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Agricultural and Biosystems Engineers Licensure Examination Renewable Energy Sources (Solar, Wind): practice questions

Agricultural and Biosystems Engineers Licensure Examination (ABELE) · Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC

Original practice questions written to the published syllabus · not past papers
10worked here, free
48written 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

Where the 48 questions fall

How our questions for Agricultural and Biosystems Engineers Licensure Examination spread across the units of Renewable Energy Sources (Solar, Wind). Difficulty: 14 easy, 24 medium, 10 hard.

Solar dryers, cookers, water heaters and stills7
PV system sizing: on-grid and off-grid6
Solar pumps and agri-voltaics5
Micro-hydro5
Wind turbines and water-pumping windmills4
Energy scenario3
Solar radiation geometry3
Power in the wind and the Betz limit3
Energy storage3
Measurement of solar radiation2
Flat-plate collectors and their efficiency2
Photovoltaic cell physics and the I-V curve2
Fill factor and MPPT2
Concentrating collectors1

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

Which of the following is a renewable (flow) energy source rather than a stock?

  1. Natural gas
  2. Hard coal
  3. Wind
  4. Uranium
Answer and working

C. Wind

Wind is continually replenished by solar heating of the atmosphere. Coal, natural gas and uranium are stocks laid down over geological time and used far faster than they form.

2 · Energy scenario · medium

In a net-metering arrangement for a farm PV system, the farmer is billed for:

  1. All energy generated, at a fixed feed-in price
  2. All energy imported from the grid at the retail tariff, with any exported energy left unpaid
  3. Net consumption: energy imported minus energy exported over the billing period
  4. Only the rated capacity of the PV system in kW
Answer and working

C. Net consumption: energy imported minus energy exported over the billing period

Under net metering the meter effectively runs backwards on export, so the bill is for net import. Selling all generation at a fixed price and buying all consumption separately is gross metering (a feed-in tariff). Exact rules and settlement prices are set by each national regulator.

3 · Energy scenario · hard

A solar pump costing 300,000 currency units replaces diesel pumping and saves 60,000 per year in fuel and maintenance. What is the simple payback period, and what is its main weakness as a decision measure?

  1. 5 years; it ignores the time value of money and savings after payback
  2. 5 years; it overstates the effect of subsidies
  3. 0.2 years; it ignores inflation
  4. 20 years; it double-counts the maintenance savings and ignores the salvage value of the pump
Answer and working

A. 5 years; it ignores the time value of money and savings after payback

Simple payback = 300,000 / 60,000 = 5 years. It treats a unit of money in year 5 as worth the same as today and ignores everything after year 5, so it is a screening tool; NPV or LCOE are used for the decision. 0.2 is the inverse ratio (60,000/300,000).

4 · Solar radiation geometry · medium

What is the solar altitude at solar noon on the June solstice (δ = +23.45°) at a site of latitude 28.6° N?

  1. 84.85°
  2. 37.95°
  3. 5.15°
  4. 61.40°
Answer and working

A. 84.85°

At noon θz = |φ − δ| = |28.6 − 23.45| = 5.15°, so altitude α = 90 − 5.15 = 84.85°. Taking θz = φ + δ gives 37.95°, a low sun wrong for midsummer; 5.15° is the zenith angle, not the altitude; 61.4° ignores declination (equinox value).

5 · Solar radiation geometry · hard

A site at 77.2° E uses a standard meridian of 82.5° E. At 12:00 standard time the equation of time is +3 min. What is the local solar time?

  1. 12:24.2
  2. 11:41.8
  3. 11:35.8
  4. 12:18.2
Answer and working

B. 11:41.8

Solar time = standard time − 4(Lst − Lloc) + E for a site west of its standard meridian in the eastern hemisphere: 12:00 − 4 × (82.5 − 77.2) min + 3 min = 12:00 − 21.2 + 3 = 11:41.8. Adding the longitude correction gives 12:24.2 (or 12:18.2 if E is also subtracted), and subtracting E gives 11:35.8. The site lies west of the standard meridian, so the sun reaches it later than the clock suggests.

6 · Measurement of solar radiation · easy

Which instrument measures beam (direct normal) solar radiation?

  1. Pyrheliometer
  2. Thermopile pyranometer
  3. Campbell–Stokes sunshine recorder
  4. Cup anemometer
Answer and working

A. Pyrheliometer

A pyrheliometer is a collimating tube pointed at the sun on a tracker, so it sees only the sun's disc and a narrow ring around it. A pyranometer sees the whole hemisphere (global radiation); the Campbell–Stokes recorder gives sunshine hours; an anemometer measures wind speed.

7 · PV system sizing: on-grid and off-grid · medium

An off-grid farmhouse uses 3 kWh per day. The design month has 5 peak sun hours, and the overall array derating factor is 0.7. What array size is needed?

  1. 600 Wp
  2. 857 Wp
  3. 420 Wp
  4. 2100 Wp
Answer and working

B. 857 Wp

Array = daily energy/(PSH × derating) = 3000/(5 × 0.7) = 857 Wp. 600 Wp ignores losses; multiplying by the derating gives 420 Wp; 2100 Wp is 3000 × 0.7, a confusion of energy with power.

8 · 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.

9 · 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.

10 · Wind turbines and water-pumping windmills · medium

A water-pumping windmill 'furls' in high winds. What does furling do?

  1. Feathers the pump valves (usually via the pump rod) so water flows back to the well
  2. Locks the rotor facing directly into the wind so it can extract maximum power during a strong gust
  3. Switches the pump from single-acting to double-acting
  4. Turns the rotor partly out of the wind (usually through the tail vane) to limit speed and loads
Answer and working

D. Turns the rotor partly out of the wind (usually through the tail vane) to limit speed and loads

Above the design speed a windmill's rotor could overspeed and overload the pump rod. Furling swings the rotor edgewise to the wind, reducing its effective area. It does not change the pump's action.

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Read the subject

Renewable Energy Sources (Solar, Wind)The full study manual, free Agricultural and Biosystems Engineers Licensure ExaminationFormat, sections and how it has changed

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Renewable Energy Sources (Solar, Wind) for other exams

ICAR AIEEA-PG · IndiaICAR AICE-JRF/SRF · IndiaGATE · IndiaASRB NET · India

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