Module 2 · Energy for size reduction
Energy laws: Rittinger, Kick and Bond
Find the power for a crushing or grinding duty from the Bond work index, or take one test result and see how far the three laws disagree when you predict a new duty. The log–log chart shows where each law is trusted.
What you're seeing
All three laws come from one differential equation for the energy \(dE\) needed to make a small change \(dL\) in the size of unit mass:
$$\frac{dE}{dL} = -\frac{C}{L^{p}}$$- \(p = 2\), Rittinger: \(E = K_R f_c\left(\dfrac{1}{L_2} - \dfrac{1}{L_1}\right)\). Energy is proportional to the new surface (surface per unit mass \(\propto 1/L\)). Holds for fine grinding, 10–1000 μm.
- \(p = 1\), Kick: \(E = K_K f_c \ln\dfrac{L_1}{L_2}\). Energy depends only on the reduction ratio; it describes elastic deformation before fracture and fits coarse crushing above about 1000 μm.
- \(p = 1.5\), Bond: \(E = 2C\left(\dfrac{1}{\sqrt{L_2}} - \dfrac{1}{\sqrt{L_1}}\right)\). Intermediate between the two; used for intermediate crushing and grinding with a measured work index.
On log–log axes the curves start at zero at the feed size and rise as the product gets finer. Fitted through the same point, Kick (slope → 0) rises most slowly and Rittinger (slope → −1) fastest; the dashed composite curve shows the Hukki idea that the local exponent changes from 1 (coarse) to 2 (fine).
Bond's law in practice
MSH form, with \(D\) in mm, \(P\) in kW and \(\dot m\) in ton/h (Eq. 29.10):
$$\frac{P}{\dot m} = 0.3162\,W_i\left(\frac{1}{\sqrt{D_{pb}}} - \frac{1}{\sqrt{D_{pa}}}\right)$$where \(D_{pa}\), \(D_{pb}\) are the sizes that 80% of the feed and of the product pass, and \(0.3162 = \sqrt{0.1}\) because \(W_i\) is defined as the energy from infinite size to 80% passing 100 μm = 0.1 mm. The Coulson & Richardson form, with \(L\) in μm and \(q = L_1/L_2\), is the same equation:
$$E = E_i\sqrt{\frac{100}{L_2}}\left(1 - \frac{1}{\sqrt q}\right)$$Work-index values are for dry crushing or wet grinding; multiply by 4/3 for dry grinding. MSH tons are short tons (2000 lb): for metric tonnes multiply the energy per ton by 1.102.
Try this
Test Kick's claim. The slides say 50 → 25 mm needs the same energy as 40 → 20 mm. Calibrate on 50 → 25 mm and predict 40 → 20 mm: Kick gives the same 1.00, Bond 1.12 times, Rittinger 1.25 times.
Fine grinding. Cement clinker from 8 mm to 75 μm at 30 ton/h. The duty falls in the Rittinger band; note how steeply the energy climbs below 100 μm.
Dry versus wet. Switch the same duty to dry grinding: the work index rises by 4/3, so the power rises by a third (420.8 → 561.0 kW).
Find p from two tests. A feed of 40 mm needs 4.0 kJ/kg to reach 10 mm and 7.3 kJ/kg to reach 5 mm. The fitted exponent comes out at p ≈ 1.50: the material follows Bond's law.
Work-index table (MSH Table 29.1 / slide Table 27-1)
| Material | Sp. gr. | Wi, kWh/ton | dry grinding ×4/3 |
|---|
Worked examples
Example 1Bond power to crush 100 ton/h of limestoneMcCabe–Smith–Harriott, 5th ed., Example 29.1
What is the power required to crush 100 ton/h of limestone if 80 percent of the feed passes a 2-in. screen and 80 percent of the product a 1/8-in. screen? Work index of limestone 12.74.
Data: \(D_{pa} = 2\) in. = 50.8 mm, \(D_{pb} = 1/8\) in. = 3.175 mm, \(W_i = 12.74\) kWh/ton, \(\dot m = 100\) ton/h. Crushing, so the table value is used as it is.
Size term: \(\dfrac{1}{\sqrt{3.175}} - \dfrac{1}{\sqrt{50.8}} = 0.5612 - 0.1403 = 0.4209\ \text{mm}^{-1/2}\)
Power:
$$P = 100 \times 0.3162 \times 12.74 \times 0.4209 = 169.6\ \text{kW}$$Check with the C&R form: \(12.74\sqrt{100/3175}\,(1 - 1/\sqrt{16}) = 1.696\) kWh/ton, the same.
Example 2Blake jaw crusher: 50 → 10 mm known, predict 75 → 25 mmBUET CHE 311, 2023-24 · Coulson & Richardson Vol. 2, Example 2.1
A material is crushed in a Blake jaw crusher from an average size of 50 mm to 10 mm with an energy consumption of 13.0 kW/(kg/s). Find the energy to crush the same material from 75 mm to 25 mm (i) by Rittinger's law, (ii) by Kick's law. Which result is more reliable, and why?
Rittinger constant: \(K_Rf_c = \dfrac{13.0}{1/10 - 1/50} = \dfrac{13.0}{0.08} = 162.5\ \text{kW·mm/(kg/s)}\)
Rittinger, 75 → 25 mm: \(E = 162.5\left(\dfrac{1}{25} - \dfrac{1}{75}\right) = 4.33\ \text{kW/(kg/s)}\)
Kick: the constant cancels in the ratio:
$$E = 13.0\,\frac{\ln(75/25)}{\ln(50/10)} = 13.0 \times \frac{1.0986}{1.6094} = 8.87\ \text{kW/(kg/s)}$$Which is reliable: a jaw crusher does coarse crushing (sizes above 1 mm), where Kick's law applies; Rittinger's law is for fine grinding and badly underestimates here. (Bond, for comparison: 6.29 kW/(kg/s).)
Example 3Crusher power for a new capacity and a finer productBUET CHE 311, 2016-17 (Rittinger's law with volume–surface mean sizes)
A crusher takes rock of volume–surface mean diameter 19 mm and discharges a product of 5 mm. The power to crush 3 kg/s is 7 kW. What power is needed if the capacity is reduced to 2.5 kg/s and the product size to 3.5 mm? Mechanical efficiency constant.
Choose the law: volume–surface mean sizes measure the specific surface, so use Rittinger, \(P/\dot m = K_R(1/\bar D_{sb} - 1/\bar D_{sa})\).
Constant: \(K_R = \dfrac{7/3}{1/5 - 1/19} = \dfrac{2.333}{0.14737} = 15.83\ \text{kW·s·mm/kg}\)
New duty:
$$P = 2.5 \times 15.83\left(\frac{1}{3.5} - \frac{1}{19}\right) = 2.5 \times 15.83 \times 0.23308 = 9.23\ \text{kW}$$Compare: Kick would give 7.39 kW and Bond 8.17 kW; in the lab, set the test energy as 7/3 = 2.333 kJ/kg.