GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersSolar 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
GATE — Agricultural Engineering (AG)
Section 3 Sources of Power: one- and two-mark numerical questions on the PV fill factor, module efficiency and area, power in the wind (½ρAV³), the Betz limit and power coefficient, and tip speed ratio. Unit discipline (kW vs kWh, rpm vs rad/s, diameter vs radius) decides marks. Solar geometry and collector efficiency appear less often but are fair game.
How our questions for GATE spread across the units of Renewable Energy Sources (Solar, Wind). Difficulty: 7 easy, 29 medium, 8 hard.
| Power in the wind and the Betz limit | 8 | |
| Solar radiation geometry | 6 | |
| Photovoltaic cell physics and the I-V curve | 5 | |
| Flat-plate collectors and their efficiency | 4 | |
| Fill factor and MPPT | 4 | |
| PV system sizing: on-grid and off-grid | 3 | |
| Solar pumps and agri-voltaics | 3 | |
| Wind turbines and water-pumping windmills | 3 | |
| Measurement of solar radiation | 2 | |
| Solar dryers, cookers, water heaters and stills | 2 | |
| Micro-hydro | 2 | |
| Concentrating collectors | 1 | |
| Energy storage | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
What is the solar altitude at solar noon on the June solstice (δ = +23.45°) at a site of latitude 28.6° N?
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).
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?
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.
What is the air mass when the sun is at a zenith angle of 60°?
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.
On a plot of flat-plate collector efficiency against (Ti − Ta)/G, the intercept and slope of the Hottel–Whillier–Bliss line are respectively:
D. FR(τα) and −FRUL
η = FR(τα) − FRUL(Ti − Ta)/G, so the intercept is FR(τα) (optical efficiency) and the slope −FRUL. Writing (τα) and UL without FR would apply only if the whole plate were at inlet temperature with perfect heat removal.
A basin-type solar still has a daily efficiency of 35%. Daily insolation is 20 MJ/m² and the latent heat of vaporisation is 2.35 MJ/kg. What is the daily distillate output per square metre?
C. 2.98 kg/m²
Output = ηH/L = 0.35 × 20 / 2.35 = 2.98 kg/m² per day, in the usual 2–3 L/m² range for basin stills. Omitting efficiency gives 8.51, far above what any still achieves; 0.041 inverts H/L while keeping the efficiency factor (ηL/H = 0.35 × 2.35/20); 10.7 multiplies the correct 2.98 by 3.6, a spurious unit conversion.
The Betz limit, the maximum fraction of wind power that any open rotor can extract, is:
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.
A micro-hydro scheme has a flow of 0.1 m³/s through a net head of 20 m with an overall efficiency of 60%. What electrical power does it produce?
D. 11.77 kW
P = ρgQHη = 1000 × 9.81 × 0.1 × 20 × 0.6 = 11772 W = 11.77 kW. Omitting efficiency gives 19.62 kW; forgetting ρg gives 1.2; treating 0.1 m³/s as 100 m³/s gives megawatts.
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