# What's the volume of water required inside the lock if

• davekardle
In summary, the conversation discusses the dimensions of a lock and the volume of water required inside the lock to prevent a ship from grounding. The question also introduces the possibility of the ship emptying its load (oil) into the lock and how that would affect the calculations. There is a debate about the details and assumptions needed to accurately solve the problem.
davekardle
What's the volume of water required inside the lock if...

The dimension of the lock is:40 long x 3.5 height x 6 width
allowing 0.5 height.

The ship weights 400T plus 300T(LOAD)
total= 700T

whats the amount of water used per lock allowing allowing 0.5 m from the height?

What if the ship empties its load on the lock ? the density of the load is 755kg/m^3.

2. Homework Equations

Archimedes principle
Mass of water displaced

3. The Attempt at a Solution

Mass of water displaced by the boat mass = 700T
VOLUME = 700M^3
Total volume inside the lock allowing 0.5

(3x40x6)m^3 - 700M^3

Is that correct ?

How about if the ship empties its load on the lock ?

You could just repeat the calculation without the load, but I'm not sure what is meant by "the volume of water required" Required to do what? To keep the lock full? To keep the ship from grounding?

haruspex said:
You could just repeat the calculation without the load, but I'm not sure what is meant by "the volume of water required" Required to do what? To keep the lock full? To keep the ship from grounding?

But how do I repeat the calculation?

the ship is 400T heavy without the oil

The amount of oil displaced by the **** is greater than the actual oil it emptied into the lock.
I don't know how to proceed

You didn't mention oil before, and when you wrote 'empties its load on the lock' I took that to mean merely that it unloaded whilst in the lock. Are you saying now the load is oil and it pours that into the lock, Exxon-style? If so, might need the density of the oil.
(Have you provided the question verbatim? If not, please do so. I get the feeling I'm missing some detail.)

haruspex said:
You didn't mention oil before, and when you wrote 'empties its load on the lock' I took that to mean merely that it unloaded whilst in the lock. Are you saying now the load is oil and it pours that into the lock, Exxon-style? If so, might need the density of the oil.
(Have you provided the question verbatim? If not, please do so. I get the feeling I'm missing some detail.)

Hey sorry, I forgot to mention some of the details yes.

The ship average displacement is 400T + 300t ( load = oil C10H22)
Oil density is 730 mg mL−1.

The ship suffers a hull fracture and empties its load (oil ) into the moving lock.

The lock dimension is: 40 long x 6 width x 3.5 height

Allowing 0.5m of the height for the water level.

So I worked out I only needed 20m^3 of water to prevent the ship grounding.

but I'm not sure how to proceed with the calculation after the load is emptied into the lock.

300T of the oil =~ 400m^3

Does the lock overflow ? what happens? I'm so lost.

If the fracture is below the water line, won't the water flow into replace lost oil, leading to the ship's foundering? If we ignore that, presumably it will stop flowing out when ... what do you think?

Thinking about this more, I don't think the question is well posed. Depends too much on details of the geometry of the ship and its cargo.
Model 1: simple hull, filled with oil from bottom up. Oil occupies ~400m3, ship+oil displaces 700m3, so 300m3 space above oil is below water line. If holed, water flows in, not oil out (until it sinks).
Model 2: Some combination of ballast and air compartments manages to lift the oil completely above the water line (somehow without tipping over). Holed in that elevated section, the oil can now flow out. But where will the oil/water boundary in the lock be in relation to the ship? Let Vw be the volume of the hull below the water line and Vo be the volume above water but below oil surface. Let ρw, ρo be the respective densities. So Vwρw+Voρo = 400T. I believe the volumes depend on hull geometry, but you could try assuming the hull is a simple box of horizontal area A and see if you can deduce anything.

haruspex said:
Thinking about this more, I don't think the question is well posed. Depends too much on details of the geometry of the ship and its cargo.
Model 1: simple hull, filled with oil from bottom up. Oil occupies ~400m3, ship+oil displaces 700m3, so 300m3 space above oil is below water line. If holed, water flows in, not oil out (until it sinks).
Model 2: Some combination of ballast and air compartments manages to lift the oil completely above the water line (somehow without tipping over). Holed in that elevated section, the oil can now flow out. But where will the oil/water boundary in the lock be in relation to the ship? Let Vw be the volume of the hull below the water line and Vo be the volume above water but below oil surface. Let ρw, ρo be the respective densities. So Vwρw+Voρo = 400T. I believe the volumes depend on hull geometry, but you could try assuming the hull is a simple box of horizontal area A and see if you can deduce anything.

Thanks for the input buddy. I think the second option is what my lecturers are really looking to see in our assumptions while doing this project. Assuming that the hull is a simple box where would the layer between oil and liquid lie ? I am trying to design a way to drain the lock up to a point where around 90% of the oil is recovered. Do you think that by removing the ship and then inserting a pipe to the lock lateral at a height where the oil/water boundary lies and then draining out all the oil layer is a good option for recovery?

davekardle said:
Assuming that the hull is a simple box where would the layer between oil and liquid lie ?
Create some variables for the knowns and unknowns, like horizontal cross-sectional area of the hull, depth of hull in water, depth of hull in oil, etc. and see what equations you can come up with.
I am trying to design a way to drain the lock up to a point where around 90% of the oil is recovered. Do you think that by removing the ship and then inserting a pipe to the lock lateral at a height where the oil/water boundary lies and then draining out all the oil layer is a good option for recovery?
Do you mean removing the ship physically? How will you do that without losing the oil?

## 1. What is the purpose of water in a lock?

The water in a lock is used to raise or lower boats to different water levels. This allows boats to pass through a canal or river that may have varying water depths.

## 2. How is the volume of water required inside a lock determined?

The volume of water required inside a lock is determined by the size of the lock, the size of the boats passing through, and the difference in water levels between the two sides of the lock.

## 3. Can the volume of water inside a lock vary?

Yes, the volume of water inside a lock can vary depending on the size and weight of the boats passing through. Additionally, if there is a large difference in water levels, more water may be required to raise or lower the boats.

## 4. Why is it important to have a specific volume of water inside a lock?

Having a specific volume of water inside a lock ensures that boats can safely pass through without hitting the lock walls or getting stuck. It also helps maintain water levels in the canal or river.

## 5. How is the volume of water inside a lock controlled?

The volume of water inside a lock is controlled by opening and closing gates at either end of the lock. These gates can be adjusted to allow more or less water to enter the lock, depending on the needs of the boats passing through.

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