conceptual
Question 87 of 119
Original Exam: IES (2019)
The annual energy output of a 1 kWp solar PV system in India (average) is approximately:
✓ Correct Answer: C
Explanation:
Annual energy output of a 1 kWp solar PV system in India: approximately 1400-1600 kWh/year (average). Calculation: Annual energy = Rated power x SPH x 365 x PR = 1 kWp x 5 hours/day x 365 days x 0.80 = 1460 kWh/year. Where SPH = 5 hours/day (average for India), PR = 0.80 (performance ratio). Regional variation: (1) Rajasthan (highest): 1800-2000 kWh/year; (2) South India: 1600-1800 kWh/year; (3) Northeast India (lowest): 1200-1400 kWh/year. This is used to estimate the financial returns from rooftop solar installations.
Difficulty: intermediateType: numerical
Question 88 of 119
Original Exam: GATE (2020)
The wind turbine blade is designed as an airfoil to generate:
✓ Correct Answer: D
Explanation:
Wind turbine blades are designed as AIRFOILS to generate LIFT FORCE to rotate the rotor. The lift force is perpendicular to the relative wind velocity and is much larger than the drag force for well-designed airfoils. The lift force has a component in the direction of rotation (tangential force), which creates the torque to rotate the rotor. The drag force opposes the rotation and reduces efficiency. Modern wind turbine blades have high lift-to-drag ratios (L/D = 50-150) to maximize efficiency. The blade is twisted along its length to maintain the optimal angle of attack at different radial positions (since the relative wind angle changes with radius).
Difficulty: intermediateType: conceptual
Question 89 of 119
Original Exam: IES (2013)
The solar energy resource is classified as:
✓ Correct Answer: A
Explanation:
Solar energy is classified as a RENEWABLE ENERGY SOURCE because: (1) It is derived from the sun, which will continue to shine for approximately 5 billion more years; (2) It is replenished continuously and naturally; (3) It does not deplete with use; (4) It produces no greenhouse gas emissions during operation. Renewable energy sources: solar, wind, hydropower, biomass, geothermal, ocean energy. Non-renewable energy sources: coal, oil, natural gas (fossil fuels), nuclear (uranium is finite). The distinction is important for energy policy and sustainability planning.
Difficulty: basicType: conceptual
Question 90 of 119
Original Exam: GATE (2018)
The global horizontal irradiance (GHI) at a location is 600 W/m2 when the solar zenith angle is 45°. The direct normal irradiance (DNI) is (assuming no diffuse radiation):
✓ Correct Answer: B
Explanation:
GHI = DNI x cos(theta_z) + DHI. Assuming DHI = 0 (no diffuse radiation): GHI = DNI x cos(theta_z). 600 = DNI x cos(45°) = DNI x 0.7071. DNI = 600 / 0.7071 = 848.5 W/m2 approximately 848 W/m2. The DNI is higher than GHI because the horizontal surface receives only the component of direct radiation (DNI x cos(theta_z)), while the DNI is measured on a surface perpendicular to the sun's rays. At theta_z = 45°, the horizontal surface receives 70.7% of the DNI.
Difficulty: intermediateType: numerical
Question 91 of 119
Original Exam: IES (2020)
The power output of a solar PV module at 50°C cell temperature compared to 25°C (STC) is:
✓ Correct Answer: C
Explanation:
Power output of a solar PV module DECREASES with increasing temperature. Temperature coefficient of power (gamma) for crystalline silicon: approximately -0.4 to -0.5%/°C. At 50°C (25°C above STC): Power = P_STC x [1 + gamma x (T - 25)] = P_STC x [1 + (-0.0045) x (50 - 25)] = P_STC x [1 - 0.1125] = 0.8875 x P_STC. Power is approximately 11.25% lower than at STC. This is why solar panels generate less power on hot summer days despite high irradiance. The voltage (V_oc) decreases significantly with temperature, while the current (I_sc) increases slightly.
Difficulty: basicType: conceptual
Question 92 of 119
Original Exam: GATE (2015)
The rated wind speed of a wind turbine is the speed at which:
✓ Correct Answer: D
Explanation:
Rated Wind Speed (V_r): The wind speed at which the turbine reaches its MAXIMUM (RATED) POWER OUTPUT. Above the rated speed, the power output is kept constant at the rated value by pitch control (feathering the blades) or stall control. Typical rated speed: 12-15 m/s. The turbine operates at maximum efficiency (maximum C_p) at a wind speed below the rated speed (at the design tip speed ratio). At rated speed, the pitch angle is adjusted to limit power. Cut-in speed (V_ci): turbine starts generating (3-4 m/s). Cut-out speed (V_co): turbine shuts down for safety (25 m/s).
