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Unit Operations in Food Processing Questions and Answers

1.

Tomato catsup with 10 Pa sn consistency coefficient and 0.8 flow behaviour index is flowing in a pipe. Generalized coefficient of viscosity of catsup, in Pa sn is

Exam Year(s)

GATE AG - 2012,

Answer: B

The generalized coefficient of viscosity is 6.93 Pa sn.

For pipe flow of a power-law fluid the Metzner-Reed approach first converts the consistency coefficient K into the pipe-flow coefficient K', then forms the generalized viscosity \(\gamma = K' 8^{\,n-1}\) that appears in the generalized Reynolds number:

\[ \begin{aligned} K' &= K\left(\frac{3n+1}{4n}\right)^{n} \\ &= 10\left(\frac{3.4}{3.2}\right)^{0.8} = 10.50 \\ \gamma &= K' 8^{\,n-1} \\ &= 10.50 \times 8^{-0.2} \\ &= 10.50 \times 0.660 = 6.93 \end{aligned} \]

Stopping at K' gives 10.5, which is not offered; the other options do not follow from K = 10 and n = 0.8 by these relations.

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2.

A packed bed of 480 kg solid particles having particle size of 0.15 mm and density of 800 kg m−3 is fluidized using air at 25 °C and 1 atmospheric pressure. If the cross section of the empty bed is 0.45 m2 and voidage at minimum fluidizing condition is 0.5, then the minimum height of the fluidized bed, in m is

Exam Year(s)

GATE AG - 2012,

Answer: C

The minimum height of the fluidized bed is about 2.7 m.

At minimum fluidization the solids occupy the fraction \((1 - \varepsilon_{mf})\) of the bed volume, so the solid mass fixes the bed height:

\[ \begin{aligned} m &= \rho_{s} A L_{mf} (1 - \varepsilon_{mf}) \\ L_{mf} &= \frac{480}{800 \times 0.45 \times 0.5} \\ &= \frac{480}{180} = 2.67\ \text{m} \end{aligned} \]

The particle size and the air temperature and pressure are needed for the minimum fluidizing velocity, not for the bed height. Ignoring the voidage gives 1.33 m, and the larger values would need a smaller area or a higher voidage than given.

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3.

Match the following reactants in Column I with the most appropriate purpose used in processing as mentioned in Column II

Column IColumn II
IVitamin EPfumigant for insect killing
IICalcium saltsQreduces shrinkage losses
IIIEdible waxesRantioxidant with vitamin A in oils
IVMethyl BromideSfirming agent in fruits

Exam Year(s)

GATE AG - 2022,

Answer: D

The correct matching is I-R, II-S, III-Q, IV-P.

  • Vitamin E (tocopherol) is an antioxidant added to oils, where it also protects vitamin A from oxidation (I-R).
  • Calcium salts such as calcium chloride are firming agents: calcium cross-links the pectin in fruit tissue and keeps canned or cut fruit firm (II-S).
  • Edible waxes coated on fruits and vegetables slow moisture loss and so reduce shrinkage losses in storage (III-Q).
  • Methyl bromide is a fumigant used to kill insects in stored grain and commodities (IV-P).

Options pairing vitamin E with fumigation or methyl bromide with antioxidant use reverse I and IV, and the option matching calcium salts to shrinkage and waxes to firming swaps II and III.

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4.

Make the correct match between specific food processing operations in Group I with their mechanism of action in Group II.

Group IGroup II
P) Ball Mill1) Compression and shear
Q) Roller Mill2) Pressure bursting
R) Flash Peeling3) Friction and shear
S) Abrasive Peeling4) Impact and shear

Exam Year(s)

GATE Life Sciences - 2012, GATE XE - 2012,

Answer: B

The match is P-4, Q-1, R-2, S-3.

  • A ball mill reduces size by the impact of falling balls and the shear between them.
  • A roller mill crushes material between rotating rolls by compression, with shear when the rolls turn at different speeds.
  • Flash (steam) peeling exposes produce to high-pressure steam and then releases the pressure suddenly, so steam under the skin bursts it off.
  • Abrasive peeling rubs the skin off against rough rollers or carborundum surfaces, by friction and shear.

The other options swap these mechanisms, for example making the roller mill act by pressure bursting.

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5.

Match the process operation in Group I with the separated constituent in Group II

Group IGroup II
P. Extraction1. Phospholipids
Q. Degumming2. Free fatty acids
R. Neutralization3. Pigments
S. Bleaching4. Crude oil

Exam Year(s)

GATE XE - 2018,

Answer: C

The match is P-4, Q-1, R-2, S-3.

