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GATE Transport Phenomena in Food Processing: Fluid Flow, Heat & Mass Transfer: practice questions

GATE — Life Sciences (XL), Section U Food Technology · India · IITs and IISc on rotation (2026: IIT Guwahati)

Original practice questions written to the published syllabus · not past papers
10worked here, free
88written to this exam in this subject
15syllabus units

What this subject covers

Mass and energy balances and the three transfer processes — momentum, heat and mass — applied to foods: pumping viscous products, heating cans and pouches, and moisture and gas movement through foods and packaging films.

In the exams: GATE XE-G Section 4 (mass and energy balance, momentum transfer, heat transfer, mass transfer); ICAR AIEEA-PG Code 18 Unit II

What GATE asks from it

GATE — Engineering Sciences (XE), Section G Food Technology

Section 4 names mass and energy balance; momentum transfer (flow rate and pressure drop relationships for Newtonian fluids in pipes, Reynolds number); heat transfer by conduction, convection and radiation and heat exchangers; and mass transfer (Fick's law, convective mass transfer, permeability through single and multilayer films). Expect numerical-answer questions on Reynolds number, Hagen–Poiseuille and Fanning pressure drop, composite-wall conduction, LMTD, Dittus–Boelter, Biot and lumped cooling, and film permeation; units and the Fanning/Darcy distinction are the common traps.

GATE — Life Sciences (XL), Section U Food Technology

Same Section 4 syllabus as XE-G; life-science candidates should practise the core balances, pipe-flow and heat-exchanger calculations, which are less familiar to them than the chemistry and microbiology sections.

Where the 88 questions fall

How our questions for GATE spread across the units of Transport Phenomena in Food Processing: Fluid Flow, Heat & Mass Transfer. Difficulty: 25 easy, 38 medium, 25 hard.

Flow in pipes: Reynolds number, friction factor and pressure drop for Newtonian fluids8
Heat exchangers: plate, tubular and scraped-surface; LMTD and effectiveness8
Units, dimensions and dimensional analysis7
Unsteady-state heat transfer: lumped systems and Heisler charts7
Material balances on mixing, concentration and drying6
Energy balances and steam tables6
Pumps for foods: centrifugal and positive-displacement6
Steady-state conduction through slabs, cylinders and composite walls6
Convection: film coefficients and dimensionless numbers6
Fluid statics and the Bernoulli equation5
Radiation heat transfer5
Convective mass transfer5
Permeability through single and multilayer films5
Flow of non-Newtonian foods in pipes4
Molecular diffusion and Fick's law4

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 · Material balances on mixing, concentration and drying · easy

5,000 kg/h of milk at 12 % total solids is concentrated to 45 % total solids. The water evaporated is closest to

  1. 1,333 kg/h
  2. 1,650 kg/h
  3. 3,350 kg/h
  4. 3,667 kg/h
Answer and working

D. 3,667 kg/h

W = F(1 − xF/xP) = 5,000 × (1 − 12/45) = 3,667 kg/h. 1,333 kg/h is the concentrate; 1,650 kg/h multiplies by 0.33 (45 − 12 as a fraction); 3,350 kg/h subtracts percentages (5,000 × 0.67).

Further reading: Singh and Heldman, Introduction to Food Engineering

2 · Energy balances and steam tables · easy

Steam is the preferred heating medium in food plants mainly because it

  1. is cheaper to make than hot air in every plant and needs no boiler or water treatment
  2. releases a large latent heat at a constant, pressure-controlled temperature
  3. heats food without any contact with a heat-transfer surface
  4. never forms condensate, so steam traps are not needed
Answer and working

B. releases a large latent heat at a constant, pressure-controlled temperature

Condensing steam delivers about 2,200 kJ/kg at a temperature fixed by its pressure, with very high film coefficients; condensate must be removed by traps.

Further reading: Singh and Heldman, Introduction to Food Engineering

3 · Fluid statics and the Bernoulli equation · medium

The Bernoulli equation for an ideal fluid states that along a streamline the sum of

  1. pressure energy and friction energy per unit mass stays constant
  2. pressure, kinetic and potential energy per unit mass is constant
  3. mass flow and heat flow is constant
  4. velocity and viscosity is constant
Answer and working

B. pressure, kinetic and potential energy per unit mass is constant

P/ρ + v²/2 + gz = constant for frictionless, incompressible, steady flow; real flows add pump work and friction terms.

