FInding distance from top of diving bell to lake surface

• Moose100
In summary, the problem involves a cylindrical diving bell being lowered into a lake until the water within the bell rises to a certain height. To determine the distance from the top of the bell to the surface of the lake, the pressure of the water and air inside the bell must be considered. Using equations such as Boyle's Law and Pascal's Law, the pressure at the interface of air and water can be calculated, which in turn gives the desired distance.
Moose100

Homework Statement

A cylindrical diving bell with open bottom and closed top 12.0m high is lowered into a lake until water within the bell rises 8.0m from the bottom end. Determine the distance from the top of the bell to the surface of the lake.

Homework Equations

I actually solved this problem through trial and error. the fundamental issue I have is defining the ratio volumes and why does this work despite it seeming to make sense or does it?
4Va/12 =8Vw/12; (Va=Volume of air, Vw=Volume of Water.)<<<Why does this work? Because it makes more sense to have 2Vw=Va

The Attempt at a Solution

PaVa=PwVw
Plugging in "equivalency" in part 2;
Pa2Vw=PwVw
2Pa=Pw
2(100kpa)=Pw
Pw=200kpa

Pw=pGh
h=Pw/pG
h=200kpa/(1000kg/m3)(9.8ms-2)
h=20.6
Height from bell surface to top of lake
20.6 - 4-=16.6m

Schaums College physics problem 16.37

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A bit hard to follow without definitions of the variables (Pw? Whereabouts?).
Clearly the air is at 3 atmospheres of pressure.
Moose100 said:
why does this work despite it seeming to make sense
Wouldn't it be more surprising if it worked but didn't make sense?

haruspex said:
A bit hard to follow without definitions of the variables (Pw? Whereabouts?).
Clearly the air is at 3 atmospheres of pressure.

Wouldn't it be more surprising if it worked but didn't make sense?

Pw=pressure of water
Pa=pressure of air
p=density of water
Va=volume air
Vw=Volume water

How is it at 3 atm?

Moose100 said:
Pw=pressure of water
How is it at 3 atm?
Yes, but pressure of water where, exactly?
How has the air volume changed since leaving the surface?

Moose100 said:
Pw=pressure of water
Pa=pressure of air
p=density of water
Va=volume air
Vw=Volume water

How is it at 3 atm?
yes it makes NO sense is what i meant ton extent.

haruspex said:
Yes, but pressure of water where, exactly?
How has the air volume changed since leaving the surface?
In the tube.

The air volume has decreased

The tube is submeged open end down and acquires an 8m influx of water

Moose100 said:
In the tube.

The air volume has decreased
The water pressure will not be the same everywhere in the bell.
In what ratio has the air volume decreased?

A third?

The pressure changes with depth

Moose100 said:
A third?

The pressure changes with depth
Right. So what will that do to the air pressure? And how would you now define your Pw?

Vw=2/3 Pressure of water decreases by 1/3

Moose100 said:
Vw=2/3 Pressure of water decreases by 1/3
Eh? The pressure of what water where decreases by a third? And why should it decrease at all? Water is virtually incompressible, and if you could reduce its volume by a third by compression the pressure would be something to be found inside a star.
How do you normally calculate the pressure at a given depth in a liquid? Why should it be any different here?

right

pressure of liquid is pgh

So volume of air decreases by a third. Pw=pgh

PaVa/3=pgh(Vw)

Moose100 said:
So volume of air decreases by a third. Pw=pgh

PaVa/3=pgh(Vw)
Not by a third.
Not sure what your equation is saying, but it looks wrong to me. Why should it be true? What principle are you applying?

The one I applied was this 4Va/12 =8Vw/12; I don't know why it is right it may not be though.

The tube is 12m long the water goes up 8m leaving 4m of empty air at the top. I figured 4/12 is air and 8/12 is water?

LOL See I told u I was lost.

Pressure of Water (Pw)= pgh
So using Boyles Law

PwVw=PaVa

pgh(Vw)=PaVa?

Hmm we can't apply boyles law really b/c like you said we can't compress water like a gas?

I know that;

Ptotal=Patm+ pgh

Moose100 said:
Hmm we can't apply boyles law really b/c like you said we can't compress water like a gas?

