Tricky Fluid Mechanics

It is a common practice in engineering fluid mechanics to express the energy of fluid flow as a quantity in units of length, called the head . The head of fluid flow consists of three (maybe five) components:

1. pressure head - head due to pressure ( $p/\gamma$ )

2. elevation head - head due to its distance from a specified datum ( $z$ )

3. velocity head - head due to the motion of fluid (its velocity) ( $v^2 /2g$ )

4. pump head - head given to the fluid by the pump ( $h_p$ )

5. turbine head - head removed from the fluid by the turbine ( $h_t$ )

These five components, assuming steady, incompressible and uniform flow, are related to each other by the energy equation as $\frac{p_1}{\gamma}+z_1+\frac{v_1^2}{2g} + h_p = \frac{p_2}{\gamma}+z_2+\frac{v_2^2}{2g} + h_t + \sum h_L$ where subscripts $1$ and $2$ are two points along the direction of fluid flow. Here, $p$ is the gage pressure, $\gamma$ is the unit weight of the fluid, and $g$ is the acceleration of free fall. The term $\displaystyle{\sum h_L}$ is called the head loss , and is the energy (in units of length) dissipated (removed) upon flow from point $1$ to point $2$ .

Energy dissipation (losses) in fluid flow occurs whenever the fluid flows through rough (and even smooth) pipes, and when the fluid encounters "obstructions" (contractions, expansions, pipe bends, valves, gates, etc.) which causes turbulence in flows. The problem deals with the latter.

Head losses due to such "obstructions" are called minor losses , minor in the sense that it is way smaller compared to pressure, elevation, velocity, pump and turbine heads. It is common practice to express minor losses as a function of the fluid's velocity head. That is, $h_{L,\mathrm{minor}} = K \frac{v^2}{2g}$ where $v$ is the velocity of flow, and $K$ is the minor loss coefficient . The value of $K$ depends on the type of "obstruction", and its value must be determined experimentally. Fortunately, almost all literature in fluid mechanics gives tables for the values of $K$ , and most manufacturers of these "obstructions" specify the $K$ values in their products.

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Now that some terms have been defined, let us go back to the problem. For a pipe entrance, we have $K=0.5$ . Therefore, the loss of head is (using $g=9.81 \mathrm{\frac{m}{s^2}}$ ), $h_{L,\mathrm{minor}} = K \frac{v^2}{2g} = 0.5 \cdot \frac{10^2}{2(9.81)} = 2.5484\text{ m}\approx \boxed{3\text{ m}}$

At first glance you might think that This is simply $L.H.=v^{2}/2g$ theoretically, a friction k must be multiplied here . The value of k is $0.5$ . So, head loss = $(Head)(k)$