Chp3_331

# Chp3_331 - FLUID STATICS 3 Hydrostatics 3.1 Hydrostatic...

This preview shows pages 1–5. Sign up to view the full content.

FLUID STATICS 3. Hydrostatics 3.1. Hydrostatic pressure Fluid mechanics is the study of fluid in motion. special case: NO motion at all. Fluid statics - determine the stress field. * The force/stress on any given surface immersed in a fluid at rest , is always perpendicular (normal) to the surface. Recall that by definition, a fluid moves and deforms when subjected to shear stress and, conversely, a fluid that is static (at rest) is not subjected to any shear stress. Otherwise it'll move τ σ * At any given point in a fluid at rest, the normal stress is the same in all directions (hydrostatic pressure) 9

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Proof : Consider a small, wedged-shaped fluid element. Fluid is in equilibrium, so F = 0 Let the element be sufficiently small so that we can assume that the stress is constant on any surface (uniformly distributed). F 1 = 1 A 1 ; F 2 = 2 A 2 ; F 3 = 3 A 3 m =   V - fluid density ; V = 1 / 2 x y z - element volume Look at the side view F x = 0 F 1 cos - F 2 = 0 1 A 1 cos - 2 A 2 = 0 1 y( z/cos  cos - 2 y z = 0 1 = 2 10
F z = 0 F 1 sin + m . g = F 3 1 A 1 sin + V g = 3 A 3 1 ( x /sin  y sin + 1 / 2 g x y z = 3 x y 1 + 1 / 2 g z = 3 Shrink the element down to an infinitesimal point, so that z 0, then 1 = 3 1 = 2 = 3 Notes : Normal stress at any point in a fluid in equilibrium is the same in all directions. This stress is called hydrostatic pressure. Pressure has units of force per unit area. F = PA [N/m 2 ] The objective is to find the stress field in a given body of fluid, namely to find the pressure at any point in a fluid at rest. 11

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3.2. Vertical distribution of pressure Look at a large tank of liquid Take a small element of fluid in a convenient shape Force balance: (P+dP) A + g A dy = P A dP/dy = - g Negative sign indicates that P decreases as y increases. For a constant density fluid, we can integrate for
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/08/2010 for the course AME 331 taught by Professor Zohar during the Fall '08 term at Arizona.

### Page1 / 14

Chp3_331 - FLUID STATICS 3 Hydrostatics 3.1 Hydrostatic...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online