Transcript Momentum Balance Steven A. Jones BIEN 501/CMEN 513 Monday, March 19, 2007 Louisiana Tech University Ruston, LA 71272

Momentum Balance
Steven A. Jones
BIEN 501/CMEN 513
Monday, March 19, 2007
Louisiana Tech University
Ruston, LA 71272
Momentum Equation
Du
 p    τ  fb
Dt
Note 1: The left hand side is the material derivative and will ultimately
give rise to the nonlinear, convective acceleration terms.
Note 2: The stresses themselves do not necessarily cause a change
in momentum (acceleration of the fluid). Spatial variations of these
stresses do.
Note 3: Remember that this equation is just a restatement of
Newton’s second law, with “ma” on the left hand side and “F” on the
right hand side.
Note 4: Remember that u is a vector and t is a 9-component tensor.
The tensor has 3 rows (one per surface) and 3 columns (one per
each velocity direction).
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Differential Form – Conservation of
Momentum
dx
Select one component of velocity
x  x
dy
vz
d  mv 
vy
vy
vx
vx
vz
dt

d
 vdV    v  v  n  dA

CV
CS
dt
We can get a differential form if we
convert the last integral to a volume
integral. The divergence theorem says:
dz

CS
 v  v  n  dA    v   v  dV
CV
dmv d
so
   vdV       vv  dV
CV
dt
dt CV
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Differential Form – Conservation of
Momentum
dx
vy
x  x
dy
vz
dmv d
   vdV       vv  dV
CV
dt
dt CV
vy
vx
vx
vz
As with conservation of mass:
d  v 
d
 vdV  
dV

CV
CV
dt
dt
dz
Combine the two volume integrals:
d  v 
d  v 
dmv

dV       vv  dV  
     vv  dV
CV
CV
CV
dt
dt
dt
d  mv 
dt
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
d  v 
dt
     vv 
Differential Form – Conservation of
Momentum
In
d  mv  d   v 

     vv 
dt
dt
The left hand term (time derivative of momentum) is equal
to the external force (by Newton’s 2nd law), so:
f ext 
d  v 
dt
     vv 
Also, d  v  / dt is the Eulerian derivative (a fixed
location in space), so:
f ext 
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  v 
t
     vv 
Differential Form for Momentum
In fluid mechanics, we write Newton’s second law
backwards, by convention, so instead of:
f ext 
  v 
t
     vv 
We write:
  v 
t
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     vv   fext
External Forces on the Differential
Cube
There are three types of external
forces on the differential cube:
1.Viscous forces   τ
2.Pressure forces (always normal) p
3.Body forces (e.g. gravity) f b
In the momentum equation, these are
expressed as force per unit volume.
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Newtonian Fluid
  1 t 12 t 13 


τ  t 21  2 t 23 
t

t

3
 31 32
 ui u j 

t ij   



x

x
i 
 j
 ui 
 (not sum m ed)
 i   p  2 
 xi 
In a general case τ   pI  S
In a New tonianfluid τ   pI  2D  S  2D
1  ui u j
Thus Dij  


2  x j xi
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


Newtonian Fluid
 u1 u2 

For exam ple t 12   

 x2 x1 
Typically one of the coordinates will be aligned with
the boundary surface. E.g., for Poiseuille flow, the
boundary is parallel to both the z and q directions.
Since velocity is zero at the wall, there is no change
with respect to the z and q directions.
Thus t zr
wall
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 ur u z 
u z
  
  

r
 z xr 
Poiseuille Flow
ur  0 everyw here
uq  0 everyw here
Thus t zr
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wall
 ur u z 
u z
  
  

r
 z xr 
Stokes Flow
ur  0 everyw here
uq  0 everyw here
Not because ur is 0, but
because ur is 0 for all q.
t rq
wall
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 uq r  1 ur
  r

r
r q

uq r 

  r
r