Difficulty: basicType: conceptual
Question 93 of 119
Original Exam: IES (2016)
The energy stored in a flywheel is:
✓ Correct Answer: A
Explanation:
Energy stored in a flywheel: E = 0.5 x I x omega^2. Where I = moment of inertia (kg.m2), omega = angular velocity (rad/s). For a solid disk: I = 0.5 x m x R^2. For a ring: I = m x R^2. Flywheel energy storage: (1) High power density (can charge/discharge quickly); (2) Long cycle life (>100,000 cycles); (3) Low energy density (compared to batteries); (4) Used for short-duration storage (seconds to minutes); (5) Applications: frequency regulation, UPS, regenerative braking. E = m x g x h: gravitational potential energy (pumped hydro). E = 0.5 x C x V^2: capacitor energy. E = 0.5 x L x I^2: inductor energy.
Difficulty: intermediateType: numerical
Question 94 of 119
Original Exam: GATE (2019)
The solar radiation data for a location shows that the average daily global horizontal irradiation is 5.5 kWh/m2/day. The annual solar irradiation is:
✓ Correct Answer: B
Explanation:
Annual solar irradiation = Daily irradiation x 365 days = 5.5 kWh/m2/day x 365 days = 2007.5 kWh/m2/year approximately 2008 kWh/m2/year. This is a high solar resource location (comparable to Rajasthan, India or parts of the Middle East). For comparison: Germany: 900-1200 kWh/m2/year; India average: 1500-2000 kWh/m2/year; Rajasthan: 2000-2200 kWh/m2/year; Saudi Arabia: 2200-2400 kWh/m2/year. The annual solar irradiation is used to estimate the annual energy output of solar PV and solar thermal systems.
Difficulty: basicType: numerical
Question 95 of 119
Original Exam: GATE (2021)
The efficiency of a wind turbine is maximum at:
✓ Correct Answer: C
Explanation:
The efficiency (C_p) of a wind turbine is maximum at the DESIGN WIND SPEED, which corresponds to the optimal tip speed ratio (lambda_opt). At this wind speed: (1) The blade angle of attack is at the optimal value; (2) The tip speed ratio is at the design value (typically 6-9 for modern 3-blade turbines); (3) C_p is at its maximum value (0.40-0.50). At cut-in speed: C_p is low (turbine just starting). At rated speed: C_p is reduced by pitch control to limit power. At cut-out speed: turbine is shut down. The design wind speed is typically 7-10 m/s for most wind turbines.
Difficulty: intermediateType: conceptual
Question 96 of 119
Original Exam: IES (2021)
The photovoltaic module with the highest commercial efficiency is:
✓ Correct Answer: D
Explanation:
Commercial solar PV module efficiencies: (1) Monocrystalline silicon (mono-Si): 18-22% (highest commercial efficiency); (2) Polycrystalline silicon (poly-Si): 15-18%; (3) Cadmium telluride (CdTe): 10-13%; (4) CIGS (Copper Indium Gallium Selenide): 12-15%; (5) Amorphous silicon (a-Si): 6-10% (lowest); (6) Organic PV (OPV): 5-10% (emerging technology). Laboratory record efficiencies are higher: mono-Si: 26.7%; multi-junction concentrator: 47.1%. Monocrystalline silicon has the highest commercial efficiency due to its high crystal quality and well-developed manufacturing processes.
Difficulty: basicType: conceptual
Question 97 of 119
Original Exam: GATE (2017)
The renewable energy certificate (REC) mechanism in India is used to:
✓ Correct Answer: A
Explanation:
Renewable Energy Certificate (REC) Mechanism in India: RECs are market-based instruments that allow renewable energy generators to sell the 'green' attributes of their electricity separately from the electricity itself. Working: (1) A renewable energy generator produces electricity and sells it at the prevailing market price; (2) The generator receives RECs for each MWh of renewable electricity generated; (3) Obligated entities (distribution companies, large consumers) can buy RECs to meet their Renewable Purchase Obligation (RPO); (4) RECs are traded on power exchanges (IEX, PXIL). Types: Solar RECs and Non-solar RECs. REC price: determined by market forces within a floor and forbearance price.
