Flow rate needed to lift object inside a vertical pipe

[alexlsu]

TL;DR Summary

How much flow rate is needed to either keep an object suspended in a large ID pipe section or to lift the object into the next section?

I have an object being forced up a vertical pipe with a slightly larger ID than the object by water. Then, the object reaches an area with a much larger ID than an object. How much flow fluid do I need to keep the object moving through the larger ID section and reach back into a small ID section? This is simplified for calculation. There are tapered areas to help guide the object into the next section. I need to know if the object will stall in the larger ID section due to fluid bypassing around it.

Smaller ID: 1.049 inches (0.027 m)
Larger ID: 1.800 inches (0.046 m)
Length of section: 24 inches (0.610 m)
Object OD: 0.955 inches (0.024 m)
Object Weight: 3 lbs (13.34 Newtons)
Use standard water as fluid: 1.0 SG
Flow rate available: up to 43 GPM (0.00276 m3/s)
Pressure is not an issue.

I tried using force calculations of the moving water against the object, and the results didn't look correct. For example, the object is 3 lbs (13.3 Newtons), and the water flow force in the large section (ID=1.8") at 43 GPM is 2.32 Newtons. So, at the max flow rate, the fluid would not be able to lift the object. Even as the fluid is moving at 1.7 m/s.

We know from practice that the object will move with no issues in the 1.049" ID pipe as the clearance between the ID and OD is very small.
Any help is appreciated.

 

[jrmichler]

I once designed a rotameter (search the term) by using the drag equation (search the term). For calculating fluid velocity, I used the flow area around the object (tube area minus cross sectional area of object).

It was a learning experience. The object, a steel ball, whirled around and wore a groove in the tube wall. Do not be surprised if your object does the same, which could cause difficulty passing the contraction.

The Mechanical Engineering forum is a better fit for this thread, so I'm moving it there.

 

[berkeman]

Welcome to PF.

Does the object have to have that exact shape? Or could it have a higher drag section near the tip to keep it oriented better in the flow direction?

TL;DR Summary: How much flow rate is needed to either keep an object suspended in a large ID pipe section or to lift the object into the next section?

There are tapered areas to help guide the object into the next section.

Could you show that in your diagram?

 

[erobz]

To hold it still you have a few forces at work, there is a thrust from the change in flow momentum, a hydrostatic pressure differential across the shuttle, and a form drag acting on its sides. All these forces add to balance the weight of the shuttle at some mass flow rate.

We can find the thrust by examining the momentum efflux:

$$ T = \dot m v_2 - \dot m v_1 $$

By continuity:

$$\dot m = \rho A_1 v_1 = \rho A_2 v_2$$

Putting everything in terms of $\dot m$:

$$ T = \frac{ \dot m ^2}{\rho}\left( \frac{1}{A_2} - \frac{1}{A_1} \right) $$

$A_1$ is full pipe area
$A_2$ is the annulus area around the shuttle
$\dot m$ is the mass flowrate
$\rho$ is the flow density

If you want to stop there for an upper bound on the flowrate just equate that with the weight of the shuttle $W$:

$$ \frac{ \dot m ^2}{\rho}\left( \frac{1}{A_2} - \frac{1}{A_1} \right) = W $$

and solve for the mass flowrate.

If you want to keep trying to trim that back, we add in the hydrostatic pressure differential across the shuttle:

$$ F_{hydro} = \rho g L A_s $$

Where $L$ is the shuttle length
$A_s$ is the shuttle cross sectional area

Then if you still want to trim some more, the last Force to add ( drag ) is more involved, but the idea will be to convert viscous dissipative force work into a pressure loss across the shuttle. I'll try to work on that tomorrow.

$$ T + F_{hydr}+F_{loss} = W $$

 

[erobz]

Looking at this in the clear light of day, I'm no longer confident in that thrust force. I tried to treat it as though the shuttle were being hit by an impinging jet. It didn't then occur to me that the fluid at the top of the shuttle is going through the opposite acceleration, returning the average flow velocity to $v_1$; meaning we can expect a net zero force from momentum change. I suppose the whole thing is just the last two terms that balance the weight. Does that sit well with any other contributors? It may require some precision fluid dynamics (that accounts for velocity and pressure distributions) to describe it, maybe @Chestermiller would share some thoughts; my foundations are shaky here.