Size Reduction and Mixing Questions and Answers

Answer: B

The work index is obtained by multiplying Bond's constant by \(\sqrt{10}\).

Bond's law gives the energy for size reduction as \(E = K_{b}\left(1/\sqrt{D_{p}} - 1/\sqrt{D_{f}}\right)\), with sizes in mm. The work index \(W_{i}\) is defined as the energy needed to reduce unit mass from a very large size (\(D_{f} \to \infty\)) to 100 µm, that is \(D_{p} = 0.1\) mm:

\[ \begin{aligned} W_{i} &= K_{b}\,\frac{1}{\sqrt{0.1}} \\ &= K_{b}\sqrt{10} \end{aligned} \]

This is the familiar relation \(K_{b} = 0.3162\,W_{i}\). Multiplying by 10, or by the cube or fourth root of 10, does not follow from the square-root form of Bond's law.

Answer: D

For Rittinger’s law, \(n = 2\).

The general size-reduction relation \(dE/dX = -C/X^{n}\) gives the three classical laws for different values of \(n\):

• \(n = 2\): Rittinger’s law. Integrating gives \(E = C(1/X_{2} - 1/X_{1})\), energy proportional to the new surface created; it suits fine grinding.
• \(n = 1\): Kick’s law, \(E = C\ln(X_{1}/X_{2})\), suited to coarse crushing.
• \(n = 1.5\): Bond’s law, \(E \propto 1/\sqrt{X_{2}} - 1/\sqrt{X_{1}}\), for intermediate sizes.

So 3/2 is Bond’s exponent, and 1/2 and \(\sqrt{2}\) correspond to none of the classical laws.

Answer: A

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

• A roller mill nips the material between rolls turning at different speeds, so it grinds by compression and shear (1).
• A ball mill tumbles grinding balls in a rotating drum; falling balls strike the material and rolling balls rub it, giving impact and abrasion (2).
• A hammer mill breaks material mainly by the impact of swinging hammers, with shear against the screen and breaker plate (4).
• An attrition mill grinds between two plates, one or both rotating, which crush and rub the material: crushing and shear (3).

The second option makes the attrition mill an impact mill, and the third and fourth give the roller mill an impact mechanism, which it does not use.

Answer: A

The energy required is about 18.17 kW.h per ton.

Bond's law gives the energy as \(E = K\left(\frac{1}{\sqrt{x_{p}}} - \frac{1}{\sqrt{x_{f}}}\right)\), with \(x_{f}\) and \(x_{p}\) the 80% passing sizes of feed and product. Find \(K\) from the first case:

\[ \begin{aligned} 10 &= K\left(\frac{1}{\sqrt{0.5}} - \frac{1}{\sqrt{6}}\right) \\ &= K(1.4142 - 0.4082) \\ K &= 9.941 \end{aligned} \]

Then the finer product:

\[ \begin{aligned} E &= 9.941\left(\frac{1}{\sqrt{0.2}} - 0.4082\right) \\ &= 9.941 \times 1.8279 = 18.17 \end{aligned} \]

A finer product must need more energy than 10 kW.h, which rules out 9.09 and 5.50 at once. 11.00 understates the effect: under Bond's law energy grows with the inverse square root of product size, so going from 0.5 mm to 0.2 mm nearly doubles it.

Answer: D

Fat globules are broken up by turbulence and cavitation.

In a homogenizer, milk is forced at high pressure through a narrow valve gap. The intense turbulence and shear in the gap, together with cavitation, the collapse of vapour bubbles as the pressure drops, tear the fat globules from about 3-4 µm down to below about 1 µm, which stops cream from rising.

Pasteurization is a separate heat treatment for microbial safety and does not break up fat globules, which rules out every option that names it. High pressure is only the means of driving milk through the valve, not the disintegrating mechanism itself.

Answer: A

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

• A hammer mill breaks material by the impact of swinging hammers.
• A burr (plate) mill grinds by attrition between two rubbing plates.
• Crushing rolls squeeze material in the nip between two rolls: compression.
• A rotary knife cuts.

The other options assign compression or cutting to the hammer mill, which works by impact.

Answer: B

The true statement is Solid particles in the suspension are disintegrated due to the high shear rate exerted by the liquid.

A high-pressure homogenizer forces the liquid at pressures of tens to hundreds of MPa through a narrow valve gap. The fluid accelerates to very high velocity, and the intense shear, turbulence and cavitation in and just after the gap break fat globules and particles into much smaller ones. This is how milk is homogenized.

It is not mere mixing, it injects no steam, and it does not compress solids into a cake; that last describes filtration or pressing.

Answer: C

The one that is not a crusher is Ball crushers.

Crushers are coarse size-reduction machines that break large lumps by compression. The standard types are jaw crushers, gyratory crushers and crushing rolls (roll crushers).

The machine with balls in a rotating drum is the ball mill, which grinds material to fine powder by impact and attrition. It is classed as a grinder or mill, not a crusher, and 'ball crusher' is not a standard machine.