Fluid dynamics and incompressible fluids

In summary: That the term in parentheses, which is negative, is the volume decrease due to the flow. Also, that the term in parentheses, which is positive, is the volume increase due to the flow.
  • #1
dunk
13
1
Homework Statement
A drop of an incompressible liquid of constant density
Relevant Equations
Navier-Stokes equation, continuity equation
Hi sorry about the way I've posted I'm new to this site. Anyway basically I've been set this question which should be attached to this post, I have attempted to do this question but I'm having trouble in forming an equation in the first place. I'm unsure where to start, I understand I need to use the equations mentioned I'm just stuck on how to form them. Any help will be very much appreciated.
 

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  • #2
At this site, you must make some attempt at solving the problem before we can help you. Please bear that in mind for the future.

For now, what is the equation of continuity for an incompressible fluid in axisymmetric flow, expressed in cylindrical coordinates? Please write the equation down.
 
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  • #3
Chestermiller said:
At this site, you must make some attempt at solving the problem before we can help you. Please bear that in mind for the future.

For now, what is the equation of continuity for an incompressible fluid in axisymmetric flow, expressed in cylindrical coordinates? Please write the equation down.
Ok I'll make sure to do that next time I post something on here.

I think it's this differential equation equation: δρ/δr+1/r(δ/δr(rρvr)+1/r(δ/δθ(ρvθ)+δ/δz(ρvz).
 
  • #4
dunk said:
Ok I'll make sure to do that next time I post something on here.

I think it's this differential equation equation: δρ/δr+1/r(δ/δr(rρvr)+1/r(δ/δθ(ρvθ)+δ/δz(ρvz).
I don't see an equal sign,, so this is not an equation. Also, if the fluid is incompressible, what can you say about ##\rho##? Also, if the system is axisymmetric, what can you say about the derivative with respect to z? Also, please read the PF tutorial on LaTex, and then use LaTex to write your equations and expressions.
 
  • #5
Chestermiller said:
I don't see an equal sign,, so this is not an equation. Also, if the fluid is incompressible, what can you say about ##\rho##? Also, if the system is axisymmetric, what can you say about the derivative with respect to z? Also, please read the PF tutorial on LaTex, and then use LaTex to write your equations and expressions.
Oh my bad this is the equation I believe: \frac{\partial\rho}{\partial t}+\frac{1}{r}(\frac{\partial}{\partial r}(r\rho v_r))+\frac{1}{r}(\frac{\partial}{\partial \theta}(\rho v_\theta))+\frac{\partial}{\partial z}(\rho v_z)=0 . For an incompressible fluid the density is constant and that the divergence of u is 0. For the derivative with respect to z, I'm not sure I think it equals 0? I thought that the derivative with respect to theta equals 0 since its a axisymmetric flow?
 
  • #6
dunk said:
Oh my bad this is the equation I believe: \frac{\partial\rho}{\partial t}+\frac{1}{r}(\frac{\partial}{\partial r}(r\rho v_r))+\frac{1}{r}(\frac{\partial}{\partial \theta}(\rho v_\theta))+\frac{\partial}{\partial z}(\rho v_z)=0 . For an incompressible fluid the density is constant and that the divergence of u is 0. For the derivative with respect to z, I'm not sure I think it equals 0? I thought that the derivative with respect to theta equals 0 since its a axisymmetric flow?
Sorry, I meant the derivative with respect to theta. Now, please write out the equation again for the case of constant density and zero derivative with respect to theta. And, again, please use LaTex.
 
  • #7
Oh ok, I assume it's a steady flow therefore the first term goes to 0. I think this is the equation: $$\frac{1}{r}(\frac{\partial}{\partial r}(r\rho v_r))+\frac{\partial}{\partial z}(\rho v_z)=0 $$
I'm using latex overleaf, I assume that works with this site?
 
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  • #8
dunk said:
Oh ok, I assume it's a steady flow therefore the first term goes to 0. I think this is the equation: $$\frac{1}{r}(\frac{\partial}{\partial r}(r\rho v_r))+\frac{\partial}{\partial z}(\rho v_z)=0 $$
I'm using latex overleaf, I assume that works with this site?
Excellent. But the density is constant, so it factors out. Please rewrite the equation without the density in there.

To answer the questions in your problem statement, you are going to have to apply dimensional analysis to your equation. Have you been learning about dimensional analysis in your studies/course?
 