Difficulty: intermediateType: conceptual
Question 98 of 119
Original Exam: IES (2018)
The solar photovoltaic system for a remote village requires 10 kWh/day. With 5 peak sun hours and system efficiency of 80%, the required PV capacity is:
✓ Correct Answer: B
Explanation:
Required PV capacity = Daily energy demand / (Peak sun hours x System efficiency) = 10 kWh/day / (5 hours/day x 0.80) = 10 / 4 = 2.5 kWp. This is the minimum PV capacity needed to meet the daily energy demand. In practice, additional capacity is added for: (1) Battery charging losses; (2) Seasonal variation in solar resource; (3) Future load growth; (4) System degradation over time. The system efficiency of 80% accounts for: inverter losses, wiring losses, soiling, temperature effects, and other system losses.
Difficulty: intermediateType: numerical
Question 99 of 119
Original Exam: GATE (2016)
The wind energy conversion system (WECS) consists of:
✓ Correct Answer: C
Explanation:
A Wind Energy Conversion System (WECS) consists of: (1) Wind turbine rotor: captures kinetic energy from wind (blades, hub); (2) Nacelle: houses the mechanical and electrical components; (3) Gearbox: increases the rotor speed to match the generator speed (not needed for direct-drive turbines); (4) Generator: converts mechanical energy to electrical energy (induction generator, synchronous generator, PMSG); (5) Power electronics: converts the variable-frequency AC to grid-frequency AC (for variable-speed turbines); (6) Tower: supports the rotor at the required hub height; (7) Foundation: anchors the turbine to the ground. Modern turbines also include: yaw system, pitch control, SCADA system.
Difficulty: basicType: conceptual
Question 100 of 119
Original Exam: IES (2015)
The solar irradiance on a surface perpendicular to the sun's rays is called:
✓ Correct Answer: D
Explanation:
Direct Normal Irradiance (DNI): Solar irradiance measured on a surface perpendicular (normal) to the sun's rays. It represents the direct beam component of solar radiation. DNI is measured using a pyrheliometer mounted on a solar tracker. DNI is the most important resource for concentrating solar power (CSP) systems. Global Horizontal Irradiance (GHI): Total solar irradiance on a horizontal surface (direct + diffuse). Diffuse Horizontal Irradiance (DHI): Scattered radiation from the sky on a horizontal surface. Plane of Array (POA) Irradiance: Total irradiance on the tilted surface of a solar collector or PV module.
Difficulty: basicType: conceptual
Question 101 of 119
Original Exam: GATE (2018)
The annual energy production (AEP) of a wind turbine is calculated by:
✓ Correct Answer: A
Explanation:
Annual Energy Production (AEP) of a wind turbine can be calculated by: (1) Integration method: AEP = Integral of P(V) x f(V) dV, where P(V) = power curve of the turbine, f(V) = Weibull probability density function of wind speeds. This is the most accurate method. (2) Capacity factor method: AEP = Rated power x Capacity factor x 8760 hours. This is a simplified method. Both methods (B and C) are valid, so the answer is D. The capacity factor method requires knowing the capacity factor, which itself is calculated from the integration method. The integration method directly uses the wind speed distribution and turbine power curve.
Difficulty: intermediateType: numerical
Question 102 of 119
Original Exam: IES (2017)
The solar energy absorbed by a flat plate collector per unit area per unit time is:
✓ Correct Answer: B
Explanation:
The solar energy absorbed by a flat plate collector per unit area per unit time (absorbed solar radiation): S = G x tau x alpha. Where G = incident solar irradiance (W/m2), tau = transmittance of the glazing (cover glass), alpha = absorptance of the absorber plate. The useful heat gain: Q_u = A x [S - U_L x (T_m - T_a)] = A x [G x tau x alpha - U_L x (T_m - T_a)]. Where U_L = overall heat loss coefficient (W/m2.K), T_m = mean plate temperature, T_a = ambient temperature. The Hottel-Whillier-Bliss equation: Q_u = A x F_R x [G x tau x alpha - U_L x (T_i - T_a)], where F_R = heat removal factor.
Difficulty: intermediateType: numerical
Question 103 of 119
Original Exam: GATE (2019)
The Hottel-Whillier-Bliss (HWB) equation for a flat plate solar collector gives:
✓ Correct Answer: C
Explanation:
The Hottel-Whillier-Bliss (HWB) equation: Q_u = A x F_R x [G x (tau x alpha) - U_L x (T_i - T_a)]. Dividing by (A x G): eta = Q_u / (A x G) = F_R x (tau x alpha) - F_R x U_L x (T_i - T_a) / G. This gives the INSTANTANEOUS EFFICIENCY as a linear function of the parameter (T_i - T_a)/G. The efficiency curve is a straight line: y-intercept = F_R x (tau x alpha) (efficiency at T_i = T_a), slope = -F_R x U_L. The stagnation temperature is when Q_u = 0: T_stag - T_a = G x (tau x alpha) / U_L. The HWB equation is the fundamental equation for flat plate collector analysis.
