Force applied on sphere by turbulent flow

[yrob]

TL;DR Summary

What is the force applied on an immobile small sphere by a uniformly flowing fluid with turbulent flow?

Hello everybody,

I am thinking of the following problem:

A sphere a radius r is in a much larger container of radius R.
In this container, a fluid continuously flows with turbulent conditions from bottom to top.
I would like to approximate the force pushing the sphere up.
Some calculations I made with F = 0.5 * Cd * rho * A * v**2 (Cd being the drag coefficient that I tried from 0.2 to 0.5, rho the fluid density, A the area of the pipe, and v the fluid vertical velocity) give me incredibly low values, below the gravity force even with enormous flows. Therefore, I believe what I calculated is only the resisting force for the sphere to move in the fluid, and not the force of the fluid on the sphere.

Would you have a hint on that problem?

Thank you.

 

[BvU]

https://en.wikipedia.org/wiki/Drag_(physics)#The_drag_equation

 

[erobz]

TL;DR Summary: What is the force applied on an immobile small sphere by a uniformly flowing fluid with turbulent flow?

Hello everybody,

I am thinking of the following problem:

A sphere a radius r is in a much larger container of radius R.
In this container, a fluid continuously flows with turbulent conditions from bottom to top.
I would like to approximate the force pushing the sphere up.
Some calculations I made with F = 0.5 * Cd * rho * A * v**2 (Cd being the drag coefficient that I tried from 0.2 to 0.5, rho the fluid density, A the area of the pipe, and v the fluid vertical velocity) give me incredibly low values, below the gravity force even with enormous flows. Therefore, I believe what I calculated is only the resisting force for the sphere to move in the fluid, and not the force of the fluid on the sphere.

Would you have a hint on that problem?

Thank you.

At the very least,
1) You have used the wrong area $A$ in your calculations. You should be using the projected area of the sphere.
2) is the sphere moving up, down, or stationary in the flow?

 

[Baluncore]

Would you have a hint on that problem?

Buoyancy may be important. Is the fluid a liquid or a gas ?
Are you boiling green peas ?

 

[berkeman]

2) is the sphere moving up, down, or stationary in the flow?

At the top of the OP, he does specify that the sphere is "immobile", so likely stationary in the flow. I guess in a practical experiment you would be using a strain gauge to measure the force of the sphere against the immobilizing strut to the side or similar.

 

[Haborix]

Alternatively, if the sphere is immobile due to force balance without being held fixed, then it seems like the force of fluid can be deduced from the sum of the forces, knowing gravity and buoyancy. Also, I don't know what "uniformly flowing fluid with turbulent flow means." Turbulence can only make this problem harder.

 

[erobz]

Alternatively, if the sphere is immobile due to force balance without being held fixed, then it seems like the force of fluid can be deduced from the sum of the forces, knowing gravity and buoyancy. Also, I don't know what "uniformly flowing fluid with turbulent flow means." Turbulence can only make this problem harder.

In practical applications the velocity distribution across the pipe in turbulent flow is approximately uniform, as opposed to the parabolic distribution found in laminar flow.

 

[yrob]

Hello and thank you for your answers.

At the very least,
1) You have used the wrong area $A$ in your calculations. You should be using the projected area of the sphere.
2) is the sphere moving up, down, or stationary in the flow?

The area was wrong, indeed, but even with the area of the sphere cross section, the force is much lower than expected. In addition, I considered the sphere stationary because it is in reality a pack of spheres, all blocked by a grid in the pipe.

Buoyancy may be important. Is the fluid a liquid or a gas ?
Are you boiling green peas ?

The buoyancy for sure should play some role, but for the sake of the example, I set the density of the sphere equal to the density of the fluid.

Alternatively, if the sphere is immobile due to force balance without being held fixed, then it seems like the force of fluid can be deduced from the sum of the forces, knowing gravity and buoyancy. Also, I don't know what "uniformly flowing fluid with turbulent flow means." Turbulence can only make this problem harder.

Sorry for the confusion. What I meant by "uniformly flowing" was that the velocity profile of the fluid (vx, vy, vz) is considered the same at all positions with vx=vy=0 and vz>0.

 

[erobz]

Hello and thank you for your answers.

