A pump is used to raise water through a height of 100m at a rate of 3.0kg/s .What is the power of the pump if the velocity of the water entering the pump can be neglected , but water is leaving the pump at 3.0m/s ?

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- Jul 7th 2009, 06:58 AMthereddevilsA physics question
A pump is used to raise water through a height of 100m at a rate of 3.0kg/s .What is the power of the pump if the velocity of the water entering the pump can be neglected , but water is leaving the pump at 3.0m/s ?

- Jul 7th 2009, 05:35 PMReferos
You know that work is the variation of mechanical energy. Consider the amount of water that flows in one second, i.e., 3kg. What's the initial mechanical energy of this amount of water? And its final mechanical energy when it reaches the top (remember to consider both gravitational potential energy and kinetic energy)? This should allow you to find the work done by the pump in one second, which is the pump's power (work over time is the definition of power).

- Jul 7th 2009, 06:29 PMphy
Variation of mechanical energy? Huh? Where did you get that idea from? Work is defined as force times distance. Or to be exact

$\displaystyle W = \int $**F***d**r**

where * is the dot product (not all that up on latex ... yet). The power being expended by the pump is the rate at which work is being done. - Jul 8th 2009, 04:05 AMskeeter
no need for integration here since the force doing work is constant ... and Referos is correct, it just might be more correct to state that the work done on the system results in a

**change**in mechanical energy of the fluid rather than a "variation" , but I'm pretty sure that is what was meant.

$\displaystyle \Delta E = mgh + \frac{1}{2}mv^2$

$\displaystyle t = \frac{h}{v}$

$\displaystyle P = \frac{\Delta E}{t}$

$\displaystyle P \approx 90 \, W$ - Jul 8th 2009, 04:14 AMthereddevils
Thanks Skeeter , but the answer says :

Power $\displaystyle =\frac{d}{dt}(mgh+\frac{1}{2}mv^2)$

$\displaystyle =\frac{dm}{dt}gh+\frac{1}{2}(\frac{dm}{dt})^2$

$\displaystyle

=2.96\times10^3W

$

I need an explaination on that . Thanks a lot . - Jul 8th 2009, 04:59 AMskeeter
- Jul 9th 2009, 06:00 AMthereddevils
- Jul 9th 2009, 03:34 PMPterid
Hi thereddevils,

The mass rate at which water is drawn into and out of the system is written as $\displaystyle \frac{dm}{dt}$. So, in a small time period $\displaystyle dt$, a small amount of water $\displaystyle dm$ is drawn into the bottom of the tower, with negligible speed.

At the same time, another (different) water element, also of mass $\displaystyle dm$, leaves the system. The element leaving has added energy:

$\displaystyle d\epsilon = dm ~ gh + \frac{1}{2}dm~v^2$

which was imparted to it during its journey through the pump. The outgoing speed is $\displaystyle v$ = 3 m / s.

Finally, the power of the pump is (energy imparted) / (time taken to impart it)

= $\displaystyle \frac{d\epsilon}{dt}$

HTH.

pterid