Difficulty: intermediateType: numerical
Question 104 of 119
Original Exam: IES (2022)
The renewable energy share in India's total electricity generation capacity (installed) as of 2022 is approximately:
✓ Correct Answer: D
Explanation:
India's installed electricity generation capacity (as of 2022): Total: approximately 400 GW. Renewable energy (excluding large hydro): approximately 160 GW (40%). Including large hydro: approximately 200 GW (50%). Breakdown: Solar: ~60 GW; Wind: ~44 GW; Large hydro: ~47 GW; Small hydro: ~5 GW; Biomass: ~10 GW; Others: ~5 GW. Thermal (coal, gas, oil): approximately 200 GW (50%). Nuclear: approximately 7 GW (2%). India has set a target of 500 GW of non-fossil fuel capacity by 2030, which would represent approximately 60-65% of total installed capacity.
Difficulty: intermediateType: conceptual
Question 105 of 119
Original Exam: IES (2020)
The solar cell with the highest theoretical efficiency potential is:
✓ Correct Answer: A
Explanation:
Multi-junction (tandem) solar cells have the highest theoretical efficiency potential. By stacking multiple p-n junctions with different bandgaps, they can capture different parts of the solar spectrum more efficiently. Theoretical efficiency limits: (1) Single-junction (Shockley-Queisser limit): ~33%; (2) Two-junction tandem: ~42%; (3) Three-junction tandem: ~49%; (4) Infinite junctions: ~68% (thermodynamic limit). Record efficiencies: (1) Single-junction Si: 26.7% (lab); (2) Three-junction concentrator: 47.1% (lab). Multi-junction cells are used in space applications and concentrating PV (CPV) systems where high efficiency justifies the high cost.
Type: conceptual
Question 106 of 119
Original Exam: GATE (2017)
The wind shear exponent (alpha) in the power law wind profile is typically:
✓ Correct Answer: B
Explanation:
Wind Shear Exponent (alpha) in the power law: V(h) = V_ref x (h/h_ref)^alpha. Typical values: (1) Open terrain (flat, few obstacles): alpha = 1/7 = 0.143; (2) Suburban terrain: alpha = 0.22-0.25; (3) Urban terrain: alpha = 0.30-0.40; (4) Offshore: alpha = 0.10-0.14. The 1/7 power law (alpha = 0.143) is the most commonly used approximation for open terrain. The logarithmic wind profile is more accurate: V(h) = (V*/kappa) x ln(h/z_0), where V* = friction velocity, kappa = von Karman constant (0.41), z_0 = roughness length. The wind shear exponent determines how much the wind speed increases with hub height.
Difficulty: intermediateType: numerical
Question 107 of 119
Original Exam: GATE (2016)
The solar irradiance on a tilted surface (I_T) can be calculated from horizontal surface data using:
✓ Correct Answer: C
Explanation:
Total irradiance on a tilted surface (isotropic sky model): I_T = I_b x R_b + I_d x (1 + cos(beta))/2 + I x rho_g x (1 - cos(beta))/2. Where: I_b = beam (direct) radiation on horizontal surface; R_b = geometric factor = cos(theta)/cos(theta_z) (ratio of beam radiation on tilted to horizontal surface); I_d = diffuse radiation on horizontal surface; beta = tilt angle; I = global horizontal radiation; rho_g = ground reflectance (albedo, typically 0.2 for grass, 0.7 for snow). The three terms represent: (1) Beam component; (2) Diffuse sky component; (3) Ground-reflected component.
Type: numerical
Question 108 of 119
Original Exam: IES (2019)
The efficiency of a geothermal power plant is typically:
✓ Correct Answer: D
Explanation:
Efficiency of geothermal power plants: typically 5-15% (much lower than conventional power plants). This is because: (1) Geothermal resources have relatively low temperatures (150-300°C for most resources); (2) The Carnot efficiency is limited by the low temperature difference; (3) For T_H = 200°C = 473 K, T_L = 30°C = 303 K: Carnot efficiency = 1 - 303/473 = 35.9%; (4) Actual efficiency is much lower due to irreversibilities. Despite low efficiency, geothermal plants have: (1) High capacity factors (>90%); (2) Low operating costs; (3) Baseload capability; (4) Small land footprint. Binary cycle plants (using lower temperature resources) have even lower efficiency (5-10%).
