Bernoulli for real flow with HGL and EGL

Derivation from Euler's equation

Consider a fluid element of area $dA$ and length $ds$ along a streamline, rising $dz$ over the length $ds$. Forces along the streamline:

• net pressure force $= p\,dA - (p + dp)\,dA = -dp\,dA$;
• weight component $= -\rho g\,dA\,ds\,\dfrac{dz}{ds} = -\rho g\,dA\,dz$.

For steady flow the acceleration is $V\,dV/ds$, so Newton's second law gives

$$-dp\,dA - \rho g\,dA\,dz = \rho\,dA\,ds\,V\frac{dV}{ds}$$

Dividing by $\rho\,dA$ gives Euler's equation:

$$\frac{dp}{\rho} + g\,dz + V\,dV = 0$$

Integrating for incompressible flow and dividing by $g$:

$$\frac{p}{\rho g} + \frac{V^2}{2g} + z = \text{constant}$$

Real flow: a real fluid is viscous, so part of the energy is lost to friction and turbulence between two sections. Also, the velocity is not uniform across a section, so a kinetic energy correction factor $\alpha$ is used. Between sections 1 and 2:

$$\frac{p_1}{\rho g} + \alpha_1\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \alpha_2\frac{V_2^2}{2g} + z_2 + h_L$$

where $h_L$ is the head loss between 1 and 2.

HGL and EGL for a horizontal, gradually expanding pipe

• The EGL (total head) always falls in the direction of flow because of the loss $h_L$.
• As the pipe expands, the velocity falls, so the velocity head $V^2/2g$ decreases. Part of it is converted to pressure head, so the HGL rises in the direction of flow (pressure recovery), staying below the EGL by $V^2/2g$.
• The gap between the EGL and HGL narrows downstream.

 head
  ^
  |  EGL (falls slowly due to losses)
  |  ___
  |   |  ''-.___
  |   | V1^2/2g ''-.___          EGL
  |   |               ''-.___
  |   |          _____..----|----- HGL
  |   |   __.--''   V2^2/2g |
  |  HGL--   (HGL rises as V falls)
  |
  |  pipe:  ==<  small  ......  large  >==
  +-----------------------------------> flow