Summative assessment for Module 6 1. An ore crusher is driven through a compound belt drive by an electric motor running at 1420 rpm. The pulley on the motor shaft is 200 mm in diameter. The driven pulley on the intermediate shaft is 400 mm in diameter and the driving pulley on the same shaft is 300 mm in diameter. The pulley on the ore crusher line shaft is 700 mm in diameter. If the crusher needs 300 kW of power and the overall drive efficiency is $$ 90 \% $$, calculate: 1.1 The motor power required. 1.2 The motor torque. 1.3 The speed of the crusher.

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### **Given Data:**
- **Motor Speed (N₁):** 1420 rpm
- **Motor Pulley Diameter (D₁):** 200 mm
- **Driven Pulley on Intermediate Shaft (D₂):** 400 mm
- **Driving Pulley on Intermediate Shaft (D₃):** 300 mm
- **Crusher Pulley Diameter (D₄):** 700 mm
- **Crusher Power Requirement (P₄):** 300 kW
- **Overall Drive Efficiency (η):** 90% (0.9)

### **1.1. Motor Power Required**

The motor must provide the necessary power to the crusher, accounting for the drive efficiency.

$$
\text{Motor Power} (P_{\text{motor}}) = \frac{\text{Crusher Power}}{\text{Efficiency}} = \frac{300\, \text{kW}}{0.9} = 333.33\, \text{kW}
$$

**Answer:** The motor requires approximately **333.33 kW** of power.

---

### **1.2. Motor Torque**

Torque can be calculated using the relationship between power and rotational speed. The formula connecting power ($$ P $$), torque ($$ T $$), and angular speed ($$ \omega $$) is:

$$
P = T \times \omega
$$

Where:
- $$ \omega $$ (in radians per second) = $$ \frac{2\pi N}{60} $$
- $$ N $$ is the rotational speed in rpm.

Rearranging to solve for torque:

$$
T = \frac{60 \times P}{2\pi N}
$$

**Calculations:**

1. **Convert Power to Watts:**
   $$
   P_{\text{motor}} = 333.33\, \text{kW} = 333,333\, \text{W}
   $$

2. **Calculate Torque:**
   $$
   T = \frac{60 \times 333,333}{2 \times \pi \times 1420} \approx \frac{20,000,000}{8933.0487} \approx 2237\, \text{Nm}
   $$

**Answer:** The motor torque is approximately **2,237 Newton-meters (Nm)**.

---

### **1.3. Speed of the Crusher**

To find the crusher's speed, we need to consider the belt drive ratios.

1. **First Belt Drive (Motor to Intermediate Shaft):**
   $$
   \text{Diameter Ratio} = \frac{D_2}{D_1} = \frac{400}{200} = 2
   $$
   $$
   N_2 = \frac{N_1}{\text{Diameter Ratio}} = \frac{1420}{2} = 710\, \text{rpm}
   $$

2. **Second Belt Drive (Intermediate Shaft to Crusher):**
   $$
   \text{Diameter Ratio} = \frac{D_4}{D_3} = \frac{700}{300} \approx 2.333
   $$
   $$
   N_4 = \frac{N_2}{\text{Diameter Ratio}} = \frac{710}{2.333} \approx 304\, \text{rpm}
   $$

**Answer:** The crusher operates at approximately **304 rpm**.

---

### **Summary of Answers:**
1. **Motor Power Required:** 333.33 kW
2. **Motor Torque:** 2,237 Nm
3. **Crusher Speed:** 304 rpm
- **Motor Power Required:** 333.33 kW - **Motor Torque:** 2,237 Nm - **Crusher Speed:** 304 rpm

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