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Formula Sheet CHE 312 Fluid Mechanics (Fall 2010)
Final Exam, December 20, 9-11 am
Print this document on a single sheet of paper and bring it to the exam; there will be no spare sheets at the exam. You are
allowed to add information on the side of the sheet you printed on; the reverse side should be blank
Newtonian fluids:
τ xy = µ
Hydrostatics
∂p
= −ρ g
∂z
Fz = V ρ f g
dvx
dy
τ xy shear stress, µ viscosity,
dvx
velocity gradient
dy
p hydrostatic pressure, ρ density, g gravitational acceleration, z points up
Fz buoyancy force, V displaced fluid volume, ρ f fluid density
dm
total mass m is a conserved quantity
= φm ,in − φm ,out
dt
Steady state mechanical energy balance (Bernoulli equation)
1
p  Wɺnf
∆  v 2 + gz +  =
− e fr ∆ is “out minus in”; Wɺnf work rate not related to flow; e fr frictional loss.
2
ρ
φ


m
Frictional loss
4A
L
In a straight pipe or channel with length L and (hydraulic) diameter Dh ≡
: e fr = 2 fU 2
.
W
Dh
Total mass balance:
U=
φV
A
: average velocity; A: cross sectional area; W: wetted perimeter.
f: Fanning friction factor, f is a function of Re =
Special cases: laminar flow in a round tube: f =
Loss coefficients K: e fr = K
ρUDh
ε
, and the relative wall roughness
.
µ
Dh
16
; turbulent flow with smooth walls: 4 f = 0.316 Re −1/ 4 .
Re
U2
.
2
Momentum balance
d
( mv ) = φm,in v in − φm,out v out + ∑ F
dt
v is the velocity vector; ∑ F is the sum of forces acting on the system.
Drag force FD on a particle moving relative to fluid
1
FD = CD ρ f A⊥ ∆v ∆v
CD drag coefficient; A⊥ area ‘seen’ by the flow; ∆v = v f − v p .
2
ρ d ∆v
CD depends on the shape of the particle and Re = f p
, d p is the particle size.
Spherical particles (diameter d p , and therefore A⊥ =
µ
π
4
d p2 )
For Re<1 Stokes law applies: FD = 3π d p µ ∆v , which is the same as CD =
(
24
.
Re
)
24
1 + 0.15 Re0.687 if Re < 1000; CD = 0.44 if Re ≥ 1000 .
Re
Ergun equation (frictional loss in a packed bed with porosity ε made of spheres all having diameter d p )
A more general correlation: CD =
vsup µ (1 − ε )
φ
1 2
e fr = 1.75 L
vsup + 150 L
vsup = V ; L is the bed length; A its cross sectional area.
3
2 3
dp
ε
ρ d pε
A
Darcy equation & permeability for flow in a porous medium
d ( p + ρ gz )
µ
=− v
k is permeability (units m2); v is superficial velocity.
dx
k
(1 − ε )
2