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Provided that the two pipe lines are of same length, same material and in the same level, is the water pressure in both the layouts same or different?

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PS: In 1st pipeline the turns are not "upward-downward" turns but "sideward-turns" in same level(height).

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The friction in the first pipe would give a very small heating effect then increasing the pressure a little bit more – huseyin tugrul buyukisik Aug 23 '12 at 16:03

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Static pressure will be the same, even if one pipe would be only half as long as the other one. That's due to Pascal's Law:

$ \Delta P = \rho \cdot g \cdot \Delta h $

So the pressure difference is proportional with difference in height, and for all points at the same height the pressure will be the same.


Dynamic pressure will be lower for the #2, as there won't be the same pressure loss due to the resistance difference.

A fluid with sufficiently high viscosity (most liquids) will have a laminar flow in pipe #2, whereas the bends in #1 will break the lamina and cause turbulence at the bends which increases flow resistance.

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Why will static pressure be the same? Are you sure for #2 dynamic pressure ($0.5 \rho v^2$) will be lower? Anyway, some argumentation based on physical laws would be nice. – Yrogirg Aug 23 '12 at 16:06
you do use some awkward notion for static an dynamic pressure. You mean by "static pressure" the pressure difference when flow rate is zero (water doesn't move), and by "dynamic pressure" the pressure on the exit when the water is flowing. But ok, it doesn't matter much. – Yrogirg Aug 23 '12 at 17:45
@Yrogirg - What terminology should I use then? – stevenvh Aug 23 '12 at 17:48
pressure drop (en.wikipedia.org/wiki/Pressure_drop) To distinct the cases just specify "water (not) flowing". Usually by static pressure meant the pressure $p$ itself at point and by dynamic pressure $\frac{1}{2} \rho v^2$ or $p + \frac{1}{2} \rho v^2$ at a point. – Yrogirg Aug 23 '12 at 17:52
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The dynamic pressure drop is not necessarily less after a turn, and turns don't break laminar flows in general. You can have a laminar turn no problem, at low enough Reynolds number, it's automatic. – Ron Maimon Aug 24 '12 at 6:27
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1st pipe: Practically speaking, there would be an appreciable change of pressure definitely at initial water flow due to the collision of water molecules on the different turns of the pipe. But as dynamic pressure comes into play, the pressure would remain constant and the water flow is even...

2nd pipe: The pressure remains the same for a horizontal pipe with no turns...

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To your understanding it is not "upward-downward" turns but "sideward-turns" in same level(height) – vivek_jonam Aug 23 '12 at 16:16
how can pressure vary for upward and downward turns whens they are in same level? It would be same pressure even if its right turn or left turn. no? – vivek_jonam Aug 23 '12 at 16:25
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So here you are talking about "vertical-vs-horizontal", in my question I talk only about horizontal with right and left bends. ok? – vivek_jonam Aug 23 '12 at 16:37
its obvious, Flows – vivek_jonam Aug 23 '12 at 16:40
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As long as the fluid is same, the pressure at same heights is always equal.

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The pressure drop in your first diagram will be substantially higher due to the bends. There is an extra pressure drop for either laminar for turbulent flow at every bend which will add to the overall pressure drop if the pipe were straight. There are standard engineering formula to use for pressure drop due to bends in the pipe.

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