Finding the pressure of water when given an area and depth only

In summary: 45x30x50 = 21000g. so 45kg x 1.47 = 58.7kg, which is too much for the person to handle and stop breathing
  • #1
murdrobe
30
0

Homework Statement


You are a secret agent deep in the jungle, escaping from the pursuing villains. You see a stream and reeds, so you cut off a reed, jump into the stream and lie about half a metre under water breathing through the reed. Assuming that the area of your chest is approx. 30x30cm, can you breathe?


Homework Equations


"pressure = hdg" is the only thing given
h= depth
d= density
g= gravity

The Attempt at a Solution


ive been trying to figure this out for days... there's many ways to work it if we knew other numbers but i don't know how to get it from this without assuming water density, which unless I am wrong is variable by temperature? we arent given a temp...

am i just blindly reading into this too much and its actually really simple?

1 thing though, PLEASE don't give me an answer, i just want a method, having an answer doesn't help if a similar question comes up in my exam.
 
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  • #2
Sometimes even very tiny differences matter, but not here. While water density varies with temperature, differences are not that large. In most cases you can safely assume 1 g/mL.
 
  • #3
so you think that in general the teacher is assuming we know an average water density and go from there with the maths? just converting the units from 1000kgm-3 into gcm-3 in order for how much it will affect the chest and then see if 30cm3 would be able to displace that much water?
 
  • #4
Where did you get 30 cm3 from?

I have a feeling this is kind of an open question - water density is obvious, but there is more information missing.
 
  • #5
sorry that should have been a 2 not a 3, it was from the area of the mans chest.
 
  • #6
just had an idea, i have been given the volume of water in a round about way. 30cm by 30cm with a depth of 50cm... given that and using an average density for water. would it be possible to use that to say that too much weight would be on the chest to be able to breath?

i think i also need to figure out a maximum depth that the person can breathe at but i have no idea where to begin with that.
 
  • #7
murdrobe said:
just had an idea, i have been given the volume of water in a round about way. 30cm by 30cm with a depth of 50cm... given that and using an average density for water. would it be possible to use that to say that too much weight would be on the chest to be able to breath?

i think i also need to figure out a maximum depth that the person can breathe at but i have no idea where to begin with that.

It's not so much of the weight on the person's chest, but more of the pressure on it. Think about it this way: what hurts more, being stepped on by a 250-lb giant wearing sneakers, or a 80-lb little lady in heels?
 
  • #8
the problem i can't get over with this question is that i don't know how much pressure it would take to stop someone breathing... it has too many variables for me to get my head around wven though i know its asking for none of them.
 
  • #9
physicsvalk said:
It's not so much of the weight on the person's chest, but more of the pressure on it. Think about it this way: what hurts more, being stepped on by a 250-lb giant wearing sneakers, or a 80-lb little lady in heels?

Completely unrelated to the problem.
 
  • #10
so i guess i should do the calculation along the lines of

50cm (depth) x 1 (density of water per cm) x 9.81, this would give me the force and then i multuiply it by the area because water affects the complete area equally?

then how would i go onto say that the person can or can't breathe when i have these numbers? guess?

google how much pressure it takes to stop someone breathing?
 
  • #11
You are right about calculating the force (that is, you need to pay attention to units, so far you are mixing meters and centimeters, that is always a possible source of problems).

You are also right about the fact this will not answer the question. Googling (or experimenting) would be a better approach than just a guess.
 
  • #12
googling the maximum pressure tells me that a diver can handle about 3 ft of water above them before it becomes difficult to breathe. 1 foot = 30.48cm 3 ft = 91.44cm
so 50cm depth = 1.64 ft. on the same website it said that 33ft of water gives off 29.4psi.

i want 0.05 of that depth to see what the maximum PsI a person could breath at, 0.05 x 29.4 = 1.47 PsI to stop a person breathing.

the weight of the water on my person is 45kg using the volume as 45000cm^3 and density as 1g/cm^3.

45kg = 99.21 lb's over the 11.81 inches^2 of the persons chest. which gives me 8.4 psi overall meaning he can't breathe?

i have definatley done something wrong because the website said 3ft is ok.
 
  • #13
i don't think i squared the 11.81, if i do that gives me 0.71 PSI which is a more realistic answer. this would also show that the person can breathe through the reed.
 
