🇮🇳 GATE Agricultural Engineering · subject

GATE Agricultural Engineering Agricultural Process Engineering Syllabus

The Agricultural Process Engineering syllabus for GATE Agricultural Engineering in full: 5 chapters, 27 topics and 9 sub-topics, every one listed below, plus 61 Agricultural Process Engineering flashcards with answers.

5Chapters
27Topics
9Sub-topics
61Flashcards

Agricultural Process Engineering syllabus: all 5 chapters

Expand any chapter to see its topics and sub-topics.

  1. Engineering properties of agriculture produce

    5 topics
    • Physical properties
    • Thermal properties
    • Frictional properties
    • Rheological properties
    • Electrical properties
  2. Evaporation and Drying

    5 topics
    • Concentration and drying of liquid foods
      • Evaporators
      • Tray dryers
      • Drum dryers
      • Spray dryers
    • Hydrothermal treatments
    • Drying and milling of cereals, pulses and oilseeds
    • Drying kinetics
    • Psychrometry
      • Properties of air-water vapor mixture
  3. Size Reduction and Material Handling

    9 topics
    • Mechanics and energy requirement in size reduction of agriculture produce
    • Particle size analysis for comminuted solids
    • Size separation by screening
    • Fluidization of granular solids
      • Pneumatic conveying
      • Bucket conveying
      • Screw conveying
      • Belt conveying
    • Cleaning and grading
    • Effectiveness of separation
    • Centrifugal separation of solids, liquids and gases
    • Homogenization
    • Filtration and membrane separation
  4. Processing of Agriculture Produce

    4 topics
    • Processing of seeds
    • Processing of spices
    • Processing of fruits and vegetables
    • Value addition of agriculture produce
  5. Storage Systems

    4 topics
    • Controlled and modified atmosphere storage
    • Perishable food storage
    • Godowns, bins and grain silos
    • Packaging material and machines

GATE Agricultural Engineering Agricultural Process Engineering flashcards — 61 questions with answers

  1. Define bulk density and how it differs from true (particle) density for agricultural produce.

    Bulk density $\rho_{b}$ is the mass of a bulk material divided by the total volume it occupies (including inter-granular air spaces): $\rho_{b} = \frac{m}{V_{bulk}}$. True/particle density $\rho_{t}$ uses only the solid volume of the particles (excluding voids). Thus $\rho_{b} < \rho_{t}$.

  2. Write the formula for porosity of a granular agricultural material in terms of bulk and true density.

    $$\varepsilon = \left(1 - \frac{\rho_{b}}{\rho_{t}}\right) \times 100\%$$ where $\rho_{b}$ is bulk density and $\rho_{t}$ is true density. Porosity is the fraction of total volume occupied by inter-granular voids.

  3. Define sphericity and give its formula for a grain.

    Sphericity $\phi$ expresses how close a particle's shape is to a sphere: $$\phi = \frac{\left(d_{a}\, d_{b}\, d_{c}\right)^{1/3}}{d_{a}}$$ where $d_{a}$ is the longest (major) dimension and $d_a,d_b,d_c$ are the three principal axes. Equivalently $\phi = \frac{d_{e}}{d_{c}}$ (diameter of equivalent-volume sphere over diameter of smallest circumscribing sphere). For a perfect sphere $\phi = 1$.

  4. What is the geometric mean diameter of a grain with principal dimensions $a$, $b$, $c$?

    $$D_{g} = (a\,b\,c)^{1/3}$$ It is the equivalent diameter of a sphere having the same volume product of the three mutually perpendicular axes.

  5. Define the coefficient of static friction and the angle of friction for a grain on a surface.

    The coefficient of static friction $\mu = \tan\phi$, where $\phi$ is the angle of friction. At the point of incipient sliding the frictional force equals $\mu N$ ($N$ = normal force). $\phi$ is the inclination of the surface at which the grain just begins to slide.

