Module 3 · Crushers
Jaw crusher and crushing rolls
Both machines break rock by compression, and both work only if the rock is gripped ("nipped") by friction. Make the nip angle too large or the rock too slippery and the lumps ride up instead of breaking.
What you're seeing
Jaw crusher. The fixed jaw (anvil) is vertical; the swing jaw is pivoted at the top (Blake type), so its biggest movement, the throw, is at the discharge. The eccentric lifts the pitman, which straightens the toggle plates and pushes the bottom of the swing jaw towards the anvil. A lump settles where the gap equals its size and is squeezed on every closing stroke; broken pieces drop lower, are squeezed again, and leave through the opening between the closed-side setting (CSS) and the open-side setting (OSS).
Why lumps slip. Each jaw presses on the lump with a normal force \(N\) and can hold it with friction up to \(\mu N\). The two normals are inclined at the nip angle \(\theta\), so they squeeze the lump upwards; friction (down the faces) must resist that. Each contact can tilt its resultant by the friction angle \(\phi = \tan^{-1}\mu\), so the lump is held only if
$$\theta \le 2\phi \quad\Longleftrightarrow\quad \tan\frac{\theta}{2} \le \mu$$With \(\mu\) = 0.3 the limit is 33.4°; wet, slimy limestone (\(\mu\) ≈ 0.2) allows only 22.6°, which is why jaws are set at 20–30° and have corrugated faces.
Capacity. Taggart's empirical formula \(T = 0.6\,L\,S\) gives short tons per hour with \(L\) (width of the feed opening) and \(S\) (open-side setting) in inches; the lab converts from mm and to metric t/h. It is an estimate: capacity also depends on speed, throw, feed size and how the feed is packed.
Crushing rolls. Two rolls turn towards each other. A particle of radius \(r\) touches both rolls when the lines from the roll centres to its centre make the angle of nip \(\alpha\) with the line of centres:
$$\cos\alpha = \frac{R + d}{R + r}$$where \(R\) is the roll radius and \(2d\) the gap. It is drawn in only if \(\tan\alpha \le \mu\) (the tangents at the two contacts meet at \(2\alpha\)). The theoretical throughput is a ribbon as wide as the face \(W\) and as thick as the gap, moving at the surface speed: \(Q = \pi D n\,W\,(2d)\,\rho\); real rolls give a fraction of this. MSH's rule of thumb \(D_{p,\max} = 0.04R + d\) is a cautious limit for most feeds.
Try this
Open the jaws too wide. Set θ = 35° with μ = 0.3: tan 17.5° = 0.315 > 0.3, so lumps ride up the jaws (red) and nothing is crushed.
Wet limestone (BUET 2023-24). At θ = 26° dry rock (μ = 0.3) is nipped; drop μ to 0.2 and it slips, because tan 13° = 0.231 > 0.2.
Feed too big for the rolls. A 70-mm lump on 1-m rolls with a 12.5-mm gap needs α = 18.9°, more than tan⁻¹0.3 = 16.7°, so it chatters on top. Smaller pieces in the same feed still go through.
Open the gap. Doubling the gap to 25 mm lets the 70-mm lump in, but the product is twice as coarse: rolls trade feed size against product size.
Small rolls. With 0.4-m rolls the largest feed falls to about 31 mm: large lumps need large rolls.
Crusher types and the forces they use
| Machine | Force | Typical duty (slides, MSH) |
|---|---|---|
| Blake jaw (pivot at top) | compression | primary; feed up to 1.8 m, large fixed gape |
| Dodge jaw (pivot at bottom) | compression | fixed discharge, uniform product, small capacity |
| Gyratory | compression | primary; continuous discharge, up to 4500 t/h |
| Smooth rolls | compression | secondary; feed 12–75 mm, product 1–12 mm |
| Toothed rolls | compression, impact, shear | softer feeds: coal, bone, soft shale |
Worked examples
Example 1Crushing rolls: largest feed and throughputCoulson & Richardson Vol. 2, Example 2.3
Crushing rolls 1 m in diameter are set so that the crushing surfaces are 12.5 mm apart, and the angle of nip is 31°. What is the largest particle that should be fed? If the actual capacity is 12% of the theoretical, find the throughput in kg/s at 2.0 Hz with a working face 0.4 m long and a feed density of 2500 kg/m³.
Data: \(R = 0.5\) m, \(d = 6.25\) mm. The 31° is the full angle between the tangents, so each roll's angle of nip is \(\alpha = 15.5^\circ\) (equivalent to \(\mu = \tan 15.5^\circ = 0.277\)).
Nip geometry:
$$R + r = \frac{R + d}{\cos\alpha} = \frac{0.50625}{\cos 15.5^\circ} = 0.52536\ \text{m} \;\Rightarrow\; r = 0.0254\ \text{m}$$Largest particle \(2r = 50.7\) mm (reduction ratio about 4).
Theoretical throughput: \(u = \pi D n = \pi(1.0)(2.0) = 6.28\) m/s;
$$Q = u\,W\,(2d)\,\rho = 6.28 \times 0.4 \times 0.0125 \times 2500 = 78.5\ \text{kg/s}$$Actual: \(0.12 \times 78.5 = 9.4\) kg/s (34 t/h).
Example 2Wet, slippery limestone in a jaw crusherBUET CHE 311, 2023-24
While crushing wet limestone with a high moisture content and a naturally slippery texture, the jaw crusher was not operating effectively. Explain the phenomenon and how it affects the crushing mechanism. (Take a nip angle of 26°, and μ = 0.3 for dry and 0.2 for wet rock.)
Condition for nipping: \(\tan(\theta/2) \le \mu\). Here \(\tan 13^\circ = 0.231\).
Dry rock: \(0.231 \le 0.3\): the lump is gripped and crushed.
Wet rock: \(0.231 > 0.2\): friction cannot resist the upward squeeze of the inclined jaws, so on each closing stroke the lump is pushed up (slippage) instead of breaking. Energy is wasted in rubbing and jaw wear, capacity falls; wet fines also pack the discharge.
Remedies: nip angle below \(2\tan^{-1}0.2 = 22.6^\circ\), corrugated jaw plates, scalp out wet fines, or use a machine that does not rely on nip (impact crusher, toothed rolls).
Example 3Jaw crusher capacity from the opening and settingExam-style, Taggart's formula
A Blake jaw crusher has a feed opening 900 mm × 1200 mm (gape × width), a closed-side setting of 125 mm, a throw of 25 mm and a nip angle of 24°. Estimate its capacity and check that dry rock (μ = 0.3) is nipped.
Open-side setting: OSS = CSS + throw = 125 + 25 = 150 mm = 5.906 in.; width \(L\) = 1200 mm = 47.24 in.
Taggart: \(T = 0.6\,L\,S = 0.6 \times 47.24 \times 5.906 = 167.4\) short ton/h \(= 151.9\) t/h.
Nip check: \(\tan 12^\circ = 0.213 \le 0.3\): nipped. Largest nip angle \(2\tan^{-1}0.3 = 33.4^\circ\).
Reduction ratio: largest feed ≈ 0.85 × 900 = 765 mm, product ≤ 150 mm, about 5 : 1.