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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.
The correct Answer is:
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:
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). 30 kg 1 min=30 kg 60 s=0.5 kg/s
The potential energy (PE) gained by lifting the water can be calculated using the formula: P E=m g h Where: - m is the mass flow rate (0.5 kg/s), - g is the acceleration due to gravity (approximately 9.81 m/s 2), - h is the height (10 m). Substituting the values: P E=0.5 kg/s×9.81 m/s 2×10 m Calculating this gives: P E=0.5×9.81×10=49.05 watts
1 horsepower is equivalent to 746 watts. To convert the power calculated in watts to horsepower, we use the following formula: Horsepower=Power in watts 746 Substituting the power we calculated: Horsepower=49.05 watts 746≈0.0657 horsepower
Thus, the minimum horsepower the engine should have is approximately: Horsepower≈0.0657 hp or 6.57×10−2 hp
The minimum horsepower required for the engine to lift water at the specified rate is approximately 6.57×10−2 hp.
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