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  • #9
Ah right yes that's true it does factor out: $$\rho(\frac{1}{r}(\frac{\partial}{\partial r}(r v_r))+\frac{\partial}{\partial z}( v_z))=0 $$
Yes, I have learned about it I just haven't been told that I can use it in this situation. Am I right in thinking I need to make a dimensional matrix for the terms $$\rho, r, v_r and v_z$$?
 
  • #10
The equation you wrote can be simplified by eliminating the density (since it is constant). Now just make an estimate of the magnitude of each term. You can do so by replacing the derivatives with finite differences in order to treat the various differentials ##dr## or ##dz## as real differences ##\Delta r##, ##\Delta z##.
Now remember that, since the fluid is incompressible, the total volume volume must be conserved.
Recall that ##V = \pi R^2 h##. This gives you the relation:
$$h \approx \frac 1 {R^2}$$
and (taking the differential):
$$d h \approx - \frac 1 {R^3} dR$$
thus
$$\Delta h \approx - \frac 1 {R^3} \Delta R$$
What can you conclude from that ?
 
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  • #11
dunk said:
Ah right yes that's true it does factor out: $$\rho(\frac{1}{r}(\frac{\partial}{\partial r}(r v_r))+\frac{\partial}{\partial z}( v_z))=0 $$
Yes, I have learned about it I just haven't been told that I can use it in this situation. Am I right in thinking I need to make a dimensional matrix for the terms $$\rho, r, v_r and v_z$$?
No. Have you not learned how to reduce an equation to dimensionless form?
 
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  • #12
Chestermiller said:
No. Have you not learned how to reduce an equation to dimensionless form?
I probably have, it's just I can't remember at the moment. I am I right in thinking you can reduce the equation to this: $$v_r \delta r+v_z \delta z=0 $$. And I believe this is dimensionless?
 
  • #13
dRic2 said:
The equation you wrote can be simplified by eliminating the density (since it is constant). Now just make an estimate of the magnitude of each term. You can do so by replacing the derivatives with finite differences in order to treat the various differentials ##dr## or ##dz## as real differences ##\Delta r##, ##\Delta z##.
Now remember that, since the fluid is incompressible, the total volume volume must be conserved.
Recall that ##V = \pi R^2 h##. This gives you the relation:
$$h \approx \frac 1 {R^2}$$
and (taking the differential):
$$d h \approx - \frac 1 {R^3} dR$$
thus
$$\Delta h \approx - \frac 1 {R^3} \Delta R$$
What can you conclude from that ?
I am right in thinking that we can right the equation like this: $$v_r \delta r+v_z \delta z=0 $$. From this and the volume, only vr depends on r meaning the derivative with respect to z doesn't affect r meaning you can consider it as negligible.
 
  • #14
dunk said:
I probably have, it's just I can't remember at the moment. I am I right in thinking you can reduce the equation to this: $$v_r \delta r+v_z \delta z=0 $$. And I believe this is dimensionless?
That's not how to do it, and isn't even correct. Here's what I'd like you to do. Start by making the following substitutions into the differential equation:
$$r=\bar{r}R(t)$$
$$z=\bar{z}H(t)$$
$$v_r=\bar{v_r}v_{r0}$$
$$v_z=\bar{v_z}v_{z0}$$
where the overbear parameters are dimensionless radius, dimensionless z coordinate, dimensionless radial velocity, and dimensionless axial velocity, and ##v_{r0}## and ##v_{z0}## are characteristic velocities in the radial- and axial directions, respectively. What do you get when you make these substitutions?
 
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  • #15
Chestermiller said:
That's not how to do it, and isn't even correct. Here's what I'd like you to do. Start by making the following substitutions into the differential equation:
$$r=\bar{r}R(t)$$
$$z=\bar{z}H(t)$$
$$v_r=\bar{v_r}v_{r0}$$
$$v_z=\bar{v_z}v_{z0}$$
where the overbear parameters are dimensionless radius, dimensionless z coordinate, dimensionless radial velocity, and dimensionless axial velocity, and ##v_{r0}## and ##v_{z0}## are characteristic velocities in the radial- and axial directions, respectively. What do you get when you make these substitutions?
Ok I did the substitution and I got this:
$$\frac{1}{\bar{r}R(t)}(\frac{\partial}{\partial r}(\bar{r}R(t)\bar{v_r}v_{r0}))+\frac{\partial}{\partial z}(\bar{v_z}v_{z0})=0$$

I am I right in thinking the $$\bar{r}R(t)$$ term cancels out with the other term? And also I wasn't sure if you could substitute in for z since its part of a partial differential?
 
