Results 1 to 8 of 8

Math Help - A physics question

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    A 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 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jan 2009
    Posts
    32
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    phy
    phy is offline
    Newbie
    Joined
    Jul 2009
    Posts
    5
    Quote Originally Posted by Referos View Post
    You know that work is the variation of mechanical energy.
    Variation of mechanical energy? Huh? Where did you get that idea from? Work is defined as force times distance. Or to be exact

    W = \int F*dr

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    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.

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

    t = \frac{h}{v}

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

    P \approx 90 \, W
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2009
    Posts
    381
    Thanks Skeeter , but the answer says :

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

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

     <br />
=2.96\times10^3W<br />

    I need an explaination on that . Thanks a lot .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by thereddevils View Post
    Thanks Skeeter , but the answer says :

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

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

     <br />
=2.96\times10^3W<br />

    I need an explaination on that . Thanks a lot .
    no idea ... I can't see how \frac{dm}{dt} is involved.

    where did this problem come from? any additional information you may have left out ... like pipe dimensions?
    Last edited by skeeter; July 8th 2009 at 06:55 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jan 2009
    Posts
    381
    Quote Originally Posted by skeeter View Post
    no idea ... I can't see how \frac{dm}{dt} is involved.

    where did this problem come from? any additional information you may have left out ... like pipe dimensions?

    It's from my Physics work book . No , i copied the question as it is .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jun 2007
    From
    Cambridge, UK
    Posts
    41
    Hi thereddevils,

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

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

    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 v = 3 m / s.

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

    = \frac{d\epsilon}{dt}

    HTH.

    pterid
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Got a physics question or interested in physics?
    Posted in the Math Topics Forum
    Replies: 16
    Last Post: February 3rd 2014, 01:56 AM
  2. Question regarding physics?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 19th 2010, 12:38 PM
  3. Physics Question
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: October 17th 2009, 10:36 PM
  4. Physics question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 29th 2008, 04:28 AM
  5. Another physics-like question...
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: October 29th 2006, 06:09 AM

Search Tags


/mathhelpforum @mathhelpforum