How Does Pressure Affect Seawater Density at Depth?

Click For Summary
SUMMARY

The discussion focuses on the effects of pressure on seawater density at a depth of 5,000 meters in the ocean. The total pressure at this depth is calculated using the equation pTotal = p_0 + rho*g*h, resulting in a pressure of 5.06×107 Pa. The fractional volume change of seawater at this pressure is determined to be -2.53×10-2 using the bulk modulus B = 0.2×1010. The misunderstanding arises in calculating the density of seawater under pressure, where the density increases due to the volume change, contrary to the initial assumption that it remains constant at 1,030 kg/m3.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically pressure and density relationships.
  • Familiarity with the concept of bulk modulus and its application in volume change calculations.
  • Knowledge of basic physics equations related to pressure, such as pTotal = p_0 + rho*g*h.
  • Ability to perform calculations involving density, volume, and pressure changes in fluids.
NEXT STEPS
  • Research the concept of bulk modulus and its significance in fluid dynamics.
  • Study the relationship between pressure, volume, and density in incompressible fluids.
  • Learn how to apply the equation dV/v = p/-B in practical scenarios involving seawater.
  • Explore the effects of depth on seawater properties using real-world oceanographic data.
USEFUL FOR

Students studying fluid mechanics, oceanographers, and anyone interested in the physical properties of seawater under varying pressure conditions.

Beginner@Phys
Messages
9
Reaction score
0

Homework Statement




Part A
What is the pressure at a depth of 5000 in the ocean? Given rho_water=1,030 kg/m^3
pTotal=p_0 + rho*g*h=5.06×107


Part B
What is the fractional volume change of seawater at this pressure? Given B=0.2*10^10
dV/v=p/-B =−2.53×10−2



Part C
What is the density of seawater at this pressure? (This is the question that I don't understand)



Homework Equations



(1)pTotal=p_0 + rho*g*h
(2)dV/v=p/-B
(3)p=F/A

The Attempt at a Solution


Part C: The liquid is incompressible. So. I guessed that the density of the liquid is still 1,030 kg/m^3 at a depth of 5,000m. And, I don't know why my answer was considered wrong.
 
Physics news on Phys.org
You determined the volume change from the bulk modulus.

But what is density? weight/volume?

So what was occupying a m³ is now occupying less volume
... so isn't the density now greater since you can now squeeze more into a m³?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
998
  • · Replies 6 ·
Replies
6
Views
779
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
3K