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Around the nozzle of a syringe (which is filled with water and the piston is being pushed in), while applying Bernoulli's equation, will the pressure just outside the syringe be zero or atmospheric pressure? Also, will the velocity of water(not volumetric flow rate) just inside the syringe be zero or some value?

Edit: This doubt stems from Irodov's problems in general physics problem 1.322. The question's solution says that pressure outside the nozzle would be zero and velocity inside would be zero, but I fail to understand how that would be correct. In another similar situation(Hole in an open water tank) pressure just outside the hole is always atmospheric pressure. Also, when a nozzle is being pushed with a constant force,i.e. continuous acceleration, shouldn't there be a buildup of velocity in the water in the syringe?

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  • $\begingroup$ What do you mean by "just outside"? In front of the nozzle, or along it? $\endgroup$ – user198207 Jul 1 '18 at 17:21
  • $\begingroup$ In front of it. Sorry if it was unclear. $\endgroup$ – Pallav Srivastava Jul 1 '18 at 17:21
  • $\begingroup$ I apologize, but this seems to me to be a homework type question, which is off limits here. If you expand your post to say what YOU think happens, and why, that might improve the chances of an answer. $\endgroup$ – user198207 Jul 1 '18 at 17:46
  • $\begingroup$ I think you will appreciate the negligibility of the non-zero fluid velocity in the syringe more when considering that this velocity in terms of mechanical energy (which Bernoulli is) shows up as a kinetic energy term proportional to the square of the velocity, i.e. $v^2$. So if $v\ll1$ as stated in @Deep's answer, consider how much smaller $v^2$ is. $\endgroup$ – nluigi Jul 3 '18 at 7:21
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Around the nozzle of a syringe (which is filled with water and the piston is being pushed in), while applying Bernoulli's equation, will the pressure just outside the syringe be zero or atmospheric pressure?

If you neglect pressure increase due to surface tension, then inside the jet issuing from the syringe the absolute pressure will be atmospheric. If you are speaking in terms of gauge pressure, then the pressure inside the jet will be zero.

Also, will the velocity of water(not volumetric flow rate) just inside the syringe be zero or some value?

The velocity of water inside the syringe ($v$) will not be zero, but it will be small compared to the velocity of the jet ($v_j$) issuing from the syringe. If $A$ is the cross-sectional area of the syringe and $A_j$ is the cross-section of the jet at the outlet of the syringe then mass conservation implies $v/v_j=A_j/A\ll1$. This justifies neglecting the velocity of the fluid inside the syringe (compared to jet velocity) as a first approximation.

In another similar situation(Hole in an open water tank) pressure just outside the hole is always atmospheric pressure.

Here Irodov is using absolute pressure in his solution, while previously he used gauge pressure. Either way will give you the same answer.

Also, when a nozzle is being pushed with a constant force,i.e. continuous acceleration, shouldn't there be a buildup of velocity in the water in the syringe?

As I said before, there will be non-zero fluid velocity inside the syringe but it will be negligible compared to jet velocity. You may include it if you want but the change in the final answer will be negligible. If further you also wish to include acceleration of fluid due to piston motion, then you must use the unsteady Bernoulli equation.

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  • $\begingroup$ "If you neglect pressure increase due to surface tension, then inside the jet issuing from the syringe the absolute pressure will be atmospheric. If you are speaking in terms of gauge pressure, then the pressure inside the jet will be zero." Right. Of course. Thank you!! I did not think of using gauge pressure in any of the questions. $\endgroup$ – Pallav Srivastava Jul 3 '18 at 8:31
  • $\begingroup$ My question about the velocity seems silly now you point out that it will negligible in comparison to the nozzle velocity, not absolutely zero. $\endgroup$ – Pallav Srivastava Jul 3 '18 at 8:34
  • $\begingroup$ @PallavSrivastava It's not silly. In mathematical modelling of physical phenomena, it is important to know how to approximate. $\endgroup$ – Deep Jul 3 '18 at 10:11

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