Difficulty: intermediateType: conceptual
Question 109 of 119
Original Exam: IES (2021)
The solar energy potential of India is estimated to be approximately:
✓ Correct Answer: A
Explanation:
India's solar energy potential: approximately 5000 GW (5 TW) based on available land area and solar irradiance. This estimate assumes: (1) 3% of India's land area (approximately 100,000 km2) is available for solar installations; (2) Average solar irradiance of 5 kWh/m2/day; (3) Solar PV efficiency of 15%. India's current installed solar capacity (2023): approximately 70 GW. India's solar target: 500 GW by 2030. The vast solar potential makes India one of the most promising countries for solar energy development. The National Solar Mission aims to make India a global leader in solar energy.
Difficulty: intermediateType: conceptual
Question 110 of 119
Original Exam: IES (2018)
The coefficient of performance (COP) of a solar-powered absorption cooling system is:
✓ Correct Answer: B
Explanation:
The COP of a solar-powered absorption cooling system is LESS THAN 1. COP = Cooling effect / Heat input from solar collector. For a single-effect LiBr-water absorption chiller: COP = 0.6-0.8. For a double-effect absorption chiller: COP = 1.0-1.4. The COP is less than 1 for single-effect systems because the heat input (from solar collector) is greater than the cooling effect produced. This is different from a vapor compression refrigeration system (COP = 2-5) which uses electrical work. Despite low COP, solar absorption cooling is attractive because it uses free solar energy and can reduce peak electricity demand.
Difficulty: intermediateType: numerical
Question 111 of 119
Original Exam: GATE (2020)
The wind turbine generator type most commonly used in modern large wind turbines is:
✓ Correct Answer: C
Explanation:
Modern large wind turbines (>1 MW) most commonly use: (1) Doubly-Fed Induction Generator (DFIG): rotor connected to grid through partial-scale power converter (30% of rated power). Variable speed operation. Most widely used type. (2) Permanent Magnet Synchronous Generator (PMSG): full-scale power converter. Direct drive (no gearbox) or with gearbox. Growing in popularity. Advantages of variable-speed operation: (1) Maximum power extraction over a wider wind speed range; (2) Reduced mechanical stress; (3) Better grid support capability. Fixed-speed induction generators (older technology) are less efficient and have limited grid support capability.
Difficulty: intermediateType: conceptual
Question 112 of 119
Original Exam: IES (2017)
The solar thermal power plant (CSP) with the highest installed capacity globally is:
✓ Correct Answer: D
Explanation:
Parabolic Trough CSP plants have the HIGHEST installed capacity globally. Global CSP installed capacity (2022): approximately 6.5 GW total. By technology: (1) Parabolic trough: ~70% of total CSP capacity; (2) Central receiver (power tower): ~20%; (3) Linear Fresnel: ~5%; (4) Parabolic dish: ~5%. Largest CSP plants: Ivanpah Solar (377 MW, power tower, USA), Noor Complex (580 MW, parabolic trough, Morocco), Ouarzazate (160 MW, parabolic trough, Morocco). Parabolic trough is the most mature CSP technology with the most operational experience.
Difficulty: intermediateType: conceptual
Question 113 of 119
Original Exam: GATE (2019)
The energy from ocean waves is proportional to:
✓ Correct Answer: A
Explanation:
Wave energy per unit surface area: E = (1/8) x rho x g x H^2. Where rho = water density (1025 kg/m3), g = 9.81 m/s2, H = significant wave height (m). Wave power per unit crest width: P = (rho x g^2 x H^2 x T) / (32 x pi). Where T = wave period (s). Wave energy is proportional to the SQUARE of wave height (H^2). This means: (1) Doubling wave height increases energy by 4 times; (2) Wave energy resources are concentrated in areas with large waves (North Atlantic, Southern Ocean, Pacific Northwest). Wave energy converters: oscillating water column (OWC), point absorbers, attenuators (Pelamis), overtopping devices.
Difficulty: intermediateType: numerical
Question 114 of 119
Original Exam: IES (2020)
The photovoltaic module degradation rate is typically:
✓ Correct Answer: B
Explanation:
Solar PV module degradation rate: typically 0.5-1% per year (power output decreases by 0.5-1% annually). After 25 years: power output = 100% - 25 x 0.7% = 82.5% of initial rated power. Most manufacturers guarantee: (1) 90% of rated power after 10 years; (2) 80% of rated power after 25 years. Causes of degradation: (1) UV-induced degradation of encapsulant; (2) Potential-induced degradation (PID); (3) Delamination; (4) Corrosion of contacts; (5) Micro-cracks in cells. The degradation rate is used to calculate the lifetime energy production and financial returns of a solar PV system.
