Module 1 · Tumbling mills

Ball mill: critical speed and the three regimes

A cross-section of a ball mill, simulated ball by ball in real time: every ball feels gravity, contact forces from its neighbours and friction from the turning shell. Change the speed and watch the charge go from cascading to cataracting to centrifuging.

Mill cross-section (rotating anticlockwise)

balls (sizes vary, as in a worn charge) tracer ball and its recent path theory: release point and parabola (outer layer) radius g/ω²: balls outside it cannot leave the wall

power draw (shape of Bond's speed factor φ(1 − 0.1/2^9−10φ)) grinding rate / product fineness (qualitative) slides 50–75% MSH 65–80%

Critical speed n_c

Operating speed n

Regime

Release angle α (outer ball)

Radius g/ω²

Power drawn by the charge (simulated)

Energy dissipated between balls

Speed band: · Simulating .

What you're seeing

The shell turns anticlockwise and lifts the balls up the right-hand side. A ball on the shell (centre at radius $R-r$) is pressed outwards by the centrifugal force $m\omega^2(R-r)$ and pulled towards the centre by the radial part of its weight, $mg\cos\alpha$. It leaves the wall where the two are equal, $\cos\alpha = \omega^2(R-r)/g = (n/n_c)^2$.

Cascading (low speed): balls are lifted only part way and roll down the sloping surface of the charge. Grinding is by attrition; impact is small.
Cataracting (about 60–100% of $n_c$): balls leave the shell high up and fly on parabolic paths (cyan dashed line) onto the toe of the charge, crushing particles by impact. This is the useful regime.
Centrifuging ($n \ge n_c$): $\cos\alpha \ge 1$, so the outer balls are never released. Every ball centred outside the radius $g/\omega^2$ (red dashed circle) rides round with the shell, so there is almost no relative motion and no grinding.

The chart below the mill shows the trend: power draw peaks a little below $n_c$ and collapses once the charge centrifuges; grinding is best inside the 50–75% (slides) or 65–80% (MSH) band. The live numbers are measured from the simulation: power is the shell torque times $\omega$, per kilogram of balls.

The simulation is two-dimensional (a slice of discs). Ball sizes vary ±20% around the slider value, the way a worn charge does. Very small balls are drawn enlarged so there are at most about 400 of them; the formulas always use the true radius.

Try this

Cascading. Run the mill at 40% of critical. The tracer ball climbs a short way and rolls back down the surface: no flight, so little impact.
Cataracting at 70%. The theory predicts release at $\alpha = \cos^{-1}0.49 = 60.7^\circ$ from the vertical. Watch the outer balls leave near the cyan line and land on the toe.
Centrifuging at 110%. With balls coloured by speed, the outer ring moves with the shell (n/n_c ≥ 1) while inner balls, inside the red circle, still fall. Go to 130% and the whole charge locks on.
Smooth liner. At 70% without lifter bars the charge slips against the shell, rises less and draws less power. This is why mills use lifters or high-friction rubber liners.
Bigger mill, slower speed. Double the radius to 1.2 m: $n_c \propto 1/\sqrt{R-r}$, so the critical speed falls from 39.9 to 27.8 rpm.

Derive the critical speed, step by step

Asked in BUET CHE 311 in 2012-13, 2016-17, 2018-19, 2019-20, 2022-23 and 2023-24. Use the forces on one ball at the shell.

Step 1 · Set-up

A ball of mass $m$ and radius $r$ is carried round by a mill of radius $R$ turning at $n$ revolutions per second, $\omega = 2\pi n$. Its centre moves on a circle of radius $R-r$. Let $\alpha$ be the angle between the radius to the ball and the vertical.

Step 2 · Radial forces

Two forces push the ball towards the centre: the radial component of its weight, $mg\cos\alpha$, and the push $N$ of the wall. Together they must supply the centripetal force for circular motion (in the rotating frame: balance the centrifugal force $m\omega^2(R-r)$):

$$N + mg\cos\alpha = m\omega^2(R-r) = 4\pi^2n^2m(R-r)$$

Step 3 · Release condition

The ball stays on the wall while $N \ge 0$. It leaves when $N = 0$, i.e. when the centrifugal force just equals the radial weight component (MSH Eq. 29.17):

$$\cos\alpha = \frac{4\pi^2n^2(R-r)}{g}$$

Step 4 · Critical speed

As $n$ rises, $\alpha$ falls. At the critical speed the ball is just held at the very top, $\alpha = 0$, $\cos\alpha = 1$:

$$\frac{4\pi^2n_c^2(R-r)}{g} = 1 \quad\Rightarrow\quad n_c = \frac{1}{2\pi}\sqrt{\frac{g}{R-r}}\qquad\text{(MSH Eq. 29.18)}$$

Step 5 · Handy forms

In rpm with diameters $D$ and $d$ in metres: $n_c = \dfrac{60}{2\pi}\sqrt{\dfrac{2g}{D-d}} = \dfrac{42.3}{\sqrt{D-d}}$ rpm. Dividing the two results gives the release angle at any speed: $\cos\alpha = (n/n_c)^2$.

