2
$\begingroup$

Someone told me: This is because the flow rate remains unchanged, so the flow rate of small diameter pipelines is faster. I don't think this explanation is right: the diameter of the pipe outlet is small, and the flow rate should be reduced. The flow rate is smaller, so the resistance of the pipeline is reduced, and the water is driven by higher pressure, so the flow rate at the outlet of the pipeline is higher. Is my explanation correct?

When the diameter of pipe outlet decreases

$\endgroup$
  • $\begingroup$ I don't understand what you're writing very well, but: wouldn't the pressure be constant because the pressure of the fluid depends only on its height, and it's the same in both diagrams? $\endgroup$ – Allure Mar 22 '19 at 3:30
  • $\begingroup$ Assuming the diagram you've drawn, and using Bernoulli's law, the mass flow rate is $\rho A u=\rho A \sqrt{2gh}$ for h the height difference of the outflow and the height of the water in the tank, $A$ the area of the jet exiting the pipe outlet, and $\rho$ the density of the fluid. So the mass flow rate changes with $A$, but the velocity of the flow is simply a function of $h$. Consequently, if you fixed the mass flow rate, and changed A, then you would change $u$, and as $u\sim A^{-1}$, then narrower orifices mean greater flow speeds. $\endgroup$ – Nick P Mar 22 '19 at 3:50
  • $\begingroup$ This can be easily understood using the continuity equation : princeton.edu/~asmits/Bicycle_web/continuity.html $\endgroup$ – himanshu Mar 22 '19 at 4:39
  • $\begingroup$ @NickP the mass flow rate when the outlet is reduced is different from that when the outlet is not reduced. $\endgroup$ – enbin Mar 22 '19 at 5:53
  • 1
    $\begingroup$ I am not in the habit of clicking weird links. Could you just put it into words? $\endgroup$ – bukwyrm Apr 4 '19 at 9:29
2
$\begingroup$

The key thing to observe is the lower level in the manometer, that vertical pipe on the left.

The larger opening has higher flow in the pipe, hence more energy loss, hence less pressure at the orifice.

The smaller opening has less flow, less loss, higher pressure.

It’s the pressure right behind/before the orifice that matters.

This effect is much more visible with a long garden hose. A wide open outlet is at low pressure as lots of flow loses energy in the hose. Put your thumb over the end, the flow decreases, and you can feel the hose get stiffer as the pressure rises.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Very well, you've found the final answer to the question. When the outlet of the pipeline is reduced (local, very short section of pipeline), the flow velocity of the non-shrinking pipeline decreases, so the loss decreases, so higher pressure is obtained at the inlet of the narrowing pipeline, so the flow velocity at the outlet of the pipeline increases. Large pressure difference is the reason for the increase of velocity. $\endgroup$ – enbin Mar 26 '19 at 14:33
1
$\begingroup$

This phenomenon is called orifice effect. What happens is the stream discharging from a nuzzle converges to a narrower diameter and faster velocity downstream at a point called Vena contracta.

There are empirical methods, based on Bernoulli's law, to calculate the discharge and velocity.

$ q = c_d A2 \frac {2 (p_1 - p_2)}{ ρ (1 - (A_2 / A_1) ^2)^{1/2} }$

$ \text{A1 is the diameter of the pipe, A2 diameter of the nuzzle, A2<A1}$,

$c_d = \frac {A{vena contracta}}{A_2}= discharge \ coeficient $

There are many chrats to find Cd. Usually it ranges around 60% to 80%, Engineering toolbox