  • #14
hmmm i think i found a better way. the college library had a physics book stating that the maximum a pair of lungs would inflate at is 20KPa and that at 2m of depth the pressure starts becomming too much.

so if i take that and then work it back. if i end up with more than 20KPa i can say that he can't breathe, or visa versa...

so P = hdg

h = 0.5 cm
d = 1000
g = 9.8

0.5x 1000 x 9.8 = 4900 pascals or 4.9 KPa meaning the guy can breathe?

am i right in thinking that because water pressure affects all of the chest at once? in which case do i also need to multiply the 4.9 by 0.3 to account for the chest?
 
  • #15
murdrobe said:
the college library had a physics book stating that the maximum a pair of lungs would inflate at is 20KPa and that at 2m of depth the pressure starts becomming too much.

Sounds like a good starting point - at least you have a source to quote.

h = 0.5 cm

m, but I assume that's just a typo.

0.5x 1000 x 9.8 = 4900 pascals or 4.9 KPa meaning the guy can breathe?

Sounds OK.

am i right in thinking that because water pressure affects all of the chest at once? in which case do i also need to multiply the 4.9 by 0.3 to account for the chest?

Why 0.3? And TBH, I have no idea what you mean.
 
  • #16
the guys chest is 30cm by 30cm, I am using m as units hence the 0.3. we were taught that water pressure affects an area all at once but I am not sure how that works, would the 4.9Pa be for 1m^3 and i need to divide it by the 0.3^2 or am i right in saying it would be 4.9 equally spread across the whole 30 by 30?
 
  • #17
What is a surface of 30 cm by 30 cm square?

What is a pressure definition?
 
  • #18
sorry I am not quite sure what your asking.
 
  • #19
@ murdrobe

Frogs breath through an active breathing system (in Biology, this is called a positive pressure breathing). This means that they actually push the air to their lungs. If we had that kind of lungs, you could use the data you found in the library. But your lungs DON'T work by pushing the air from the atmosphere as frogs do.

Humans use a negative pressure breathing. We suck the air from the atmosphere because when the diaphragm moves, it reduces the pressure in the thorax. Since gases move from the area of higher pressure to a lower pressure area, air will passively enter our lungs. There's no pushing of the air - it simply is sucked into our lungs.

From that reasoning, it's possible to realize that the pressure in your thorax (and, by extension, in your lungs) cannot be higher than the atmospheric pressure, or you wouldn't be able to breathe. Thus, it will be impossible to breathe through a reed if the pressure in the water is higher than the atmospheric pressure.

From now on, the reasoning is pretty straightforward. The pressure in the water will be equal to the atmospheric pressure at a depth d = 10 m (you may get this using P = pho*g*d, and using the fact that the atmospheric pressure is roughly 10^5 Pa). So, at a depth of 10 m it is absolutely impossible to breath using a reed. If someone wants to go deeper than that, he or she MUST use a pressurized air cylinder. Since the pressure in the cylinder is higher than the water pressure on your thorax, he will be able to breathe normally, using the human negative pressure respiratory system.

Now, as you found in other books, it gets fairly difficult to breathe below some 2 to 3 meters, because the water pressure is already considerable at those depths.
 
  • #20
thanks for that, the book was talking about a person so I am fine with using that as the quote. the thing I am stuck on is that I am not sure how water pressure works exactly. would it be:
4KPa over the whole area of his chest
or 4KPa times the area of his chest
or have i worked out that it would be 4KPa for 1m^2 and i need to divide the answer?
 

1. How do you calculate the pressure of water when given an area and depth?

The formula for calculating pressure is pressure = density x gravity x depth. To find the pressure of water, you will need to know the density of water and the acceleration due to gravity, which is typically 9.8 m/s^2. Once you have this information, you can plug in the values for the area and depth to calculate the pressure.

2. What is the unit of measurement for pressure?

The unit of measurement for pressure is typically expressed in pascals (Pa) or newtons per square meter (N/m^2). Other common units include pounds per square inch (psi) and atmospheres (atm).

3. How does the pressure of water change with depth?

The pressure of water increases with depth due to the weight of the water above it. For every 10 meters of depth, the pressure increases by approximately 1 atmosphere.

4. What is the effect of area on water pressure?

The area has no direct effect on the water pressure. The pressure is determined by the depth and density of the water, regardless of the size of the area. However, a larger area may distribute the pressure over a larger surface, resulting in a lower force per unit area.

5. How does the density of water affect its pressure?

The density of water has a direct impact on its pressure. The higher the density, the greater the pressure will be at a given depth. This is because denser objects have more mass, which results in a greater force exerted on the surface below it.

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