What actually is static pressure?

  • #1
adjurovich
53
11
I’ve been confused about the term static pressure for quite some time. Different sources use very different definitions. From the problems perspective, it’s usually some external pressure. For example we are having a pool with tiny hole on the bottom which makes water level decrease and it flows through this tiny whole with some velocity.
Bernoulli equation (for this example) would go like this:

ρgh = 1/2ρv²

We are getting this result because the static pressures are equal to atmospheric so we can eliminate these two from equation, and the pressure on some elevation h (which isn’t actually hydrostatic pressure) is bigger than zero while on the ground (where this tiny hole is), h=0 so it is equal to zero. We can simply ignore the dynamic pressure on elevation h because the water level decreases much much slower than it flows through this tiny hole.

So my conclusion is that static pressure is actually sum of all external pressures (atmospheric pressure, piston pressure, etc) that push on fluid and cause its motion? Please correct me if I’m wrong.

Also, what happens to the ACTUAL hydrostatic pressure since ρgh is technically potential energy of fluid rather than its hydrostatic pressure.
 
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  • #2
The quantity ##P_z = P+\rho g z ## is called the piezometric pressure. The quantity ## \rho gz## is the hydrostatic pressure. i.e. pressure from the weight the fluid above the datum specifically. ##P## is the static pressure; an initial condition for some point in a fluid. You aren't always going to be analyzing from atmosphere to atmosphere.
 
  • #3
I used an example with atmospheric pressure because it’s the simplest one. My question is very clear: is static pressure basically pressure by some external factor, like atmosphere? Or something else. Would this kind of pressure be present if some gas was hermetically sealed? By my knowledge of thermodynamics there must be a pressure as long as there is non-absolute zero temperature inside the container
 
  • #4
adjurovich said:
I used an example with atmospheric pressure because it’s the simplest one. My question is very clear: is static pressure basically pressure by some external factor, like atmosphere? Or something else. Would this kind of pressure be present if some gas was hermetically sealed? By my knowledge of thermodynamics there must be a pressure as long as there is non-absolute zero temperature inside the container
Yeah, I would say that sounds correct. It's a pressure acting on the system in question (some arbitrary collection of mass) caused by something external to that arbitrarily selected system.
 
  • #5
adjurovich said:
... So my conclusion is that static pressure is actually sum of all external pressures (atmospheric pressure, piston pressure, etc) that push on fluid and cause its motion? Please correct me if I’m wrong.
Welcome, @adjurovich ! :smile:

What about external heat that increases the kinetic energy of the molecules of that fluid?
Think of a pressure cooker.
 
  • #6
AFAIK static pressure is simply pressure in a static situation. It has the same definition as pressure.
 
  • #7
Static pressure is not an external factor. Within the bulk of an inviscid fluid, static pressure is the force per unit area the parcels of liquid exert on their neighboring parcels at the fictitious surface separating one parcel from another. At a boundary with the atmosphere, the static pressure is equal to the atmospheric pressure.
 
  • #8
Thermodynamically speaking, the cause of it should be the fact that air exerts some force on the surface of water? In order for water to not get compressed (like gases do when piston acts on them with some force) the water must have the same pressure as air? But if it’s the case, why then is the actual hydrostatic pressure not included in Bernoulli’s equation? The term ρgh is rather the “potential energy” of fluid. This is what confuses me the most. Let’s say we are talking about the total pressure of some cask with a small hole on the bottom, what happens to the real hydrostatic pressure in that case?
 
  • #9
adjurovich said:
Thermodynamically speaking, the cause of it should be the fact that air exerts some force on the surface of water? In order for water to not get compressed (like gases do when piston acts on them with some force) the water must have the same pressure as air? But if it’s the case, why then is the actual hydrostatic pressure not included in Bernoulli’s equation? The term ρgh is rather the “potential energy” of fluid. This is what confuses me the most. Let’s say we are talking about the total pressure of some cask with a small hole on the bottom, what happens to the real hydrostatic pressure in that case?
Bernoulli's ( valid for inviscid - steady flow ), states the Energy per unit Volume of a fluid along a streamline is constant.

$$\overbrace{P_1}^{\text{static pressure}} + \overbrace{ \rho g z_1}^{\text{hydrostatic pressure}} + \overbrace{\frac{1}{2}\rho V_1^2}^{ \text{kinetic/dynamic pressure}} = const. = P_2 + \rho g z_2 + \frac{1}{2}\rho V_2^2 $$

So what do you mean when you say its not included?
 
  • #10
adjurovich said:
.. This is what confuses me the most. Let’s say we are talking about the total pressure of some cask with a small hole on the bottom, what happens to the real hydrostatic pressure in that case?
For the small volume instantaneously contained within the small hole, the hydrostatic plus the atmospheric pressures push from above, while only the atmospheric pressure pushes from below.
Therefore, that small volume is motivated to move downwards.

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/14-1-fluids-density-and-pressure/

https://en.wikipedia.org/wiki/Torricelli's_law#Discharge_experiment,_coefficient_of_discharge

800px-BernoulliHydrodynamicaFig28.jpg



CNX_UPhysics_Figure_14_01_VarDensity.jpg
 
  • #11
erobz said:
Bernoulli's ( valid for inviscid - steady flow ), states the Energy per unit Volume of a fluid along a streamline is constant.

$$\overbrace{P_1}^{\text{static pressure}} + \overbrace{ \rho g z_1}^{\text{hydrostatic pressure}} + \overbrace{\frac{1}{2}\rho V_1^2}^{ \text{kinetic/dynamic pressure}} = const. = P_2 + \rho g z_2 + \frac{1}{2}\rho V_2^2 $$

So what do you mean when you say its not included?
The first question one would ask is “what causes hydrostatic pressure”? And the answer would be: the weight of the water. The thing I want to say is: hydrostatic pressure increases with depth. Which is quite intuitive. But if we derive Bernoulli’s equation from the law of conservation of energy, one can clearly see that the therm ρhz comes from potential energy. The potential energy (gravitational potential energy) must increase with height. Height and depth are much different so how can we consider ρgz hydrostatic pressure at all? It simply can’t be the same thing
 
  • #12
adjurovich said:
The first question one would ask is “what causes hydrostatic pressure”? And the answer would be: the weight of the water. The thing I want to say is: hydrostatic pressure increases with depth. Which is quite intuitive. But if we derive Bernoulli’s equation from the law of conservation of energy, one can clearly see that the therm ρhz comes from potential energy. The potential energy (gravitational potential energy) must increase with height. Height and depth are much different so how can we consider ρgz hydrostatic pressure at all? It simply can’t be the same thing
The datum for zero gravitational potential energy is arbitrary. It’s usually set to be zero at one of the end points on the streamline for computational convenience, but it could be anywhere on it.

The Static Pressure is an initial condition from the integration of the hydrostatic differential equation:

$$ \frac{dP}{dz} = - \rho g \implies P = -\rho g z + C $$

##C## is just an arbitrary boundary condition(or initial condition); Meaning to explicitly solve it we need to say what ##P## is at some particular ##z##.
 
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