Derivation of the Navier-Stokes equations - DOC - DOC by zT0ZIQ49

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									                                                                                              Juster, GLY5932

Reduced pressure: If density is constant the Navier-Stokes equations can be simplified somewhat. We can
rearrange the equation of hydrostatics as
               
       0  gk  Ps
where the subscript s on P means that the pressure is hydrostatic. We then subtract this equation from the
full Navier-Stokes equation:
                   
        Dv                          
               gk  P   2 v
         Dt
                     
      0   gk  Ps
         
       Dv                          
             ( P  PS )   2 v
        Dt
             ˆ
If we define P = P – PS the Navier-Stokes equations can then be written as:
         
       Dv       ˆ        
           P   2 v
       Dt
 ˆ
 P is variously called the ―dynamic pressure‖ or ―reduced pressure‖, but is a very common concept in fluid
mechanics. It obviously simplifies the Navier-Stokes equations and makes their solutions easier to
              ˆ
understand. P can be understood as the pressure in excess of hydrostatic that is responsible for creating the
pressure gradient forces that actually cause the fluid to move. (Recall that the hydrostatic pressure gradient
just keeps the fluid steady by balancing its weight.)
The hydrostatic pressure can be written as:
             z
     PS   g  dz
                  ˆ
             0

This expression is for gage pressure; i.e., P(z=0) = 0. Since it is a definite integral evaluated between two
limits we need to be careful about the symbols—z represents the ultimate elevation at which the pressure is
                              ˆ
to be evaluated so I’ve used z to represent the ―dummy variable of integration‖. The ―dummy variable‖
represents the general variation in pressure while the variable z in this case means the specific elevation at
which the integral is to be evaluated. For example, when ρ is constant we integrate this expression to:
              z            z
     PS   g  dz   g  dz   gz 0
                                        z
                  ˆ           ˆ       ˆ
             0            0

                                            ˆ
The last part means that the expression -ρg z is to be evaluated between z (a specific elevation) and 0
(another specific elevation). This evaluates as - ρg (z – 0) = - ρgz, which we knows is correct. Therefore, if
density is constant the hydrostatic gage pressure at any elevation z is given by:
     PS   gz
                          ˆ
and the reduced pressure, P = P – PS, as
     ˆ
     P  P  gz




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                                                                                              Juster, GLY5932

Steady flow in a crack: Consider the general case of a crack of
width W with a general orientation in a vertical plane, and assume
the crack is infinite into the page in the horizontal direction. We
then choose a coordinate system with x parallel to the crack and z
perpendicular to the crack. If the crack is not horizontal then
neither the x nor z axes will be vertical, but this doesn’t matter—we
just have to be careful how we handle the body force due to gravity.
As it turns out, by using the reduced pressure we don’t have to do
anything at all.
The flow in the crack is steady, so the acceleration term vanishes regardless or whether or not we can ignore
the inertial term, so the Navier-Stokes equations in the x- and z- directions are:
         ˆ
        P        2v  2vx 
    0        2x 
                 x         
        x            z 2 
        Pˆ      v2
                        v 
                         2
    0        2z  2z 
                 x
        z             z  
                                    ˆ
We have used the reduced pressure P so there is no gravity term in either equation. Because flow is entirely
in the x-direction we can conclude:
      ˆ
     P
         0;
     z
                2vz  2vz
    vz  0          2 0
               2x    z
                                 v x
and because the flow is steady        = 0 as well. Thus the Navier-Stokes equations reduce to:
                                 x
           ˆ
          dP    d 2vx
    0      
          dx     dz 2
There is only one equation to solve because the equation for the z-direction completely vanishes! Notice I
                                                                        ˆ
have replaced the partial derivatives with normal derivatives, because P ( x) only and vx(z) only. The fluid is
driven through the crack by the reduced pressure gradient in x, which must be constant if the flow is steady,
and the flow is opposed by viscous forces that exactly balance the pressure-gradient drive. In other words,
      ˆ
     dP
         constant
     dx
The Navier-Stokes equation is a second-order (2nd-derivative) differential equation in z. To solve we
rearrange and integrate twice, producing a constant of integration each time:




