Reynolds number calculation

  • Thread starter MonroeX
  • Start date
  • #1
5
0

Main Question or Discussion Point

Hello,

The particle Reynolds number makes me confused, and I hope someone can help me with this, please! Reynolds number is calculated by Re = ρfUD/μ where U is particle velocity, D can be particle diameter, and ρf and μ are density and viscosity of the continuous fluid. How I can Reynolds number if the particle velocity U is unknow? Is it correct to calculate first the particle terminal velocity Ut = (gD2pf))/(18μ) , (where ρp is the density of the particle), then to use the calculated Ut to calculate the Reynolds number? I need to calculate the particle's terminal velocity but stuck with the Re number.

Thank you in advance.
 

Answers and Replies

  • #2
boneh3ad
Science Advisor
Insights Author
Gold Member
3,071
741
I am not at all clear what you are trying to ask here. You want to calculate the Reynolds number but you don't have enough information to do so?

What is the overarching goal here? What is the application?
 
  • #3
5
0
I want to calculate the particle settling velocity. I don't have any velocities for Reynolds number. Reynolds number is needed in order to find whether we are in laminar (Rep < 20) or in turbulent flow (Rep > 100) to use the correct equation to find settling velocity.

UPD I want to calculate the required flowrate to lift the proppant after frac. The well is new, hasn't been flowing yet, so velocities are unknown.
 
Last edited:
  • #4
19,921
4,096
I want to calculate the particle settling velocity. I don't have any velocities for Reynolds number. Reynolds number is needed in order to find whether we are in laminar (Rep < 20) or in turbulent flow (Rep > 100) to use the correct equation to find settling velocity.

UPD I want to calculate the required flowrate to lift the proppant after frac. The well is new, hasn't been flowing yet, so velocities are unknown.
It looks like you are trying to determine the terminal velocity. The key to this is knowing that the drag coefficient is a function of the Reynolds number. You have graphs of this functionality. So, you are going to have to solve the non-linear equations involved for the velocity. The drag equation is $$\frac{4F}{\pi D^2}=\frac{1}{2}\rho v^2 C_D(Re)$$
This is solved in conjunction with the force balance on the sphere.
 
  • #5
boneh3ad
Science Advisor
Insights Author
Gold Member
3,071
741
This sounds to me like a simple situation where you have two equations and two unknowns and just need to solve the system.
 
  • #6
5
0
Chestermiller, boneh3ad, could you please explain how to find particle's velocity and the Reynolds number, I don't get how to obtain them if they are dependent from each other
 
  • #7
19,921
4,096
Let's neglect buoyancy temporarily, and do a force balance on the sphere when terminal velocity has been attained. Then you have that the drag force is equal to mg. So, $$\frac{4mg}{\pi D^2}=\frac{1}{2}\rho v^2 C_D$$where ##C_D## is a known (graphical) function of the Reynolds number. Multiplying this equation by ##\rho D^2/\mu^2## gives:$$\frac{8mg\rho}{\pi \mu^2}=C_DRe^2$$Both sides of this equation are dimensionless. So you make a graph of ##C_DRe^2## vs Re, and find the value of Re at which the left hand side of the above equation matches.
 
  • #8
5
0
Let's neglect buoyancy temporarily, and do a force balance on the sphere when terminal velocity has been attained. Then you have that the drag force is equal to mg. So, $$\frac{4mg}{\pi D^2}=\frac{1}{2}\rho v^2 C_D$$where ##C_D## is a known (graphical) function of the Reynolds number. Multiplying this equation by ##\rho D^2/\mu^2## gives:$$\frac{8mg\rho}{\pi \mu^2}=C_DRe^2$$Both sides of this equation are dimensionless. So you make a graph of ##C_DRe^2## vs Re, and find the value of Re at which the left hand side of the above equation matches.
My inputs are: density of particle is 2700 kg/m3, diameter of particle is 0.001 m, viscosity of liquid is 0.002 Pa⋅s, and the ##C_DRe^2=23.8383##.
How do I build the graph ##C_DRe^2## vs Re?

I found one method described in the book "Introduction to Practical Fluid Flow" by R.P. King (2002). Photos are attached. The example is on the last two pages. He is using similar method, ##C_DRe^2## vs ##C_D##, and this ##C_DRe^2##=[4/3⋅(ρs - ρf)⋅ρf⋅(g/μ2)]d3. Needed to find the second solution to confirm the result.
 

