Transom Drag

I thought the water falling from the wash is pushing the boat forwards??

 

I wouldn't say that it is really a hydraulic jump as such.

Attached is a short paper that shows experiments and predictions for transom
unwetting and the effects on resistance. The hulls are very simple parabolic
hulls with rectangular cross-sections, so they can be easily reproduced as input
for CFD and other codes.

Good luck!

 

Both are true.

The water pressure at the air-water interface (surface) is the same on both the air and the water side - and it is equal to the atmospheric pressure, regardless of the boat speed. So the interface will lower down towards the transom edge as the pressure at the transom edge approaches the atmospheric one.

On the other hand, at a given speed, the hull will set itself at such trim as to permit the water to leave the transom smoothly, tangent to the bottom surface. This is true for all regimes, wet or dry transom.
Without bothering you with the math (but I can, if you want), the tangency of the exiting flow lines to the buttock lines at the transom edge stems from the impossibility of the water to perform too-tight turns, as it would require strong pressure gradients (tending to infinity as the turn radius tightens to zero). But the available pressure gradient around the transom is actually rather mild, due to the fact that the minimum pressure is atmospheric and the maximum possible pressure is the total stagnation pressure (never actually achieved in the stern area).

So the water, being unable to turn upwards around the transom bottom corner, leaves the transom smoothly and lets some other fluid fill the above volume. That other fluid can be either recirculating water in case of a wet transom, or the air in case of a dry transom.

In the case of the hull examined with the CFD at very low speed, this is an interesting visualization of what happens behind the transom:

 

So what is the difference in water pressure of the turbulent water against the transom, and normal hydrostatic pressure on the transom ?

 

Yes. If you keep the transom draft constant, then the wider the transom, the smaller the transverse flow components.
On the other hand, the 2-D case has no transverse flow components.
Hence, transoms with high B/T ratio are more accurately rappresented by the 2-D model then narrow ones.

 

What is the "normal hydrostatic pressure"?

 

What the transom would experience at rest, at the same immersion. I expect it has to be less, but...... ?

 

When in motion, each point of the transom is subject to the sum of the hydrostatic pressure and the static pressure (pressure due to the fluid motion) acting at that point.

In the case of a simple step model (post #12) with assumed flat water surface, on the virtual forward side of the body the pressure is:

Pfwd = rho g z + Patm​

where Patm is the reference atmospheric pressure acting on the water surface, z is the draft of the point being considered.

At the transom:

Ptran = rho g z + Patm + Phydro​

The difference is then:

Ptran - Pfwd = Phydro​

In other words, the difference is only due to the pressure of the wake behind the transom.
Since the net result is a drag force, it means that the mean pressure Phydro acting on the transom is less than zero (it is the gauge pressure, so can be negative - a "suction").
See the picture attached to the post #48. The blue zone is a negative gauge pressure - hence suction area, hence drag.

As a general rule, whenever you see vortices like those over a body moving through a fluid - that is the area of suction.

This is valid only for very low speeds, where the draft z is assumed constant. When z starts to decrease due to transom ventilation, the things get complicated because wave elevations around the entire hull have to be taken in consideration.

 

I wonder if the air drag of trying to fill the void is greater than the possible wet transom that might or might not be creating more drag as the speed goes up?

 

As the speed goes up, if the air wasn't able to fill the volume behind the transom, the vortex system behind the wet transom would create a tremendous suction (and drag).
But it cannot, because the air will displace the water, keeping the pressure behind the transom slightly less than the atmospheric (because the air wake is also at a slightly negative pressure) and decreasing the drag component due to the transom.

Think of how the steps of stepped hulls are made. Without air channels, the vortex system behind the steps would create a tremendous drag. Instead, the air rushes in as the pressure behind the steps decrease, decreasing the drag due to steps. It is the same basic physical principle.

 

deep transom vs max wavemaking

The guys from Australia have posed so much info since yesterday that is is all hard to absorb and generates several questions -- but I went back into the thread in search of an earlier statement by MrEfficiency referenced in one of these posts -- and ran across the one quoted above --

Can someone read the above referenced post and explain it to me?? I have no argument with the idea that transom goes dry at around the time of max wavemaking -- but if hull stays displacement it would seem to me that wavemaking might continue to grow but at slower rate -- so the statement could imply when wavemaking growth begins to level out. And I guess it makes sence that wavemaking could or should be the major retardant to boat's progress at this point --

BUT -- the statement I don't understand is --- "the resistance would be much higher if it wasn't' for the deep transom in that mode" --- This seems to be telling us that max wavemaking resistance might be higher in a hull with canoe stern than in a "similar" hull with deep transom --- can this be so??

 

Back to the original question

1st part of Barry's post describes hydrostatic transom pressure and it's gradual loss as transom goes dry when hull speeds up -- This yields the loss of forward pressure which is usually referred to as "transom drag". My original question was per possibility of drag higher than hydrostatic being produced by transom of a hull moving below dry transom speed & a Van Oossnan reference as to as much as 2x the full hydrostatic pressure (apparently relating to vortex action in the violent wake at speeds below dry transom speed) -- Daiquiri's CFD analysis posts seem to support this idea --

As to Barry's question as to cause of transom going dry, I feel that transom goes dry due to inertia / momentum causing water to separate from hull at the transom edge but I guess that is just another way of saying what he states as the water's inability to fill the voad (wake hollow) -- which seems to me to have a surface exposed to ambient air pressure (I don't know what the internal water pressures might be in this area)

 

Daiquiri -- Great Responses

Daiquiri -- I think your responses are simply great and really appreciate the effort you put into helping to make these things understandable. We have all heard of suction at aft end and stories of sinking a canoe by towing it too fast. You have clearly shown that a partially wet transom, when underway can exhibit more drag than that which is associated with the total loss of hydrostatic pressure (forward) when hull goes dry, that this increase can be a major factor of low speed drag, and that it is due to suction action of vortex flow at partially wet transom.

I can understand how Bernouli effect can reduce pressure on hull bottom area behind max beam if buttock lines, and cross section provide convex curved surface to the hull in this area, but I never could understand how any suction can act against a transom in an area so close to ambient air -- is the speed of everything taking place allowing suction to perform it's effect on drag before it can dissipate??

 

with no offense intended, this question doesn't make much sense. Or at least I am unable to see the sense of it.

 

"in other words the difference is only due to the pressure of the wake behind the transom"

Agreed

"Since the net results is drag force, it means that the pressure can be negative, a suction (below Atmospheric)"

The pressure behind the transom cannot be less than atmospheric or the transom would merely ventilate to atmosphere.

"as a general rule, when you see vortices like those over a body moving through a fluid, that is an area of suction"

In your explanation, you said that suction is a negative gauge pressure, ie less than atmospheric.

I think the statement should be "as a general rule, when you see vortices, there will be an area of LOWER PRESSURE THAN THE SURROUNDING AREA

( and you do not even require a pressure differential to create vortices, even two opposing stream, moving parallel will create vortices)

You do not need to get to a negative pressure (suction) below atmospheric pressure to create vortices.

I have been looking at transom drag as what is the pressure acting on the transom say in psi and that this is planing hull with a sharp transition line.

At 0.00001 knots , the pressure is a maximum, and the drag is minimal. Say at 3 knots the pressure is declining but still positive in the direction of travel, the drag is increasing. As speed increases, pressure diminishes and when the transom ventilates, the hydrostatic pressure goes to zero and this is the point of maximum drag, from zero knots to transom ventilation speed in knots