The area was wrong, indeed, but even with the area of the sphere cross section, the force is much lower than expected. In addition, I considered the sphere stationary because it is in reality a pack of spheres, all blocked by a grid in the pipe.

Pack of spheres? Can you give us a complete diagram of the experimental setup and data?

 

[yrob]

Pack of spheres? Can you give us a complete diagram of the experimental setup and data?

Sure, here is it. One assumption I make, to make it a bit simpler, is that the "push" from the fluid is the same for each sphere and assume a "void fraction" to calculate the velocity (v_assumed = v_fluid / void_fraction). I also attached a dummy example I made, the parameters are:
- Density of the fluid: 2000 kg/m^3
- Density of the sphere: 1800 kg/m^3
- Mass flow rate: 1000 kg/s (huge)
- Radius of sphere: 5 cm
- Radius of pipe: 100 cm
- Void fraction = 1

The drag seems unbelievably low.

 

[erobz]

Sure, here is it. One assumption I make, to make it a bit simpler, is that the "push" from the fluid is the same for each sphere and assume a "void fraction" to calculate the velocity (v_assumed = v_fluid / void_fraction). I also attached a dummy example I made, the parameters are:
- Density of the fluid: 2000 kg/m^3
- Density of the sphere: 1800 kg/m^3
- Mass flow rate: 1000 kg/s (huge)
- Radius of sphere: 5 cm
- Radius of pipe: 100 cm
- Void fraction = 1

The drag seems unbelievably low.

Radius of each sphere is 5 cm? the spheres go 50 cm across the pipe? Is that red line above them holding them from accelerating upward?

 

[yrob]

Radius of each sphere is 5 cm? the spheres go 50 cm across the pipe? Is that red line above them holding them from accelerating upward?

The radius of each sphere is indeed in this example 5 cm.
And the red line can be thought of a grid which prevents spheres to go further.

 

[erobz]

Ok, so the red line is applying a force $N$ to the spheres in the same direction as their weight. Can you write a force balance for the spheres so we can see what we can figure out, or try to determine what information is necessary to (perhaps) fill in the knowledge gap ( i.e. $N$)?

 

[yrob]

Thank you for your patience with me.

I see that there is a lot of confusion with this problem. Let me disclose some more information.
These spheres are packed naturally with buoyancy. However, now, the flow in the pipe is turned on (let's say, with a pump). The packing should be tighter. I am trying to apply an extra force on each sphere to make that happen. In practice they are not immobile, since I need them to have a tighter packing, but since the displacement is probably very slight, and that the velocity will be much lower than the velocity of the fluid, I was assuming fixed spheres.

 

[erobz]

Thank you for your patience with me.

I see that there is a lot of confusion with this problem. Let me disclose some more information.
These spheres are packed naturally with buoyancy. However, now, the flow in the pipe is turned on (let's say, with a pump). The packing should be tighter. I am trying to apply an extra force on each sphere to make that happen. In practice they are not immobile, since I need them to have a tighter packing, but since the displacement is probably very slight, and that the velocity will be much lower than the velocity of the fluid, I was assuming fixed spheres.

Thats not the problem I am having though. The spheres without the grid above them would accelerate upwards in the flow. They are precisely NOT accelerating upwards in the flow because the grid (which is attached to the walls of the pipe) is applying a force to them...

The external force balance on the body ( collection of spheres) is:

$$ F_D + F_b - N - W = 0 $$

As I hope you can see, we have 2 unknowns; the restraining force $N$, and the drag force $F_D$. We need more information. ( i.e. the pressure on both sides, and assumed velocity distributions on both sides). With that we can apply "The Momentum Equation" in fluid mechanics for the control volume to determine $N$ (in a round about way).

 

[yrob]

I see now what you mean. We could consider a "perfect" grid which would not induce any pressure loss. Therefore, we could assume a v_A ~= v_B and P_A ~= P_B.

However, what would probably better to take a step back here. Since the solution I am looking for is (obviously) very simplified, maybe a better representation would be the following: a sphere that can move is pushed, unrestrained, upwards, by a flow of constant velocity everywhere. The forces I am expecting to consider for the sphere are probably the gravity, buoyancy, the push from the fluid, and the friction resisting the motion. Would that make it simpler?

 

[erobz]

I see now what you mean. We could consider a "perfect" grid which would not induce any pressure loss. Therefore, we could assume a v_A ~= v_B and P_A ~= P_B.

