# Drag in water - object in a tunnel/tube

• Maximusflash
In summary, the drag equation for a sphere traveling at 2m/s under water with a radius of 1m can be calculated using the formula: 0.5 x 1000 x 2² x 0.47 x π x 1² = 3kN. However, this equation has limitations and may not be applicable in certain situations such as when the sphere travels through a tunnel or pipe. In these cases, other factors such as friction and the size of the tunnel may affect the drag force. Additionally, for objects with a high Reynolds number, the use of lubrication theory may be more appropriate in calculating drag coefficients. Further research is needed to determine the most accurate equations for these scenarios.
Maximusflash
Hi everyone

I was reading about drag, and didn't quite find what I was looking for.
A general drag equation is : 0,5 x ρ x u² x Cd x A, where there are some drag coefficients that have been found during laboratory experiments.
Let's say a sphere with a radius of 1m is traveling at 2m/s under water, with buoyancy and gravity being equal.
0,5 x 1000 x 2² x 0,47 x π x 1² = 3kN
Not much, but at the same time speed is low.

But I guess that this equation has it's limitations: a free body of water with no restrictions?

What if this sphere were to travel trough a tunnel or pipe? A train going trough a tunnel at 200km/h there are a lot of factors of friction, and the size of the tunnel being a big part of it (as well as aerodynamic shape of the train).
If the underwater-tunnel is to small, even at slow speed, the sphere would be pushing the water in stead of going trough. But is it enough if the free area around the sphere is as big as the sphere, or are there other rule of thumbs here?

Maximusflash
It's easier to find what you are looking for when you know what's it called :-)

So, spending a lot of hours reading different reports and discussions online I did come to some understanding on the topic (or at least I think I did).
I did find one report very close to my question. It defined a factor β as dp/DT (particle width / pipe width), and an equation for fw (wall factor) based on either Newton or Stokes regime. (β<0,5 => Newton, β>0,5 Stokes).
With this they managed to come ut with a revised equation for fw = 1,092 / (1+A x Re^β).
But, as I have learned, this is only valid in Reynolds number 0,38<Re<310,7, witch is relatively low Reynolds number in the scales of what I am looking for.
Reynolds number for a particle moving through fluid is Re = ρ x Vp x Dp / μ
My original thought was to find an estimate of forces of drag for a small submarine going through a underwater tunnel. So ρ would be fluid density 998 kg/m³, Vp (particle velocity) of 1m/s, Dp (diameter particle) 2m and μ (fluid viscosity) of 0,001 kg/m-s. But this leads to a rather high Reynolds number.. 2 x 10^6. I've been trying to find an equation to use with large Re, but haven't really found any. Maybe because it's not that straight forward?
I have found different graphs showing a smaller Cd for bigger Re, but haven't really found this wall effect with high Re.

Anyone with any insight as to where I need to look?

Maximusflash said:
It's easier to find what you are looking for when you know what's it called :-)

So, spending a lot of hours reading different reports and discussions online I did come to some understanding on the topic (or at least I think I did).
I did find one report very close to my question. It defined a factor β as dp/DT (particle width / pipe width), and an equation for fw (wall factor) based on either Newton or Stokes regime. (β<0,5 => Newton, β>0,5 Stokes).
With this they managed to come ut with a revised equation for fw = 1,092 / (1+A x Re^β).
But, as I have learned, this is only valid in Reynolds number 0,38<Re<310,7, witch is relatively low Reynolds number in the scales of what I am looking for.
Reynolds number for a particle moving through fluid is Re = ρ x Vp x Dp / μ
My original thought was to find an estimate of forces of drag for a small submarine going through a underwater tunnel. So ρ would be fluid density 998 kg/m³, Vp (particle velocity) of 1m/s, Dp (diameter particle) 2m and μ (fluid viscosity) of 0,001 kg/m-s. But this leads to a rather high Reynolds number.. 2 x 10^6. I've been trying to find an equation to use with large Re, but haven't really found any. Maybe because it's not that straight forward?
I have found different graphs showing a smaller Cd for bigger Re, but haven't really found this wall effect with high Re.

Anyone with any insight as to where I need to look?
If that is the case, then drag on a sphere is not going to be a good approximation. You should be looking for drag coefficients on torpedo type shapes, which is definitely available on-line. You might also consider looking at a case where you have a small clearance between the object and the wall so that lubrication theory might be applicable in that limit.

