# Speed of liquid being blocked at end of pipe

How fast would water go if at the end of of a 1 inch diameter pipe was closed by a valve? The system is as follows:

5 meter high source of water that feeds a 1 in pipe. The pipe goes straight down from the middle of a 25 square meter box. (1 high x 5 wide x 5 deep)

At the end of of the pipe, a valve is attached.

The valve is opened instantly.

So far I have worked out the velocity of the water as if there were no blockage.

$v = (2gh)^{1/2}$

$v = 9.9 m/sec$

Does the presence of the blockage change the velocity of the system when it is removed? Diagram: Sorry for poor quality, in a hurry because of school

• You better post a diagram, as the concept of a "block" is unclear, as well as the orientation of the pipe and the volume of your source of water. – Carl Witthoft Mar 31 '14 at 19:34

The Torricelli Formula you used is certainly a good way to start if the level of the water source is stationary. In addition there will be losses which depend on geometry of your flow restriction and on Reynolds Number. Loss coefficients $\zeta_{loss}$ for free jet discharge, valves, etc. can be found in literatute, for example here: http://books.google.com/books/about/Handbook_of_Hydraulic_Resistance.html?id=GrcZPQAACAAJ&redir_esc=y

The equation then expands basically as follows:

$\rho gh-1/2\rho v^2=\Delta p_{loss}=\zeta_{loss}/2 \rho v^2$

so

$v= \sqrt{\frac{2gh}{\zeta_{loss} +1}}$

• Sorry, bit confused by loss coefficient concept. Is the loss coefficient based on the fluid or the valve? – Ryan Sinclair Mar 31 '14 at 21:42
• And what do you mean by stationary? – Ryan Sinclair Mar 31 '14 at 21:50
• @RyanSinclair In general it depends on both. But sometimes values based on geometry only are given as well (valid for a cartain range of Reynolds). – Mirc Breitschuh Mar 31 '14 at 21:52
• @RyanSinclair It does not change – Mirc Breitschuh Mar 31 '14 at 22:05
• so the water does not need time to accelerate? – Ryan Sinclair Mar 31 '14 at 22:11