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Suppose we have a water reservoir at some elevation say z and the flow is frictionless then by Bernoulli's equation we have the exit velocity (2gz)^0.5 . This means that the static pressure in the middle of horizontal pipe would always be atmospheric. Now what if we split the pipe into two at the exit, would the velocity at each exit be the same (2gz)^0.5 ? If so, then this implies that the pressure in the middle of the same horizontal pipe is negative (gauge). Does this make sense ?

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  • $\begingroup$ What about the pressure created by the height of the liquid? $\endgroup$ – Solar Mike Apr 25 '19 at 16:03
  • $\begingroup$ The height of the liquid at the base of the tank (hydrostatic pressure) is converted into dynamic pressure upon the exit in the first case but how would it happen in the second case ? $\endgroup$ – Yousuf Khan Apr 25 '19 at 16:11
  • $\begingroup$ What about the liquid at the top of the tank? $\endgroup$ – Solar Mike Apr 25 '19 at 16:12
  • $\begingroup$ Elevation or potential head, right ? $\endgroup$ – Yousuf Khan Apr 25 '19 at 16:14
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Yes.

Conservation of mass dictates that the velocity at an exit with double the area of the upstream pipe is 1/2 the velocity in the upstream pipe. For an ideal fluid in an ideal (horizontal) expansion transition, there is no change in entropy. So Bernoulli's equation gives you the relationship for calculating the pipe's static pressure given that the exit is at atmospheric pressure. This pressure will be the same throughout the pipe, and will be less than atmospheric pressure. We are also assuming an ideal bell transition from the reservoir to the pipe.

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  • $\begingroup$ Thanks Phil, I agree but if we take a hypothetical condition in which one of the exit pipes is raised above the tank level, then the pressure in that pipe will be atmospheric. Now let us start cutting the pipe gradually reducing its height, would we get the flow out of that ? If so, at what height ? Isn't it physically misleading ? $\endgroup$ – Yousuf Khan Apr 28 '19 at 4:22

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