Question: pressure inside an object submerged in water

  • #1
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Main Question or Discussion Point

I'm doing a question relating to the pressure inside on object submerged in water. Here is the question:

A tube, height 1.2m, is submerged vertically in the ocean where the waters density is 10^3 kg/m^3. A diver initially holds the tube vertically directly on top of the water. He then dives to a certain depth where the water reaches a height 40cm, from the bottom of the tube. The tube is open at the bottom and closed at the top. Calculate the depth to which he dived.

The relevant equations would be P = ro*g*h where h is the depth to which he dived. But I'm unable to calculate the pressure inside the tube and relate it to the pressure outside.
 

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  • #2
Merlin3189
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How does pressure change with volume for a fixed quantity of air?
 
  • #3
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How does pressure change with volume for a fixed quantity of air?
If the volume increases pressure decreases and if volume decreases pressure increases?
 
  • #4
Merlin3189
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Yes. And you know a formula for that (Boyle's law)
You know how much the volume has changed and the starting pressure.
 
  • #5
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Yes. And you know a formula for that (Boyle's law)
You know how much the volume has changed and the starting pressure.
Sorry, I'm not sure what the starting pressure and change in volume is. Is the starting pressure the atmospheric pressure?
 
  • #6
Merlin3189
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Yes. He starts at the surface with air at atmospheric pressure and the full volume of the tube. As he dives the pressure increases and the volume decreases.
Since no one mentions the air pressure, I think you just assume standard atmospheric pressure. (Since the water is 1000kgm-3 they're using rough figures, so don't worry about precision.)
 
  • #7
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Okay so then what's the inital and final volumes? I don't know the dimensions of the tube, so is there any other way to calculate it? And how does this relate to the pressure outside?
 
  • #8
Merlin3189
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You only need the ratio of the volumes. It is a cylinder whose length is changing.

If the tube is open, then the pressure inside must be the same as the pressure outside. If it were lower, more water would come in: if it were higher, the air would push some water out.
Since we are looking at the trapped air, the pressure in question is at the surface of the water inside the tube.
 
  • #9
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Okay so then P1*V1 = P2*V2
Then P2 = P1*(V1/V2)
If you subsititue this into p= r*g*h and solve for h I get 1cm... That's not correct, what am I doing wrong?
 
  • #10
Merlin3189
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Well let's see this calculation and see what's going wrong.

Edit: Maybe do it in bits? Post any parts you can do:
How much has the volume changed?
How much has the pressure changed? What is the new pressure? What is the extra pressure due to the water?
What depth of water corresponds to that .

Edit2: Substituting one formula into another is efficient and good maths, but IMO is more prone to error than putting the numbers in and working out each formula separately.
 
Last edited:
  • #11
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Well let's see this calculation and see what's going wrong.

Edit: Maybe do it in bits? Post any parts you can do:
How much has the volume changed?
How much has the pressure changed? What is the new pressure? What is the extra pressure due to the water?
What depth of water corresponds to that .

Edit2: Substituting one formula into another is efficient and good maths, but IMO is more prone to error than putting the numbers in and working out each formula separately.
The ratio in volumes is h1/h2 right? Which is 1.2/0.4? So the new pressure should be 101.325*1.2/0.4? Wait what do you mean by extra pressure due to water??
 
  • #12
Merlin3189
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The ratio in volumes is h1/h2 right? Which is 1.2/0.4? So the new pressure should be 101.325*1.2/0.4?
which equals?
But no.( your answer would make you a lot deeper than 1cm), your 1.2 and 0.4 are not the volumes (or lengths) of the air in the tube. If the tube had an area of A then the original volume of air is 1.2 x A. The new volume of air is not 0.4 x A, so what is it?