These are the steps of edible oil processing, in order. Extraction (pressing or solvent extraction) separates the crude oil from the seed. Degumming with water or acid hydrates and removes the phospholipids (gums). Neutralization with alkali converts free fatty acids into soap, which is separated. Bleaching with adsorbent earth removes pigments such as chlorophyll and carotenoids.

The other options swap two of these, for example giving bleaching the free fatty acids or degumming the pigments.

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6.

Identify an example of a classical diffusional mass transfer process without involving heat, among the following.

Exam Year(s)

GATE XE - 2018,

Answer: B

The example is carbonation of beverages.

In carbonation, carbon dioxide dissolves into the chilled beverage and diffuses into the liquid, driven by its concentration difference under pressure. It is a gas absorption, a pure mass transfer process, with no heat supplied to drive it.

Drying of food grains and concentration of fruit juice remove water by evaporation, which needs heat for the latent heat of vaporisation. Distillation of alcohol separates components by boiling and condensing, so heat transfer is essential to all three.

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7.

Match the edible oil refining stages given in Column I with their respective functions in Column II

Column IColumn II
P. Degumming1. Separation of waxes
Q. Neutralization2. Removal of pigments
R. Bleaching3. Removal of phosphatides
S. Winterization4. Removal of free fatty acids

Exam Year(s)

GATE Life Sciences - 2022, GATE XE - 2022,

Answer: C

The match is P-3, Q-4, R-2, S-1.

  • Degumming removes phosphatides (gums) by hydration with water or acid.
  • Neutralization (alkali refining) converts free fatty acids into soaps that are separated.
  • Bleaching with adsorbent earths removes pigments such as chlorophyll and carotenoids.
  • Winterization chills the oil to crystallize and filter off waxes and high-melting glycerides.

The other options misplace neutralization or bleaching, for example making neutralization remove pigments.

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8.

Make the correct pair of food packaging technology given in Column I with operating principle or description in Column II.

Column IColumn II
P. Aseptic packaging1. Control of the concentration of O2 and CO2 inside the package
Q. Active packaging2. Create a skin tight package wall
R. Modified atmosphere packaging3. Independent sterilization of food and packaging material and packaging under sterile environment
S. Vacuum packaging4. Makes non-passive contribution to product development

Exam Year(s)

GATE Life Sciences - 2022, GATE XE - 2022,

Answer: A

The match is P-3, Q-4, R-1, S-2.

  • Aseptic packaging sterilizes the food and the packaging separately and fills them in a sterile environment.
  • Active packaging contains components such as oxygen scavengers or antimicrobials that take a non-passive part in keeping the product.
  • Modified atmosphere packaging sets the concentrations of O2 and CO2 inside the pack.
  • Vacuum packaging removes air so the film pulls tight around the product, forming a skin-tight wall.

The other options confuse vacuum packaging with active packaging, or modified atmosphere with aseptic packaging.

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9.

Which one of the following steps is used to remove phosphatides from crude oil in the refining process?

Exam Year(s)

GATE Life Sciences - 2024, GATE XE - 2024,

Answer: C

Phosphatides are removed by degumming.

Crude oils, especially soybean oil, contain phospholipids (phosphatides, the gums) that hydrate and settle out. In degumming the oil is treated with water or dilute acid; the hydrated gums are separated by centrifugation and are the source of commercial lecithin.

Neutralization removes free fatty acids as soaps, bleaching removes pigments with adsorbent earths, and deodorization strips volatile odour compounds by steam distillation under vacuum.

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10.

Identify the gas composition (in percent) suitable for packaging cured meat under MAP conditions.

Exam Year(s)

GATE XE - 2025,

Answer: A

The suitable mixture is \(O_{2} = 0; CO_{2} = 50; N_{2} = 50\).

The pink colour of cured meat comes from nitrosyl myoglobin, which fades quickly in the presence of oxygen and light. Oxygen also promotes rancidity. Cured meats are therefore packed with oxygen excluded:

  • Carbon dioxide inhibits aerobic spoilage bacteria and moulds.
  • Nitrogen is an inert filler that prevents pack collapse as CO2 dissolves into the meat.

Mixtures with 50% oxygen suit fresh red meat, where oxygen keeps oxymyoglobin bright red, but they would discolour cured meat. Pure nitrogen excludes oxygen but gives no antimicrobial action, so it offers less protection than the CO2/N2 mix.

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