  • #16
dunk said:
Ok I did the substitution and I got this:
$$\frac{1}{\bar{r}R(t)}(\frac{\partial}{\partial r}(\bar{r}R(t)\bar{v_r}v_{r0}))+\frac{\partial}{\partial z}(\bar{v_z}v_{z0})=0$$

I am I right in thinking the $$\bar{r}R(t)$$ term cancels out with the other term? And also I wasn't sure if you could substitute in for z since its part of a partial differential?
I've decided it would have been better to normalize the spatial coordinates in terms of the maximum radius and maximum height at time zero. Thus, with this change, we would have:
$$r=\bar{r}R(0)$$
$$z=\bar{z}H(0)$$

And the continuity equation would become:
$$\frac{1}{\bar{r}R^2(0)}(\frac{\partial}{\partial \bar{r}}(\bar{r}R(0)\bar{v_r}v_{r0}))+\frac{1}{H(0)}\frac{\partial}{\partial \bar{z}}(\bar{v_z}v_{z0})=0$$Since the characteristic velocities and dimensions are taken as constants, they can be factored out of the derivatives, to yield:$$\frac{v_{r0}}{R(0)}\frac{1}{\bar{r}}\frac{\partial}{\partial \bar{r}}(\bar{r}\bar{v_r})+\frac{v_{z0}}{H(0)}\frac{\partial}{\partial \bar{z}}(\bar{v_z}
)=0$$or, equivalently,
$$\frac{1}{\bar{r}}\frac{\partial}{\partial \bar{r}}(\bar{r}\bar{v_r})+\left[\frac{v_{z0}R_(0)}{v_{r0}H(0)}\right]\frac{\partial}{\partial \bar{z}}(\bar{v_z}
)=0$$

OK so far?
 
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  • #17
Yes I think that makes sense, so in a way your just relating the r and z terms ( and making them dimensionless) with the parameters within this problem?
 
  • #18
dunk said:
Yes I think that makes sense, so in a way your just relating the r and z terms ( and making them dimensionless) with the parameters within this problem?
You'll see how this works soon.

So far, we have specified the characteristic lengths in the radial and thickness directions, but we haven't said anything about the characteristic velocities. We can now deal with that by, with no loss of generality, setting the dimensionless group in brackets in the final equation of post #16 equal to unity: $$\frac{v_{z0}R_(0)}{v_{r0}H(0)}=1\tag{1}$$such that our dimensionless continuity equation now becomes simply:
$$\frac{1}{\bar{r}}\frac{\partial}{\partial \bar{r}}(\bar{r}\bar{v_r})+\frac{\partial}{\partial \bar{z}}(\bar{v_z}
)=0\tag{2}$$If you use Eqn. 1 to express the characteristic velocity in the thickness direction ##v_{z0}## in terms of the other characteristic parameters for the system, what do you get? What does this tell you about the magnitude of the characteristic velocity in the thickness direction in comparison with the magnitude of the characteristic velocity in the radial direction? (This provides the answer to part (a) of your problem)
 
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  • #19
Ok I think I understand it now, I rearranged to get this:
$$v_{z0}=\frac{H(0)}{R(0)}v_{r0}, v_{r0}=\frac{R(0)}{H(0)}v_{z0}$$
And since H(0)<<R(0), it means that the magnitude of the vz component is negligible compared with the vr term.
 
  • #20
dunk said:
Ok I think I understand it now, I rearranged to get this:
$$v_{z0}=\frac{H(0)}{R(0)}v_{r0}, v_{r0}=\frac{R(0)}{H(0)}v_{z0}$$
And since H(0)<<R(0), it means that the magnitude of the vz component is negligible compared with the vr term.
Correct. Ready to tackle part (b) using analogously methodology?
 
