# Bernoulli's principle

In steady flow, the air that flows faster causes a lower pressure, the lower pressure causes it to move faster, or exactly both at the same time?

Also how is it that we're talking about the pressure of the fluid, if in reality that term for pressure comes from a pressure that's outside the immediate system (according to the derivation; see below).

$$dW=p_1A_1ds_1 - p_2A_2ds_2$$

• $dW$ is work done on the system
• $p_1A_1ds_1$ is work caused by a pressure that's outside the fluid and in the same direction as the flow
• $p_2A_2ds_2$ is work caused by a pressure at the other ending of the system that opposes the direction of the flow.

• If this is not immediately clear, I'm considering a closed pipe and my system is an element of fluid within that pipe.

One last thing: how do you solve any relativistic effect for the Bernoulli equation? Like, suppose I'm in a car with a certain velocity, is the pressure inside my car greater or less than the pressure outside? Well that depends on the frame of reference, etc.

EDIT: The car is not sealed.

• The pressure inside a car is very close to the outside pressure since there are several ways for the air to enter the compartment.
– LDC3
Commented May 25, 2014 at 0:51
• The pressure difference is relative to the "same air at rest" which is typically air that is far away from the area where you are measuring. Inside the (sealed) car there is no connection to "stationary" air so the basis for the equation (work done by the air to get the air moving) does not apply Commented May 25, 2014 at 3:14
• The car isn't sealed. Suppose i'm smoking a cigarette inside, where does the smoke go?
– DLV
Commented May 25, 2014 at 4:05
• The air in the car will be in equilibrium with the air outside the car - so pressure will drop slightly and some smoke will escape. More interesting (given shape of car) is that pressure will be different on front and rear windows, so if you crack both open there will be a circulation of air in the car Commented May 25, 2014 at 13:12

Confusion abounds in conversation relating to Bernoulli's principle, so I will do my best to dispel some of the more pernicious misconceptions.

(1) Bernoulli's principle, in general, applies only to a single streamline, with unique stagnation properties. While many aerodynamic/hydrodynamic flows (e.g. irrotational flows) do exhibit uniform stagnation pressure everywhere, this is generally not to be expected. Case in point, even though the static pressure just inside and just outside your car window are equal, the stagnation properties will be very different (higher outside, substantially lower within the confines of the car, where the flow is a swirling welter of separated flow). Faster velocity does not always mean lower pressure, contrary to what many are led to believe. You must look at the flow as a whole, especially the upstream conditions.

(2) Stagnation properties do in fact depend on the reference frame, while static properties do not. That is to say, stagnation properties are relative. In the stationary (ground) reference frame the ambient air is not moving, so the static and stagnation properties are identical. However, in the car's reference frame, the flow far upstream is moving rather quickly, so it's stagnation pressure, temperature, and density will necessarily be higher than the corresponding static values.

(3) Pressures, or more accurately, pressure gradients cause accelerations in an ideal flow, not the other way around; forces cause accelerations, not vice versa. This is a big source of confusion when talking about aerodynamic lift, vorticity distributions, and the Biot-Savart Law, but just keep in mind the difference between a physical principle and a merely useful mathematical tool/concept. The phenomena are coupled via conservation of mass and conservation of linear momentum, but ultimately the imposed pressure field causes the observed velocity distribution.

(4) It's best not to construe Bernoulli's principle as an expression of conservation of energy, as technically it's not (it completely neglects internal energy). No, Bernoulli's principle is best understood as an integrated expression of the conservation of linear momentum, $F=ma$, where nominal force is replaced with force per unit volume (which is equal to the negative local pressure gradient), and mass is replaced by mass per unit volume (which is the fluid density). In these terms, Newton's Second Law becomes:

$\frac{Force}{Unit Volume}=\left(\frac{Mass}{Unit Volume}\right)(Acceleration)$

which gives us

$-\frac{dP}{ds}=\rho v\frac{dv}{ds}$

which is rearranged as

$\frac{dP}{ds}+\rho v\frac{dv}{ds}=0$,

or

$\frac{dP}{\rho}+vdv=0$,

which when integrated gives us the classic incompressible Bernoulli Eqn.:

$\boxed{P+\frac{1}{2}\rho v^2=P_0}$,

The best reference on this topic (like most topics concerning fluid mechanics) is the NCFMF series entry entitled "Pressure Fields and Fluid Acceleration," hosted by the legend himself, Ascher H. Shapiro. He clears up a lot of the confusion surrounding this concept in a readily intelligible and fascinating way.

• Thanks a lot. I was wondering though, in the Venturi effect, what has caused this pressure gradient? Also, In the simplified model of lift, what caused the pressure gradient so as to create more rapid flow on top of the wing? I understand that as flow hits the top bump in the wing, it wants to keep a linear trajectory, and because the wing has curvature this will then create some kind of partial vacuum, and then this causes the pressure gradient. Am I right?
– DLV
Commented May 28, 2014 at 0:03
• And one last thing: I can see the case for steady flow if I consider the wing to be stationary with respect to the air, but if I switch reference frames where the wing is moving and the air is stationary, I just can't picture it as steady flow. Do you have trouble with this? Or should I just keep thinking about it, because it's not really a problem? Thanks.
– DLV
Commented May 28, 2014 at 0:04
• The best way to answer this question is to convey the idea that the flow must be consistent with conservation of mass, linear momentum, and energy all simultaneously. There is only one stable solution that satisfies all of these criteria, and it is the one we actually observe. This may not be the most satisfying answer I could give you, but it may be the only one that's true. No correct discussion of lift can be had without discussion of viscosity and flow separation. The best online reference for how airfoils work is included below. av8n.com/how/htm/airfoils.html Commented May 28, 2014 at 15:17
• You may have an improper understanding of how reference frames work. The wing is moving relative to the air (and vice versa). However, only in the reference frame of the wing are all of the time derivatives equal to zero (steady flow). When a car drives by you, there is initially no breeze but after it passes there is a breeze in the direction that the car was travelling. In your reference frame the velocity increased over time, so the flow is not steady in this frame. In the car's frame, however, the flow is steady because at a given distance away from the car the flow is always the same. Commented May 28, 2014 at 15:26