  6. Differentiate the angle of repose (static vs dynamic) for granular agricultural materials.

    The angle of repose is the angle the free surface of a heaped bulk material makes with the horizontal. The static (emptying/filling) angle of repose is formed when grain is poured into a heap; the dynamic angle of repose is formed when grain flows out of an orifice (during emptying). The dynamic angle is generally smaller than the static angle.

  7. Define specific heat and give the units used in food/grain thermal property work.

    Specific heat $c_{p}$ is the heat required to raise the temperature of unit mass by one degree: $Q = m\,c_{p}\,\Delta T$. Units: $\mathrm{J\,kg^{-1}\,K^{-1}}$ (or $\mathrm{kJ\,kg^{-1}\,^{\circ}C^{-1}}$).

  8. State Siebel's equation for the specific heat of a food above its freezing point.

    Above freezing: $$c_{p} = 0.837 + 3.349\,X_{w}\quad (\mathrm{kJ\,kg^{-1}\,K^{-1}})$$ where $X_{w}$ is the mass fraction of water. Below freezing the second coefficient is reduced (ice term): $c_{p} = 0.837 + 1.256\,X_{w}$.

  9. Define thermal conductivity and thermal diffusivity, and give the relation between them.

    Thermal conductivity $k$ measures the rate of heat conduction (Fourier's law $q = -k\,\frac{dT}{dx}$, units $\mathrm{W\,m^{-1}\,K^{-1}}$). Thermal diffusivity $\alpha$ measures how fast temperature changes propagate: $$\alpha = \frac{k}{\rho\,c_{p}}\quad (\mathrm{m^{2}\,s^{-1}})$$

  10. What is the latent heat of vaporization of water at $100\,^{\circ}\mathrm{C}$, a key value in drying and evaporation calculations?

    $\lambda \approx 2257\ \mathrm{kJ\,kg^{-1}}$ (about $2.26\ \mathrm{MJ\,kg^{-1}}$) at $100\,^{\circ}\mathrm{C}$ and atmospheric pressure.

  11. Define the dielectric constant and dielectric loss factor relevant to electrical heating of agricultural produce.

    The complex permittivity is $\varepsilon^{*} = \varepsilon' - j\varepsilon''$. The dielectric constant $\varepsilon'$ is the material's ability to store electrical energy; the dielectric loss factor $\varepsilon''$ governs conversion of electrical energy into heat (dielectric/microwave heating). The loss tangent is $\tan\delta = \frac{\varepsilon''}{\varepsilon'}$.

  12. Give the equation for power dissipated per unit volume during dielectric (microwave/RF) heating.

    $$P = 2\pi f\,\varepsilon_{0}\,\varepsilon''\,E^{2}$$ where $f$ is frequency, $\varepsilon_{0}=8.854\times10^{-12}\ \mathrm{F\,m^{-1}}$, $\varepsilon''$ is the dielectric loss factor and $E$ is the electric field strength. Power absorbed rises with frequency, loss factor and the square of field strength.

  13. Classify the rheological behaviour of fluid foods using the power-law (Ostwald-de Waele) model.

    $$\tau = K\,\dot{\gamma}^{\,n}$$ where $\tau$ = shear stress, $\dot{\gamma}$ = shear rate, $K$ = consistency index, $n$ = flow behaviour index. $n=1$: Newtonian; $n<1$: pseudoplastic (shear-thinning); $n>1$: dilatant (shear-thickening).

  14. Distinguish a Bingham plastic fluid from a Newtonian fluid rheologically.

    A Newtonian fluid obeys $\tau = \mu\,\dot{\gamma}$ and flows at any nonzero stress. A Bingham plastic requires a finite yield stress $\tau_{0}$ before flow begins: $$\tau = \tau_{0} + \mu_{p}\,\dot{\gamma}\quad (\tau > \tau_{0})$$ where $\mu_{p}$ is the plastic viscosity. Examples: tomato paste, ketchup, chocolate melt.