You could consider the grid to induce negligible pressure loss, but the spheres are responsible for the eddy currents(pressure loss) and velocity distribution downstream. Neglecting everything leaves you with nothing to figure out.

However, what would probably better to take a step back here. Since the solution I am looking for is (obviously) very simplified, maybe a better representation would be the following: a sphere that can move is pushed, unrestrained, upwards, by a flow of constant velocity everywhere. The forces I am expecting to consider for the sphere are probably the gravity, buoyancy, the push from the fluid, and the friction resisting the motion. Would that make it simpler?

If a sphere(s) are unrestrained and are not accelerating in the flow, then the force of drag can be found immediately.

 

[yrob]

In that simplified moving case then, the chart with dummy values I sent would be correct? I am highly surprised that 1 ton per second flowing in this pipe would not be enough to fight gravity (if we do not account for buoyancy, with equal fluid and sphere densities).

 

[erobz]

In that simplified moving case then, the chart with dummy values I sent would be correct? I am highly surprised that 1 ton per second flowing in this pipe would not be enough to fight gravity (if we do not account for buoyancy, with equal fluid and sphere densities).

Water floats in water. There will be no drag force necessary if the spheres are the same density as the water.

 

[yrob]

Maybe I am misunderstanding your answer, I apologize.
Let us say that now the sphere is slightly denser than the fluid.
The push that I am calculating (from the drag formula above) is almost negligible, and would need several tons of fluid flowing every single second just to fight gravity. I feel like there is something missing.

 

[erobz]

Maybe I am misunderstanding your answer, I apologize.
Let us say that now the sphere is slightly denser than the fluid.
The push that I am calculating (from the drag formula above) is almost negligible, and would need several tons of fluid flowing every single second just to fight gravity. I feel like there is something missing.

You are blundering computationally if you find anything other than 0 flow = 0 drag force, for spheres of water immersed in...water.

 

[erobz]

imagine a sphere of not water, but with the same density as water in the unconstrained case. Write the force balance:

$$F_D + \overbrace{-W + F_b}^{ =0} = 0 $$

In the case of equal density sphere and water the buoyant force and the weight force add to zero.

and you are left with $F_D = 0$. There is no possibility of a unbalanced drag force in the unconstrained case for non accelerating sphere of water density.

In the case where there is restraint and neutral buoyancy you get the same thing but with $N$:

$$ F_D + \overbrace{-W + F_b}^{ =0} -N = 0 $$

$$ \implies F_D = N $$

Again, not enough to determine either variable as is.

 

[yrob]

You are blundering computationally if you find anything other than 0 flow = 0 drag force, for spheres of water immersed in...water.

I agree with you, but I was trying to simplify the example, and did not say 0 flow.
We start from a sphere immersed in a pipe filled with a fluid (we could assume we insert it at z=0), then turn on a pump which will make the fluid flow vertically. If the formula I am using is correct and using a slightly denser sphere than the fluid (1900 vs 2000 kg/m^3), the amount of fluid I need to fight gravity is 2.3 tons per seconds, 8280 tons per hour, this cannot be correct, right?

 

[erobz]

I agree with you, but I was trying to simplify the example, and did not say 0 flow.
We start from a sphere immersed in a pipe filled with a fluid (we could assume we insert it at z=0), then turn on a pump which will make the fluid flow vertically. If the formula I am using is correct and using a slightly denser sphere than the fluid (1900 vs 2000 kg/m^3), the amount of fluid I need to fight gravity is 2.3 tons per seconds, 8280 tons per hour, this cannot be correct, right?

Is the sphere constrained of not? If it is not, is the sphere accelerating? If it is not accelerating is it moving at some non-zero constant velocity or is it completely stationary w.r.t the inertial frame? Share the derived equation of motion, computations, assumptions state clearly! I'm not a mind reader.

 

[yrob]

I see now what you mean. We could consider a "perfect" grid which would not induce any pressure loss. Therefore, we could assume a v_A ~= v_B and P_A ~= P_B.

However, what would probably better to take a step back here. Since the solution I am looking for is (obviously) very simplified, maybe a better representation would be the following: a sphere that can move is pushed, unrestrained, upwards, by a flow of constant velocity everywhere. The forces I am expecting to consider for the sphere are probably the gravity, buoyancy, the push from the fluid, and the friction resisting the motion. Would that make it simpler?

