Calculating Pressure at Depth in Oceans/Lakes

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Discussion Overview

The discussion revolves around calculating the pressure at a given depth in oceans or lakes, specifically examining the pressure equation P = ρgh and its implications in different unit systems. Participants explore the relationship between depth, pressure, and atmospheric contributions without reaching a consensus on the best approach to unit conversion.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that 1 atmosphere supports a column of water 32 feet high and proposes dividing the lake's depth by this height to calculate pressure in atmospheres.
  • Another participant confirms the pressure equation P = ρgh is valid in any arbitrary system of units, emphasizing the flexibility of unit choice.
  • A third participant states that the pressure under water is given by P = ρgh + Patm and notes that this is measured in Pascals.
  • A later reply elaborates on the unit independence of the pressure equation, explaining that substituting numerical values can lead to unit dependence if not handled correctly, and discusses the implications of using different unit systems.

Areas of Agreement / Disagreement

Participants express differing views on the implications of unit choice in the pressure equation, with some emphasizing unit independence while others highlight potential pitfalls in unit conversion. No consensus is reached regarding the best method for calculating pressure at depth.

Contextual Notes

Participants do not fully resolve the implications of unit conversion, and there are assumptions about the applicability of the pressure equation across different systems of measurement that remain unexamined.

cragar
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if I want to calculate the pressure at a given depth in the ocean or a lake
1 atmosphere of pressure will support a column of water 32 feet high right?
so if the lake is 20,000 feet deep can we just divide 20,000
by 32 and get 625 atmospheres at the depth?

i see that the pressure equation is p=(rho)gh is this measured in Pascal’s .
This is not a homework question.
 
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Yes, but

P = rho g h

is an equation that is valid in any arbitrary system of units. The whole point of units is that you can choose whatever units are most convenient for you.
 
The pressure under water P = \rho gh + P_{\text{atm}} is measured in Pascal.
 
fatra2 said:
The pressure under water P = \rho gh + P_{\text{atm}} is measured in Pascal.

This equation is valid in any arbitrary units. If you were to take this equation and then substitute numerical values in it, like
g = 9.81 m/s^2, then it is still unit independent, because the meters and seconds in this equation will transform covariantly when changing to other units (e.g. to miles and weeks from meters and seconds).

Only if you were to substitute numerical values like g = 9.81 ommitting the meters/second^2, then the equaion becomes unit dependent, it will only be valid in SI units. You are then effectively putting meter = second = kilogram = 1, analogous to what we do in theoretical physics when we put hbar = c = G = 1.

The answer you get for P will then be a number and then the unit Pascal has to be inserted by hand, it doesn't come out of the equation itself. You can also say that since we have put meter = second = kilogram = 1, you can multiply by any arbitrary combination of meters, seconds and kilograms, but only one combination is dimensionally correct. So, the correct unit, Pascal, was set to 1. We can thus multiply by Pascal as that is equal to 1 without changing anything. But then the equation has become dimensionally correct again, and therefore this is the correct way to put back the units.
 

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