pressure question ?????

Hi this is messing with my head n I can't figure out what im actually supposed to be doing pahah

heres the question:

A capsule, containing air at a pressure of 150 kPa, is sealed by a valve which can withstand a pressure difference of 650 kPa. How deep in sea water, of density 1020 kg m-3, can the capsule be lowered if no water is to enter? [Pressure]

thanksssss :smile:

Reply 1

GabiAbi84

You are told the maximum difference in pressure the valve can withstand -so what is the maximum pressure the water can be at before it breaks?

Then use that number in the equation for pressure at a depth to work out how deep it can go.

Reply 2

17hcooper

Original post by GabiAbi84

im sorry im really dumb, but when it says pressure difference, does it mean the difference in pressures between inside the capsule (the air bit) and the water outside ?

Reply 3

GabiAbi84

Original post by 17hcooper

im sorry im really dumb, but when it says pressure difference, does it mean the difference in pressures between inside the capsule (the air bit) and the water outside ?

Yes

And you’re not dumb.

Reply 4

17hcooper

Original post by GabiAbi84

Yes

And you’re not dumb.

so would the maximum pressure the water could be at be 650 000 - 150 000 which would be 500000 Pa ?

and thank you ahah, I just find it hard to get my head around the numbers lmao

Reply 5

GabiAbi84

Original post by 17hcooper

so would the maximum pressure the water could be at be 650 000 - 150 000 which would be 500000 Pa ?

and thank you ahah, I just find it hard to get my head around the numbers lmao

Not quite:

Inside is 150kPa -outside can be no more than 650kPa More than that

The difference would be

Outside pressure - inside pressure = pressure diff

Reply 6

17hcooper

Original post by GabiAbi84

Not quite:

Inside is 150kPa -outside can be no more than 650kPa More than that

The difference would be

Outside pressure - inside pressure = pressure diff

OHHHHHHHH okayyyy I think I get it

so outside pressure would = 150000 + 650000 = 800000

and then you put that into the equation:

800000 = 1020 x 9.81 x h

so h = 800000/(1020 x 9.81) = 80 m

Reply 7

GabiAbi84

Original post by 17hcooper

OHHHHHHHH okayyyy I think I get it

so outside pressure would = 150000 + 650000 = 800000

and then you put that into the equation:

800000 = 1020 x 9.81 x h

so h = 800000/(1020 x 9.81) = 80 m

Remember it is at sea and not in a closed container so you have to take into account the atmospheric pressure also.

Reply 8

17hcooper

Original post by GabiAbi84

Remember it is at sea and not in a closed container so you have to take into account the atmospheric pressure also.

oh god what does that mean I need to do ?

Reply 9

GabiAbi84

Original post by 17hcooper

oh god what does that mean I need to do ?

What level are you at?

Reply 10

17hcooper

Original post by GabiAbi84

What level are you at?

Well it says that the pressure at the surface of the sea is 101 kPa and I calculated that the maximum pressure of the water is 800 kPa, but I’m not sure how to calculate it without ignoring atp

Reply 11

GabiAbi84

Original post by 17hcooper

Well it says that the pressure at the surface of the sea is 101 kPa and I calculated that the maximum pressure of the water is 800 kPa, but I’m not sure how to calculate it without ignoring atp

(I meant what education level is it for but that’s ok 😉)

Okay so if that 800kPa -101kPa is due to the air, the rest is due to the water.
So calculate how much is due to the water and Then use the equation as before.