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Archived Fluid Mechanics - Piezometric Head

  1. Jun 9, 2012 #1
    1. The problem statement, all variables and given/known data

    3qK3j.png

    2. Relevant equations

    Provided above

    3. The attempt at a solution

    Piezometric head is defined as the pressure head plus the elevation head, correct? But how does having a high piezometric head tell you about the fluid flow?

    Since the diameters are equal, the area's are equal, hence velocities are equal. But how do I show the direction of flow without putting in any numerical values?
     
    Last edited: Jun 9, 2012
  2. jcsd
  3. Jul 15, 2016 #2
    First, let's assume point 1 is upstream from point 2. It makes sense that way, as the head gain from the pump (hp) appears in the LHS of the equation, and the head losses from the turbine and due to friction appear in the RHS. They can be thought of as inputs and outputs in an energy balance, well, that is exactly what the equation is. There are no pumps or turbines in the system and the diameter is constant, so
    [tex]h_p = 0[/tex]
    [tex]h_t = 0[/tex]
    [tex]V_1 = V_2[/tex]
    Now, the energy equation is reduced to
    [tex]\left( \frac{p_1}{\gamma} + z_1 \right) - \left( \frac{p_2}{\gamma} + z_2 \right) = h_L[/tex]
    Piezometric head is defined as
    [tex]h = \frac{p}{\gamma} + z[/tex]
    So we have
    [tex]h_1 - h_2 = h_L[/tex]
    Now, head loss is always a positive number, therefore
    [tex]h_1 > h_2[/tex]
    So the fluid flows from the point with higher piezometric head to the point with lower piezometric head.
     
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