Calculating Power of Pump to Remove Water from Basement

In summary, LawrenceC and I were both trying to solve the problem of pumping water using the power of the pump, and we got different results because we were not accounting for the efficiency of the pump.
  • #1
Hiche
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Homework Statement



Water is pumped steadily out of a flooded basement at a speed of 5.5 m/s through a uniform hose of radius 1.0 cm. The hose passes out through a window to a street ditch 2.8 m above the waterline. What is the power of the pump?

Homework Equations



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The Attempt at a Solution



We find the flow which is the area multiplied by the speed. Then we find the power which is the flow x the height (2.8 m) x gravitational acceleration x 1000 (fluid density). At least that's what my friend told me I should do. I had an answer of 47.4129 W but the answer is seemingly wrong. Can someone help me here, please?
 
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  • #2
I compute 47.4 watts. The problem is not accounting for the length of the hose so there is no hose friction to overcome. I don't see what is wrong with your answer.
 
  • #3
There is confusion and uncertainty if you just use equations that you get from someplace without understanding them.

Power is just energy/time

the energy needed to lift a mass m through height h is mgh - like you knew all along.
if it take time T to pump 1kg then the power is P = gh/T
so far so easy?

You are pumping a volume of 550 x pi ccs per second.
1cc of water is 1g, so you are pumping 0.55pi kg of water per second.

this is mass/time ... so we can finesse the previous formula:
P = gh(m/T) = 9.81x2.8x0.55pi = 47.461W

So I agree with LawrenceC - what makes you think this is the wrong answer?
 
  • #4
Thank you for the reply and for the explanations.

Well, our instructor posts our homework on wileyplus.com and it gave me an incorrect answer after plugging in the value.

I was checking through the problems on wileyplus and found this method:

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The values are different, by the way (use the values from the problems statement I posted). I plugged the values and it gave me something around ~55 (or slightly more). The answer was also wrong. I don't think I my calculations are wrong - I triple-checked my computations.

I'm new to fluids as we did not cover it in high school.
 
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  • #5
Ah - that method also takes account that the pump has to accelerate the water, so gives it some kinetic energy. Your numbers are fine - 55.78J I get.

I think you need to look through your class notes for the definition you are given for the power of a pump. LawrenceC and I (and you, at first) have been working out just the rate of work against gravity alone.

Is there anything left out of the problem you've given to us?
Efficiency of the pump perhaps?

If that is everything then you want to check with other students to see how they did it (did your friend get the right answer?) and, if this is common, confront your instructor: explain how you did it and ask what you missed out.
 
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FAQ: Calculating Power of Pump to Remove Water from Basement

1. How do I calculate the power of a pump needed to remove water from my basement?

To calculate the power of a pump, you will need to know the flow rate of the water and the total head or height that the water needs to be pumped. You can use the following formula to calculate power: Power (W) = (Flow rate (m3/s) x Total head (m) x Specific gravity) / Pump efficiency. The specific gravity is the density of the liquid being pumped and the pump efficiency can be found in the manufacturer's specifications.

2. What is the flow rate of water in my basement?

The flow rate of water in your basement will depend on the size and severity of the flooding. You can measure the flow rate by using a flow meter or by timing how long it takes for a bucket to fill up with water. If you do not have access to a flow meter, you can estimate the flow rate by measuring the dimensions of the basement and calculating the volume of water that has entered.

3. How do I determine the total head or height that the water needs to be pumped?

The total head or height is the vertical distance that the water needs to be pumped from the basement to the desired discharge point. This can be calculated by measuring the depth of the water in the basement and adding it to the height of the discharge point. Keep in mind that the pump will also need to overcome any friction losses in the pipes or hoses.

4. What is the specific gravity of water?

The specific gravity of water is 1.0. This value is used in the formula for calculating power because it represents the density of the liquid being pumped. If the water in your basement is contaminated with other substances, the specific gravity may be slightly different and should be adjusted accordingly in the formula.

5. How do I determine the efficiency of my pump?

The pump efficiency can be found in the manufacturer's specifications or by testing the pump. If you do not have access to this information, you can estimate the efficiency by using the following values: 50% for small centrifugal pumps, 65% for larger centrifugal pumps, and 80% for positive displacement pumps. Keep in mind that the efficiency may also decrease over time with wear and tear, so it is important to regularly maintain and service your pump.

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