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Velocity of a Cart accelerated by a Jet stream

  1. May 8, 2017 #1
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    This is a problem for a fluid dynamics class I'm in. My current approach is to use a conservation of mass approach and say that the d/dt(momentum in the cart) = momentum into the cart. This leads to (u is speed of the cart, V is volume J is jet velocity, A is cross sectional area of the jet)

    d/dt(Vρu) = ρ dV/dt J

    where the left hand term represents the derivative of the current momentum of the cart, and the right hand term represents to momentum flux into the cart.

    I found that the rate at which water from the jet enters the cart will be

    dV/dt = A (J-u), i.e. the faster the cart is going the less water that actually enters it. Going back, and cancelling density,

    d/dt(Vu) = dV/dt J
    V' u + u' V = V' J, where prime is a time derivative.
    u' = V'(J-u)/V

    a = u' = A (J-u)^2/V

    but I cant proceed from here. Not only is the differential equation nonlinear, but the volume in the denominator will be an integral depending on u,

    V(t) = Vo + ∫(0 to t) V'(τ)dτ = Vo + A J t - A∫(0 to t) u(τ)dτ

    Since I couldn't solve it analytically I used numerical integration to get the u vs t curve and it looks as you'd suspect, with asymptotic behavior near u=J

    My question is whether or not there is a symbolic answer, and if so how is it obtained?
     
    Last edited: May 8, 2017
  2. jcsd
  3. May 8, 2017 #2

    scottdave

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    Hi Will. Thanks for your interest. Here is I found out, shortly after joining this site. The questioners are expected to pose what they know about the problem, and any attempts at a solution. Then people will offer guidance to help the asker along and gain further understanding.

    Do you understand what laws would be applicable to this type of problem? I noticed that you tagged it with momentum. What can you say about the momentum of the cart? Momentum of the water?

    Since there are not any numbers, you will have an answer as an expression with a bunch of variables. What assumptions will you need to make? I hope this helps you get started.
     
  4. May 8, 2017 #3
    I'm new here, updated the post
     
  5. May 8, 2017 #4

    boneh3ad

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    Is there any reason you decided to rely solely on conservation of mass for a problem involving forces and acceleration?
     
  6. May 8, 2017 #5
    I like everything you did. So now you have two coupled differential equations

    V' = Ak
    k' = -A k2/V

    And you know the initial conditions: V0 = 0 and k0 =J

    What do you get when you integrate V'?
     
  7. May 8, 2017 #6
    I used a conservation of mass approach (we are covering the momentum tensor in my continuum mechanics class right now and I am quite sure this is the approach eh wanted us to use). The conservation of mass only plays a role since the mass of the cart varies with time.
     
  8. May 8, 2017 #7
    this is exactly where I am stuck. If i try to integrate V the expression involves an integral of k,

    V(t) = Vo + A ∫(0 to t) k dτ

    but I dont see how that helps. I did try doing something along the lines of

    V'' = A k'

    k' =V''/A ,
    k' = -A k2 /V

    V''/A = -A k2 /V

    V'' = -A2k2/V
    V * V'' = V'2

    but this is nonlinear and not easy to solve.

    I threw your coupled diff eqs into wolfram alpha and they game me fairly simple analytic solutions though. using the same IC's I used when i numerically integrated it also gave me the same results. Im just unsure how to solve these Diff Eqs
     
  9. May 9, 2017 #8

    TSny

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    Shouldn't you allow for the fact that the empty cart has some mass?
     
  10. May 9, 2017 #9
    This can be accounted for by having an initial condition that allows for the initial volume of the fluid to have the same mass of the cart.

    i.e. if it said "the initial mass of the cart and water at time 0 is 100 lbs..." you could set Vo = 100lbs/(density of water). We treat the IC as cart of mass 0 and initial water of mass 100 lbs, which has the same effect.
     
  11. May 9, 2017 #10

    TSny

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    OK. Sounds good.

    Note that you can integrate d/dt(Vu) = dV/dt J immediately. ("Cancel" the dt on each side.)
    Use the result to help solve dV/dt = A (J-u).
     
    Last edited: May 9, 2017
  12. May 9, 2017 #11

    scottdave

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    Yes I think that could be the key, right there.
     
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