  • #21
Yes I think so, I have made an attempt on part b using a similar method we just used by using dimensionless variables into the Navier Stokes equation:
$$ \rho(\frac{\partial u}{\partial t} + u \cdot \nabla u)= \rho g- \nabla p+ \mu \nabla^2u$$
$$\hat{u}= \frac{u}{U}, \hat{x}=\frac{x}{L}, \hat{t}= \frac{t}{T}, \hat{p}= \frac{pL}{\mu U}$$
Rearranging and substituting them back into the Navier Stokes equation and knowing the Reynolds number, Froude number and frequency parameter:
$$F_r=\frac{U}{\sqrt{gL}}, \beta=\frac{L^2}{vT}, Re=\frac{UL}{v}, v=viscosity$$
The equation reduces to:
$$\beta \frac{\partial \hat{u}}{\partial \hat{t}}+Re\hat{u} \cdot \hat{\nabla} \hat{u}= -\hat{\nabla}\hat{p}+\hat{\nabla^2}\hat{u}+\frac{Re}{F_r^2}\frac{\hat{g}}{g}$$
Where ##\frac{\hat{g}}{g}## is a unit vector expressing the direction of the body force.
In order for the left hand side to be negligible Re<<1 and ##\beta<<1##, then by reverting back to dimensional variables we find:
$$-\nabla p+\mu \nabla^2u +\rho g=0$$
I think I've done this in the correct way, let me know what you think.
 
  • #22
dunk said:
Yes I think so, I have made an attempt on part b using a similar method we just used by using dimensionless variables into the Navier Stokes equation:
$$ \rho(\frac{\partial u}{\partial t} + u \cdot \nabla u)= \rho g- \nabla p+ \mu \nabla^2u$$
$$\hat{u}= \frac{u}{U}, \hat{x}=\frac{x}{L}, \hat{t}= \frac{t}{T}, \hat{p}= \frac{pL}{\mu U}$$
Rearranging and substituting them back into the Navier Stokes equation and knowing the Reynolds number, Froude number and frequency parameter:
$$F_r=\frac{U}{\sqrt{gL}}, \beta=\frac{L^2}{vT}, Re=\frac{UL}{v}, v=viscosity$$
The equation reduces to:
$$\beta \frac{\partial \hat{u}}{\partial \hat{t}}+Re\hat{u} \cdot \hat{\nabla} \hat{u}= -\hat{\nabla}\hat{p}+\hat{\nabla^2}\hat{u}+\frac{Re}{F_r^2}\frac{\hat{g}}{g}$$
Where ##\frac{\hat{g}}{g}## is a unit vector expressing the direction of the body force.
In order for the left hand side to be negligible Re<<1 and ##\beta<<1##, then by reverting back to dimensional variables we find:
$$-\nabla p+\mu \nabla^2u +\rho g=0$$
I think I've done this in the correct way, let me know what you think.
I think you did it correctly. I would have done it a little differently (by working with the equations in component form), but you definitely have the right approach.
 
  • #23
Ok great, thank you for all your help I appreciate it.
Chestermiller said:
I think you did it correctly. I would have done it a little differently (by working with the equations in component form), but you definitely have the right approach.
 

Related to Fluid dynamics and incompressible fluids

1. What is fluid dynamics?

Fluid dynamics is the study of the motion of fluids, including liquids and gases. It involves the analysis of how fluids flow and interact with their surroundings, and is an important field in understanding many natural and man-made phenomena.

2. What is an incompressible fluid?

An incompressible fluid is a fluid that has a constant density, meaning that its volume does not change when subjected to pressure. This is in contrast to compressible fluids, such as gases, which can change in volume when subjected to pressure.

3. What is the difference between laminar and turbulent flow?

Laminar flow is a smooth, orderly flow of fluid, where the layers of fluid move in parallel without mixing. Turbulent flow, on the other hand, is characterized by chaotic, irregular motion of fluid particles, with mixing and eddies occurring. The transition from laminar to turbulent flow is determined by the Reynolds number, which takes into account factors such as fluid velocity, density, and viscosity.

4. How is Bernoulli's principle related to fluid dynamics?

Bernoulli's principle states that as the speed of a fluid increases, its pressure decreases. This principle is important in fluid dynamics as it helps explain the behavior of fluids in motion, such as the lift force on an airplane wing or the flow of water through a pipe. It is also used in the design of many engineering applications, such as pumps and turbines.

5. What are some real-world applications of fluid dynamics?

Fluid dynamics has many practical applications in various fields, including engineering, meteorology, and oceanography. It is used to design and optimize the performance of aircraft, ships, and cars. It also plays a crucial role in understanding weather patterns and climate change. In addition, fluid dynamics is used in the design of water and sewage systems, as well as in the development of medical devices such as blood flow meters and respiratory equipment.

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