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Mechanical energy equation for flow b/n 2 points

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data
    a) Write down the steady-state mechanical energy equation for flow from
    point (1) (located at the free surface of the creek) to point (2) located at the free surface of
    the tank. For the purposes of this problem you can assume the pressures at both points
    are atmospheric.

    b)The steady-state mechanical energy balance will still apply at every
    moment of time if terms in the equation adjust much faster than changes in tank water
    level. Under such conditions, the unsteadiness in Darcy’s system can be modeled by
    taking the time derivative of both sides of the mechanical energy equation obtained from
    Part 1. Write down this unsteady equation.

    2. Relevant equations
    Mechanical energy equation:
    [itex]{\frac{\Delta P}{\rho}}+{\frac{\Delta V^2}{2}}+g\Delta z + Ws + F = 0 [/itex] between two points.
    Where v = average velocity, z = height, Ws = shaft work, and F = work done per unit mass against friction between points 1 and 2.

    3. The attempt at a solution
    So for part a) I'm not really sure what assumptions are reasonable given the attached diagram.

    Can the [itex]V^2[/itex] term be canceled due to continuity? i.e. area is the same along the pipe and so velocity is the same?

    The question also says that pressure is atmospheric at P1 and P2, so does the pressure term disappear as well? I would have thought we needed to include the weight force = [itex]\rho gz[/itex] which increases as the pump transfers water from the creek to the tank, reducing the volumetric flow rate with time. So does it just mean we can ignore the atmospheric pressure component?

    Also can [itex]\Delta z[/itex] just equal z2, taking z1 as reference point? i.e. z1 = 0?

    So would my equation look something like this?
    [itex]{\frac{P_{atm}}{\rho}} = {\frac{(P_{atm} + \rho gz)}{\rho}}+z+Ws+F[/itex]

    Then the left hand side term disappears and would become:
    [itex]0 = gz+z+Ws+F[/itex]

    I could then divide the equation by g to make the mech energy eq. in head form, allowing me to substitute [itex]{\frac{W_s}{g}}[/itex] and [itex]{\frac{F}{g}}[/itex] for others with more appropriate variables. E.g. Ws/g in terms of mass flow rate and actual power transferred. "g" would also disappear from the "gz" term.

    But then I'm missing a time variable because the following step requires me to take the time derivative and I can't in its current form..

    Any help is greatly appreciated, thanks!
     

    Attached Files:

    Last edited: May 5, 2014
  2. jcsd
  3. May 5, 2014 #2
    attached diagram????

    Chet
     
  4. May 5, 2014 #3
    Sorry Chet - original post has been updated with the relevant diagram.

    Thanks!
     
  5. May 5, 2014 #4
    OK. Here are some thoughts:

    1. The tank velocity is much lower than the velocity in the pipe. The two are related by the continuity equation. The velocities at points 1 and 2 need to be included. The velocity at point 2 is also the rate of increase of depth in the tank.

    2. The difference in elevations should be included in the gz term. But shouldn't be included twice (as you did).

    3. The ΔP is indeed zero.

    Chet
     
  6. May 6, 2014 #5
    Hi Chet, thanks again for the reply.

    I've taken another look at the problem and my equation, can't we assume the velocity term is 0, because the points are taken at the surface level and we assume that the tank and creek are large enough that the surface velocity is unaffected by the flow?

    The flow velocity is still considered, however as a part of the shaft friction term - i.e.:
    [itex]F={\frac{2fL{V^2}}{D}}[/itex]
    Where f = fanning friction factor, V = average velocity, L = pipe length, D = diameter of pipe

    So taking into consideration your above thoughts, my equation becomes:
    [itex]g\Delta z+W_s+F=0[/itex]
    or
    [itex]\Delta z+{\frac{W_s}{g}}+{\frac{F}{g}} = h_p[/itex]

    Then for part b) taking the time derivative, I think I should end up with:
    [itex]{\frac{\partial z}{\partial t}} = - {\frac{\partial}{\partial t}}({\frac{W_s}{g}}) - {\frac{\partial}{\partial t}}({\frac{F}{g}})[/itex]

    Thanks again!
     
  7. May 6, 2014 #6
    Yes. My mistake.

    Looks good. I must admit, I had trouble figuring out what they were driving at in part (b), but this looks like it is probably what they were looking for.

    Chet
     
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