Step 6 · Comment on the parameters

$n_c$ depends only on $R-r$: not on the mass or density of the balls, nor on the material. Larger mills must turn more slowly ($n_c \propto 1/\sqrt{R-r}$); larger balls raise $n_c$ slightly. The derivation assumes no slip at the wall (lifters or friction hold the ball). Mills run at 50–75% of $n_c$ (slides) or 65–80% (MSH, lower for viscous wet grinding); above $n_c$ the balls centrifuge and grinding stops.

Worked examples

Example 1Recommended speed for a 1200-mm mill with 75-mm ballsMcCabe–Smith–Harriott, 5th ed., Problem 29.6

What rotational speed, in revolutions per minute, would you recommend for a ball mill 1200 mm in diameter charged with 75-mm balls?

Data: $R = 0.600$ m, $r = 0.0375$ m, so $R - r = 0.5625$ m.
Critical speed:
$$n_c = \frac{1}{2\pi}\sqrt{\frac{9.81}{0.5625}} = 0.665\ \text{r/s} = 39.9\ \text{rpm}$$
Operating range: MSH 65–80% of $n_c$ → 25.9–31.9 rpm; slides 50–75% → 19.9–29.9 rpm.
Choose about 72% of $n_c$ = 28.7 rpm, inside both bands. At this speed the outer balls leave the shell at $\cos\alpha = 0.72^2 = 0.518$, $\alpha = 58.8^\circ$.

Answer$n_c = 39.9$ rpm; run at about 28–30 rpm (lower end for viscous wet grinding).

Example 2A 1.2-m mill run at 0.8 Hz is not grinding: why?Coulson & Richardson Vol. 2, Problem 2.5

A ball mill 1.2 m in diameter is run at 0.8 Hz and it is found that the mill is not working satisfactorily. Should any modification in the conditions of operation be suggested?

Critical speed (ball radius small, $R - r \approx R = 0.6$ m):
$$n_c = \frac{1}{2\pi}\sqrt{\frac{9.81}{0.6}} = 0.643\ \text{Hz}\ (38.6\ \text{rpm})$$
Compare: $n/n_c = 0.8/0.643 = 1.24$. The mill runs 24% above critical: the balls centrifuge, so there is no cascading or cataracting and almost no grinding.
Remedy: slow the mill to 50–75% of $n_c$: 0.32–0.48 Hz (19–29 rpm), say 0.45 Hz. Then check the ball charge (30–50% of the mill volume) and the feed level.
In the lab the balls are given a 10-mm radius, which raises $n_c$ to 0.649 Hz, so 0.8 Hz is 123% of critical. The charge locks onto the shell.

Answer0.8 Hz = 1.24 $n_c$: the mill is centrifuging. Reduce the speed to about 0.32–0.48 Hz.

Example 32.4-m mill with 80-mm balls: speeds, release angle, ball sizeExam-style (gvisualizer practice P10a), based on the slides and MSH Eqs. 29.17–29.18

A ball mill has an internal diameter of 2.4 m and is charged with 80-mm balls. (i) Find the critical speed and an operating speed of 70% of critical. (ii) At what angle from the vertical does a ball in contact with the shell leave the wall? (iii) The balls were chosen for a 25-mm feed; what size would you use for a 9-mm feed?

(i) $R - r = 1.2 - 0.04 = 1.16$ m:
$$n_c = \frac{1}{2\pi}\sqrt{\frac{9.81}{1.16}} = 0.463\ \text{r/s} = 27.8\ \text{rpm},\qquad n = 0.70\,n_c = 19.4\ \text{rpm}$$
(ii) $\cos\alpha = (n/n_c)^2 = 0.70^2 = 0.49$ → $\alpha = 60.7^\circ$ from the vertical on the rising side. The ball then falls on a parabola onto the toe.
(iii) The optimum ball diameter is roughly proportional to the square root of the feed size (slides): $d_b = 80\sqrt{9/25} = 48$ mm → use 50-mm balls.

Answer$n_c = 27.8$ rpm, run at 19.4 rpm; release at about 61° from the vertical; about 50-mm balls for a 9-mm feed.