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Change in gauge pressure using viscous flow through a pipe.

  1. Jul 26, 2015 #1

    KAC

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    1. The problem statement, all variables and given/known data
    Water flows at 0.25 L/s through a 9.0-m-long garden hose 2.0 cm in diameter that is lying flat on the ground. The temperature of the water is 20 ∘C. What is the gauge pressure of the water where it enters the hose?

    Side question: does the velocity of the water flow need to be converted to m/s rather than L/s? Upon conversion this becomes m^3/s, and I am not sure if this is suitable units for the velocity.

    R = .01m
    n = 1.0*10^-3 Pa*s
    L = 9m
    v= 0.25L/s

    2. Relevant equations
    Q = (pi*R^4*p)/8*n*L (p = pressure change; n = coefficient of viscosity, in this case 1.0*10^-3 due to the temperature of the water; L= length of the tube, 9m; R = .01m)
    Q = v(avg)*A (A = cross sectional area, of a tube is pi*r^2)
    v(avg) = (R^2/8nL)/p.

    3. The attempt at a solution

    I am unsure of where to begin. The first equation I would think to use is v(avg) = (R^2/8nL)*p so that I could plug this into Q = v(avg)A but the first equation requires p and that is what I am solving the problem for. To start, would it be suitable to use the velocity given and change the units to m^3/s? If I take this route, I get:

    v = 0.25L/s
    1000L = 1m^3
    0.25L/s * 1m^3/1000L = 0.00025 m^3/s = v

    Q = v*A; A = pi*R^2
    A = pi* 0.01^2; A = 0.000314 m^2
    Q = 0.00025m^3/s * 0.000314 m^2; Q = 7.85 * 10^-8

    Q = (pi*R^4*p)/8nL
    p= [(8nL)*Q]/pi*R^4
    p = [(8 * (1.0*10^-3 Pa*s)) * 9m]/pi * 0.01^4
    p = (0.072)/3.14*10^-8
    p = 2292993 or 2.293 * 10 ^6.

    This answer is incorrect and I am weary of the steps I took to get there. Can someone please point me in the correct direction? The first concern I have is converting L/s for velocity. Should I be leaving velocity in terms of L/s or changing them into m^3/s as I have done here? Also, is this velocity able to be used at all? The formula calls for average velocity, but with the formula for average velocity, I am unable to solve because it calls for the use of p, which I do not have.

    Any help is appreciated. Thank you in advance.
     
  2. jcsd
  3. Jul 26, 2015 #2
    What do you think the units of velocity are?

    Chet
     
  4. Jul 26, 2015 #3

    KAC

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    Chet,

    I think that the units of velocity should be m/s. Since it is a fluid and not an object in this case, I am assuming that m^3/s is suitable to use?
     
  5. Jul 26, 2015 #4

    SteamKing

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    Staff Emeritus
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    Are m3/s the same as m/s?

    You're mixing rate of flow with flow velocity, which are two different concepts.

    If you had carried the units in the calculation of Q = A * v as you should have, this would also tell you that rate of flow cannot be substituted for flow velocity.
     
  6. Jul 26, 2015 #5
    So, out of the whole list of the symbols you have shown, which one does the 0.25 l/ s correspond to?
     
  7. Jul 26, 2015 #6

    KAC

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    If I convert L to m3, it corresponds to Q, the flow rate of the fluid. So in this case, I don't actually need to solve for Q because this is already given?
    After conversions Q = 2.5*10-4; if I plug this into my equation:

    p= [(8nL)*Q]/pi*R^4
    p=[(8*(1.0*10-3 Pa*s)*9m)*2.5*10-4 m3/s]/pi*0.01m4
    p= 1.8*10-5/3.14*10-8
    p = 573.2 Pa

    Does my form appear to be correct here?
     
  8. Jul 26, 2015 #7
    I haven't checked your arithmetic, but this now looks right. You should check the reynolds number to confirm that the flow is laminar. If it is turbulent, then a different equation prevails, and pressure drop is larger.

    Chet
     
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