Tank outlet velocity exit pipe cross-section area dependence

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SUMMARY

The discussion centers on the dependence of outlet velocity on the cross-section area of an exit pipe, utilizing Torricelli's Law and Bernoulli's equation. The observed phenomenon indicates that a partially closed exit pipe results in a higher velocity compared to a fully open pipe, despite theoretical calculations suggesting otherwise. The participants emphasize the importance of accurately measuring flow velocity and consider factors such as the continuity equation and the relationship between flow rate and cross-sectional area. The conclusion drawn is that real-world fluid dynamics often deviate from idealized equations due to various influencing factors.

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Dileep Ramisetty
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Sir, actually in a tank by the toricelli's law and also from bernoulli equation, we have outlet velocity as V= (2*g*H)^(1/2). In the case 2, I have closed the pipe exit partially with hand and observed a higher velocity than case 1, in practical. but when I applied the bernoulli equation at the surface of tank and pipe's outlet in the both cases, I am getting velocity V=(2*g*H)^(1/2) by energy conservation. But in practical the velocity is high in case 2.
so, someone please solve me this, by considering the bernoulli energy equation at 1.the surface of tank and 2.at pipe exit
Thank you
 
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The mean depth at the outlet is greater in case 2 than in case 1 .

This should only make a noticeable difference though if the nominal mean depth is only equal to a small number of outlet pipe diameters .

You don't say how you measured the velocity . Simple visual estimates of flow velocity can be very inaccurate and you may have just overestimated the difference in flow velocity for the two cases .
 
Last edited:
Lets check both figures and let's forget for a while the pipes at the bottom of the tanks:

Neglecting everything about fluid dynamics, and centering only on basic physics ...

The fluid inside the tank will flow down at a speed of:

V= sqrt(dens g h). If the surface area of the free surface is "S", the flowrate will be: Q= S x VLets now have a look to the output pipe at the bottom, and let's assume, the height of the exit, in both cases to be the same.

If the pipe section area is given by "A" and the output speed by "Ve", the output flowrate will be: Qo= Ve x AContinuity equation (and common sence) says that in "ideal" conditions, Flowrate at which the tank level goes down, must be equal to the output flowrate, hence ...

Q= Qo
S x V= Ve x A

Ve= V x S/A


Meaning that ... The less output section area "A", the higher output speed "Ve".The real world is not so "perfect". Fluids do not tend to behave so simply and many variables have to be taken into account.
 

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