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 of calculating the minimum horsepower the engine should have to lift water from a level 10 m below the ground at a rate of 30 kg/min, we can follow these steps:

### Step 1: Convert the mass flow rate from kg/min to kg/s
The mass flow rate is given as 30 kg/min. To convert this to kg/s, we divide by 60 (since there are 60 seconds in a minute).

\[ \frac{30 \text{ kg}}{1 \text{ min}} = \frac{30 \text{ kg}}{60 \text{ s}} = 0.5 \text{ kg/s} \]

### Step 2: Calculate the gravitational potential energy change per second
The potential energy (PE) gained by lifting the water can be calculated using the formula:

\[ PE = mgh \]

Where:
- \( m \) is the mass flow rate (0.5 kg/s),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)),
- \( h \) is the height (10 m).

Substituting the values:

\[ PE = 0.5 \text{ kg/s} \times 9.81 \text{ m/s}^2 \times 10 \text{ m} \]

Calculating this gives:

\[ PE = 0.5 \times 9.81 \times 10 = 49.05 \text{ watts} \]

### Step 3: Convert the power from watts to horsepower
1 horsepower is equivalent to 746 watts. To convert the power calculated in watts to horsepower, we use the following formula:

\[ \text{Horsepower} = \frac{\text{Power in watts}}{746} \]

Substituting the power we calculated:

\[ \text{Horsepower} = \frac{49.05 \text{ watts}}{746} \approx 0.0657 \text{ horsepower} \]

### Step 4: Final Result
Thus, the minimum horsepower the engine should have is approximately:

\[ \text{Horsepower} \approx 0.0657 \text{ hp} \quad \text{or} \quad 6.57 \times 10^{-2} \text{ hp} \]

### Summary
The minimum horsepower required for the engine to lift water at the specified rate is approximately \( 6.57 \times 10^{-2} \) hp.

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