Ten litre of water per second is lifted from well through 20m and delivered with a velocity of 10 m/s, then the power of the motor is :

Options

A. 1.5 kW
B. 2.5 kW
C. 3.5 kW
D. 4.5 kW

Answer

To solve the problem step by step, we need to calculate the power of the motor required to lift 10 liters of water per second from a well through a height of 20 meters and deliver it with a velocity of 10 m/s.

### Step 1: Convert the volume of water to mass
Given that the flow rate of water is 10 liters per second, we need to convert this volume into mass. The density of water is approximately 1000 kg/m³.

\[ \text{Mass flow rate} (m) = \text{Volume flow rate} \times \text{Density} \]
\[ m = 10 \, \text{liters/second} \times 1000 \, \text{kg/m}^3 = 10 \, \text{kg/s} \]

### Step 2: Calculate the potential energy (PE)
The potential energy gained by the water when lifted to a height \( h \) is given by the formula:

\[ PE = mgh \]
Where:
- \( m = 10 \, \text{kg/s} \)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 20 \, \text{m} \)

Substituting the values:

\[ PE = 10 \times 10 \times 20 = 2000 \, \text{J/s} = 2000 \, \text{W} \]

### Step 3: Calculate the kinetic energy (KE)
The kinetic energy of the water delivered at a velocity \( v \) is given by the formula:

\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m = 10 \, \text{kg/s} \)
- \( v = 10 \, \text{m/s} \)

Substituting the values:

\[ KE = \frac{1}{2} \times 10 \times (10)^2 = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J/s} = 500 \, \text{W} \]

### Step 4: Calculate the total energy per second
The total energy per second (total power) is the sum of the potential energy and kinetic energy:

\[ \text{Total Power} = PE + KE = 2000 \, \text{W} + 500 \, \text{W} = 2500 \, \text{W} \]

### Step 5: Convert power to kilowatts
To express the power in kilowatts:

\[ \text{Power in kilowatts} = \frac{2500 \, \text{W}}{1000} = 2.5 \, \text{kW} \]

### Final Answer
The power of the motor is **2.5 kW**.

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