  15. Define apparent viscosity for a non-Newtonian fluid.

    Apparent viscosity is the ratio of shear stress to shear rate at a given shear rate: $$\eta_{app} = \frac{\tau}{\dot{\gamma}} = K\,\dot{\gamma}^{\,n-1}$$ (for a power-law fluid). It is not constant but varies with shear rate, unlike the true viscosity of a Newtonian fluid.

  16. What is thixotropy, and how does it differ from rheopexy?

    Thixotropy is a time-dependent decrease in apparent viscosity under constant shear, with recovery on rest (e.g. starch gels, mayonnaise). Rheopexy (negative thixotropy) is a time-dependent increase in viscosity under constant shear. Both are reversible time effects, in contrast to permanent shear breakdown.

  17. List the main purposes of concentrating liquid foods before drying.

    Concentration removes part of the water to: reduce mass/volume (cheaper transport and storage), lower water activity for preservation, reduce energy needed for subsequent drying, and improve product properties. It is usually achieved by evaporation, freeze concentration, or membrane (reverse osmosis) concentration.

  18. Write the steady-state mass balance for total solids in a single-effect evaporator concentrating a feed.

    For feed $F$ at solids fraction $x_{F}$ giving product $P$ at $x_{P}$ and vapour $V$: Total: $F = P + V$. Solids: $F\,x_{F} = P\,x_{P}$ (solids leave only in product). Thus $P = \dfrac{F\,x_{F}}{x_{P}}$ and $V = F - P$.

  19. Define the steam economy of an evaporator.

    Steam economy is the mass of water vapour evaporated per unit mass of heating (live) steam supplied: $$\text{Steam economy} = \frac{\text{kg water evaporated}}{\text{kg steam used}}$$ A single-effect evaporator has economy $\lesssim 1$; an $N$-effect system has economy approaching $N$.

  20. How does the number of effects relate to steam economy in multiple-effect evaporation?

    In an $N$-effect evaporator the vapour from one effect becomes the heating medium for the next, so roughly $1\ \mathrm{kg}$ of steam evaporates about $N\ \mathrm{kg}$ of water. Steam economy $\approx N$ (slightly less due to losses), while the total heat-transfer area and capital cost rise roughly in proportion to $N$.

Show the other 41 Agricultural Process Engineering flashcards
  1. What is boiling-point elevation (BPE) in evaporators and why does it matter?

    BPE is the increase in the boiling point of a solution above that of pure water at the same pressure, caused by dissolved solids (a colligative effect). It reduces the effective temperature-difference driving force $\Delta T$ available for heat transfer, lowering evaporator capacity. It is accounted for using Dühring's rule.

  2. Why are falling-film and rising-film evaporators preferred for heat-sensitive liquid foods?

    They give thin liquid films with short residence times and high heat-transfer coefficients, allowing low-temperature operation (often under vacuum). This minimizes thermal damage, fouling and quality loss in heat-sensitive foods such as fruit juices and milk.

  3. Describe the operating principle of a tray (cabinet) dryer.

    A tray dryer is a batch convective dryer: product is spread on perforated or solid trays stacked in an insulated chamber; heated air is circulated (by fans, parallel or through-flow) over/through the trays. Moisture is carried away by the air. It is simple, low-cost, used for small batches of fruits, vegetables and spices, but gives non-uniform drying between trays.

  4. Describe the construction and use of a drum (roller) dryer.

    A drum dryer consists of one or two slowly rotating, internally steam-heated metal drums. A thin film of liquid/slurry food is applied to the hot drum surface, dries within part of a revolution, and is scraped off by a doctor blade as flakes/sheet. Used for milk, mashed potato, starch, baby foods; gives short contact time but high surface temperatures.

  5. Explain the working principle of a spray dryer.

    A spray dryer atomizes a liquid feed into fine droplets (by rotary atomizer, pressure nozzle, or two-fluid nozzle) into a chamber of hot air. The huge surface area gives almost instantaneous drying; dried powder falls and is separated (cyclone/bag filter) from the exhaust air. Used for milk powder, instant coffee, eggs; very short residence time protects heat-sensitive products.

