calculate the momentum deficit correction factor.

Control Volume (Open Systems)

  eeiiCVCV vmvmFdt )vm(d 

 Control Mass (Closed Systems)

 Fdt vd

m 

Pressure Forces

pFd 

Ad 

AdpFp 



BODY FORCES:SURFACE FORCES

Viscous Forces

visc w ˆF =-i τ dA



     

2

du N τ =μ

dy m

Shear Stress due to friction acting in the opposite direction of flow velocity, u.

μ=viscosity coefficient(kg/m‐s)

gmWFgrav 



Weight

( ) d

F mv dt

  

CONSERVATION OF MOMENTUM – FLUID STATICS (FLUIDS AT REST)

Applications: Hydrostatic equation

g dz dp

 If density (and g) uniform:

gYpp )zz(gp)z(p

topbottom

refref

 

z

zref=0

Example: The human eardrum will rupture under an absolute pressure of about 2.4atm. How deep can you dive before you go deaf? (Also see Example 6.1, p418.)

Manometry: Manometers (U‐tubes) are used to determine gage pressures when one leg is connected to a known pressure or measure pressure difference between two locations. The manometer fluid is usually denser than the fluid to which it is connected, commonly mercury, water and oils.

'B B

B T

p p p p gY

manometer fluid density 



   

Applications: Buoyancy Force fluid displaced

fluid displacedB VgF

 

gVF airB 

Helium

V

mgW 

Air gVF OHB 2

mgW 

A piece of wood (=465kg/m3) floats at rest partially submerged in water. Develop an  expression for the depth of submersion, h, as a function of the given dimensions. Ignore  air effects.

H

h

L

D

CONSERVATION OF MOMENTUM – FLUID STATICS (FLUIDS AT REST)

Example: How do birthday Helium balloons work? What is the force the clown needs to exert to hold the 30cm‐diameter balloon if the balloon material mass is 4.25gm and the string’s mass is 0.75gm? (gm=grams)