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ICAR AIEEA-PG Renewable Energy Sources (Solar, Wind): practice questions

ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
74written to this exam in this subject
14syllabus units

What this subject covers

Solar thermal, solar PV, wind and micro-hydro for farms, from sun geometry to sizing a solar pump.

In the exams: GATE AG Section 3 (solar PV in 5 of 8, wind in 5 of 8 papers 2019–26); ICAR AIEEA-PG Unit V; ICAR AICE subject 42 Units 4, 5 and 7; NCEES PE energy sources

What ICAR AIEEA-PG asks from it

ICAR AIEEA-PG — Agricultural Engineering & Technology

Unit V 'renewable energy': definitions and principles — solar radiation instruments, flat-plate collector parts and efficiency, solar dryers, cookers, stills and water heaters, PV cell characteristics, wind turbine types and the Betz limit — with short calculations.

Where the 74 questions fall

How our questions for ICAR AIEEA-PG spread across the units of Renewable Energy Sources (Solar, Wind). Difficulty: 31 easy, 43 medium, 0 hard.

Measurement of solar radiation8
Solar dryers, cookers, water heaters and stills7
Photovoltaic cell physics and the I-V curve7
Solar radiation geometry6
Flat-plate collectors and their efficiency6
Fill factor and MPPT6
Energy storage6
Energy scenario5
Solar pumps and agri-voltaics5
Wind turbines and water-pumping windmills5
Concentrating collectors4
Power in the wind and the Betz limit4
Micro-hydro3
PV system sizing: on-grid and off-grid2

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Energy scenario · easy

One kilowatt-hour of energy is equal to:

  1. 1 kJ
  2. 3.6 MJ
  3. 3.6 kJ
  4. 36 MJ
Answer and working

B. 3.6 MJ

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J = 3.6 MJ. 3.6 kJ is the energy of 1 W for an hour (1 Wh), a factor-of-1000 slip. 36 MJ is a factor-of-ten slip.

2 · Solar radiation geometry · medium

What is the air mass when the sun is at a zenith angle of 60°?

  1. 0.5
  2. 1.5
  3. 1.15
  4. 2.0
Answer and working

D. 2.0

Air mass m ≈ 1/cos θz = 1/cos 60° = 1/0.5 = 2.0. 0.5 is cos θz itself; 1.5 is the standard PV rating spectrum (θz ≈ 48°); 1.15 is 1/cos 30°, confusing zenith with altitude.

3 · PV system sizing: on-grid and off-grid · easy

Anti-islanding protection in a grid-connected PV inverter:

  1. Keeps the farm supplied from the PV array during grid outages by forming a separate local island network
  2. Disconnects the inverter when the grid fails, so the system does not energise lines being repaired
  3. Prevents the PV array from being shaded
  4. Limits battery charging current
Answer and working

B. Disconnects the inverter when the grid fails, so the system does not energise lines being repaired

An 'island' is a section of network kept live by local generators after the grid has gone down; it endangers line workers. Grid-tied inverters must shut down on loss of grid, which is why an ordinary on-grid system without battery backup gives no power in an outage.

4 · Solar pumps and agri-voltaics · easy

In most farm solar pumping systems, energy for cloudy periods is stored mainly as:

  1. Charge in large lead-acid batteries
  2. Compressed air in the borewell
  3. Water in an elevated tank or reservoir
  4. Heat in the PV modules
Answer and working

C. Water in an elevated tank or reservoir

Storing water is far cheaper and longer-lived than storing electricity, so most solar pumps are battery-less and directly coupled through a controller, with a tank or the soil profile buffering supply.

5 · Power in the wind and the Betz limit · easy

The Betz limit, the maximum fraction of wind power that any open rotor can extract, is:

  1. 8/27 ≈ 0.296
  2. 2/3 ≈ 0.667
  3. 1/3 ≈ 0.333
  4. 16/27 ≈ 0.593
Answer and working

D. 16/27 ≈ 0.593

Momentum theory gives Cp = 4a(1 − a)², maximised at a = 1/3, where Cp = 16/27 ≈ 0.593. 2/3 is the velocity ratio at the rotor at that condition and 1/3 is the induction factor, not the power coefficient.

6 · Micro-hydro · medium

In a run-of-river micro-hydro scheme, the forebay is:

  1. A tank at the penstock head that settles sediment and gives a steady water level at the penstock inlet
  2. The steep pressure pipe that carries water down from the settling basin to the turbine in the powerhouse
  3. The channel that returns water to the river after the turbine
  4. The dam that stores a season's flow
Answer and working

A. A tank at the penstock head that settles sediment and gives a steady water level at the penstock inlet

Water from the headrace enters the forebay, where sediment settles and a stable level prevents air being drawn into the penstock. The penstock is the pressure pipe, the tailrace returns water, and run-of-river schemes have no seasonal storage dam.

7 · Energy storage · medium

1000 m³ of water is pumped into a reservoir 50 m above the turbine. With a round-trip efficiency of 75%, how much electrical energy can be recovered?

  1. 136.2 kWh
  2. 181.7 kWh
  3. 102.2 kWh
  4. 0.102 kWh
Answer and working

C. 102.2 kWh

Stored energy = ρgVh = 1000 × 9.81 × 1000 × 50 = 490.5 MJ = 136.25 kWh. Recovered = 0.75 × 136.25 = 102.2 kWh. Ignoring efficiency gives 136.2; forgetting the water density (using volume as mass) gives 0.102 kWh. 181.7 kWh divides by the efficiency instead of multiplying.

8 · Concentrating collectors · medium

Which of these concentrators is a line-focus system that uses rows of nearly flat mirror strips?

  1. Paraboloidal dish
  2. Central receiver tower
  3. Linear Fresnel reflector
  4. Compound parabolic concentrator with one curved mirror
Answer and working

C. Linear Fresnel reflector

A linear Fresnel reflector approximates a trough with many flat or slightly curved strips focusing on a fixed linear receiver above them. Dishes and towers are point-focus; a CPC uses two parabolic segments, not flat strips.

9 · Measurement of solar radiation · medium

The clearness index KT is defined as:

  1. The ratio of diffuse radiation to global radiation on a horizontal surface, averaged over the same period
  2. The ratio of sunshine hours to day length
  3. The ratio of global horizontal radiation to extraterrestrial horizontal radiation over the same period
  4. The fraction of time the sky is cloud-free
Answer and working

C. The ratio of global horizontal radiation to extraterrestrial horizontal radiation over the same period

KT = H/H0. It is used to correlate the diffuse fraction. n/N is the relative sunshine duration used in Ångström–Prescott, a different quantity.

10 · Fill factor and MPPT · medium

In the perturb-and-observe MPPT method, if a small increase in operating voltage produces a decrease in array power, the controller should next:

  1. Increase the voltage further
  2. Hold the voltage at open-circuit
  3. Disconnect the array
  4. Decrease the operating voltage
Answer and working

D. Decrease the operating voltage

A power drop after a voltage increase means the operating point has passed the maximum and is on the right side of the peak, so the perturbation direction is reversed. The method then oscillates around the maximum power point.

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Read the subject

Renewable Energy Sources (Solar, Wind)The full study manual, free ICAR AIEEA-PGFormat, sections and how it has changed

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Renewable Energy Sources (Solar, Wind) for other exams

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