A water pump lifts water from a level 10 m below the ground. Water is pumped at a rate of 30 kg /minute with negligible velocity.Caculate the minimum horsepower the engine should have to do this.

Explanation

To solve the problem, we need to calculate the minimum horsepower required for a water pump that lifts water from a depth of 10 meters below the ground at a rate of 30 kg/minute.

### Step-by-Step Solution:

1. **Convert the mass flow rate to kg/s**:
- The water is pumped at a rate of 30 kg/minute.
- To convert this to kg/s, we divide by 60 (since there are 60 seconds in a minute):
\[ \text{Mass flow rate} = \frac{30 \text{ kg}}{60 \text{ seconds}} = 0.5 \text{ kg/s} \]

2. **Identify the gravitational potential energy formula**:
- The gravitational potential energy (GPE) gained by the water when lifted to a height \( h \) is given by:
\[ \text{GPE} = mgh \]
- Where:
- \( m \) = mass (in kg)
- \( g \) = acceleration due to gravity (approximately \( 10 \text{ m/s}^2 \))
- \( h \) = height (in meters)

3. **Calculate the power required**:
- The power (P) required to lift the water is the energy per unit time. Since we are lifting water at a rate of 0.5 kg/s, the power can be calculated as follows:
\[ P = \text{mass flow rate} \times g \times h \]
- Substituting the values:
\[ P = 0.5 \text{ kg/s} \times 10 \text{ m/s}^2 \times 10 \text{ m} = 50 \text{ watts} \]

4. **Convert power from watts to horsepower**:
- We know that 1 horsepower (hp) is equivalent to 746 watts. To convert watts to horsepower, we use the formula:
\[ \text{Horsepower} = \frac{\text{Power (in watts)}}{746} \]
- Substituting the power we calculated:
\[ \text{Horsepower} = \frac{50 \text{ watts}}{746} \approx 0.067 \text{ hp} \]
- This can also be expressed in scientific notation:
\[ \text{Horsepower} \approx 6.7 \times 10^{-2} \text{ hp} \]

### Final Answer:
The minimum horsepower the engine should have to lift the water is approximately **0.067 hp** or **6.7 x 10^-2 hp**.