ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersTurning crop residues, dung and oils into fuel: briquettes, gasifiers, biogas plants and liquid biofuels, and auditing farm energy.
In the exams: ICAR AIEEA-PG Unit V; ICAR AICE subject 42 Units 1–3 and 8; ICAR AICE subject 39 Unit 9; GATE AG biogas sizing (2 of 8 papers)
ICAR AICE-JRF/SRF (PhD) — Renewable Energy Engineering (subject 42) and Farm Machinery & Power (subject 39)
PG-depth questions on biomass chemistry and thermal decomposition, gasification reactions and equilibrium trends, equivalence ratio effects, reactor sizing, Buswell stoichiometry and COD-based methane yield, digestion kinetics, cellulosic ethanol pretreatment and SSF, transesterification chemistry and FFA handling, and energy-in-agriculture indicators.
How our questions for ICAR AICE-JRF/SRF spread across the units of Bio-energy Systems: Design & Applications. Difficulty: 1 easy, 60 medium, 29 hard.
| Gasification: updraft and downdraft gasifiers | 9 | |
| Bio-ethanol | 9 | |
| Biodiesel and transesterification | 9 | |
| Energy audit and energy ratio in agriculture | 9 | |
| Anaerobic digestion stages | 8 | |
| Biomass characterisation: proximate and ultimate analysis | 7 | |
| Biogas plant sizing and gas yield | 7 | |
| Calorific value | 6 | |
| Pyrolysis | 6 | |
| Producer gas and dual-fuel engines | 6 | |
| Combustion of biomass | 5 | |
| Biogas plant types: floating drum and fixed dome | 5 | |
| Briquetting and pelleting | 4 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A sample of crop residue has, on an as-received basis, moisture 12%, volatile matter 62% and ash 8%. Its ash content on a dry basis is closest to:
C. 9.09%
Dry-basis value = as-received value ÷ (1 − moisture fraction) = 8 ÷ (1 − 0.12) = 9.09%. Leaving the 8% unchanged ignores the basis; dividing by (1 − moisture − ash) = 0.80 (10.00%) is the dry ash-free conversion, which cannot be applied to ash itself. Multiplying by (1 + M), 8 × 1.12 = 8.96%, is a common but wrong shortcut.
A dry biomass has a higher heating value of 18.5 MJ/kg and contains 6% hydrogen (dry basis). Taking the latent heat of water as 2.442 MJ/kg, its lower heating value (dry basis) is:
D. 17.18 MJ/kg
Water formed = 9 × H = 9 × 0.06 = 0.54 kg per kg fuel. LHV = HHV − 2.442 × 0.54 = 18.5 − 1.32 = 17.18 MJ/kg. Forgetting the factor 9 (1 kg H2 forms 9 kg H2O) gives 18.35 MJ/kg; adding instead of subtracting (19.82 MJ/kg) reverses the definition; 8.51 MJ/kg multiplies the HHV by (1 − 9H) = 0.46 instead of subtracting the latent heat.
A fast-pyrolysis unit converts dry biomass (HHV 19 MJ/kg) to bio-oil at a yield of 65% by mass; the bio-oil HHV is 17 MJ/kg. The fraction of feed energy recovered in the bio-oil is:
B. 58.2%
Energy recovery = 0.65 × 17 ÷ 19 = 58.2%. Bio-oil has a lower HHV than the dry feed largely because it carries 15–30% water and is rich in oxygen, so energy recovery is below the mass yield. 65.0% is the mass yield; 72.6% inverts the heating-value ratio (0.65 × 19 ÷ 17); 89.5% is the heating-value ratio alone (17 ÷ 19).