Further reading: Geankoplis, Transport Processes and Separation Process Principles

4 · Flow in pipes: Reynolds number, friction factor and pressure drop for Newtonian fluids · easy

Milk (ρ = 1,030 kg/m³, μ = 2.0 mPa s) flows at 1.5 m/s in a 38 mm tube. The Reynolds number is closest to

  1. 29,355
  2. 29
  3. 29,355,000
  4. 2,936
Answer and working

A. 29,355

Re = ρvD/μ = 1,030 × 1.5 × 0.038/0.002 = 29,355, turbulent. 29 and 29,355,000 come from unit slips in μ (Pa s vs mPa s); 2,936 uses D = 3.8 mm.

Further reading: Singh and Heldman, Introduction to Food Engineering

5 · Steady-state conduction through slabs, cylinders and composite walls · medium

In a heat exchanger with a thin stainless-steel wall, the wall usually contributes little to the overall resistance because

  1. metal conductivity is high compared with the fluid films
  2. stainless steel is a perfect insulator at food temperatures
  3. the fluid films always have zero resistance
  4. heat passes around the wall rather than through it
Answer and working

A. metal conductivity is high compared with the fluid films

A 1–2 mm steel wall has resistance far below the convective films and fouling layers.

Further reading: Incropera and DeWitt, Fundamentals of Heat and Mass Transfer

6 · Convection: film coefficients and dimensionless numbers · hard

Milk with Re = 29,000 and Pr = 13 is heated in a 38 mm tube (k = 0.55 W/m K). Using Dittus–Boelter for heating, Nu = 0.023 Re^0.8 Pr^0.4, the film coefficient is closest to

  1. 2,669 W/m² K
  2. 3,450 W/m² K
  3. 238 W/m² K
  4. 16.5 W/m² K
Answer and working

B. 3,450 W/m² K

Nu = 0.023 × 29,000^0.8 × 13^0.4 = 238; h = Nu k/D = 238 × 0.55/0.038 = 3,450 W/m² K. The 0.3 exponent is for cooling; 238 is Nu itself; 16.5 inverts k/D.

Further reading: Incropera and DeWitt, Fundamentals of Heat and Mass Transfer

7 · Convection: film coefficients and dimensionless numbers · medium

Natural (free) convection is driven by

  1. a pump or fan moving the fluid
  2. radiation from the hot surface
  3. density differences caused by temperature differences
  4. pressure differences created by a compressor in the circuit
Answer and working

C. density differences caused by temperature differences

Warm fluid is less dense and rises; forced convection uses pumps or fans. The Grashof number characterises it.

Further reading: Incropera and DeWitt, Fundamentals of Heat and Mass Transfer

8 · Unsteady-state heat transfer: lumped systems and Heisler charts · hard

A food slab (k = 0.5 W/m K, ρ = 1,000 kg/m³, cp = 3,600 J/kg K) of half-thickness 20 mm is heated for 30 min. The Fourier number is about

  1. 0.156
  2. 0.01042
  3. 0.62
  4. 2.5
Answer and working

C. 0.62

α = k/ρcp = 1.39e-07 m²/s; Fo = αt/L² = 1.39e-07 × 1,800/0.02² = 0.62. 0.156 uses the full thickness 40 mm; 0.01042 uses 30 s; 2.5 uses 10 mm.

Further reading: Incropera and DeWitt, Fundamentals of Heat and Mass Transfer

9 · Heat exchangers: plate, tubular and scraped-surface; LMTD and effectiveness · hard

A counter-current heat exchanger has equal heat capacity rates (Cr = 1) and NTU = 2.0. Its effectiveness is

  1. 0.865
  2. 0.667
  3. 0.500
  4. 1.000
Answer and working

B. 0.667

For Cr = 1, ε = NTU/(1 + NTU) = 2/3 = 0.667. 0.865 is 1 − e^−2, valid only when Cr = 0; 0.500 is the co-current limit; 1.000 would need infinite NTU.

Further reading: Incropera and DeWitt, Fundamentals of Heat and Mass Transfer

10 · Convective mass transfer · hard

Water evaporates from a 0.5 m² wet surface with km = 0.02 m/s; vapour concentrations are 0.017 kg/m³ at the surface and 0.008 kg/m³ in the air. The evaporation rate is

  1. 0.17 g/s
  2. 0.08 g/s
  3. 9.0 g/s
  4. 0.09 g/s
Answer and working

D. 0.09 g/s

N = kmA(Cs − C∞) = 0.02 × 0.5 × 0.009 = 9.00e-05 kg/s = 0.09 g/s. 0.17 g/s uses Cs only; 0.08 g/s uses C∞ only; 9.0 g/s is a unit slip.

Further reading: Geankoplis, Transport Processes and Separation Process Principles

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Transport Phenomena in Food Processing: Fluid Flow, Heat & Mass TransferThe full study manual, free GATEFormat, sections and how it has changed

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