Attachments

  • #9
19,921
4,096
My inputs are: density of particle is 2700 kg/m3, diameter of particle is 0.001 m, viscosity of liquid is 0.002 Pa⋅s, and the ##C_DRe^2=23.8383##.
You can't know this yet because you don't know the terminal velocity.
How do I build the graph ##C_DRe^2## vs Re?
You use the first figure you presented. You pluck points off the graph, and for each combination of Re and Cd, you calculate ReCd^2. You then plot a graph (preferably with log-log scales).
I found one method described in the book "Introduction to Practical Fluid Flow" by R.P. King (2002). Photos are attached. The example is on the last two pages. He is using similar method, ##C_DRe^2## vs ##C_D##, and this ##C_DRe^2##=[4/3⋅(ρs - ρf)⋅ρf⋅(g/μ2)]d3. Needed to find the second solution to confirm the result.
This method is basically what I indicated. I don't know what you mean about "a second solution."

I've given you enough hints for now. Together with the reference you cited, it's now up to you to work this out.
 
  • #10
JBA
Science Advisor
Gold Member
1,540
459
It appears to me that the appropriate Reynolds number, and whether the well flow is laminar or turbulent, would be a function of the well fluid bulk properties and and tubing roughness factor not the properties of a particle in the fluid.
 
  • #11
19,921
4,096
It appears to me that the appropriate Reynolds number, and whether the well flow is laminar or turbulent, would be a function of the well fluid bulk properties and and tubing roughness factor not the properties of a particle in the fluid.
What well and what tubing? This is just a sphere falling though a liquid (of infinite extent).
 
  • #12
JBA
Science Advisor
Gold Member
1,540
459
UPD I want to calculate the required flowrate to lift the proppant after frac. The well is new, hasn't been flowing yet, so velocities are unknown.
The above post clearly indicates that the base issue is flushing an a newly fractured oil or gas well of excess formation proppants, generally sand, pumped into the the well formation fractures during a fracturing process. The well is flowed and well fluid is used to lift the proppants to the surface thru the well tubing for disposal or recovery following a well formation fracturing process. This is standard procedure in the completion process for a newly drilled well.

One other issue is that there is that there will generally be hundreds to thousands proppant particles in the fluid during this process so the composite density of the bulk fluid varies during this process.
 
  • #13
19,921
4,096
The above post clearly indicates that the base issue is flushing an a newly fractured oil or gas well of excess formation proppants, generally sand, pumped into the the well formation fractures during a fracturing process. The well is flowed and well fluid is used to lift the proppants to the surface thru the well tubing for disposal or recovery following a well formation fracturing process. This is standard procedure in the completion process for a newly drilled well.

One other issue is that there is that there will generally be hundreds to thousands proppant particles in the fluid during this process so the composite density of the bulk fluid varies during this process.
Ahh. I forgot about that post. Thanks for refreshing my memory.

Earlier he had me confused when I though he was referring to the placement of the proppant particles in the frac to keep it open (rather than the recovery of the proppant fluid).
 
  • #14
5
0
The above post clearly indicates that the base issue is flushing an a newly fractured oil or gas well of excess formation proppants, generally sand, pumped into the the well formation fractures during a fracturing process. The well is flowed and well fluid is used to lift the proppants to the surface thru the well tubing for disposal or recovery following a well formation fracturing process. This is standard procedure in the completion process for a newly drilled well.

One other issue is that there is that there will generally be hundreds to thousands proppant particles in the fluid during this process so the composite density of the bulk fluid varies during this process.
How the calculations change taking this into account?
 
  • #15
19,921
4,096
I think the point is that, in your analysis, you are inherently assuming that each particle is surrounded by an infinite ocean of fluid, and does not interact fluid dynamically with other particles. This is valid if the volume fraction of particles is very low (say, < 5%). If you have a more concentrated suspension, then there will be interactions of the fluid dynamics with adjacent particles. This will reduce the terminal velocity (i.e., the relative velocity of the particles with respect to the fluid).
 

Related Threads on Reynolds number calculation

  • Last Post
Replies
1
Views
2K
Replies
9
Views
4K
  • Last Post
Replies
14
Views
10K
  • Last Post
Replies
2
Views
2K
Replies
3
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
6
Views
9K
  • Last Post
Replies
8
Views
14K
Replies
1
Views
471
Top