Food Process Engineering

Bond's law size reduction

Bond's law relates the energy of size reduction to the feed and product sizes through a material property called the work index. Bond defined the work index W_i as the specific energy, in kWh per tonne, needed to reduce a material from an infinitely large feed to a product of which 80% passes 100 µm. For a feed with 80% passing size F_{80} and a product with 80% passing size P_{80}, both in micrometres, the specific energy is W = 10\,W_i\left(\dfrac{1}{\sqrt{P_{80}}} - \dfrac{1}{\sqrt{F_{80}}}\right) kWh/t. Dividing the measured power by the throughput gives W, and the work index follows by rearranging this expression. A larger W_i means a harder material.

Because W_i is a constant of the material, the energy for a different product size can be found by comparing the two brackets, since the ratio of energies equals the ratio of the brackets. The bracket grows quickly as P_{80} becomes smaller, so fine grinding is much more energy-intensive than coarse grinding.

Bond's law is one of three classical size-reduction laws. Kick's law assumes that the energy is proportional to the logarithm of the reduction ratio, E = K_K \ln(D_1/D_2), and it is applied mainly to coarse crushing. At a fixed feed rate the power follows the same logarithmic dependence, so a power value can be scaled to another product size by the ratio of logarithms. Rittinger's law assumes that the energy is proportional to the new surface area created, so it suits fine grinding. Bond's law lies between the two, since its size exponent is intermediate between theirs.

Janssen equation for bin wall pressure

Janssen's theory explains why the pressure on the walls of a grain bin does not grow without limit with depth. Part of the weight of the grain is carried by friction on the wall, so the vertical pressure approaches a limiting value. Let \gamma be the bulk weight density (bulk density times g), R the hydraulic radius (cross-sectional area divided by wetted perimeter), \mu = \tan\phi_w the wall friction coefficient, K the ratio of horizontal to vertical pressure, and z the depth. The vertical pressure is p_v = \dfrac{\gamma R}{\mu K}\left(1 - e^{-\mu K z/R}\right), and the lateral pressure is p_h = K p_v = \dfrac{\gamma R}{\mu}\left(1 - e^{-\mu K z/R}\right). The same exponential form is used for silos in Airy's theory.

Because the exponential term tends to zero, increasing the fill depth raises the lateral pressure by a factor smaller than the depth ratio, and the size of that factor depends on \mu K/R.

For a circular bin the hydraulic radius is R = D/4, so the equivalent diameter D_e = 4R equals the actual diameter of the bin. This is the reason the hydraulic radius, rather than the diameter, is the parameter that enters the wall-pressure expression.

Multiple effect evaporator mass and energy balance

An evaporator concentrates a solution by boiling water off. Three balances describe it: a total mass balance, a solids balance, and an energy balance. Since solids leave only in the concentrate, the solids balance is F x_F = P x_P, where F and P are the feed and product flow rates and x the solids fraction. The total mass balance is F = P + V, with V the vapour evaporated. In the energy balance the heat supplied equals the sensible heat needed to raise the feed to the boiling temperature plus the latent heat of the evaporated water, Q = F c_p (T_b - T_F) + V\lambda.

If the concentration ratio of product solids to feed solids is known and all solids leave in the product, the product mass is the feed mass divided by that ratio, and the evaporated water follows from the difference. The feed temperature does not enter the solids balance.

Steam economy is the ratio of the total vapour evaporated to the mass of steam supplied, \text{SE} = \dfrac{\sum V}{S}. In a forward-feed multiple-effect system, the vapour from one effect heats the next, so the steam is reused and the economy is greater than one. Ideally it approaches the number of effects, so a triple-effect unit can reach roughly three.

