# Is the value of the discharge coefficient depends on the depth of opening and the pre

Discussion in 'Boat Design' started by xichyu, Sep 26, 2016.

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### xichyuJunior Member

When a ship was damaged. A opening was made. I am sure that the value of the discharge coefficient(flooding water) depends on the size and shape of the opening. Is the value of the discharge coefficient depends on the depth of opening and the pressure of the air pocket of damaged compartment?

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### rwatsonSenior Member

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### xichyuJunior Member

The paper，Cd(discharge coefficient) is semi-empirical，which assumed to be 0.6. The value is not precise if air can not escape from the damaged compartment, so I need to take into account also the air pressure inside the tank. And I also want to know the factors that influence the value of discharge coefficient.

TANSL TANSL :
The pressure varies with the difference of the heights of the liquid inside and outside the tank. So, effectively, as the compartment is filled, or the draft of the ship changes for any circustance, pressure changes, and therefore the "Cv" varies. Perhaps you could use in your calculations mean value of Cv, half-sum of the initial and final values (final value = 0).
I do not know what you want to calculate but I'm sure it will be much more complicated than just determine the value of "Cv".

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### gonzoSenior Member

This calculator may help you. However, holes from groundings and collisions are usually irregular, so they are difficult to calculate exactly. You can approximate a size and shape.
http://www.lmnoeng.com/TankDischarge.php

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### xichyuJunior Member

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### fredrosseUSACE Steam

Discharge Coefficient

Two things are being mixed up here.

The discharge coefficient for a "sharp edged orifice" (think of a clean hole drilled through a relatively thin hull plate) will typically be close to 0.6 . If the hole has a rounded inlet form (think of a circular spear piercing a hull plate, causing a rounded form before the spear pierces the hull), then the discharge coefficient will approach high value, in the vicinity of 0.9 to 0.98 . In the example photographs, flow from the face of the sign to the backside would enjoy a discharge coefficient considerably greater than 0.6 and flow in the reverse direction would have a Cd well below 0.6

The hole in the ship will have a discharge coefficient somewhere between these values depending on the shape of the fluid entrance.

The discharge coefficient describes the ratio of actual flow to the flow that would exist with a "perfect orifice", having a maximum value discharge coefficient of unity.

The flow however is entirely a function of the fluid conditions (differential pressure, fluid density,, etc.) upstream and downstream of the orifice, as mentioned on previous posts. This differential pressure can vary considerably and is generally a transient phenomenon for the conditions of a hole in a sinking ship. For example, if the hole leaks into a sealed compartment where air cannot escape, then as sea flows into this compartment the air pressure will rise until the flow across the hole stops.

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### xichyuJunior Member

You are out of sight!

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### gonzoSenior Member

The calculator allows for the type of edge.

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### PetrosSenior Member

as I recall from engineering school, the coefficient was empirically determined on standardized openings. That means it would only be valid for similar openings, a round sharp edged hole vs. a round radius entry hole. Presumably they would have to be round holes to be valid.

There is no valid "calculation" you can do to approximate an irregular jagged hull breach, the fluid mechanics would be very complex and even a simulation would not be much better than a Scientific Wild-*** Guess (SWAG). fluid velocity, differential pressure, turbulence at the opening, Reynolds number... I can not see how even a simulation would be any better than a guess.

I suppose you can guesstimate what the coefficient would be for an irregular torn hole, though it could be off by quite a lot without some testing. It would be nothing more than a SWAG, without some kind of testing on a "typical" (if there is such a thing) random breach in a hull.

Good luck.

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### gonzoSenior Member

For a flooding calculation, assuming a smooth edge, gives the worst case scenario. Any irregular shapes, jagged edges and other obstructions will slow the water flow by causing more turbulence. It will give a conservative estimate, which is a good thing for calculating sinking time.

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### baeckmoHydrodynamics

To cope with the flow through an irregular opening, we use the concept of the "Hydraulic diameter":

Dh = (4 x A)/P;

where Dh is hydraulic diameter, A is actual opening area and P is the total wetted perimeter.

The flow characteristics of an irregular opening with hydraulic diameter Dh will be similar to a circular opening with the diameter Dh, which for the round hole is equal to its diameter. This means that you can substitute the irregular opening (with Dh) with a circular opening of the same Dh when calculating the flow.The influence of the remaining geometric features (rounded, sharp aso) do not change. Note that Reynolds number in all cases refer to the Dh.

That said, and with the use of conventional hydraulic calculations, the opening discharge coefficients are NOT the main sources of uncertainty. Instead it is the final pressure differential that is the critical unknown; ie at what depth will the opening be when inflow starts and stops?

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### TANSLSenior Member

Amazing.

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### fredrosseUSACE Steam

Hydraulic Diameter

".....To cope with the flow through an irregular opening, we use the concept of the "Hydraulic diameter":

Dh = (4 x A)/P;

where Dh is hydraulic diameter, A is actual opening area and P is the total wetted perimeter....."

While the concept of "Hydraulic diameter" is useful in many flow situations, the use of the equations can be far from truth if applied without several relevant caveats and boundary conditions.

For example, consider the attached figure, a circular hole (having a basic diameter of 1.0), and the same hole, with a narrow crack extending for a length of 2.0 The "Hydraulic diameter" calculation shows a "Hydraulic diameter" for the basic hole as 1.0, yet for the same hole, with crack, the "Hydraulic diameter" is only 0.44 Common sense tells that either flow arrangement is virtually the same, flow character is no different from the simple hole to the same hole with crack, yet the "Hydraulic diameter" calculation yields far different results.

While this is an extreme example, one can envision many flow configurations where the simple application of "Hydraulic diameter" equations are far from the real flow character. Unfortunately many references neglect to mention the limitations associated with the concept.

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### gonzoSenior Member

Does that method assume a fixed flow coefficient?

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### xichyuJunior Member

Discharge coefficient is bound up with flooding quantity of damaged compartment. And is there any other use about discharge coefficient?

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