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ My question: The outlet of pinched pipeline and the outlet of non-pinched pipeline have higher velocity. Why? $\endgroup$ – enbin Mar 23 '19 at 22:57
  • $\begingroup$ @enbin zheng, a rough explanation is $ PA = 1/2 \dot mv^2 $ . Because the pressure decreases relatively slowly we can assume for short time interval as constant. And so by narrowing the opening we have to increase the speed to maintain the equilibrium. But air friction to the jet and viscosity of liquid passing through smaller orifice make this relationship nonlinear. The speed of ejection will go up to a certain point then will decrease reducing to droplets. That curve has much to do with the shape of the orifice, its edges, the head of pressure. hence the empirical measures. $\endgroup$ – kamran Mar 24 '19 at 2:14
  • $\begingroup$ @enbin zheng, and by "pinched pipeline and the outlet of non-pinched pipeline " have higher velocity you mean the velocity of the horizontal pipe with no nuzzle, the answer is the same as above. $\endgroup$ – kamran Mar 24 '19 at 3:49
  • $\begingroup$ Q=π·r4·Δp/(8ηL) Should the flow rate decrease when the pressure difference remains constant and the diameter decreases? $\endgroup$ – enbin Mar 24 '19 at 6:05
0
$\begingroup$

Why can water be sprayed farther when the pipe outlet is narrowed?

Your assumption that the water will go further (i.e., velocity will be greater) when the pipe outlet is narrowed is incorrect. For an ideal fluid (assumptions below) the velocity will remain the same and the flow rate will decrease based on the Bernoulli equation and the continuity equation.

See diagram below.

Bernoulli equation (conservation of energy):

$P_{1}+\frac{ρV_{1}^2}{2}+ρgz_{1}= P_{2}+\frac{ρV_{2}^2}{2}+ρgz_{2}$

Continuity (conservation of mass) equation:

$A_{1}V_{1}= A_{2}V_{2}$

Assumptions:

  1. Steady flow
  2. Incompressible fluid
  3. No or negligible viscosity
  4. No friction losses
  5. The diameter at the top of the tank (assumed circular) is much greater than the diameter of outlet.

Then given

$P_{1}=P_{2}$= 1 atm

$A_{1}>>A_{2}$ therefore, $V_{1}<<V_{2}$

$z_{1}-z_{2}= h$

The velocity at the outlet is

$$V_{2}=\sqrt{2gh}$$

Note that this is identical to the velocity of an object initially at rest that falls to the ground from a height $h$, due to the conversion of potential to kinetic energy. It is independent of the diameter of the pipe outlet. If the velocity is the same and the cross sectional area of the pipe is decreased, then by the continuity equation the flow rate decreases.

Hope this helps.

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ but $P_2 \ne P_1$. $P_2=P_1+ \rho A_1 (z_1 - z_2) g$ $\endgroup$ – Paul Childs Mar 27 '19 at 2:07
  • $\begingroup$ @PaulChilds P1 and P2 are the static pressures OUTSIDE the fluid above the tank and to the left of the outlet, respectively, and are both atmospheric pressure. $\endgroup$ – Bob D Mar 27 '19 at 5:16
  • $\begingroup$ But your Bernoulli equation is not about the air but the fluid. $\endgroup$ – Paul Childs Mar 27 '19 at 5:50
  • $\begingroup$ You have also said the OP is incorrect when clearly they have experimentally demonstrated it and are now asking why. $\endgroup$ – Paul Childs Mar 27 '19 at 5:53
  • $\begingroup$ @PaulChilds I read the OP question and saw no statement the question was based on an experiment the OP did. So I assumed it was a theoretical question. For what it's worth, I did conduct an, admittedly crude, experiment. Took a large plastic cup, about 4.5 in tall, 4.5 in. dia at top, 2.5 in. dia at bottom. I drilled a 1/16 in dia hole one inch from the bottom on one side and a 1/4 in dia hole at the same height on the opposite side of the cup. Filled it with water and held it above and over and the middle of the sink. Both streams went the same distance. $\endgroup$ – Bob D Mar 27 '19 at 10:52
0
$\begingroup$

It is assumed that the pipeline consists of an unshrinking pipeline L1 and a shrinking pipeline L2. L1 is much larger than L2. pipeline L1 and L2 As the outlet of the pipeline shrinks and the flow is hindered, the velocity of L1 in the pipeline decreases, so the pressure drop also decreases. According to the following formula, the pressure drop is proportional to the square of the velocity, so when the velocity decreases, the pressure drop will decrease greatly.

$$h_f=λ \frac{l}{d} \frac{v^2}{2g}$$

Because the pressure drop is greatly reduced, pipeline L2 will get higher pressure, so the flow rate of pipeline L2 will increase.

