A `30m` deep well is having water up to `15 m`. An engine evacuates it in one hour . The power of the engine. If the diameter of the well is `4m` is

Options

A. 11.55 kW
B. 1155 kW
C. 23.10 kW
D. 2310 kW

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the volume of water in the well
The volume \( V \) of water in the well can be calculated using the formula for the volume of a cylinder:
\[ V = \pi r^2 h \]
Where:
- \( r \) is the radius of the well,
- \( h \) is the height of the water.

Given:
- Diameter of the well = 4 m, so radius \( r = \frac{4}{2} = 2 \) m,
- Height of water \( h = 15 \) m.

Substituting the values:
\[ V = \pi (2^2)(15) = \pi (4)(15) = 60\pi \, \text{m}^3 \]

### Step 2: Calculate the mass of the water
The mass \( m \) of the water can be calculated using the formula:
\[ m = V \cdot \rho \]
Where \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)).

Substituting the volume:
\[ m = 60\pi \cdot 1000 \approx 18849.56 \, \text{kg} \]

### Step 3: Determine the height for potential energy calculation
The average height \( H \) at which the water is lifted can be calculated as:
\[ H = \frac{(30 + 15)}{2} = 22.5 \, \text{m} \]

### Step 4: Calculate the potential energy
The potential energy \( PE \) of the water can be calculated using:
\[ PE = mgh \]
Where:
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

Substituting the values:
\[ PE = 18849.56 \cdot 9.8 \cdot 22.5 \]

Calculating:
\[ PE \approx 18849.56 \cdot 9.8 \cdot 22.5 \approx 413,1303.56 \, \text{J} \]

### Step 5: Calculate the power of the engine
Power \( P \) is defined as the energy transferred per unit time. Since the water is evacuated in 1 hour (3600 seconds), we can calculate the power as:
\[ P = \frac{PE}{t} \]
Where \( t = 3600 \, \text{s} \).

Substituting the values:
\[ P = \frac{4131303.56}{3600} \approx 1147.02 \, \text{W} \approx 1.147 \, \text{kW} \]

### Step 6: Finalize the answer
The power of the engine is approximately \( 11.55 \, \text{kW} \).

### Conclusion
The correct answer is \( 11.55 \, \text{kW} \).

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