Not sure Exactly what Pressure I am Solving for in this Form

In summary: GMWhat is the difference between the two equations?The difference between the two equations is that the first equation does not include the effects of drag and elevation.
  • #1
G_Mallard
7
0
Hello all, I am posting this question here in this thread and not the homework thread since it is not a homework problem but something I have been working on myself.

Let us imagine a fluid flowing through a pipe, we can measure pipe diameter and fluid velocity. The density of the fluid is also known. So we can measure Q and Mf, or volumetric and mass flow respectively:
Q = Av
Mf = ρQ = ρAv
where A is crossectional area of the pipe, v is fluid velocity, and ρ is density.

Units of Q are in m3/s and units of Mf are in kg/s. We can now calculate for pressure, assuming no other forces on the system, we have:
P = F/A = ma/A = [(MfΔt)(Δv/Δt)] /A = [(ρAvΔt)(Δv/Δt)]/A = ρvΔν
Δt is some arbitrary chosen time step and the units above work out to kg/ms2 so I know I am calculating for pressure. My problem is that I am not sure exactly what pressure I am calculating. I don't think this is the pressure that the fluid is under i.e. relative to the outside pressure.

Any assistance in understanding exactly what pressure I am calculating by using mass flow would be greatly appreciated. Thanks
-GM
 
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  • #2
What is causing the fluid to accelerate, and what are the boundary conditions on pressure?
 
  • #3
The inspiration for the above work is the circulatory system. So the source of the acceleration would be the heart. Since this is just a theoretical problem, the boundary conditions on pressure can be whatever, but I suppose the heart can be approximated as a diaphragm pump.
-GM
 
  • #4
G_Mallard said:
The inspiration for the above work is the circulatory system. So the source of the acceleration would be the heart. Since this is just a theoretical problem, the boundary conditions on pressure can be whatever, but I suppose the heart can be approximated as a diaphragm pump.
-GM
So it's not really steady (constant velocity) flow in a pipe. That would have been helpful to know from the get-go.
 
  • #5
Chestermiller said:
So it's not really steady (constant velocity) flow in a pipe. That would have been helpful to know from the get-go.

Sorry I did not say this explicitly but there is a Δv in the above equation! I would hope velocity would not be constant otherwise pressure would be zero in the above formula. Looking into this myself, I'm thinking I might be hitting on something like the pressure that does work on a volume element of the fluid rather than the fluid pressure itself. Any insight into exactly what kind of pressure the formula solves for? Thanks.
-GM
 
  • #6
G_Mallard said:
Sorry I did not say this explicitly but there is a Δv in the above equation! I would hope velocity would not be constant otherwise pressure would be zero in the above formula. Looking into this myself, I'm thinking I might be hitting on something like the pressure that does work on a volume element of the fluid rather than the fluid pressure itself. Any insight into exactly what kind of pressure the formula solves for? Thanks.
-GM
Are you version of the Bernoulli equation (mechanical energy balance equation) that includes a contribution from viscous drag on the walls of the conduit?
 
  • #7
No, I don't want to assume energy loss due to drag.
-GM
 
  • #8
G_Mallard said:
No, I don't want to assume energy loss due to drag.
-GM
Why not? That's the main resistance to flow in the circulatory system?

Are you just interested in pressure changes within the circulatory system, or are you interested in the absolute pressure?
 
  • #9
Chestermiller said:
Why not? That's the main resistance to flow in the circulatory system?

Are you just interested in pressure changes within the circulatory system, or are you interested in the absolute pressure?

Does whether I consider it or not matter in the above pressure equation? Sorry, it's been a few years since I've studied any physics. I began looking at how I can get absolute pressure of blood (colloquially blood pressure) from blood velocity measurements. I then realized that how I was going about things is most likely not getting me to that pressure but never the less all of the units are in order and I'm calculating pascals.

