Answer :
Final answer:
The power required to pump 50 m3 of water in 15 minutes against a head of 5 m, with a pump efficiency of 60%, is approximately 2.72 kW.
Explanation:
To calculate the power required to pump water, we can use the formula Power = Flow x Pressure. In this case, the pressure can be calculated as pgH, where p is the density of water, g is the acceleration due to gravity, and H is the head in meters.
Given that the pump efficiency is 60%, we need to take this into account by dividing the power by 0.6.
First, let's convert the given flow rate from m3/minute to m3/hour. Since there are 60 minutes in an hour, the flow rate is 50 m3 / (15/60) hours = 200 m3/h.
Next, we can calculate the pressure using the formula Pressure = pgH. The density of water, p, is approximately 1000 kg/m3, and the head, H, is given as 5 m. Therefore, the pressure is (1000 kg/m3) x (9.8 m/s2) x (5 m) = 49,000 N/m2.
Now, we can calculate the power using the formula Power = Flow x Pressure. The flow rate is 200 m3/h and the pressure is 49,000 N/m2. Therefore, the power required is (200 m3/h) x (49,000 N/m2) = 9,800,000 Nm/h = 9,800,000 J/h.
Finally, we need to convert the power from joules per hour to kilowatts. Since 1 watt is equal to 1 joule per second, we can divide the power by 3,600,000 (the number of seconds in an hour) to get the power in kilowatts. Therefore, the power required to pump the water is 9,800,000 J/h / 3,600,000 = 2.72 kW.
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