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ICAR AICE-JRF/SRF Post-Harvest Engineering of Cereals, Pulses & Oilseeds: practice questions

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 papers
10worked here, free
40written to this exam in this subject
15syllabus units

What this subject covers

Moisture, drying, parboiling, milling, cleaning, size reduction and conveying of grains, pulses and oilseeds.

In the exams: GATE AG Section 6 (~14 marks a year 2019–26; psychrometry in all 8 papers, rice milling in 6 of 8); ICAR AIEEA-PG Unit VI; ICAR AICE subject 41; NCEES PE processing; PRC ABE bioprocess; ASRB NET ASPE

What ICAR AICE-JRF/SRF asks from it

ICAR AICE-JRF/SRF (PhD) — Processing and Food Engineering (subject 41)

Units on post-harvest unit operations, process technology and machinery and design of processing machinery ask for PG depth: sorption models (GAB, BET, modified Henderson), thin-layer and diffusion models and effective diffusivity, deep-bed models, dryer design and performance indices, parboiling kinetics and quality, extraction and refining steps, and design relations for conveyors and mills.

Where the 40 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Post-Harvest Engineering of Cereals, Pulses & Oilseeds. Difficulty: 1 easy, 20 medium, 19 hard.

Thin-layer drying models7
Equilibrium moisture content and sorption isotherms6
Size reduction: Rittinger, Kick and Bond laws5
Drying theory: constant and falling rate periods4
Oil expression and solvent extraction4
Psychrometry and properties of air2
Cleaning, grading and screening effectiveness2
Conveying: belt, bucket, screw and pneumatic2
Seed processing2
Moisture content: wet and dry basis1
Grain dryers: bin, LSU, recirculating, fluidised-bed and solar1
Parboiling and hydrothermal treatment1
Paddy and rice milling: husking, whitening, polishing and yields1
Wheat milling1
Pulse milling1

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 · Moisture content: wet and dry basis · hard

One tonne of maize at 30% moisture (dry basis) is dried to 12% moisture (wet basis). The water removed is closest to

  1. 180 kg
  2. 231 kg
  3. 205 kg
  4. 126 kg
Answer and working

D. 126 kg

Dry matter = 1000/(1 + 0.30) = 769.2 kg. Final mass at 12% wb = 769.2/0.88 = 874.1 kg. Water removed = 1000 − 874.1 = 125.9 kg. 180 kg subtracts percentages on mixed bases; 231 kg is the total water initially present, as if dried to zero moisture. 205 kg treats the initial 30% as wet basis.

Further reading: Chakraverty, Post Harvest Technology of Cereals, Pulses and Oilseeds

2 · Drying theory: constant and falling rate periods · hard

A tray holds 50 kg of dry solid on 2 m² and dries at a constant rate R_c = 1.5 kg/m²·h down to the critical moisture X_c = 0.6 (db); the rate then falls linearly to zero at X_e = 0.05. The falling-rate time from 0.6 to 0.15 kg/kg (db) is closest to

  1. 15.6 h
  2. 25.6 h
  3. 9.2 h
  4. 12.7 h
Answer and working

A. 15.6 h

t_f = [W_s(X_c − X_e)/(A R_c)] ln[(X_c − X_e)/(X − X_e)] = (50 × 0.55/3) × ln(0.55/0.10) = 9.167 × 1.7047 = 15.63 h. Dropping the log term gives 9.2 h; using total rather than free moisture in the log gives 12.7 h; 25.6 h is the total including the constant-rate period.

Further reading: Earle, Unit Operations in Food Processing

3 · Thin-layer drying models · hard

Treating a grain kernel as a sphere of radius 2 mm, the late-period plot of ln MR against time (s) has a slope of −1.2 × 10⁻⁴ s⁻¹. The effective moisture diffusivity is closest to

  1. 4.80e-10 m²/s
  2. 1.95e-10 m²/s
  3. 4.86e-11 m²/s
  4. 1.53e-10 m²/s
Answer and working

C. 4.86e-11 m²/s

First term of the sphere solution: ln MR = ln(6/π²) − (π²D/r²)t, so D = |slope|·r²/π² = 1.2 × 10⁻⁴ × (2 × 10⁻³)²/9.8696 = 4.86e-11 m²/s, within the usual 10⁻¹¹–10⁻¹⁰ range for grains. Omitting π² gives 4.80e-10; using the diameter instead of the radius gives a value four times too large.

Further reading: Brooker, Bakker-Arkema and Hall, Drying and Storage of Grains and Oilseeds

4 · Thin-layer drying models · hard

Effective diffusivity of paddy is 3.0 × 10⁻¹¹ m²/s at 40 °C and 6.0 × 10⁻¹¹ m²/s at 60 °C. The Arrhenius activation energy is closest to

  1. 0.692 kJ/mol
  2. 13.1 kJ/mol
  3. 69.2 kJ/mol
  4. 30.1 kJ/mol
Answer and working

D. 30.1 kJ/mol

E_a = R ln(D₂/D₁)/(1/T₁ − 1/T₂) = 8.314 × 0.6931/(1/313.15 − 1/333.15) = 8.314 × 0.6931/1.9171e-04 = 30061 J/mol ≈ 30.1 kJ/mol. Using Celsius temperatures gives a nonsense small value; multiplying by 2.303 treats the natural log as a common log. 13.1 kJ/mol uses log₁₀ 2 instead of ln 2; 0.692 kJ/mol is the Celsius result; 69.2 kJ/mol multiplies by 2.303.