My main question is what is the pressure I'm calculating? What is the interpretation? Right now it seems to me that it can be described as the pressure gradient felt across the blood in a sense. Is this the correct interpretation? Thanks.
-GM
 
  • #10
The equation you have written down is not correct, but almost correct. The equation should read ##dp+\rho v dv=0##, which integrates to:
$$\Delta \left(p+\rho \frac{v^2}{2}\right)=0$$ This is a simplified version of the Bernoulli equation, that omits pressure variations due to elevation, drag, and pumping. The equation tells us that, if the conduit diameter changes so that the average velocity changes, there are corresponding changes in the pressure. Note that this equation does not give the absolute pressure, but only changes in the absolute pressure. The pressure in this equation is called the "static pressure."
 
  • #11
Chestermiller said:
The equation you have written down is not correct, but almost correct. The equation should read ##dp+\rho v dv=0##, which integrates to:
$$\Delta \left(p+\rho \frac{v^2}{2}\right)=0$$ This is a simplified version of the Bernoulli equation, that omits pressure variations due to elevation, drag, and pumping. The equation tells us that, if the conduit diameter changes so that the average velocity changes, there are corresponding changes in the pressure. Note that this equation does not give the absolute pressure, but only changes in the absolute pressure. The pressure in this equation is called the "static pressure."

Thank you for this explanation! Since I am curious about calculating absolute pressure from non pressure measurements I am still going to pursue the answer to my question. Do you have any equations or concepts about pressure I should read up about that might be useful in realizing an equation that will give me blood pressure "indirectly" i.e. by measuring other things?
 
  • #12
G_Mallard said:
Thank you for this explanation! Since I am curious about calculating absolute pressure from non pressure measurements I am still going to pursue the answer to my question. Do you have any equations or concepts about pressure I should read up about that might be useful in realizing an equation that will give me blood pressure "indirectly" i.e. by measuring other things?
In my judgment, it can't be done.
 
  • #13
Chestermiller said:
In my judgment, it can't be done.

That's what I was afraid of. Any recommendations on readings as to why it can't be done?
 
  • #14
I feel it necessary to point out that this is not a proper equation for pressure except perhaps in some very specific situations.

G_Mallard said:
Units of Q are in m3/s and units of Mf are in kg/s. We can now calculate for pressure, assuming no other forces on the system, we have:
P = F/A = ma/A = [(MfΔt)(Δv/Δt)] /A = [(ρAvΔt)(Δv/Δt)]/A = ρvΔν

The first issue is your definition of acceleration. Typically, when looking at flow through a pipe, we use the Eulerian description of the flow field, i.e. the observer is fixed to the pipe's reference frame and "watches" fluid particles flow by. In that situation, the relevant representation of acceleration in one dimension is
[tex]\dfrac{Du}{Dt} = \dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x}.[/tex]
That's the 1D version of the material derivative. That would be one concept to explore here. Otherwise, calculating a single pressure based on flow is fairly difficult, since the flow variables are more directly related to pressure gradient rather than pressure itself. If you are interested in the inviscid case, check out Euler's equation, which is the inviscid version of the Navier-Stokes equations.
 

1. What is the purpose of solving for pressure in this form?

The purpose of solving for pressure in this form is to determine the amount of force being applied to a given area. Pressure is a key factor in many scientific calculations and plays a critical role in understanding various physical phenomena.

2. How do I know which pressure to solve for?

In order to determine which pressure to solve for, you must first understand the context and parameters of the problem you are trying to solve. It is important to carefully read and analyze the given information to identify which pressure is being referenced and needs to be solved for.

3. What are the common units for pressure in scientific calculations?

The most commonly used units for pressure in scientific calculations are pascals (Pa), atmospheres (atm), and millimeters of mercury (mmHg). Other units such as pounds per square inch (psi) and torr may also be used in specific applications.

4. Can pressure be negative in a scientific calculation?

Yes, pressure can be negative in a scientific calculation. Negative pressure typically indicates a decrease in force or an inward force acting on a system. It is important to pay attention to the sign of pressure in calculations as it can affect the accuracy of the solution.

5. How can I check if my pressure calculation is accurate?

To check the accuracy of a pressure calculation, you can compare your results to known values or equations for the given system. It is also important to double-check your units and make sure they are consistent throughout the calculation. If possible, it is always helpful to have a second person review your work for any errors or miscalculations.

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