Force on an object that is stationary in flowing water

[MrNewton]

Hello there,

I was wondering, and i can't seem to really find the answer to my question, i found something that looks like it, but I am not sure if i understand correctly, that why I am turning to PF.com

1) What is the basic formula of the force on a object in flowing water when we assume that the object is tied to a point that is upstream. So the rope between the object and his anchorpoint upstream is always tight.

2) What is the influence of the shape of the object and the depth (surface increased) of the object in the water on the formula.

I assume that their are a couple factors:
-The surface of the object that is in the water
-The drag of the object (aerodynamica)
-The speed of the moving water
-anything else?

The reason i ask:
I want to calculate roughly how much power is needed to hold a ship on the same location (relative to the shore) when its pointed upstream. (and I am not yet taking the efficiency of the prop in account, let's say the efficiency is 100% at this time)

 

[PeroK]

What a drag!

https://en.wikipedia.org/wiki/Drag_coefficient

 

[MrNewton]

Thanks you! That was the word i was searching for. I adjusted it in the post

 

[MrNewton]

Okay, so i found this

Where A is the surface
Cw is dragcoefficient
p is density of the liquid
v is the speed

So with this formula i can calculate the friction between the water and the object in Newton.
But now i want to know how to calculate the power that is needed to maintain the objects position without the anchorpoint to the shore. I assume that the weight becomes a great factor now?

 

[PeroK]

Power in this case should be $Fv$. The mass of the object is not relevant (unless you want to accelerate it).

 

[kuruman]

I assume that the weight becomes a great factor now?

The weight becomes a factor only to keep it from sinking, not if it is neutrally buoyant.

 

[Nugatory]

But now i want to know how to calculate the power that is needed to maintain the objects position without the anchorpoint to the shore. I assume that the weight becomes a great factor now?

This is the same problem as pushing the object at a constant speed $v$ through still water, and the standard energy and power relationships are all you need: $W=Fd$, $d=vt$, $P=W/t$. The weight doesn’t show up in the calculation (unless the object is supported by hydrodynamic forces, in which case the problem is waaaaay more complicated).

 

[MrNewton]

Thanks for the replies.
But how is it possible that the weight of the object has no place in the formula?
If you have an object A with a very large weight and another object B with very little weight, but the same surface and shape as A, it will require less power to move object B then A right?

 

[MrNewton]

Power in this case should be $Fv$. The mass of the object is not relevant (unless you want to accelerate it).

Yes i do want to accelerate is :)

 

[PeroK]

Yes i do want to accelerate is :)

What is the required acceleration?

If you have an object A with a very large weight and another object B with very little weight, but the same surface and shape as A, it will require less power to move object B then A right?

What do you mean by "move"?

 

[MrNewton]

What is the required acceleration?What do you mean by "move"?

Before i answer those questions i think i have to change the question a little bit. As Nugatory said: "This is the same problem as pushing the object at a constant speed v through still water"
Now the ship has no propulsion. It is floating in the water with 0 speed. Now it wants to move up to 5 km/u and continue with that speed.

I think i understand the answers to the original questions, thanks!

What is the required acceleration?

From 0 to 5 km/u through the water. After that is has to keep going at 5km/h

What do you mean by "move"?

Ah, good question. I mean:
Move the object from a stationairy 0 km/h to a 5km/h and keep it there.
Initially the friction between the water and the ship is 0, but as the speed increased, the resistance between water and ship increases.

 

[PeroK]

Before i answer those questions i think i have to change the question a little bit. As Nugatory said: "This is the same problem as pushing the object at a constant speed v through still water"
Now the ship has no propulsion. It is floating in the water with 0 speed. Now it wants to move up to 5 km/u and continue with that speed.

I think i understand the answers to the original questions, thanks!
From 0 to 5 km/u through the water. After that is has to keep going at 5km/uAh, good question. I mean:
Move the object from a stationairy 0 km/u to a 5km/u and keep it there.
Initially the friction between the water and the ship is 0, but as the speed increased, the resistance between water and ship increases.

In English it's $km/h$ rather than $km/u$.

If you have the power to sustain $5 km/h$, then you have the power to accelerate to that speed, for all practical purposes. That speed will effectively be your terminal velocity through the water.

 

[Arjan82]

The general formula for the drag is a bit inaccurate for a ship. It only concerns the part of the resistance that scales with the dynamic pressure.

There are normally three components in ship resistance:

Frictional resistance.
Just the friciton of the water alongside the hull. Normally you can compute this with a surface which has the length of the ship, and a width that is chosen such that the surface area is equal to that of the submerged surface area of the ship. You then need a frictional line to relate the surface area, ship velocity, viscosity and density of the water (i.e. the Reynolds number) to an actual resistance, see here for example.

Viscosity induced pressure resistance.
This has to do with pressure recovery at the aft body of the ship. At the fore body a high pressure is generated because of the water hitting the bow. Then the pressure is lowered as the flow is accelerated alongside the ship hull. Then, it should be the case that the aft-body of the ship 'recovers' this pressure by increasing it again. This can be done by shaping the aft body gently such that the flow is slowly decelerated again without any separation. For potential flow theory the pressure recovery is always a 100%, but due to the formation of a boundary layer (a viscous effect) this is not true for real flows. This is why it is called viscosity induced.
Normally this is taken into account by a 'form factor' This is a number higher than 1, which is the increase in frictional resistance due to this viscosity induced pressure resistance. For a well designed ship this is in the order of 1.2 (so this resistance is about 20% of the frictional resistance). But actually this is very dependent on the quality of the hull shape.

Wave resistance.
This is the resistance that is due to the fact that a ship makes waves. Generating a wave costs energy, which results in extra drag. There is no easy formula for the wave resistance but it depends on the Froude number. A Froude number below 0.5 means you have a 'displacement ship' for which frictional resistance is often dominant, between 0.5 to 1.0 is a semi-planing ship for which wave making resistance becomes most important and Froude > 1.0 means the ship hull is lifted from the water. Wave making is very important here as well. For displacement ships you can use this formula:

as explained here. This is a semi-emperical formula based on many tests and statistics. It is only valid for certain types of ships.

The weight of the ship is an important parameter for the resistance of the ship, but only indirectly. The weight must of course be compensated with the buoyancy. The buoyance is directly related to the volume of water that is displaced. That on its turn is related to the surface area and thus to friction. But a bigger displaced volume also means a higher wave resistance.