I'm Studying the streamline flow, specifically the continuity equation Bernoulli's Principle.

Consider the following system where a liquid flows through the pipe of a uniform area of cross section A, from high pressure P2 to low pressure P1, both of which are maintained.

<--------length = r----------->
P2                           P1
           flow ->

Based on the assumptions that

  1. Difference in pressure is what causes the liquid to flow $P_2 > P_1$
  2. The continuity equation $A_1v_1 = A_2v_2$ holds
  3. The Bernoulli's Principle holds

Then my analysis is as follows

The Bernoulli's equation gives $$ P_2 + \frac{1}{2}\rho v_2^2 = P_1 + \frac{1}{2} \rho v_1^2$$ Therefore, the speed of liquid element will be more at $P_1$ and less at $P_2$ since $P_2 > P_1$

But then

$$A_1v_1 = A_2v_2$$

and since the Area of cross section is uniform, the velocities must be same, which is contradicting the result from Bernoulli's equation!

There is something wrong with my analysis or assumptions, but I can't figure out what it is


3 Answers 3


In the form you've stated it ($A_1 v_1 = A_2 v_2$), the continuity equation is only holds for incompressible fluids. So what you've found is that the type of accelerated flow you're describing cannot happen for an incompressible fluid in a pipe of uniform cross-section. The more general form for the continuity equation is based on conservation of mass (i.e., mass per time entering = mass per time exiting), and states $$ Q_m = \rho_1 A_1 v_1 = \rho_2 A_2 v_2, $$ where $Q_m$ is the mass flow rate (i.e., mass per time). This means that if the fluid increases in velocity, it must decrease in density.

An analogy here would be cars on a highway. Suppose you have a highway that leads from the center of a small town out into the country. Suppose further that the drivers are all perfectly safe drivers and obey the two-second rule, i.e., the cars pass a given point in the highway at a rate of one car every two seconds. If the speed limit in the town is low, then the cars will be more closely spaced, since two seconds corresponds to less distance. Thus, the density of cars is higher at this point. When the cars get out of town and the speed limit increases, the cars get further apart in distance (since two seconds now corresponds to a longer distance), and so the density decreases.

Bernoulli's equation, meanwhile, doesn't hold so simple a form for compressible fluids. Rather, you have to define a pressure potential $w(P)$ from the equation of state $\rho(P)$: $$ \frac{1}{2} v^2 + gy + w(P) = \mathrm{const.}, $$ where $$ w(P) = \int \frac{\mathrm dP}{\rho(P)}. $$ For the case of an "incompressible" fluid, $\rho(P) = \rho$ is a constant, $w(P) = P/\rho$, and the familiar form of Bernoulli's Law is recovered. But for a compressible fluid, the equation may look quite different.

  • $\begingroup$ For an incompressible fluid, the volume flow rate must remain the same and thus the velocities of two ends should be the same, which makes sense to me, yes. But then what I'm unable to comprehend is what causes the flow of liquid if not a difference in pressure at two ends? If there weren't difference in pressure, the liquid would not move, correct? Or is it the case that the difference in pressure does cause the liquid to flow, but Bernoulli's equation doesn't apply in the setup as I stated above (with incompressible fluid) and some other law will give an indication of speed of flow? $\endgroup$ Commented Mar 21, 2016 at 17:14
  • $\begingroup$ @PeeyushKushwaha: It is perfectly consistent for the fluid to be moving without any change in pressure from one end of the pipe to the other. When the pressures and the areas are the same on either side of a "slice" of water in the pipe, this means that the forces from either side cancel out; but this just means that the water is moving at a constant velocity, not necessarily zero velocity. (Newton's First Law!) You're right that you would need a difference in pressure initially to get the flow started; but once it's moving, no difference in pressure is needed to maintain the flow. $\endgroup$ Commented Mar 21, 2016 at 19:24
  • $\begingroup$ @PeeyushKushwaha: Just to complete my thought: Bernoulli's equation and the continuity equation (even in the form I wrote above) implicitly assume that nothing is depending on time, i.e., the velocity at a given point in the pipe isn't changing. So you can't really use either one to address how the flow gets started; they only really deal with the steady-state flow, once everything is moving nice and smoothly throughout the system. $\endgroup$ Commented Mar 21, 2016 at 19:28
  • $\begingroup$ That settles my confusion. I was for some reason thinking that there needs to be a constant difference of pressure between the two ends of a pipe to "drive" the flow, without which the flow must stop. $\endgroup$ Commented Mar 22, 2016 at 6:09
  • $\begingroup$ @PeeyushKushwaha: You're not entirely wrong; in the "real world", fluids have viscosity, which causes an extra force between the pipe walls and the fluid. If you want to take this into account, you end up with something like Poiseuille's equation instead, where a pressure difference is required to maintain a steady flow. Bernoulli's equation is really an approximation in which we treat the viscosity as negligible. It's not unlike how we talk about "frictionless tables" in introductory mechanics — such a thing doesn't really exist, but in many situations the effects are negligible. $\endgroup$ Commented Mar 22, 2016 at 12:45

here is the diagram

The question is very good but the assumption you have made is wrong!

I agree that the water is moving because of the pressure difference between water and air

As I have shown point $P_1$ and $P_2$ in the diagram.

Now the water moves because of pressure difference between this $P_1$ and $P_2$ so as the water reaches the end of the tube(according to the diagram) the pressure in the water is still $P_1$ and pressure just next to it (I mean pressure of air next to it) is still $P_2$

Now the mistake you have done is you have taken Bernoulli's equation for air and water which is completely wrong. Bernoulli's equation is valid only for one fluid at a time, but you have taken two separate fluids which is totally wrong.

You may say why the pressure has to be equal at the starting point and at the end point. The reason is fluid is at horizontal level; more over it has uniform cross-section and according to equation of continuity it must have equal velocity. So it has to satisfy Bernoulli's equation that's why pressure should be equal.(because Bernoulli's equation is a property of streamline flowing liquid)


  • Bernoulli's equation can be applied for one fluid at a time.
  • If fluids are at same horizontal level and have same cross-section then they have same velocity always (streamline flowing liquid only).
  • $\begingroup$ Where did OP invoke water and air? $\endgroup$
    – Kyle Kanos
    Commented Mar 21, 2016 at 17:03
  • $\begingroup$ @KyleKanos OP? i didn't get this $\endgroup$ Commented Mar 21, 2016 at 17:05
  • $\begingroup$ OP is common parlance for original poster, i.e. the person who wrote the question. $\endgroup$
    – Kyle Kanos
    Commented Mar 21, 2016 at 17:06
  • $\begingroup$ he has invoked water and air by saying because of pressure difference water is moving and the pressure difference is between water and air only not between water at starting and a end point $\endgroup$ Commented Mar 21, 2016 at 17:07
  • 2
    $\begingroup$ well, this seemed the quite logical reason for me! feel free to down-vote :) $\endgroup$ Commented Mar 21, 2016 at 17:18

In pipe flow two terms are added to Bernoulli's equation. Major loss and minor loss terms. Major loss is due to wall friction and minor losses are due to shape change in the pipe. (Valves, elbows etc.)

In the system that you draw the pressure difference overcomes the wall friction. The wall friction increases with velocity of the fluid, therefore, increasing pressure difference increases the fluid velocity.


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