BTW you might like to put some units, so that I can check those. 101.325 sounds like kN per sq metre, which is fine as long as you remember the kilo bit !?
Wait what do you mean by extra pressure due to water??
Ok, you start at atmospheric pressure and say you went down 10m the pressure would increase to about 2x atmospheric pressure. But only half of that is due to the weight of the water, the other half is the original atmospheric pressure on the surface of the water. For your rgh formula you are calculating the pressure provided by the water.
 
  • #13
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which equals?
But no.( your answer would make you a lot deeper than 1cm), your 1.2 and 0.4 are not the volumes (or lengths) of the air in the tube. If the tube had an area of A then the original volume of air is 1.2 x A. The new volume of air is not 0.4 x A, so what is it?

BTW you might like to put some units, so that I can check those. 101.325 sounds like kN per sq metre, which is fine as long as you remember the kilo bit !?

Ok, you start at atmospheric pressure and say you went down 10m the pressure would increase to about 2x atmospheric pressure. But only half of that is due to the weight of the water, the other half is the original atmospheric pressure on the surface of the water. For your rgh formula you are calculating the pressure provided by the water.
ohhh is the new volume (1.2m-0.4m)*A? so that means the ratio in volumes is 1.2/0.8m? Yep its 101.325kN/m^2. So that means then that the pressure provided by the water is 101.325*1.2/0.8? Is that right? So does that meant the total pressure is the original atmospheric pressure + the pressure provided by the water?
 
  • #14
Merlin3189
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Yes
 
  • #15
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Yes
So then we can substitute 101.325 + 101.325*1.2/0.8 kPa as the value for P in P = rgh and solve for h? I did this and the answer still isnt right.
 
  • #16
Merlin3189
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Can we start by getting some answers (and the calculations that give them?)
I think you said
Psubmerged =Patmospheric x 1.2/0.8
and that would equal 1.5 x atmospheric pressure (which you call 101.325 no units!)

Then you go a bit wrong. This has calculated the pressure of the air in the tube when it is submerged. That is equal to the water pressure around it at that depth.

So if you have an expression for the water pressure at that depth, you can indeed equate them.

You say, "we can substitute 101.325 + 101.325*1.2/0.8 kPa as the value for P in P = rgh" but I don't agree.

If "101.325*1.2/0.8 kPa" is the pressure of the air inside the tube when submerged,
then "101.325 + 101.325*1.2/0.8 kPa" is more than the pressure inside the tube.

"the value for P in P = rgh" is not the pressure outside the tube, nor the pressure inside the tube. What is it?
 
  • #17
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Can we start by getting some answers (and the calculations that give them?)
I think you said
Psubmerged =Patmospheric x 1.2/0.8
and that would equal 1.5 x atmospheric pressure (which you call 101.325 no units!)

Then you go a bit wrong. This has calculated the pressure of the air in the tube when it is submerged. That is equal to the water pressure around it at that depth.

So if you have an expression for the water pressure at that depth, you can indeed equate them.

You say, "we can substitute 101.325 + 101.325*1.2/0.8 kPa as the value for P in P = rgh" but I don't agree.

If "101.325*1.2/0.8 kPa" is the pressure of the air inside the tube when submerged,
then "101.325 + 101.325*1.2/0.8 kPa" is more than the pressure inside the tube.

"the value for P in P = rgh" is not the pressure outside the tube, nor the pressure inside the tube. What is it?
Im not sure what that pressure is for. But is an expression for the water pressure at that depth Pwater = pwater * g * h, where h is the depth of the tube(distance from surface of water to top of tube)? And as we just found out the pressure of the water outside the tube, substituting it in gives the value 0.0155m. Adding that to the length of the tube, it means the diver is at a depth of 1.2155m...
 
  • #18
Merlin3189
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OK, the P=ρhg gives the pressure due to the weight of water.
Then there are two approaches to making the equation:
The pressure inside the submerged tube is 1.5 atm. It started at 1 atm and increased by 0.5 atm as it was submerged.
So you could say the increase in pressure due to the water is 0.5 atm and that is equal to ρhg.
Otherwise you could say, the pressure outside the tube is 1 atm plus ρhg due to the water, then equate the total inside and outside pressures,
1.5 atm = 1atm + ρhg
Once atmospheric pressure is arithmetically eliminated, they are both the same.

