Calculating Pressure at Depth in Oceans/Lakes

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SUMMARY

The pressure at a given depth in oceans or lakes can be calculated using the equation P = ρgh, where P is pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. For example, at a depth of 20,000 feet, the pressure can be calculated by converting the depth to appropriate units and applying the equation. The pressure is measured in Pascals, and while the equation is unit-independent, specific numerical values must adhere to SI units for accuracy.

PREREQUISITES
  • Understanding of fluid density (ρ) and its units
  • Knowledge of gravitational acceleration (g) and its standard value (9.81 m/s²)
  • Familiarity with the concept of atmospheric pressure (P_atm)
  • Basic understanding of unit conversion in physics
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  • Research the implications of pressure changes at varying depths in different fluids
  • Learn about the relationship between pressure and depth in various environments, such as oceans and lakes
  • Explore the concept of hydrostatic pressure and its applications in engineering
  • Investigate unit conversion techniques for pressure measurements in different systems
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Students, engineers, and scientists interested in fluid dynamics, environmental science, and anyone involved in calculating pressure in aquatic environments.

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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