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                                                                                                Juster, GLY5932

     d 2 v x 1  dP    ˆ
                         
      dz 2        dx 
                          
     dv x 1  dP     ˆ
                       z  C1
      dz        dx 
                       
                     ˆ
             1  dP  2
     vx                z  C1 z  C 2
            2   dx 
                       
This is a general solution to the differential equation. To get a specific solution this needs to be evaluated at
two boundary conditions (since we have two constants), and these are the no-slip conditions for a viscous
fluid at a hard boundary:
     v x  0 at x  0;
     v x  0 at x  W
The first boundary condition yields C2 = 0. Plugging in for the second boundary condition we get
                 ˆ
           1  dP  2
     0           W  C1W
          2   dx 
                  
                   ˆ
             1  dP 
     C1            W
            2   dx 
                    
Inserting this in the general solution and reducing using algebra the final
expression for the steady flow of fluid in a crack is obtained:
                   ˆ
             1  dP  2
     vx        
            2   dx 
                       
                      z  Wz    
                    
The velocity distribution is in the form of a parabola, with the fastest velocity in
the center of the crack and friction causing the velocity to decrease outwards.
For comparison with Darcy’s Law it is useful to convert this equation into an expression for the discharge, Q
= dV/dt, and the specific discharge q = Q/A. Specific discharge has the dimensions of a velocity, is
sometimes called the ―Darcy velocity‖, and indeed can be considered an ―average‖ velocity that fluid has to
produce the required discharge Q. The general relationship between Q and velocity is one that was derived
earlier:
            dV d ( Ax)    dx
     Q                A  Av
            dt    dt      dt
where A is the cross-sectional area through which all the flow passes. Because the velocity varies across the
crack, we must integrate the velocity over the area; i.e., we need to compute
     Q   vdA

The expression for the flow in a crack derived above is for an infinite crack, so we must define a discharge
per unit length (otherwise the discharge would be infinite!). This is equivalent to assuming the crack has a
length L = 1 so that area is defined as A = L × W = 1 × W = W, and dA = dw. The discharge Q is then found
from

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                                                                                                 Juster, GLY5932

                      W
     Q   v x dA   v x dw
                      0

Substituting in the expression for velocity and integrating:
                   ˆ
              1  dP  2
                               
           W
     Q     2  dx  z  Wz dw
           0
                
                
                     
                     
                  ˆ W 2
            1  dP 
       
           2   dx  
                         
                     z  Wz dw 
                   0
                                    W
                ˆ
          1  dP   z 3 Wz 2 
                         
         2   dx   3
                       2 0
                                    W
                ˆ
          1  dP   W 3 W 3 
                        
         2   dx   3
                       2 0
                             
After a little re-arranging, this yields the final expression for the discharge per unit-length of a crack of width
W:
                      ˆ
               W 3  dP 
     Q               
               12  dx 
                       
The specific discharge q = Q/A is found simply by dividing by the cross-sectional area of the crack A = W:
                      ˆ
               W 2  dP 
     q               
               12  dx 
                       


Steady flow in an inclined tube: The next simple geometry we will solve is for
the steady flow of fluid in an inclined tube. Because the tube is circular in cross-
section we will use a cylindrical coordinate system, with the x-direction parallel
to the flow, the r-direction perpendicular to flow with radial symmetry, and z
directed upwards, as usual. The cylindrical coordinate system necessitates a
change in the Navier-Stokes equation in the x-direction (which can be easily
derived and is included in any physics or fluids book):

               P             v  
     0           g x   r x  
               x         r  r  r  
                                       
where gx is the component of the gravitational force in the x-direction.
Notice that because the tube can be inclined the weight of the fluid may
augment or oppose the pressure gradient in x that is driving the flow.
From geometry, we can see that gx = gsinφ, where φ is the angle between x
and the horizontal.
The first two terms include the pressure-gradient force and the
gravitational (body) force, and can be combined as follows:

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                                                                                               Juster, GLY5932

       P         
           g x  P  gx sin  
       x         x
But x sin  = z, the elevation, so we can write:

     P                       ˆ
         g x  P  gz  
                              dP
     x         x            dx
        ˆ
where P is the reduced pressure if density ρ is constant. Just as in the case with the flow in the crack, we
can eliminate the body force from the Navier-Stokes expression by using the reduced pressure—that’s why it
was invented! Also notice that I have replaced the partial derivative with a regular derivative because the
reduced pressure only varies in the x-direction.
The tube flow is then reduced to a single 2nd-order differential equation in r:
               ˆ
              dP   d  dv x  
     0               r      
              dx r  dr  dr  
                            
Although the derivative terms seem messy, they are really nothing more difficult than what we’ve seen
before. To solve we separate and integrate once:
      d  dv x  1  dP    ˆ
     dr r dr     dx r
                             
