fluid mechanics -...
TRANSCRIPT
Fluid Mechanics
Lecture #8
Stability• Center of the buoyancy and center of the solid body can be different.
• When?– The solid material is not homogeneous.– The solid body is not fully submerged.
Fully submerged body• Assume an equilibrium state (W=FB)
c: center of buoyancy; CG: center of solid mass
Consider a moment around CG.
Stable equilibrium
Unstable equilibrium
Stable vs. Unstable Equilibrium
Floating Body
Applications
Fluid with Acceleration• Fluid is always under influence of the acceleration due to gravity (az=‐g)
• Often, there exists acceleration in another direction (ax,ay≠ 0)
1p gk a
1px
ax, 1py
0, 1pz
g az (ax, az are constant)
px
ax, py
0, pz
(az g)
dp px
dx py
dy pz
dz
( ) 0x za dx a g dz dzdx
ax
az g
X
Z
At the water surface, p=patm (constant)
PROVE IT YOURSELFKEYS1. Total differential2. Pressure is constant at the water
surface: dp=0
dzdx
ax
az g
Physical Meaning
At the free surface, dx/dzdenotes the free surface slope.
X
Z
Review of Chapter I• Fluid cannot sustain a shear stress.• Fluid viscosity gives rise to a shear stress.• For Newtonian fluids, • Surface tension is caused by the property of fluids that minimizes their surface area.
• Unit is Force/Length [N/m]
dudy
Review of Chapter II• Pressure at a point is independent of directions.
• Understanding of total differential• For fluids with a zero shear stress, the equation for the pressure is
• The above equation can be integrated to obtain pressure distributions.
1p gk a
Review of Chapter II• In compressible fluids, one should consider density variations for integrating the pressure equations.
• In incompressible fluids, density variation is neglected.
• Methods for calculating the hydrostatic force– Inclined plane wall (Ixx Ixy)–Pressure prism method– Free‐body diagram method
Homework #2• Due date: April 7th• 2.95, 2.108, 2.109, 2.143, 2.148, 2.149, 2.151
• Plus 8 additional problems (check the website)