I know that;

Ptotal=Patm+ pgh
That's better. But what would be a useful value for h in the present context, i.e. whereabouts would you like to know what the pressure in the water is?

The top of the water inside the bell to the surface of the ocean?

Oops! DOuble post!

Moose100 said:
Top most of the water in the bell to the top of the ocean?
Yes. And what can you say that pressure will be equal to?

haruspex said:
Yes. And what can you say that pressure will be equal to?

Pressure of water= pgh
We already know the atm pressure as given by the problem 1.00atm or 100kpa

Hmm still got to get total pressure

Ptotal= 1atm + 1000kg/m3)(9.8m/s-2)(h)

So that Ptotal is also equal to the air pressure in the tube at depth h?

Moose100 said:
Pressure of water= pgh
We already know the atm pressure as given by the problem 1.00atm or 100kpa

Hmm still got to get total pressure

Ptotal= 1atm + 1000kg/m3)(9.8m/s-2)(h)
Yes, but I mean what equation can you write equating this pressure to another?

Boyles law? As it relates to the changes in volume before and after submerging.

Moose100 said:
So that Ptotal is also equal to the air pressure in the tube at depth h?
Right. And how does that relate to atmospheric pressure (according to Boyle)?

haruspex said:
Right. And how does that relate to atmospheric pressure (according to Boyle)?
I edited above

Ok I was thinking on this earlier and had it but wasnt confident I figured it out with your help...It think

My mistake was using boyles law on water as if it was a gas. Theres a before and after volume for the air in the tube. One that we can find with Boyles law. But we can also find the after air pressure using Pascual(?)

since they are equal we solve for h to find the pressure (basically the interface of air and water.), and subtract that from the column of air to get the distance between top of tube and surface of lake.

Also I was focusing too much on the inside of the tube and missed that that interface sat at a certain height in the lake overall.

I solved it before with no idea about this. I guessed and missed all this thanks for the help!

One last thing. Is it corret to say that the pressures at that interface are the same?

Moose100 said:
Is it corret to say that the pressures at that interface are the same?
Yes. Action and reaction, equal and opposite.

Ok. So the air portion in the tube has a uniform pressure correct? While the fluid portion varies with depth?

This is why there is a partition. The Gas exerts the same pressure on the liquid as it does on the gas.

Moose100 said:
Ok. So the air portion in the tube has a uniform pressure correct? While the fluid portion varies with depth?

This is why there is a partition. The Gas exerts the same pressure on the liquid as it does on the gas.
Yes. Strictly speaking, the pressure also varies with depth in the air, but the density of air is so low that you can ignore that over a height of a few metres.

RIght because gravity does affect it but so much that it's negligible so. You can use pascual on gases so to speak but the affect or calulation would be super small..

I guess another way to look at it is that the gas has to be equal because it stops the water from moving(equal and opposite).

1. How is the distance from the top of a diving bell to the lake surface measured?

The distance from the top of a diving bell to the lake surface is typically measured using a depth gauge or a sonar device. The depth gauge measures the pressure exerted by the water on the diving bell, which can then be converted to distance. Sonar devices use sound waves to determine the distance between the diving bell and the lake surface.

2. What factors can affect the accuracy of the distance measurement?

The accuracy of the distance measurement can be affected by factors such as water currents, temperature, and visibility. These factors can cause the diving bell to move or affect the readings from the depth gauge or sonar device. It is important to take these factors into consideration when measuring the distance.

3. Can the distance from the diving bell to the lake surface change over time?

Yes, the distance from the diving bell to the lake surface can change over time. This can be due to changes in water level, tides, or other environmental factors. It is important to regularly monitor and update the distance measurement to ensure accuracy.

4. How is the distance from the diving bell to the lake surface used in scientific research?

The distance from the diving bell to the lake surface is an important measurement in scientific research, particularly in studies related to underwater habitats and ecosystems. It can also be used to monitor changes in water levels and to study the effects of human activities on the underwater environment.

5. Are there any safety precautions that should be taken when measuring the distance from the diving bell to the lake surface?

Yes, there are several safety precautions that should be taken when measuring the distance from the diving bell to the lake surface. These include ensuring proper training and equipment for the divers, monitoring weather and water conditions, and having a backup plan in case of emergency. It is important to prioritize safety when conducting any scientific research in a diving bell.