  6. Compare drum drying and spray drying on contact temperature, product form and residence time.

    Drum drying: conductive, high surface temperature, product as flakes/sheet, residence a few seconds to tens of seconds, lower capital cost. Spray drying: convective, lower particle temperature (evaporative cooling), product as free-flowing powder, residence very short (seconds), higher capital cost but better quality for heat-sensitive liquids.

  7. What is parboiling of paddy and what are its main objectives?

    Parboiling is a hydrothermal treatment of paddy involving soaking, steaming and drying before milling. Objectives: gelatinize starch, increase head (whole) rice yield by reducing breakage, drive water-soluble vitamins/minerals from bran into the endosperm (nutrient retention), and harden the kernel. It is a classic hydrothermal treatment of cereals.

  8. List the three sequential operations of paddy parboiling.

    (1) Soaking — paddy is hydrated in warm water to raise moisture (~30% wb); (2) Steaming/heating — gelatinizes the starch; (3) Drying — moisture is reduced for safe storage and milling. Subsequent milling then removes husk and bran.

  9. Differentiate the husking and polishing/whitening steps in rice milling.

    Husking (dehusking/shelling) removes the outer husk (hull) to give brown rice, typically with rubber-roll or disc shellers. Polishing/whitening (milling) removes the bran layers and germ from brown rice by abrasion/friction to give white (polished) rice, optionally followed by polishing for shine.

  10. What is the degree of milling (DOM) in rice processing?

    Degree of milling is the extent to which bran and germ are removed from brown rice, usually expressed as the percentage of bran removed or the percentage weight reduction of brown rice during whitening. Higher DOM gives whiter rice but lower yield and reduced nutrients.

  11. Distinguish dry milling and wet milling of cereals such as maize.

    Dry milling tempers grain to a moderate moisture then breaks it to separate germ, bran and endosperm, yielding grits, meal and flour. Wet milling steeps grain (often with $\ce{SO2}$), then mills and separates components in water to give relatively pure starch, gluten, germ (oil) and fibre. Wet milling gives purer fractions; dry milling is simpler and cheaper.

  12. Define moisture content on wet basis and dry basis, and give the conversion between them.

    Wet basis: $M_{wb} = \dfrac{m_{w}}{m_{w}+m_{d}}$. Dry basis: $M_{db} = \dfrac{m_{w}}{m_{d}}$. Conversions: $$M_{db} = \frac{M_{wb}}{1 - M_{wb}}, \qquad M_{wb} = \frac{M_{db}}{1 + M_{db}}$$ where $m_w$ is water mass and $m_d$ is dry-matter mass.

  13. Define equilibrium moisture content (EMC) and free moisture content in drying.

    EMC ($M_{e}$) is the moisture content a material attains in equilibrium with air of a given temperature and relative humidity (no further net moisture exchange). Free moisture content is the moisture available for removal: $$M_{free} = M - M_{e}$$ where $M$ is the current moisture content (dry basis).

  14. Sketch and label the phases of a typical drying-rate curve.

    Plotting drying rate vs free moisture: (1) an initial transient (warming-up) period; (2) the constant-rate period, where rate is limited by surface evaporation and moisture is freely available; (3) the falling-rate period(s), beginning at the critical moisture content, where internal moisture diffusion controls and rate declines to zero at EMC.

  15. What is the critical moisture content in drying?

    The critical moisture content $M_{c}$ is the free moisture content at which drying changes from the constant-rate period to the falling-rate period. Above $M_c$ the surface stays saturated; below it the surface begins to dry and internal moisture movement limits the rate.

  16. Write the constant-rate-period drying equation for heat-controlled surface evaporation.

    $$R_{c} = \frac{h\,(T_{a} - T_{s})}{\lambda}$$ where $R_c$ = constant drying rate per unit area, $h$ = convective heat-transfer coefficient, $T_a$ = air temperature, $T_s$ = wet-surface (wet-bulb) temperature, and $\lambda$ = latent heat of vaporization. Equivalently mass-transfer form: $R_{c} = k_{y}(H_{s} - H_{a})$.

