Tank outlet velocity exit pipe cross-section area dependence

AI Thread Summary
The discussion centers on the relationship between outlet velocity and pipe cross-section area in fluid dynamics, referencing Torricelli's law and Bernoulli's equation. It highlights an observed discrepancy where a partially closed pipe exit resulted in a higher velocity than predicted by theoretical calculations. The conversation emphasizes the importance of accurately measuring flow velocity, as visual estimates can lead to inaccuracies. It also notes that the flow rate from the tank must equal the output flow rate, implying that a smaller outlet area increases the exit speed. Overall, the complexities of real-world fluid behavior challenge the simplicity of theoretical models.
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 .
 
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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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