I've no idea which approach you may have been taught to use, but it does not matter so long as you are clear in your own mind (and ideally. write down in your method) what each number or expression represents.

If it's any help, the depth comes out around 5m and I would not worry too much about the precision of atmospheric pressure, as the question only gives the density of seawater to within 2 or 3% . Atmospheric pressure should be good enough at 100kPa.
 
  • #19
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OK, the P=ρhg gives the pressure due to the weight of water.
Then there are two approaches to making the equation:
The pressure inside the submerged tube is 1.5 atm. It started at 1 atm and increased by 0.5 atm as it was submerged.
So you could say the increase in pressure due to the water is 0.5 atm and that is equal to ρhg.
Otherwise you could say, the pressure outside the tube is 1 atm plus ρhg due to the water, then equate the total inside and outside pressures,
1.5 atm = 1atm + ρhg
Once atmospheric pressure is arithmetically eliminated, they are both the same.

I've no idea which approach you may have been taught to use, but it does not matter so long as you are clear in your own mind (and ideally. write down in your method) what each number or expression represents.

If it's any help, the depth comes out around 5m and I would not worry too much about the precision of atmospheric pressure, as the question only gives the density of seawater to within 2 or 3% . Atmospheric pressure should be good enough at 100kPa.
okay i think i understand now. So basically the pressure is conserved? So the initial atm + pressure due to water = 1.5atm. So then the pressure due to water is 0.5atm. And thats the value you subsitute for P in P = pgh. so 0.5atm = p*g*h and yep i got around 5m as well for h. Thank you!
 
  • #20
Merlin3189
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I'd be more convinced if I saw it! But if you're happy now, that's good.

Since the question was a bit approximate, no doubt that's good enough. But just as a little rider for you, if you want, which part of the tube is at the depth you calculated? Is it the top (closed end), the bottom (open end) or what? (I assume the tube must be glass, perspex or something you can see through, so that you can see how much the water has risen.)
 
  • #21
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I'd be more convinced if I saw it! But if you're happy now, that's good.

Since the question was a bit approximate, no doubt that's good enough. But just as a little rider for you, if you want, which part of the tube is at the depth you calculated? Is it the top (closed end), the bottom (open end) or what? (I assume the tube must be glass, perspex or something you can see through, so that you can see how much the water has risen.)
i think the depth i calculated was the top of the tube. so the actual depth of the tube would be 5m + the length of the tube, 1.2m so 6.2m?
 
  • #22
as a theoretical physicist
At the beginning I must state that we are modeling a system that is isothermal at the same temperature in the initial and final state + that the density of air inside the tube is uniform in both states + that the gravitational field is uniform and constant + that there is no exchange of matter between the air inside the tube and the water during the experiment, + that the air behaves like a perfect gas … and may be many other assumptions that are quite reasonable near our oceans.
(Well, I’m not sure that water vapor pressure is negligible in this experiment).
Within this simplifying assumptions,
if
Pmsl = 101,325 Pa = pressure at mean sea level
ρ = 1,000 kg/cubic meter = density
g = 9.80665 m/s2
H = 120 cm = tube heigth
h = 40 cm is = final heigth of water inside the tube
x = depth to which the diver dived = distance of the bottom of the tube from the ocean surface
then
we have to solve the system of two equations:
(1) Pmsl*H = P*(H - h)
(2) P = Pmsl + (x - h)*ρ*g

from (1) you get P = Pmsl*H/(H-h)
and from (2) x = Pmsl*(H/(H-h)-1)/ρ/g+ h

numerically: x = 5.566 meters ( 1 cm more than the engineer )
 