  17. Give the thin-layer (exponential) drying model for the falling-rate period.

    Assuming drying rate proportional to free moisture (first-order): $$\text{MR} = \frac{M - M_{e}}{M_{0} - M_{e}} = e^{-k t}$$ where MR is the moisture ratio, $M_0$ the initial moisture, $M_e$ the equilibrium moisture, $k$ the drying constant and $t$ time. (Page model: $\text{MR}=e^{-k t^{n}}$.)

  18. Define psychrometry and the dry-bulb temperature of air.

    Psychrometry is the study of the thermodynamic properties of moist air (air-water-vapour mixtures) and their use in processes that change air temperature and humidity. The dry-bulb temperature is the air temperature measured by an ordinary thermometer shielded from radiation and moisture.

  19. Define absolute humidity (humidity ratio) of moist air and give its formula.

    Absolute humidity (humidity ratio) $W$ is the mass of water vapour per unit mass of dry air: $$W = \frac{m_{v}}{m_{a}} = 0.622\,\frac{p_{v}}{P - p_{v}}$$ where $p_{v}$ is the partial pressure of water vapour, $P$ is total pressure, and $0.622$ is the ratio of molar masses ($18/29$).

  20. Define relative humidity and contrast it with absolute humidity.

    Relative humidity is the ratio of the actual vapour partial pressure to the saturation vapour pressure at the same temperature: $$\mathrm{RH} = \frac{p_{v}}{p_{vs}} \times 100\%$$ It is a dimensionless percentage that depends on temperature, whereas absolute humidity (humidity ratio) is the actual mass of vapour per kg dry air and is temperature-independent.

  21. Distinguish dew-point temperature and wet-bulb temperature of moist air.

    Dew-point temperature is the temperature to which moist air must be cooled at constant pressure and humidity ratio to reach saturation (condensation begins). Wet-bulb temperature is the steady temperature reached by a wetted thermometer in moving air, reflecting simultaneous heat and mass transfer (evaporative cooling). For unsaturated air: $T_{dry} > T_{wet} > T_{dew}$.

  22. Define the enthalpy of moist air per kg of dry air.

    $$h = c_{pa}\,T + W\,(\lambda_{0} + c_{pv}\,T)$$ where $c_{pa}\approx1.005\ \mathrm{kJ\,kg^{-1}\,K^{-1}}$ (dry air), $c_{pv}\approx1.88\ \mathrm{kJ\,kg^{-1}\,K^{-1}}$ (vapour), $\lambda_{0}\approx2501\ \mathrm{kJ\,kg^{-1}}$ (latent heat at $0\,^{\circ}\mathrm{C}$), $W$ = humidity ratio, $T$ = dry-bulb temperature in $^{\circ}\mathrm{C}$.

  23. Define humid (specific) volume of moist air.

    Humid volume $v_{H}$ is the volume of moist air per unit mass of dry air: $$v_{H} = \left(\frac{1}{29} + \frac{W}{18}\right)\frac{R\,T}{P}$$ approximately $v_{H} = (0.00283 + 0.00456\,W)\,T\ (\mathrm{m^{3}\,kg^{-1}\ dry\ air})$, with $T$ in kelvin. It increases with temperature and humidity ratio.

  24. On a psychrometric chart, what process is represented by a horizontal line, and what by a constant-enthalpy line?

    A horizontal line (constant humidity ratio $W$) represents sensible heating or cooling — only dry-bulb temperature changes. A constant wet-bulb / constant-enthalpy line (sloping down to the right) approximates an adiabatic saturation/evaporative-cooling process, where dry-bulb falls as humidity rises at nearly constant enthalpy.

  25. State Rittinger's law for energy in size reduction and the property it assumes.

    Rittinger's law: energy required is proportional to the new surface area created, i.e. proportional to the increase in specific surface (inverse of particle size): $$E = K_{R}\left(\frac{1}{d_{2}} - \frac{1}{d_{1}}\right)$$ where $d_1,d_2$ are feed and product sizes. Best suited to fine grinding.