Further reading: Brooker, Bakker-Arkema and Hall, Drying and Storage of Grains and Oilseeds

5 · Thin-layer drying models · medium

In the Page thin-layer model MR = exp(−k tⁿ), the exponent n

  1. is an empirical shape parameter letting the curve depart from a simple exponential
  2. is the activation energy divided by RT
  3. must equal 1 for grains
  4. is the number of terms retained from the series solution of Fick's second law of diffusion
Answer and working

A. is an empirical shape parameter letting the curve depart from a simple exponential

Page added n to Lewis's exponential to improve fit; n = 1 reduces it to the Lewis model. It has no direct physical meaning, unlike k, which follows Arrhenius behaviour.

Further reading: Brooker, Bakker-Arkema and Hall, Drying and Storage of Grains and Oilseeds

6 · Size reduction: Rittinger, Kick and Bond laws · hard

Smooth crushing rolls of 500 mm diameter are set with a 10 mm gap and fed 30 mm particles. The minimum coefficient of friction needed for the particles to be nipped is closest to

  1. 0.283
  2. 0.615
  3. 0.197
  4. 0.272
Answer and working

A. 0.283

cos b = (R + d)/(R + r) = (250 + 5)/(250 + 15) = 0.9623, so the half nip angle b = 15.79°, and μ ≥ tan b = 0.283. Using the full nip angle gives 0.615; using the full gap instead of half the gap gives 0.197. 0.272 is sin b instead of tan b.

Further reading: McCabe, Smith and Harriott, Unit Operations of Chemical Engineering

7 · Size reduction: Rittinger, Kick and Bond laws · medium

Ground feed has a fineness modulus of 3.0. Using the Henderson–Perry relation D = 0.104 × 2^FM, the average particle size is closest to

  1. 0.62 mm
  2. 0.0328 mm
  3. 0.94 mm
  4. 0.83 mm
Answer and working

D. 0.83 mm

D = 0.104 × 2³ = 0.832 mm. 0.0041 × 2^FM gives inches (0.0328 in), not millimetres; the other options misapply the exponent. 0.62 mm multiplies 2 by FM instead of raising it (0.104 × 6); 0.94 mm squares FM (0.104 × 3²).

Further reading: Henderson and Perry, Agricultural Process Engineering

8 · Conveying: belt, bucket, screw and pneumatic · hard

A screw conveyor has a 0.30 m flight diameter, 0.06 m shaft, 0.30 m pitch, runs at 60 rpm with a loading factor of 0.30, and handles grain of 750 kg/m³. Its capacity is closest to

  1. 16.5 t/h
  2. 0.27 t/h
  3. 17.2 t/h
  4. 55.0 t/h
Answer and working

A. 16.5 t/h

Q = (π/4)(D² − d²)·p·N·ρ·φ = 0.7854 × (0.09 − 0.0036) × 0.30 × 60 × 750 × 0.30 = 274.8 kg/min = 16.5 t/h. Ignoring the shaft gives 17.2 t/h; ignoring the loading factor gives 55.0 t/h (a full trough). 0.27 t/h is the 274.8 kg/min read as t/min and mislabelled.

Further reading: Sahay and Singh, Unit Operations of Agricultural Processing

9 · Conveying: belt, bucket, screw and pneumatic · medium

In horizontal dilute-phase pneumatic conveying of grain, if the air velocity falls below the saltation velocity,

  1. the cyclone stops separating
  2. the grain accelerates and moves faster than the conveying air in the pipe
  3. particles drop out of suspension and form deposits or dunes in the pipe
  4. the system switches to positive pressure
Answer and working

C. particles drop out of suspension and form deposits or dunes in the pipe

Saltation velocity is the minimum horizontal velocity to keep particles suspended; below it they settle and can block the line. The vertical equivalent is the choking velocity.

Further reading: McCabe, Smith and Harriott, Unit Operations of Chemical Engineering

10 · Seed processing · hard

By Harrington's rules of thumb, lowering seed moisture by 2 percentage points and storage temperature by 10 °C would multiply expected storage life by roughly

  1. 16
  2. 8
  3. 6
  4. 4
Answer and working

A. 16

Moisture: 2 × 2 = 4; temperature: each 5 °C fall doubles life, so 10 °C gives 4; combined 4 × 4 = 16. Adding the factors (4 + 4 = 8) is the common slip. 6 adds the moisture factor to a single temperature doubling (4 + 2).

Further reading: Harrington's rules for seed storage

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Post-Harvest Engineering of Cereals, Pulses & OilseedsThe full study manual, free ICAR AICE-JRF/SRFFormat, sections and how it has changed

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