Difficulty: basicType: conceptual
Question 115 of 119
Original Exam: GATE (2014)
The solar energy stored in a solar water heater (SWH) system is used for:
✓ Correct Answer: C
Explanation:
A Solar Water Heater (SWH) system is used for DOMESTIC HOT WATER HEATING. It uses flat plate or evacuated tube collectors to heat water to 50-80°C. Applications: (1) Domestic hot water (bathing, cooking, cleaning); (2) Swimming pool heating; (3) Industrial process heat (below 100°C). SWH systems are the most widely deployed solar thermal technology in India. India has approximately 10 million m2 of solar water heater collector area installed. The Bureau of Energy Efficiency (BEE) and MNRE promote SWH through subsidies and mandatory installation requirements in new buildings.
Difficulty: basicType: conceptual
Question 116 of 119
Original Exam: IES (2016)
The renewable purchase obligation (RPO) in India requires:
✓ Correct Answer: D
Explanation:
Renewable Purchase Obligation (RPO) in India: Distribution companies (DISCOMs) and large consumers (open access consumers, captive power plants) are required to purchase a MINIMUM PERCENTAGE of their total electricity consumption from renewable energy sources. RPO targets (as of 2022): Total RPO = 43.33% by 2030 (including 6.94% solar, 2.82% wind, 0.81% hydro, etc.). RPO is enforced by State Electricity Regulatory Commissions (SERCs). Non-compliance results in penalties. RPO can be met by: (1) Purchasing renewable energy directly; (2) Purchasing Renewable Energy Certificates (RECs); (3) Setting up captive renewable energy plants.
Difficulty: intermediateType: conceptual
Question 117 of 119
Original Exam: GATE (2021)
The solar irradiance on a horizontal surface at solar noon on a clear day at a location with latitude 28°N on summer solstice (declination = 23.45°) is approximately:
✓ Correct Answer: A
Explanation:
At solar noon on summer solstice at latitude 28°N: Solar altitude angle = 90° - (latitude - declination) = 90° - (28° - 23.45°) = 90° - 4.55° = 85.45°. Zenith angle = 90° - 85.45° = 4.55°. Air mass = 1/cos(4.55°) = 1/0.9968 = 1.003 (approximately AM1). At AM1 on a clear day, the direct normal irradiance is approximately 950-1000 W/m2. Global horizontal irradiance = DNI x cos(theta_z) + DHI = 950 x cos(4.55°) + 100 = 950 x 0.9968 + 100 = 947 + 100 = 1047 W/m2. However, accounting for atmospheric losses at AM1: GHI approximately 950 W/m2.
Type: numerical
Question 118 of 119
Original Exam: IES (2021)
The efficiency of a solar cell increases when:
✓ Correct Answer: B
Explanation:
Solar cell efficiency INCREASES when: (1) Temperature DECREASES: V_oc increases, P_max increases (temperature coefficient of power is negative: -0.4 to -0.5%/°C for Si); (2) Irradiance INCREASES: I_sc increases proportionally, V_oc increases logarithmically, P_max increases. Therefore, efficiency increases when temperature decreases AND irradiance increases. This is why solar cells perform best on cold, sunny days (e.g., winter in high-altitude locations). The efficiency is highest at low temperatures and high irradiance. In practice, high irradiance often comes with high temperature (summer), which partially offsets the irradiance benefit.
Difficulty: intermediateType: conceptual
Question 119 of 119
Original Exam: GATE (2022)
The global renewable energy share in total electricity generation as of 2022 is approximately:
✓ Correct Answer: C
Explanation:
Global renewable energy share in electricity generation (2022): approximately 30% (including all renewables: hydro, wind, solar, biomass, geothermal). Breakdown: (1) Hydropower: ~15%; (2) Wind: ~7%; (3) Solar PV: ~4%; (4) Bioenergy: ~3%; (5) Other renewables: ~1%. The renewable share has been growing rapidly: 2010: ~20%; 2015: ~23%; 2020: ~29%; 2022: ~30%. The IEA projects renewable energy to provide 60% of global electricity by 2030 under the Net Zero scenario. Solar and wind are the fastest-growing energy sources.