# Fluid dynamics and incompressible fluids

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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Chestermiller
Mentor
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.

sysprog and vanhees71
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).

Chestermiller
Mentor
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.

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?

Chestermiller
Mentor
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.

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?

sysprog
Chestermiller
Mentor
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?

sysprog
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 $$\rho, r, v_r and v_z$$?

dRic2
Gold Member
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 ?

sysprog
Chestermiller
Mentor
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?

sysprog
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?

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.

Chestermiller
Mentor
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?

Last edited:
sysprog
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?

Chestermiller
Mentor
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?

sysprog
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?

Chestermiller
Mentor
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)

dRic2 and sysprog
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.

Chestermiller
Mentor
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?

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.

Chestermiller
Mentor
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.

Ok great, thank you for all your help I appreciate it.
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.