Let's solve the problem **step-by-step**: --- ## **Given:** - Initial Sauter diameter, \( d_{1} = 500\, \mu m \) - First final diameter, \( d_{2} = 100\, \mu m \) - Energy consumption for this change, \( E_{1} = 0.5 \) kWh/ton We are asked: - What is the energy consumption if crystals are ground to \( d_{3} = 50\, \mu m \)? - **(a) According to Rittinger's Law** - **(b) According to Kick's Law** - **(c) Comment on energy consumption values** --- # **Step 1: Rittinger's Law** Rittinger's Law states that **energy required for size reduction is proportional to the new surface area created**: \[ E \propto \left( \frac{1}{d_{final}} - \frac{1}{d_{initial}} \right) \] So, for two grinding steps, the ratio of energies is: \[ \frac{E_2}{E_1} = \frac{\left( \frac{1}{d_3} - \frac{1}{d_1} \right)}{\left( \frac{1}{d_2} - \frac{1}{d_1} \right)} \] Where: - \( E_1 \) = energy for reducing from \( d_1 \) to \( d_2 \) - \( E_2 \) = energy for reducing from \( d_1 \) to \( d_3 \) --- ### **Plug in the values:** - \( d_1 = 500 \mu m \) - \( d_2 = 100 \mu m \) - \( d_3 = 50 \mu m \) - \( E_1 = 0.5 \) kWh/ton Calculate numerator and denominator: \[ \frac{1}{d_3} - \frac{1}{d_1} = \frac{1}{50} - \frac{1}{500} = \frac{10 - 1}{500} = \frac{9}{500} = 0.018 \] \[ \frac{1}{d_2} - \frac{1}{d_1} = \frac{1}{100} - \frac{1}{500} = \frac{5 - 1}{500} = \frac{4}{500} = 0.008 \] So, \[ \frac{E_2}{E_1} = \frac{0.018}{0.008} = 2.25 \] \[ E_2 = 2.25 \times E_1 = 2.25 \times 0.5 = 1.125 \text{ kWh/ton} \] --- ### **Final answer (a):** \[ \boxed{1.125\ \text{kWh/ton}} \] is the energy required to grind from 500 µm to 50 µm **according to Rittinger's Law**. --- # **Step 2: Kick's Law** Kick's Law states that **energy required is proportional to the size reduction ratio**: \[ E \propto \ln \left( \frac{d_{initial}}{d_{final}} \right) \] So, \[ \frac{E_2}{E_1} = \frac{\ln \left( \frac{d_1}{d_3} \right)}{\ln \left( \frac{d_1}{d_2} \right)} \] Plug in the numbers: \[ \ln \left( \frac{d_1}{d_3} \right) = \ln \left( \frac{500}{50} \right) = \ln(10) \approx 2.3026 \] \[ \ln \left( \frac{d_1}{d_2} \right) = \ln \left( \frac{500}{100} \right) = \ln(5) \approx 1.6094 \] So, \[ \frac{E_2}{E_1} = \frac{2.3026}{1.6094} = 1.431 \] \[ E_2 = 1.431 \times 0.5 = 0.7155 \text{ kWh/ton} \] --- ### **Final answer (b):** \[ \boxed{0.72\ \text{kWh/ton}} \] (Rounded to two decimal places) is the energy required to grind from 500 µm to 50 µm **according to Kick's Law**. --- # **Step 3: Comment on energy consumption values (c)** **Rittinger's Law** is most applicable for **fine grinding** (creating large surface area, small particle sizes), while **Kick's Law** is more appropriate for coarse crushing (large particles, small surface area created). - **Rittinger's Law predicts higher energy requirement** for finer grinding because as particle size decreases, the increase in surface area becomes very large, demanding more energy. - **Kick's Law underestimates** energy for fine grinding since it doesn't account for the increasing surface area at small sizes. For this case, **grinding from 500 µm to 50 µm** is already in the fine grinding range, so Rittinger's Law is more appropriate. The calculated values: - Rittinger: **1.13 kWh/ton** - Kick: **0.72 kWh/ton** Thus, **actual energy required will likely be closer to Rittinger's value**. --- ## **Final Summary Table** | Law | Energy for 500 µm → 50 µm (kWh/ton) | |--------------|-----------------------------| | Rittinger | **1.13** | | Kick | **0.72** | --- ### **Short Answers** **a. According to Rittinger's Law:**  **1.13 kWh/ton** **b. According to Kick's Law:**  **0.72 kWh/ton** **c. Comment:**  Rittinger's Law predicts higher energy, as it considers surface area increase, and is more appropriate for fine grinding. Kick's Law underestimates energy for fine grinding situations. For reducing sugar crystals to fine powder, Rittinger's value is more realistic.