Deriving Pressure on a Fluid Surface at Rest with Constant Density

Click For Summary
SUMMARY

The discussion focuses on deriving the pressure distribution P(z) in a stationary fluid with constant density under the influence of gravity, represented by the force vector F = (0, 0, -g). The solution utilizes Euler's equation, leading to the conclusion that P(z) = P_a + ρg(h_0 - z), where P_a is the atmospheric pressure at z = h_0. The pressure at the surface z = 0 is determined to be P(0) = P_a + ρgh_0, demonstrating the relationship between pressure and depth in a fluid at rest.

PREREQUISITES
  • Understanding of Euler's equation in fluid dynamics
  • Knowledge of pressure gradients in fluids
  • Familiarity with boundary conditions in fluid mechanics
  • Basic concepts of hydrostatics
NEXT STEPS
  • Study the derivation of pressure equations in stationary fluids
  • Learn about boundary conditions in fluid dynamics
  • Explore hydrostatic pressure calculations in different fluid scenarios
  • Investigate the implications of constant density assumptions in fluid mechanics
USEFUL FOR

Students studying fluid mechanics, engineers working with fluid systems, and anyone interested in the principles of hydrostatics and pressure distribution in fluids.

coverband
Messages
170
Reaction score
1

Homework Statement


Consider a stationary fluid (u=0) with constant density and take F= (0,0,-g). Find P(z) which satisfies [tex]P=P_a[/tex] on [tex]z=h_0[/tex], where z is measured positive upwards. What is the pressure on z=0?


Homework Equations



Euler's equation: [tex]\frac{Du}{Dt}=-\frac{1}{\rho}\nabla P + F[/tex]

The Attempt at a Solution


[tex]\frac{1}{\rho}\nabla P = (0,0,-g)[/tex] Gives the answer in the back of the book as:
then [tex]P = P_a + \rho g(h_0-z); P(0) = P_a + \rho g h_0[/tex]. How did they get this? Thanks
 
Physics news on Phys.org
As they say u=0, then as you correctly wrote down:
[tex] \frac{1}{\rho}\nabla P=(0,0,-g)[/tex]
Which means that:
[tex] \frac{\partial P}{\partial x}=0,\quad\frac{\partial P}{\partial y}=0,\frac{\partial P}{\partial z}=-\rho g[/tex]
Which shows that the pressure in independent of both x & y. so you are left to solve:
[tex] \frac{\partial P}{\partial z}=-\rho g[/tex]
Can you solve this? What are the boundary conditions that you need to use?
 
Thanks
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K