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A tank contains sea water of density $1030 \frac{kg}{m^3}$ The pressure at point M is $60 kPa$ more than atmospheric pressure. Calculate the depth of M below the surface of the water.

My attempt: $P = \rho gh \therefore (1030)(10)(h) - 101kPa = 60kPa$

However the markscheme gives the equation $(1030)(10)(h) = 60kPa$

I feel that this equation does not account for the information the question gives, that the pressure is $60 kPa$ more than atmospheric pressure. Is there some concept that I am missing in my attempt?

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    $\begingroup$ I can't see where point M is, sorry. $\endgroup$
    – Kenshin
    Commented Apr 1, 2016 at 4:18

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Since we are talking about the difference in pressure between two levels in a single liquid, to avoid confusion, it is more appropriate to use the formula $\Delta P=\rho \times g \times h,$ where it means exactly what it is meant to. The question is concerned with this change in pressure rather than the absolute pressure you are concerned with. This fact is indeed signalled in the statement, "The pressure at point M is 60kPa more than atmospheric pressure;" the word more than is the key to understand that we are dealing with relative change rather than absolute value.

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  • $\begingroup$ Thank you for your answer, but isn't that exactly what I'm doing? I say $(1030)(10)(h)−101kPa$ and this is $\Delta P$, isn't it? Furthermore since it's a closed tank, atmospheric pressure is not experienced at the surface, but rather somewhere below. Hence, calculating $\Delta P$ relative to this point would not give the desired result, which is relative to the surface of the water. $\endgroup$
    – Airdish
    Commented Apr 1, 2016 at 4:32
  • $\begingroup$ Actually no. $\Delta P$ is the first term just by itself where you can also see it as $P_{low} - P_{up}$. Here atmospheric pressure is embedded in both terms $P_{low}$ and $P_{up}$ which is cancelled when taken together (one with positive contribution and the other with negative contribution). This is why the atmospheric pressure is completely washed away so long as we are concerned with the Change in Pressure. $\endgroup$
    – Benjamin
    Commented Apr 1, 2016 at 4:37
  • $\begingroup$ Oh so what you're saying is $P_{low} = P_{atm} + P_{water}$? $\endgroup$
    – Airdish
    Commented Apr 1, 2016 at 4:41
  • $\begingroup$ Exactly, you got it. But I would put it this way, $P_{low} = P_{atm} + \Delta P$ where $\Delta P = \rho \times g \times h$. $\endgroup$
    – Benjamin
    Commented Apr 1, 2016 at 4:43
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I am not sure about the answer , but i think it has something related to the closed tank structure , there should be some air under that and also on the water of course so , you shouldn't put the atmospheric pressure in your calculations.

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