Maximusflash
Yes, a round sphere wasn't initially the optimum design, and it would be better to have a more streamline design. But it seemed to be a lot of experiments and documentation around the sphere design. The Cd could be changed with the design, but it was more in the lines of the wall effect a was trying to look in to. The hypothetical passenger transport submarine would need to be in the venosity of 2m i diameter, and would be transported via an external pulley system through the tunnel. Meaning it would need some clearance, but "as small as would be needed", as underwater tunnel making isn't your everyday activity.
So a sphere-like object of diameter 2m, where as the tunnel would probably need to be at least 3m a cross (which would mean that water areal around the sphere is grater that the area of the sphere).
Seeing as this is way out of my theoretical background, I was wondering if there were a set of equations with their own limits so one could get as close to the answer using the right type of equations. The reports I have read do reference to a set of equations, but not in the Re numbers this kind of particle would be.
I am guessing (without knowing) that lubrication theory means a very small clarence, and not it the 0,5 x particle size range? Probably more in the "torpedo leaving torpedo-hole" kind of clarence?

Once you get down to small radial clearances you will approach the condition of having a piston in a cylinder . The fluid in the cylinder does not then pass over the piston (except in a small amount) but rather gets pushed ahead of the piston and out of the cylinder at some remote location . New fluid may or may not enter the cylinder behind the piston depending on the practical arrangement of the system .

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Well, I would start out with something simple like axial movement of a solid cylinder within a tube. If the cylinder were very long compared to its diameter, then most of the flow resistance would be in the annular region, and not at the leading and trailing edges. I would also look at shear flow between infinite parallel plates, where the Reynolds number would be defined in terms of the plate velocity and channel opening (rather than based on an object diameter). I would see if there was any information on turbulent flow for this situation.

BvU
So, Max, what is the purpose of the tunnel ?

The purpose of the tunnel is (hypothetical) to use it as a transportation system, either as an attraction or an alternative means of getting from A to B through a solid mass without having to leave the water.
It could be done in a single person submarine (sphere like form) og a multiple person submarine (more of a bullet shape). It would be pulled along the way by a pulley system, and the question is how much force is needed to move said submarine at a speed of (let's say..) 2m/s.
Well, we do have the drag equation 0,5 x ρ x u² x Cd x A, but this isn't open water. In a restricted area in a tube (tunnel) the distance from the sphere to the walls will give a increased drag force due to the wall effect. I have only managed to find equations for low Reynolds number (in laminar flow) and not for high Re (turbulent flow).
The case being that I am only trying to find "ballpark numbers". And I was hoping for an equation in relations with sphere width / tube width. The distance from the sphere to the walls will need to be in the lines of 0,5m to 1,0m at least to make rom for pulley system and so on.

Maximusflash said:
It would be pulled along the way by a pulley system
Still don't understand the need for a tunnel. If the pulley system is in place, what more do you need ? Unless this is about a transport system on dry land -- but then: why bring in the water ? A vacuum would be much more sensible !

The under-water-tunnel would be there to not need to convert the watercraft to a land craft and then back to a water craft. And in stead of use the one-board propulsion-system to drive it through the tunnel, the one-board system could power down to save energy and in stead be pulled trough the tunnel on a system that is permanently installed in the infrastructure (tunnel). Further I would assume a propeller of some sort on this sub would contribute to the drag force from the turbulence (?). Figure:

When going over or around is a unwanted situation.
So the drag force in the sphere will determine what is needed to pull this along.

There is plenty of information in the literature about the drag coefficient on a sphere at high Reynolds number without the surrounding walls present. This would give a lower bound to the actual drag.

Maximusflash
Yes Chestmiller, this was my initial thought as well. I was more curious as if the wall effect would x2, x10 or more to the drag force. But I have come to a understanding that this topic in high Re isn't straight forward. It's heavy physics well beyond my understanding of the subject.
So to be on the conservative side of thing, any gut feelings as to how much the drag would increase due to walls? In my head, as I have noted before, one should at least have a equivalent area around the sphere, so water could pass around it without functioning as a piston.

I would use the high Reynolds number result for a sphere without the wall effect, and then combine it with the wall correction factor for the low Re case. At least I would look at this.

## 1. What is drag in water?

Drag in water is the force that opposes the motion of an object as it moves through water. It is caused by the resistance of water molecules to the movement of the object.

## 2. How does the shape of an object affect drag in water?

The shape of an object can greatly affect the amount of drag it experiences in water. Objects with streamlined shapes, such as a fish or submarine, experience less drag than objects with flat or irregular shapes.

## 3. What factors affect the amount of drag an object experiences in water?

The amount of drag an object experiences in water is affected by several factors, including its size, shape, speed, and the density and viscosity of the water.

## 4. Can drag be reduced in water?

Yes, drag can be reduced by changing the shape of the object, decreasing its speed, or using special coatings or materials that reduce friction and turbulence in the water.

## 5. How is drag in water calculated?

The calculation of drag in water involves complex equations and varies depending on the specific properties and conditions of the object and the water. However, it is generally calculated using principles of fluid dynamics and mathematical models.

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