A downdraft gasifier consumes 50 kg/h of wood chips and 75 kg/h of air. The stoichiometric air–fuel ratio of the wood is 6 kg/kg. The equivalence ratio is:
B. 0.25
Actual A/F = 75 ÷ 50 = 1.50 kg/kg. ER = actual ÷ stoichiometric = 1.50 ÷ 6 = 0.25, within the usual gasification range of about 0.2–0.4. 1.50 is the actual air–fuel ratio, not the ER; 4.00 is its inverse. 0.67 is the fuel–air ratio (50 ÷ 75).
Producer gas contains, by volume, CO 20%, H2 18%, CH4 2%, the rest CO2 and N2. The stoichiometric air requirement (air 21% O2 by volume) is:
A. 1.10 m³ air per m³ gas
O2 needed: CO + ½O2 → CO2 gives 0.2 × 0.5; H2 + ½O2 → H2O gives 0.18 × 0.5; CH4 + 2O2 → CO2 + 2H2O gives 0.02 × 2. Total O2 = 0.10 + 0.09 + 0.04 = 0.23 m³/m³. Air = 0.23 ÷ 0.21 = 1.10 m³/m³. Using 0.232 mixes the mass fraction into a volume calculation. The ratio is close to 1:1, which is why producer-gas mixers admit roughly equal volumes of gas and air.
A producer gas of LHV 5.18 MJ/Nm³ needs 1.10 Nm³ of air per Nm³ for stoichiometric combustion. The energy content of the stoichiometric gas–air mixture, and the main consequence for an engine converted from petrol (whose stoichiometric mixture holds roughly 3.5–3.8 MJ/m³), are:
D. 2.47 MJ/m³ of mixture; the engine is derated because each cylinder charge holds less energy
Mixture energy = gas LHV ÷ (1 + air/gas) = 5.18 ÷ (1 + 1.10) = 2.47 MJ/m³. A cylinder fills with mixture, not with gas, so the energy per stroke is roughly 65–71% of that on petrol (2.47 ÷ 3.8 to 2.47 ÷ 3.5) before efficiency effects. Producer-gas engines are therefore derated unless compression ratio or turbocharging compensates; the high knock resistance of CO and CH4 permits a higher compression ratio.
Biogas that is to be compressed and used as a vehicle fuel must first be upgraded, which mainly means:
B. Removing CO2 and impurities
Raw biogas is about 50–70% CH4 with 30–50% CO2, traces of H2S and saturation water. Upgrading by water scrubbing, pressure-swing adsorption, membranes or chemical absorption removes CO2 and impurities to give biomethane with over about 90–97% methane, depending on the specification. H2S must also be removed because it is toxic and corrosive.
For a family biogas plant, the hydraulic retention time chosen at the design stage should be:
A. Longer in cold climates and seasons, because gas production per day falls as temperature drops
Microbial activity slows sharply with temperature, so unheated plants in cooler regions need longer HRT (bigger digesters) for the same degradation. National programmes therefore specify longer HRTs for hill and cold regions than for warm plains. Very short HRTs risk washout of slow-growing methanogens, which is the opposite problem.
A distillery ferments 1 t of molasses containing 45% fermentable sugars (expressed as glucose). With 0.511 kg ethanol per kg glucose theoretically and a fermentation efficiency of 90%, the ethanol produced (density 0.789 kg/L) is about:
B. 263 L
Sugar = 1000 × 0.45 = 450 kg. Ethanol = 450 × 0.5114 × 0.9 = 207.1 kg = 207.1 ÷ 0.789 = 263 L. 207 is the mass in kg; applying the yield to the whole molasses mass ignores its water and non-fermentable solids. Distillation losses reduce the product further. 292 L ignores the 90% fermentation efficiency (230.1 ÷ 0.789).
A wheat crop has a total energy input of 15000 MJ/ha and an energy output (grain + straw) of 90000 MJ/ha. The energy ratio and the net energy are:
D. 6.0 and 75000 MJ/ha
Energy ratio = 90000 ÷ 15000 = 6.0. Net energy = 90000 − 15000 = 75000 MJ/ha. The inverse (0.17) is input per unit output.
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