                            
        dv  1  dP    ˆ
     d r x              rdr
        dr    dx 
                           
       dv          ˆ
              1  dP  r  2
     r x             C
        dr      dx  2
                     
Even before we integrate again, we can solve for the first constant by using the fact that the derivative of the
velocity at r = 0 must vanish; i.e.,
     dvx
                  0
     dr    r 0

Plugging this in immediately shows that C = 0. We then separate and integrate once more:
                    ˆ
               1  dP 
     dv x            rdr
              2  dx 
                     
                ˆ
           1  dP  r 2         ˆ
                          1  dP  2
     vx          C          r  C
          2  dx  2
                       4   dx 
                                 
To evaluate this constant we use the no-slip boundary at the edge, where vx = 0 at r = R:
                  ˆ
            1  dP  2
     0            R C
           4   dx 
                   
                  ˆ
            1  dP  2
    C             R
           4   dx 
                   

                                                                                                     Page 5 of 7
                                                                                                   Juster, GLY5932

Finally, plugging this expression for C into the general solution yields the final answer, the expression for the
velocity of steady flow in an inclined tube:
                  ˆ
             1  dP  2
     vx       
            4 
                       
                dx  r  R
                            2
                                     
                    
This is called the Poiseuille’s equation, after the French physicist J.L.M. Poiseuille
who first solved it. Like the solution for a crack, it describes a parabolic velocity
distribution with the fastest velocity down the center and zero velocity (no-slip
condition) at the outside edge of the tube.
Once again we can determine the discharge Q by integrating the velocity over the area. For the circular tube,
     A   2r  dr

and the discharge Q is found from
                                ˆ
                           1  dP  2                      ˆ
                                                       dP   R 3                 
                                            
            R                 R                                               R
     Q   v x 2r dr           r  R 2 2r dr            r dr  R 2  rdr 
         0              0
                          4  dx 
                                                   2  dx   0
                                                                            0     
After evaluating at the limits of integration and a little re-arranging, we get the final expression:

             R 4      ˆ
                      dP 
     Q                  
             8       dx 
                         
The get the specific discharge, q, we divide A by the cross-sectional area = πR2:

         Q      R2            ˆ
                             dP 
     q                        
        R 2    8           dx 
                                


A quick introduction to Darcy’s Law: In 1856 Henry
Darcy experimentally determined the relationship between
discharge Q and head, h, for flow through an inclined tube:
                 dh
     Q   KA
                 ds
where K is the hydraulic conductivity, A is the cross-
sectional area of the tube, and s is the direction parallel to
the tube. Head is the height that water rises to in a
manometer, measured relative to an arbitrary datum. We
will discuss the meaning of head later, but for now what is
important is the form of Darcy’s Law. The specific
discharge q can quickly be seen as:
          Q      dh
     q      K
          A      ds                                                 Darcy’s experiment. Figure from Middleton &
                                                                    Wilcock (1994).
Specific discharge is sometimes called the ―Darcy velocity‖
because it has the dimensions of velocity (L/T) and because

                                                                                                          Page 6 of 7
                                                                                                  Juster, GLY5932

it represents the average velocity at water flows through the porous medium in order to produce the observed
discharge Q.
The hydraulic conductivity K is a property of both the fluid and the medium. Since the two pertinent fluid
properties are its density ρ and viscosity μ, we can say that K (  ,  , ) , where Φ is a variable that depends
only on the porous medium and includes measures of the abundance, distribution, geometry, and size of
channels through which the water flows. Just to be clear, then:
                            dh
     q   K (  ,  , )                                                                                       (5)
                            ds


Darcy’s Law from first-principles: Because the geometry of the pore network is random and hopelessly
complicated it’s impossible to derive Darcy’s Law from first-principles using the Navier-Stokes equations.
We can, however, get a sense of where Darcy’s Law comes from by comparing it to the solutions for both
the infinite crack and the tube. They are reproduced here just for convenience in comparing:

            Darcy’s Law                            Crack flow                               Tube flow

                                 dh                           ˆ
                                                       W 2  dP                               R2      ˆ
                                                                                                      dP 
        q   K (  ,  , )                     q                                   q              
                                                       12  dx                               8     dx 
                                 ds                                                                    
The similarity of these equations is instructive: all demonstrate the flow is proportional to the gradient of a
           ˆ
function, P in the case of the crack and tube flow, and h in the case of Darcy’s Law. (We will show how
these are related soon.) By comparing like terms it is apparent that the hydraulic conductivity




                                                                                                         Page 7 of 7

								
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