  26. State Kick's law of size reduction.

    Kick's law: the energy required is proportional to the logarithm of the size-reduction ratio (energy per unit mass is constant for a given reduction ratio): $$E = K_{K}\,\ln\!\left(\frac{d_{1}}{d_{2}}\right)$$ where $d_1$ and $d_2$ are initial and final particle sizes. Best for coarse crushing.

  27. State Bond's law and define the work index.

    Bond's law: $$E = K_{B}\left(\frac{1}{\sqrt{d_{2}}} - \frac{1}{\sqrt{d_{1}}}\right)$$ The work index $W_{i}$ is the energy (kWh per tonne) needed to reduce material from theoretically infinite size to 80% passing $100\ \mu\mathrm{m}$. Bond's law is intermediate between Kick's and Rittinger's, used for intermediate grinding.

  28. Show that Kick's, Rittinger's and Bond's laws are special cases of the general (Walker) differential equation $dE = -C\,\frac{dx}{x^{n}}$.

    Integrating $dE=-C\,x^{-n}\,dx$: for $n=1$ gives Kick's law (energy $\propto \ln(d_1/d_2)$); for $n=2$ gives Rittinger's law (energy $\propto 1/d_2 - 1/d_1$); for $n=1.5$ gives Bond's law (energy $\propto 1/\sqrt{d_2}-1/\sqrt{d_1}$). Thus the exponent $n$ distinguishes the three laws.

  29. Define the reduction ratio in a size-reduction (comminution) operation.

    Reduction ratio is the ratio of a characteristic feed size to the corresponding product size: $$R = \frac{d_{feed}}{d_{product}}$$ Often based on 80%-passing sizes ($d_{80}$). It measures how much the material has been reduced in one pass.

  30. Define mean particle diameters used in particle-size analysis of comminuted solids: the surface-volume (Sauter) mean.

    The Sauter (surface-volume) mean diameter is $$D_{sv} = \frac{\sum n_{i} d_{i}^{3}}{\sum n_{i} d_{i}^{2}} = \frac{1}{\sum (x_{i}/d_{i})}$$ where $x_i$ is the mass fraction in size interval $i$ with mean diameter $d_i$. It is the diameter of a sphere having the same surface-to-volume ratio as the whole sample, important for surface-area-controlled processes.

  31. Define the fineness modulus and uniformity index obtained from sieve analysis of ground feed.

    Fineness modulus (FM) is obtained by summing the cumulative percentages of material retained on a standard set of sieves and dividing by 100; higher FM means coarser product. The uniformity index expresses the proportions of coarse, medium and fine fractions (often as a three-figure ratio) describing the spread of the size distribution.

  32. In screening, define screen effectiveness (efficiency).

    Screen effectiveness measures how well a screen separates oversize from undersize. A common overall form is the product of recovery of undersize in the underflow and rejection of oversize: $$E = E_{o}\,E_{u}$$ where each term compares actual mass of correctly sorted material to the amount in the feed. Higher feed rate generally lowers effectiveness.

  33. Write the material balance for an ideal screen separating feed $F$ into overflow $D$ (oversize) and underflow $B$ (undersize).

    Total: $F = D + B$. Component (fraction of material finer than the cut, $x$): $$F\,x_{F} = D\,x_{D} + B\,x_{B}$$ Solving gives $\dfrac{D}{F} = \dfrac{x_{F}-x_{B}}{x_{D}-x_{B}}$ and $\dfrac{B}{F} = \dfrac{x_{D}-x_{F}}{x_{D}-x_{B}}$.

  34. What is the relationship between mesh number and aperture size in standard screens?

    Mesh number is the number of openings per linear inch of screen. As mesh number increases, the aperture (opening) size decreases, so higher mesh numbers pass finer particles. Aperture also depends on wire diameter: $\text{aperture} = \frac{1\ \text{inch}}{\text{mesh}} - (\text{wire diameter})$.

