An installation consisting of an electric motor driving a water pump left 75 L of water per second to a height of 4.7 m . If the motor consumes a power of 5 k W , then efficiency of the installation is

Options

A. `39%`
B. `69%`
C. `93%`
D. `96%`

Answer

To solve the problem, we need to calculate the efficiency of the installation consisting of an electric motor driving a water pump. The efficiency can be calculated using the formula:

\[ \text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power}} \times 100 \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Volume of water pumped per second, \( V = 75 \, \text{L/s} \)
- Height to which water is pumped, \( h = 4.7 \, \text{m} \)
- Power consumed by the motor, \( P_{\text{input}} = 5 \, \text{kW} = 5000 \, \text{W} \)

2. **Convert Volume to Mass:**
- Since 1 liter of water has a mass of approximately 1 kg, the mass flow rate (\( \dot{m} \)) can be calculated as:
\[ \dot{m} = 75 \, \text{kg/s} \]

3. **Calculate the Output Power:**
- The output power can be calculated using the formula for gravitational potential energy:
\[ P_{\text{output}} = \dot{m} \cdot g \cdot h \]
- Where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).
- Substituting the values:
\[ P_{\text{output}} = 75 \, \text{kg/s} \cdot 9.8 \, \text{m/s}^2 \cdot 4.7 \, \text{m} \]
\[ P_{\text{output}} = 75 \cdot 9.8 \cdot 4.7 = 3467.25 \, \text{W} \approx 3.467 \, \text{kW} \]

4. **Calculate the Efficiency:**
- Now, we can calculate the efficiency using the output and input power:
\[ \text{Efficiency} = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100 \]
\[ \text{Efficiency} = \frac{3467.25 \, \text{W}}{5000 \, \text{W}} \times 100 \approx 69.35\% \]

5. **Final Answer:**
- The efficiency of the installation is approximately \( 69.35\% \).