How deep can you dive to get 3x pressure? (101,3 kPa)

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To achieve a total pressure of 3 times the air pressure at the surface (101.3 kPa), one must dive to a depth where the pressure increases to 303.9 kPa. The fluid pressure formula p = ρgh is used, leading to the calculation of depth h = p/(ρg). The correct depth is determined to be approximately 20 meters, not 30 meters, due to the distinction between "3 times as large" and "2 times larger" than the surface pressure. This confusion stems from the interpretation of pressure increments and the need to account for the surface pressure in the calculations. Understanding these terms clarifies the solution to the problem.
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I'm very new to physics and I haven't learned exactly
what everything is called yet. Sorry if I confuse you.

1. Homework Statement


How deep do you have to dive in a lake to get
the total pressure 3 times as large as the airpressure
at the surface of the water (101,3 kPa)?

Normal pressure at sea surface: 101,3 kPa = 101,3*(10^3) Pa
Water density: 998 kg/m3
Gravity in Sweden: 9.82 N/kg

p = 101,3 kPa
ρ = 998 kg/m3
g = 9.82 N/kg

Homework Equations


Fluid pressure formula: p = ρgh
Get deep (h): h = p/(ρ*g)

The Attempt at a Solution


h = (101,3*(10^3))/(998*9.82)
h = 10.34 m

Total pressure three times as large as the airpressure at the sea surface

h = 10.34 * 3
h = 31 m

Correct answer: 20 m
 
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Hi,

your calculations are basically correct. Imagine you are at 30 m and the pressure is about 300 kPa. Now you get up to the surface. What pressure is there, if you had 300 kPa at 30 m?
 
I don't get it because if you get up to the sea surface then the pressure would be 101,3 kPa again according to the exercise.

But if I am at 30 m and the pressure is 300 kPa wouldn't it be something like:

30 m = 300 kPa
20 m = 200 kPa
10 m = 100 kPa
0 m = 0 kPa

I don't really get what you mean and how I'm supposed to solve it by that.
 
DecoratorFawn82 said:
But if I am at 30 m and the pressure is 300 kPa wouldn't it be something like:

30 m = 300 kPa
20 m = 200 kPa
10 m = 100 kPa
0 m = 0 kPa

That's exactly the point. At the surface the pressure isn't 0. If you rewrite this table starting at 0 m (surface) with the surface pressure and add the pressure difference every ten metres...
 
Sorry if I'm slow but I still don't get why the answer is 20 m instead of closely 30 m.
 
I think you are dealing with a terminology issue. They are asking for the depth at which the pressure is 3x as large as the pressure at the surface. That means that it is 2x larger than the pressure at the surface.
 
Yeah, but why is the pressure 2x larger than the pressure at the surface and not 3x?
 
DecoratorFawn82 said:
Yeah, but why is the pressure 2x larger than the pressure at the surface and not 3x?
The term 3x as large as is semantically the same thing as the term 2x larger.
 
I mean that if you have for example an rectangle. And you can extend it. Then 3x and 2x isn't the same thing because if x = 4 then:

3 * 4 = 12
2 * 4 = 8
 
  • #10
DecoratorFawn82 said:
I mean that if you have for example an rectangle. And you can extend it. Then 3x and 2x isn't the same thing because if x = 4 then:

3 * 4 = 12
2 * 4 = 8
I mean how the terms "as large as" and "larger than" are used in the English language. If they asked how deep do you have to go so that the pressure is 2x larger than at the surface, what would your answer have been?

Chet
 
  • #11
p = 101,3 kPa
p = 101,3 kPa * 2
p = 202,6 kPa

h = 10,34 m
h = 10,34 m * 2
h = 20,68 m

That's what my answer would be?
 
  • #12
DecoratorFawn82 said:
p = 101,3 kPa
p = 101,3 kPa * 2
p = 202,6 kPa

h = 10,34 m
h = 10,34 m * 2
h = 20,68 m

That's what my answer would be?
Yes. That's the depth at which the pressure is 2x larger than the pressure at the surface and 3x as large as the pressure at the surface.

Part of your problem might also be that your starting equation should have been:
$$p=p_0+ρgh$$
where p0 is the pressure at the surface.

Chet
 
  • #13
DecoratorFawn82 said:
Sorry if I'm slow but I still don't get why the answer is 20 m instead of closely 30 m.

The table starting from sea surface and getting deeper could look like this:

h0 = 0⋅10 m / p0 = ... kPa (=p0⋅1)
h1 = 1⋅10 m / p1 = ... kPa (=p0⋅(1+...))
...
hn = n⋅10 m / pn = ... kPa (=p0⋅(1+...))
 
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