Video Interpretation of Sprinkler Speed

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

This question has been bothering me for a while, and here's what I've come up with.

Bernoulli's equation is clearly the starting point to think about this problem: $p + \frac{1}{2} \rho v^2 + \rho g h = E$ where $E$ is a quantity they is conserved throughout the fluid, basically its energy per unit volume. In other words, the sum of the kinetic energy, potential energy and pressure is the same everywhere.

With atmospheric pressure roughly equal at the surface of the tank and at the exhaust, all of the potential energy at the surface of the tank should be transformed into kinetic energy at the exhaust such that $v_{\text{exhaust}} = \sqrt{2gh}$.

Note that the diameters of the pipe or of the exhaust aren't in this formula, so under conditions where the Bernoulli equation applies, the narrowing in the nozzle doesn't change anything.

So, since we know this isn't true... what gives?

An effect that's not captured by Bernoulli's equation are friction losses in the pipe: friction in viscous fluids is generally proportional to $v^2$ (see Navier-Stokes equation) and velocity is proportional to the cross-sectional area of the pipe so that doubling the diameter of the pipe while keeping the exhaust diameter the same would reduce the friction losses by a factor of 16 (not accounting for the change in exhaust velocity, but you get the idea...)

These losses have the form $\alpha l v^2_{\text{pipe}}$ where $\alpha$ is some coefficient which would include things like the viscosity of water and $l$ is the length of the pipe since it makes sense that the losses would scale linearly with that.

Finally, call $\beta$ the ratio of the exhaust diameter to pipe diameter. From this we get $v_{\text{pipe}} = \beta^2 v_{\text{exhaust}}$.

Now equate fluid energy at the exhaust (including losses from the pipe) to energy at the top of the tank (using Bernoulli's equation the $p$ terms since they are almost equal on both ends and dividing by $\rho$). You get:

$\frac{1}{2} v_{\text{exhaust}}^2 + \alpha \beta^4 l v_{\text{exhaust}}^2 = g h$

Solving for $v_{\text{exhaust}}$, we get: $v_{\text{exhaust}}=\sqrt{\frac{2gh}{1 + 2 \alpha \beta^4 l}}$

Note that taking pipe length $l \rightarrow 0$, or friction coefficient $\alpha \rightarrow 0$ or diameter of the exhaust $\beta \rightarrow 0$ yields the friction-less Bernoulli formula.

(I'm not a physicist so I look forward to replies with corrections/confirmation)

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

The pressure is the same in either case, but as pressure equals force divided by area, you are ejecting less water at higher force in order to maintain the same pressure, hence the droplet spray goes further. Flow rate doesn't (have to) come into it.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I agree, but maybe throw in an equation for good measure $\endgroup$ – InertialObserver Mar 22 '19 at 5:33
  • $\begingroup$ Can you write the equation? $\endgroup$ – enbin Mar 22 '19 at 6:29
  • $\begingroup$ Anything beyond just $F = P A$ is useless when dealing with non-laminar flow. Bernouli only deals with laminar flow. $\endgroup$ – Paul Childs Mar 24 '19 at 19:40
  • $\begingroup$ What cross-section is your pressure? $\endgroup$ – enbin Mar 25 '19 at 21:33
  • $\begingroup$ The cross sectional area A is taken wherever you want to measure the force; typically at the point it is ejected from the pipe. $\endgroup$ – Paul Childs Mar 26 '19 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.