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Homework Help: Mass conservation in fluid mechanics

  1. Jan 21, 2006 #1
    Part of the mechanics course I'm taking this semester are also fluids, but the material our teacher gave us to this topic is very poor (but unfortunately I haven't found a better source). The problem is that there are many "magic formulas", which come just out of nowhere - without any explanation.
    I have a problem with the very first chapter and that is the mass conservation in fluids.
    So if we have a fluid in a pipe with constant area S,
    fluid has density [tex]\varrho(x,t)[/tex] (depending on time and location), and velocity v(x,t) then the mass of the fluid in volume [tex]V = S*\Delta x[/tex] is [tex]m = S*\Delta x * \varrho[/tex]
    And now the formula I don't understand occurs:
    [tex]\frac{\partial (\varrho*S*\Delta x)}{\partial t} = \varrho(x,t)*v(x,t)*S - \varrho(x+ \Delta x,t)*v(x+ \Delta x,t)*S[/tex]
    Could someone plaese explain me how can we come to that equation?
     
    Last edited: Jan 21, 2006
  2. jcsd
  3. Jan 21, 2006 #2

    arildno

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    1.Mass conservation implies that the only way in which the mass within the pipe region of length [itex]\bigtriangleup{x}[/itex] and cross-sectional area S is that mass either leaves or enters the region (there is no mass source or sink within the region).
    2. The change of mass within the region from time t to [itex]t+\bigtriangleup{t}[/itex] is clearly given by:
    [tex]\bigtriangleup{m}=(\rho(x,t+\bigtriangleup{t})-\rho(x,t))S\bigtriangleup{x}\approx\frac{\partial\rho}{\partial{t}}S\bigtriangleup{t}\bigtriangleup{x}[/tex]
    3. However, the amount of mass entering the region from the "x"-side during the same time interval is given by:
    [itex]\rho(x,t)v(x,t)S\bigtriangleup{t}[/itex]
    whereas the amount of mass leaving at the [itex]x+\bigtriangleup{x}[/itex]-side is given by:
    [itex]\rho(x+\bigtriangleup{x},t)v(x+\bigtriangleup{x},t)S\bigtriangleup{t}[/itex]
    Thus, we have an alternative expression for [itex]\bigtriangleup{m}[/itex]:
    [tex]\bigtriangleup{m}=\rho(x,t)v(x,t)S\bigtriangleup{t}-\rho(x+\bigtriangleup{x},t)v(x+\bigtriangleup{x},t)S\bigtriangleup{t}\approx{-}\frac{\partial}{\partial{x}}(\rho{v})S\bigtriangleup{x}\bigtriangleup{t}[/tex]
    4. Setting these expressions for the change in mass equal to each other yields finally:
    [tex]\frac{\partial\rho}{\partial{t}}=-\frac{\partial}{\partial{x}}(\rho{v})[/tex]
     
    Last edited: Jan 21, 2006
  4. Jan 22, 2006 #3
    thanks a lot, this was really helpful.

    But unfortunately I am stuck again - now with turbulent flow:
    [tex]\frac {\partial \rho v_{x}}{\partial t} + \frac {\partial \rho v_{x}^2}{\partial x} = -\frac {\partial p}{\partial x} + \rho g_{x} - \frac {\lambda \rho V_{x}^2}{2D}[/tex]

    there are some symbols that are unknown to me - D is probably hydraulic diameter, but I don't know what [tex]V_{x}[/tex] and [tex]\lambda[/tex] stand for. And another thing - further in the text when solving some problems the left side of the equation is 0 - why?
    Any ideas?
     
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