Centrifugal separation principles

In a centrifuge the gravitational acceleration is replaced by the centrifugal acceleration \omega^2 r, where \omega is the angular speed and r the radius. For a small particle obeying Stokes' law in this field, the terminal settling velocity is v = \dfrac{d^2(\rho_p - \rho_f)\omega^2 r}{18\,\mu}, so v \propto d^2\omega^2 at a fixed radius. Because the diameter enters squared, halving the globule diameter reduces the velocity to one quarter, while tripling the speed multiplies it by nine. The two changes combine as a product of their factors.

The separation factor is the ratio of the centrifugal to the gravitational acceleration, S = \dfrac{\omega^2 r}{g} = \dfrac{v_t^2}{rg}, where v_t is the tangential velocity. It does not depend on the particle diameter, so it describes the strength of the field and not the settling of a particular particle.

In a rotating liquid the radial pressure gradient is \dfrac{dp}{dr} = \rho\,\omega^2 r. Integrating between two radii gives the pressure difference \Delta p = \tfrac{1}{2}\rho\,\omega^2 (r_2^2 - r_1^2). This relation is used in basket centrifuges and liquid-liquid separators to find the pressure needed to hold an interface at a given radius. The angular speed is \omega = 2\pi N/60 with N in rpm.

Mass balance in food drying

In drying, the dry solids are conserved, and water is the only component that leaves the material. Therefore the solids in the feed equal the solids in the product, m_{in}(1 - M_{in}) = m_{out}(1 - M_{out}) on a wet basis, which is the usual starting point for a drying mass balance. When a pretreatment such as peeling removes material before drying, the loss must be subtracted from the input mass before the solids balance is applied. The product yield is then the product mass expressed as a percentage of the original fresh mass.

The energy for drying is found from the water evaporated, which is the difference between the feed water and the product water. The water content of the product is obtained from the dry solids and the final moisture content. The energy has two parts: the latent heat of vaporisation of the evaporated water at the drying temperature, and the sensible heat needed to raise the material to that temperature, Q = m_w \lambda + m_{s} c_p \Delta T.

• wet basis — moisture expressed as a fraction of the total wet mass.
• total solids — the complement of moisture, expressed on the same basis.
• yield — the mass of product obtained per unit mass of fresh raw material.

Ball mill critical speed

The critical speed of a ball mill is the rotational speed at which a ball is just held against the shell at the top of its path, so that the centrifugal force equals the weight. Setting \omega^2 (R - r) = g gives N_c = \dfrac{60}{2\pi}\sqrt{\dfrac{g}{R - r}} rpm, where R is the inside radius of the mill and r the radius of the ball. When the ball size is negligible compared with the mill, the common approximation is N_c \approx 42.3/\sqrt{D} rpm with D the mill diameter in metres.

Because a larger ball reduces R - r, it raises the critical speed. The mill is operated below N_c, usually at a fixed fraction of it, so that the balls cascade rather than centrifuge.

For a given mill, the critical speed is computed from the inside radius and the ball radius only, so the mill diameter and ball diameter are the two data needed.

Moisture content basis conversion

Moisture content can be expressed on a wet basis or a dry basis. On a wet basis, M_{wb} = \dfrac{W}{W + D}, where W is the mass of water and D the mass of dry matter. On a dry basis, M_{db} = \dfrac{W}{D}. The two are related by M_{db} = \dfrac{M_{wb}}{1 - M_{wb}} and M_{wb} = \dfrac{M_{db}}{1 + M_{db}}. Because the dry basis uses dry matter in the denominator, its value can exceed 100%. A wet-basis moisture of 70% already gives a dry-basis value above 200%.

If the ratio of wet-basis to dry-basis moisture is known, the ratio of dry matter to water follows directly, since \dfrac{M_{wb}}{M_{db}} = \dfrac{D}{W + D}. Solving for D/W gives the ratio of dry matter to water in the grain.

• dry basis — water mass per unit dry matter mass, which can be above 100%.
• wet basis — water mass per unit total mass, which is always below 100%.

