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I Bernoulli equation derivation with added mechanical work

  1. May 26, 2017 #1
    I´m able to derive Bernoulli equation from the force bilance of the flowing element (particle):
    [tex]dm\cdot \left (\frac{D \vec{v}}{D \tau } \right )_{flowing particle}=\vec{dF}_{g}+\vec{dF}_{p}[/tex]
    where dFg is the gravitational force acting on the particle, dFp is the pressure force acting on the particle. From now on I neglect the action of friction forces. For such a flowing particle I´m able to derive The Euler equation:
    [tex]\vec{v}\cdot grad\vec{v}+\frac{\partial \vec{v}}{\partial \tau} =\vec{g}-\frac{1}{\rho }\cdot gradp[/tex]
    Further I consider steady state (∂v/∂τ=0). By multiplying the Euler equation with vector of elementary path ds I get:
    [tex]\left [\tau \right ]:\vec{v}\cdot d\vec{v}(dx,dy,dz)=dU(dx,dy,dz)-\frac{dp(dx,dy,dz)}{\rho }[/tex]
    where dU(dx,dy,dz) is the differential of gravitational potential. Integrating such equation through the whole streamline from the beginning 1 to the ending 2 I get the Bernoulli equation:
    [tex]\frac{v_{2}^{2}-v_{1}^{2}}{2}+g(z_{2}-z_{1})+\frac{p_{2}-p_{1}}{\rho }=0[/tex]
    Problem is I find this derivation correct also when the mechanical work is applied to the flowing particle (eg. from the agitator), but according to the literature there has to be another member e in the equation.

    What is the change in this derivation when there is a mechanical work from eq. agitator acting on the particle?
  2. jcsd
  3. May 26, 2017 #2
    If there is mechanical work from an agitator, then the streamlines are not steady. They move all over the place.

    Are you (a) familiar with the open system (control volume) version of the first law of thermodynamics, or (b) able to integrate the equation of motion dotted with the velocity vector over a control volume (even if some of the walls of the control volume are moving internally)?
  4. May 27, 2017 #3
    All these things look little bit strange. Bernoulli equation is a consequence from the Euler equations of inviscid fluid. The Euler equation in its part follows from general equations of continuous medium. All these theory is completely standard and mathematically rigorous. Why do not you use a textbook instead of developing such an informal and irregular argument?
  5. Jun 1, 2017 #4
    I don't know what are you talking about.
  6. Jun 1, 2017 #5
    That's right! I can imagine that at some place (x,y,z) near the agitator the velocities are changing with time.
  7. Jun 2, 2017 #6
    Let's have a fluid flowing through a pipe with a pump.
    Let's have a control volume bordering the pipe and the pump. The fluid enters the control volume in surface 1 and exits the control volume in surface 2. The control volume is always full of fluid. The control volume does not move relative to the inertial coordinate system.
    Let's consider the gravitational field as the only force field acting in the control volume. The gradient of the gravitational potential is parallel to the z coordinate axis.
    Then the balance of energy of the system is:
    where u is specific internal energy, v is velocity, ρ is density, n0 is unit normal vector of the control surface.
    According to the first thermodynamic law this is equal to the heat power transfered through control surface plus work power done on the control surface:

    From this balance I'm not able to derive something similar to this equation.
    As Chestermiller said there is no steady state in the control volume with the moving pump/agitator. Then the volume integral (in energy balance) is not constant in time thus its time derivation is nonzero.
    How to get rid of this volume integral?
  8. Jun 2, 2017 #7
    OK. In this equation, the rate of doing work ##\dot{W}## in this equation is going to be split into two parts, the rate of work done by pressure forces exerted by the fluid ahead and behind, and the remainder of the work, which is referred to as shaft work ##\dot{W}_s##. The shaft work refers to the rate of work done by the blades of the turbine or pump. $$\dot{W}=\int_{A_{Outlet}}{p(\vec{v}\centerdot \vec{n_0})dA}-\int_{A_{Inlet}}{p(\vec{v}\centerdot \vec{n_0})dA}+ \dot{W_s}$$

    What do you get when you substitute this? Also, next assume that the flow velocity is uniform over the exit and entrance cross sections. What does this give?
  9. Jun 2, 2017 #8
    Ok but my question was what happens with this part:
    which is the rate of change of energy inside the control volume. With a pump this part is nonzero. Since it is missing in the equation on the picture I suppose it was simply neglected.
  10. Jun 2, 2017 #9
    At steady state, it is zero. Even though the internal energy per unit mass may be changing locally within the pump at the various locations, the time average of the internal energy integrated over the volume is virtually constant, and any small temporal fluctuations that may be occurring in the total internal energy will take place over a time scale equal to the period of rotation divided by the number of blades (i.e., the time between blades sweeping past the inlet and outlet ports, which is very short).
  11. Jun 3, 2017 #10
    Since the velocities v in some point (x,y,z) near agitator are changing in time, I'm not sure the part:
    is equal to zero.
  12. Jun 3, 2017 #11
    It fluctuates slightly with time, but, of course, on average it is zero.
  13. Jun 4, 2017 #12
    But the derivation must be also possible with the control volume whose inlet and outlet walls are not moving - then the work done by pressure forces of entering and leaving fluid is simply zero. In that case I receive (assuming incompressible fluid, a steady state, well defined inlet and outlet, neglecting the volume integral):
    $$\dot{q}+\dot{w}=u_{out}+\frac{v_{out}^{2}}{2}+g\cdot z_{out}-u_{in}-\frac{v_{in}^{2}}{2}-g\cdot z_{in}$$
    With the definition of enthalpy $$u=h-\frac{p}{\rho }$$
    I get:
    $$\dot{q}+\dot{w}=h_{out}-\frac{p_{out}}{\rho }+\frac{v_{out}^{2}}{2}+g\cdot z_{out}-h_{in}+\frac{p_{in}}{\rho }-\frac{v_{in}^{2}}{2}-g\cdot z_{in}$$
    In this equation I'm just worried about the sign of pressure difference Δp - it's opposite to the other terms. Is it fine?
  14. Jun 4, 2017 #13
    You need to go back and review the derivation of the open system (control volume) version of the first law of thermodynamics. Even though the inlet and outlet walls of the imaginary control volume are not moving, the actual fluid at the inlet and outlet is moving, and work is being done on the fluid at the inlet and outlet.
    The correct version of this equation should be:
    $$\dot{q}+\dot{w}_{shaft}+\dot{m}\left(\frac{p_{in}}{\rho}-\frac{p_{out}}{\rho}\right)=\dot{m}[(u+\frac{v^{2}}{2}+g\cdot z)_{out}-(u-\frac{v^{2}}{2}-g\cdot z)_{in}]$$where ##\dot{m}## is the mass flow rate. In terms of enthalpy, this equation reads:
    $$\dot{q}+\dot{w}_{shaft}=\dot{m}[(h+\frac{v^{2}}{2}+g\cdot z)_{out}-(h-\frac{v^{2}}{2}-g\cdot z)_{in}]$$
  15. Jun 5, 2017 #14
    Ok so it's the basic definition of mechanical work for an open system, that inlets and outlets also do mechanical work on the system.

    Thank you for your effort, it helped me a lot!
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