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.
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$\begingroup$ Any pipe, orifice or valve generates a pressure loss due to friction. $\endgroup$– Solar MikeCommented Oct 14 at 17:20
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$\begingroup$ I don't think that the pipe end should generate any significant local pressure loss when there is discharge of water into atmospheric air. $\endgroup$– Tomáš LétalCommented Oct 14 at 20:30
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$\begingroup$ gauge pressure at the exit is zero. Every water hose works this way, and has friction head loss. Delta p is fluid height pressure minus zero. $\endgroup$– Tiger GuyCommented Oct 15 at 2:32
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$\begingroup$ Do you have a nozzle at the end of the drain pipe? Are you discharging into Air or liquid? $\endgroup$– Sami SafariniCommented Oct 15 at 11:31
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$\begingroup$ @SamiSafarini there is no nozzle. Discharging to air. $\endgroup$– remusconnorCommented Oct 15 at 18:02
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1 Answer
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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!