  35. Name and distinguish the common industrial screening equipment by motion.

    Grizzly: static inclined parallel bars for coarse scalping. Trommel: rotating cylindrical screen. Vibrating screen: high-frequency vibration for efficient sizing. Gyratory/oscillating screen: gyratory horizontal motion for fine separation. They differ in the motion imparted (none, rotation, vibration, gyration) and the size range handled.

  36. What is the terminal velocity of a particle and its role in pneumatic/aspiration separation of grains?

    Terminal velocity is the constant velocity a particle attains in a fluid when gravity balances drag and buoyancy: $$v_{t} = \sqrt{\frac{4\,g\,d\,(\rho_{p}-\rho_{f})}{3\,C_{D}\,\rho_{f}}}$$ Particles with different terminal velocities are separated in an air stream (aspiration), e.g. cleaning grain from chaff and light impurities.

  37. In the constant-rate drying period, why does the surface stay at the wet-bulb temperature of the drying air?

    During the constant-rate period the surface is fully wetted, so all sensible heat supplied by the air is consumed as latent heat of evaporating free water. This dynamic equilibrium between convective heat gain and evaporative cooling holds the surface at the air's wet-bulb temperature until the critical moisture content is reached.

  38. What is hysteresis in moisture sorption isotherms of grains?

    Sorption hysteresis is the phenomenon where the equilibrium moisture content at a given relative humidity is higher along the desorption (drying) isotherm than along the adsorption (wetting) isotherm. The two curves form a loop, attributed to capillary condensation, pore-shape ('ink-bottle') effects and structural changes.

  39. Give the Brunauer (BET) interpretation relevant to sorption isotherms of foods.

    The BET equation describes multilayer moisture sorption and is used to estimate the monolayer moisture value $M_{0}$: $$\frac{a_{w}}{(1-a_{w})M} = \frac{1}{M_{0}C} + \frac{C-1}{M_{0}C}\,a_{w}$$ where $a_{w}$ is water activity, $M$ moisture content, and $C$ a constant. The monolayer value indicates the moisture for maximum storage stability.

  40. Define water activity and its significance in drying and storage of foods.

    Water activity $a_{w} = \dfrac{p_{v}}{p_{vs}} = \dfrac{\mathrm{ERH}}{100}$ is the ratio of the vapour pressure of water in the food to that of pure water at the same temperature (equal to equilibrium relative humidity/100). It governs microbial growth, enzymatic and chemical reactions; lowering $a_w$ by drying improves shelf stability (most bacteria need $a_w>0.9$).

  41. Differentiate the through-flow and cross-flow (parallel) air patterns in tray dryers.

    In cross-flow (parallel-flow) tray dryers heated air passes horizontally over the surface of the product on the trays. In through-flow tray dryers air is forced vertically through perforated trays and the bed of product, giving greater contact area, faster and more uniform drying, but higher fan power.

GATE Agricultural Engineering Agricultural Process Engineering FAQ

What is in the GATE Agricultural Engineering Agricultural Process Engineering syllabus?

GATE Agricultural Engineering Agricultural Process Engineering is split into 5 chapters: Engineering properties of agriculture produce, Evaporation and Drying, Size Reduction and Material Handling, Processing of Agriculture Produce and Storage Systems. Together they contain 27 topics and 9 sub-topics.

How many chapters are in GATE Agricultural Engineering Agricultural Process Engineering?

5 chapters, containing 27 topics and 9 sub-topics. The largest chapter is Size Reduction and Material Handling, with 9 topics.

Are there flashcards for GATE Agricultural Engineering Agricultural Process Engineering?

Yes — a 61-card Agricultural Process Engineering deck, printed in full in the flashcards section of this page and available in the Examius app with spaced repetition. The free plan covers one exam of your choice, including its flashcards; Examius Pro, a one-time purchase, unlocks every other exam and deck.