Calculating Pressure Drop from Water Pipe Discharging to Atmosphere

Question

I have an application at work where I am draining a water tank. The outlet valve is a flush mount, and I know the Cv value from manufacturer website. I can calculate the pressure drop through this valve and piping. The piping goes to an atmospheric drain which is 5 ft lower than the tank bottom. Will there be a discharge pressure drop at the very end of the drain pipe that I must take into account? The end goal is I want to calculate the free-drain flow rate at various tank water levels. A derivation/proof that there is in fact a pressure drop there would be especially helpful.

Answer 1

In general, the Loss coefficient(K) in calculation for the inlet and outlet of piping is considered to be 0.5 and 1.0. These values are exemplified in Crane Tech. Report 410 or in the Fluid Mechanics book, which is a textbook for universities.

This is a value that is used based on experimental results, and the derivative formula of the component itself is unknown.

Answer 2

Using Bernoulli equation:

$$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 $$

• $P_1$ and $ P_2 $ are the pressures at the surface and the orifice, respectively.
• $v_1=0 \ $ The water velocity at the surface, can be ignored if the tank's surface is much bigger than the orifice.
• $v_2 $ is the velocity of water at the orifice$\ =\sqrt{2gh}.\ $ We ignore the valve friction, in this case, it's irrelevant.

Pressure drop at the orifice

$$P_1 + \rho gh +0= P_2 + \frac{1}{2} \rho v^2 $$ We set $P_1=P_2= 1atm$

$$\Delta P = P_1 - P_2 = \rho gh - \frac{1}{2} \rho v^2$$ Substituting $$v = \sqrt{2gh}\\ \rightarrow \quad \Delta P = \rho gh - \frac{1}{2} \rho (2gh) = \rho gh - \rho gh = 0$$

There is no pressure loss at the discharge. This roundabout was just to repeat the obvious: when the discharge and tank are both exposed to the atmosphere, they are at the same pressure!