  • #23
as an engineer I observe that
before diving the volume of air in your tube is proportional to 1.2
and after it is proportional to 0.8 =1.2 - 0.4
then from the equation
P*V = constant at constant temperature
you get:
Pmsl*1.2 = P*0.8
the the pressure of water at the free surface inside the tube is
P = Pmsl*1.2/0.8 = 1.5*Pmsl = Pmsl + water column
as Pmsl is approximately = 10.3 meters of water, then the water column is nearly 5.15 meters.
This is the height of the column of water between the sea surface and the free surface inside the tube.
You have to add 0.4 meters to find the depth of the tube bottom.
The answer of an engineer is 5.15 + 0.4 = 5.55 meters
a warning: the density of oceanic water is not 1000 kg/cubic meter
 
  • #24
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as an engineer I observe that
before diving the volume of air in your tube is proportional to 1.2
and after it is proportional to 0.8 =1.2 - 0.4
then from the equation
P*V = constant at constant temperature
you get:
Pmsl*1.2 = P*0.8
the the pressure of water at the free surface inside the tube is
P = Pmsl*1.2/0.8 = 1.5*Pmsl = Pmsl + water column
as Pmsl is approximately = 10.3 meters of water, then the water column is nearly 5.15 meters.
This is the height of the column of water between the sea surface and the free surface inside the tube.
You have to add 0.4 meters to find the depth of the tube bottom.
The answer of an engineer is 5.15 + 0.4 = 5.55 meters
a warning: the density of oceanic water is not 1000 kg/cubic meter

as a theoretical physicist
At the beginning I must state that we are modeling a system that is isothermal at the same temperature in the initial and final state + that the density of air inside the tube is uniform in both states + that the gravitational field is uniform and constant + that there is no exchange of matter between the air inside the tube and the water during the experiment, + that the air behaves like a perfect gas … and may be many other assumptions that are quite reasonable near our oceans.
(Well, I’m not sure that water vapor pressure is negligible in this experiment).
Within this simplifying assumptions,
if
Pmsl = 101,325 Pa = pressure at mean sea level
ρ = 1,000 kg/cubic meter = density
g = 9.80665 m/s2
H = 120 cm = tube heigth
h = 40 cm is = final heigth of water inside the tube
x = depth to which the diver dived = distance of the bottom of the tube from the ocean surface
then
we have to solve the system of two equations:
(1) Pmsl*H = P*(H - h)
(2) P = Pmsl + (x - h)*ρ*g

from (1) you get P = Pmsl*H/(H-h)
and from (2) x = Pmsl*(H/(H-h)-1)/ρ/g+ h

numerically: x = 5.566 meters ( 1 cm more than the engineer )
your working as the engineer is the same as merlin3189. So i was wrong, you are meant to add the height of water that entered the tube..
 
  • #25
Merlin3189
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Yes, impressive. And a very good example to OP (and most PF questioners) - a list of variables, their names, what they mean clearly stated, their units and then formulae, expressions and equations with explanations of where they come from and what they mean. If people worked like that, 9 times out of 10 they would not need to ask questions, because the working explains itself to them. (And on the tenth occasion, they'd get a solution in the first reply, because people would be able to see exactly what was wanted.)

I'm still trying to see where your non-engineer calculation differs, but with the detail you give, that should be easy enough. (I just noticed your post when I popped in from gardening to have a tea break.)
Yes, I didn't bother with Pascals and just used half an atm. I had that as 10.2m water so got 5.1m to the surface of the water in the tube. (To be honest, I had to look up the 10.2m. I still remember it as 30ft! Which must be wrong, as that's a lot less than 10.2m.) If that's the engineers way, count me on the side of the engineers! I'm in awe of the mathematical wizardry of the PF physicists, but I often come into threads after them and try to explain things in simpler ways to people who seem to me to have been befuddled by the obscure maths treatment. (Incidentally, the best mathematician I know - in terms of being able to solve problems that others can't, and get reliable results - is an engineer.)

When I saw the 1000kg/m3 for seawater, I thought 101325Pa for 1atm was a bit OTT.
 

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