I was referring to this "step back" we took.

 

[erobz]

I see now what you mean. We could consider a "perfect" grid which would not induce any pressure loss. Therefore, we could assume a v_A ~= v_B and P_A ~= P_B.

However, what would probably better to take a step back here. Since the solution I am looking for is (obviously) very simplified, maybe a better representation would be the following: a sphere that can move is pushed, unrestrained, upwards, by a flow of constant velocity everywhere. The forces I am expecting to consider for the sphere are probably the gravity, buoyancy, the push from the fluid, and the friction resisting the motion. Would that make it simpler?

The friction is not "resisting" motion, if anything it is providing it. Ideally if the sphere is at rest in the inertial frame than we just have:

$$ \uparrow^+ \sum F = F_d + F_b -W = 0 $$

$F_b$ and $W$ you know. There are still two unknowns in $F_d$(drag force), $C_d$ and $v$.

If you are doing the experiment and controlling/measuring the flowrate, you are controlling $v$ such that the acceleration of the sphere is zero. Then...you may compute $C_d$. If you are asking what $v$ is without experimentation, then you either need $C_d$ (typically a function of the velocity) or other information about pressure and velocity distribution before and after the "obstruction" (sphere) with further analysis.

So show me exactly how you calculated both $C_d$ and $v$?

 

[erobz]

For a single sphere the cumulative data was compiled and a correlation was proposed by Clift and Gauvin for the drag coefficient which is reported to deviate from the data by -4% to 6% for $\text{Re} < 3 \cdot 10^5$ according to [1]

$$C_d (\text{Re}) = \frac{24}{\text{Re}}\left( 1 + 0.15~{\text{Re}}^{0.687} \right) + \frac{0.42}{1+ 4.25 \cdot 10^4 ~ {\text{Re}}^{-1.16}}$$

Where the Reynolds number is given by:

$\text{Re} = \frac{\rho_w v D}{\mu}$

Using some general variables and Newtons Second Law we can write the following:

$$ \frac{4}{3}Dg \left( \frac{\rho_s}{\rho_w}-1 \right) = C_d(\text{Re}) v^2 $$

$D$ sphere diameter
$\rho_s$ density of sphere
$\rho_w$ density of fluid (water)
$\mu$ viscosity of fluid(water)

For a single sphere you could use this relationship to solve for $v$ iteratively, but this says nothing about the drag coefficient for your group of packed spheres which have an entirely different coefficient function (likely unknown), so predicting your bodies drag coefficient without properly conducted experimentation is unlikely.Reference:
[1] Engineering Fluid Mechanics, Crowe, Elger, Williams, Roberson-9th Edition.

 

[erobz]

Just an FYI. I get $v \approx 4.3 ~\rm{ \frac{m}{s}}$ using the method above for a single steel sphere of diameter $10 ~\rm{cm}$. So, In a pipe that is $100 ~\rm{cm}$ in diameter ( that's a LARGE pipe ), it would require a mass flow rate of about $3.4 \cdot 10^3 ~\rm{\frac{kg}{s}}$.

I'd recommend decreasing the diameter of your pipe, and using smaller spheres!

 

[jrmichler]

it is in reality a pack of spheres, all blocked by a grid in the pipe.

- Density of the fluid: 2000 kg/m^3
- Density of the sphere: 1800 kg/m^3

First of all, those spheres have density lower than the fluid. They will float, and the fluid velocity needs to be downward to keep them down.

Second, that's some interesting fluid. Liquid metals have specific gravity higher than that. Ordinary liquids mostly have specific gravity less than 1.5. I have seen slurries with specific gravity as high as 2.2, but those had relatively high viscosity.

Third, the drag on a packed bed is not calculated from the drag on a single sphere. Good search terms are porous media flow, fluidization, and packed bed flow. You can also add pressure drop to those search terms. My old copy of Chemical Engineers' Handbook has a section on pressure drop in packed beds, but it mostly applies to distillation column packings. The fluidized bed people are mostly interested in the flow and pressure drop to lift a porous media, but those calculations can be used in your case. Since your spheres are restrained, just use their calculations to get pressure drop vs flow.