Osmotic dehydration and water activity

For an ideal solution, Raoult's law gives the water activity as the mole fraction of water, a_w = x_w = \dfrac{n_w}{n_w + n_s}, where n_w and n_s are the moles of water and of solute. Osmotic dehydration works because water moves from the food, whose water activity is higher, to a solution whose water activity is lower. The most effective solution is therefore the one with the lowest water activity below that of the food tissue, since this gives the largest driving force.

Adding a salt to a sugar solution lowers its water activity more than the same mass of sugar would. If NaCl is treated as dissociating into two ions, its effective moles double, and the solution's water activity falls further. A balance on water and solute can then be solved for the salt needed to reach a target water activity.

• water activity — ratio of the vapour pressure of the food to that of pure water at the same temperature.
• mole fraction — moles of a component divided by total moles in the solution.

Sieve analysis mean particle diameter

Sieve analysis gives the mass fraction x_i retained between sieves whose average opening is D_{pi}. The mass-weighted average particle size is \bar{D} = \dfrac{\sum m_i d_i}{\sum m_i}, where m_i is the mass of the ith fraction and d_i its representative size. This is the mean size most often reported for a flour or powder.

The volume-surface (Sauter) mean diameter weights each fraction by its surface and volume. A fraction of mass x_i contains a number of particles proportional to x_i/D_{pi}^3. The ratio of total volume to total surface then gives \bar{D}_{vs} = \dfrac{1}{\sum \left(x_i / D_{pi}\right)}. For example, two equal mass fractions of sizes 1 mm and 2 mm give \bar{D}_{vs} = 1/(0.5 + 0.25) = 1.33 mm, which is smaller than the arithmetic mean of 1.5 mm.

Cake filtration specific resistance

In cake filtration, solids deposit on the medium and form a cake whose resistance controls the rate. The specific cake resistance \alpha (m/kg) is defined per unit mass of dry cake per unit filter area, so it is a property of the solids and their compressibility, not of the cake thickness. The total resistance is the medium resistance R_m plus the cake resistance \alpha w, where w is the mass of cake solids per unit area.

At constant pressure the time to collect a filtrate volume V is t = \dfrac{\mu}{A\,\Delta P}\left(R_m V + \dfrac{\alpha c V^2}{2A}\right), where \mu is the viscosity, A the filter area and c the mass of solids per unit volume of filtrate.

The pressure drop across the cake is \Delta P_c = \mu\,\alpha\,w\,u, where u is the superficial velocity of the filtrate through the cake. This is Darcy's law written with the specific resistance. The pressure drop is therefore proportional to viscosity, to specific resistance, to the cake mass per unit area and to the filtration rate.

Parboiling of paddy (hydrothermal treatment)

Parboiling is a hydrothermal treatment in which paddy is soaked, steamed and dried before milling. The steaming gelatinises the starch and the grain becomes harder, so it resists breakage during milling. The hardened endosperm also needs a longer time to cook to the same softness as raw rice.

During soaking, water-soluble vitamins of the bran migrate into the endosperm, so parboiled rice retains more vitamin B than raw milled rice. The bran of parboiled paddy is also reported to contain a higher oil content than that of raw paddy.

To raise the moisture content of paddy from one dry-basis value to a higher one, the water required equals the dry matter multiplied by the change in dry-basis moisture, W = m_{d}\,(M_{2,db} - M_{1,db}). The dry matter is obtained from the given wet mass and the initial dry-basis moisture, m_d = m/(1 + M_{1,db}).

Rheological behavior of food materials

Food materials are classified by their flow curves, which show shear stress against shear rate. A Newtonian fluid has a constant viscosity, so shear stress is proportional to shear rate. A Bingham plastic does not flow until the shear stress exceeds a finite yield stress, after which its shear rate is proportional to the excess stress. A pseudoplastic (shear-thinning) fluid has a viscosity that decreases as the shear rate increases, while a dilatant (shear-thickening) fluid has a viscosity that increases with shear rate.