Friction factor
 
The friction factor is a dimensionless number, which provides an idea about the magnitude of shear stress produced by a solid boundary as fluid flows. This is defined as the ratio of shear stress at the wall and the kinetic energy head of the fluid,. Here, ρ is the density and is the average velocity of fluid. The friction factor is thereby defined as
 
where, is the shear force per unit area on the wall of the tube. This may be calculated as shown below
 
Here, first minus sign is used as the inside surface of the tube wall has outer normal in the negative r direction and second minus sign is used because is treated here as momentum flux. If is treated as actual shear force then positive sign would have to be taken. For fully developed laminar flow, the velocity profile is parabolic and is given by
 
Evaluating the velocity gradient at the wall (r=R), we have
 
Thus, the shear stress considered as momentum flux is given by
 
 
or
 
 
The friction factor may now be calculated as shown below
 
 
or
 
 
Equation (11.7) shows that the friction factor in laminar flow region depends only on the Reynolds number. Clearly, the friction factor is also a dimensionless number.
 
Friction factor in turbulent flow
 
Fig 11.1 Smooth and rough surface of pipe
 
In turbulent flow, the friction factor also depends on the surface of the pipe. A rough pipe leads to higher turbulence than a smoother pipe, so that the friction factor for smoother pipes is less than that for rougher pipes. The ratio of surface roughness height (∈) to pipe diameter (D) is used to quantify the “roughness” of the pipe surface. In practice, the shear stress on the wall may be calculated by measuring the pressure drop across the pipe for a given flow rate. Thus, friction factor may be calculated as the function of Reynolds number and plotted on a log-log plot for a given surface roughness. The curves are different for different surface roughness as shown in figure. (11.2). The collection of these f-Re plots is called Moody Chart as shown in figure below, and can be used for estimating the friction factor for given flow parameters.
 
Ref: http://www.brighthub.com/engineering/civil/articles
 
Solution of some more fluid flow problems by shell momentum balance approach
 
In this section, we solve a few more fluid mechanics problems in simple geometries using the shell momentum balance approach. The detail procedure, which was also used in previous example, is outlined below.
 

1) Make a diagram of the flow geometry with the appropriate coordinate system

 

2) Specify all necessary assumptions

 
3) Intuitively assume the velocity profile
 
This is an important step for solving these problems. In laminar flow, the fluid flows in parallel layer without mixing. Thus, it is easy to guess the non-zero components of velocities by intuition.
 
4) Apply of the equation of continuity to modify the velocity profile
 
5) Determine the non-zero shear stress component(s)
 
Since the shear stress components depend on the velocity profile, the non-zero shear stress components may now be determine.

 

6) Determine control volume and make shell momentum balance for the control volume
 

Draw control volume in system diagram according to system shape, size and problem statement. The selection of proper control volume is very important to solve problem correctly. The control volume should be select in such way that it can be easily integrated for whole system. The differential length of control volume should be taken in direction of changing velocity.

Write momentum balance equation for the control volume. The shear stress may be considered as shear force or as momentum flux, both provide the same results as shown in previous example . Write down all surface and body force acting on the fluid carefully. Finally obtain a appropriate differential equation and integrate.

 
7) Boundary conditions
 
Use appropriate boundary conditions which help us to determine the constant of integration in above step.
 

Falling film on an inclined flat surface

 
An inclined surface of length L and width W is situated at an angle Β to the vertical direction as shown in Fig. (11.3). A Newtonian fluid is freely falling on the surface as a film of thickness δ. Assuming the flow to be laminar, determine the velocity profile, flow rate and shear force on the surface by the fluid.
 
Solution
Fig 11.3 Laminar flow on an inclined surface
 
Assumptions
 
  • Constant density, viscosity
  • Steady state
  • Laminar flow (simple shear flow)
  • Fully developed flow
  • Newton's law of viscosity is applicable
 
Assume velocity profile

The fluid is flowing in the z direction, hence only the z component of velocity is non-zero. Thus, we may assume
 
, and
 
We may further assume that vz does not depends upon y coordinate. Since the flow is steady, vz does not depend on time. Thus,
 
 
Using the equation of continuity in the cartesian coordinates for constant fluid density, we have
 
 
which reduces to
 
 
Equation (11.11) indicates that vz does not depend on the z coordinate. Thus,
 
 
There are nine components of the shear stress as shear force or momentum flux, namely
 
 
Since vz is only the non-zero velocity, and also it is the function of x coordinate , is the only significant component of shear stress and we need to write momentum balance only in z direction. Because the pressure is same at both ends of the inclined plane, there is no pressure force on the fluid. Now, we can solve this problem by assuming shear stress as a shear force or shear stress as momentum flux.
 
Assuming as momentum flux
Draw a control volume of length L, width W and differential thickness dx.
 
Fig 11.3 Control volume for falling film problem
 
Momentum balance in x direction
 
Rate of momentum flux entering CV due to viscous transport at
 
 
Rate of momentum flux leaving CV due to viscous transport at
 
 
Gravity force acting on fluid in z direction
 
 
Rate of momentum flux entering in CV due to convective transport
 
 
Rate of momentum flux leaving from CV due to convective transport
 
 
Now, when above terms are substituted for z-momentum balance, we obtain
 
 
Since the velocity vz does not depends on z coordinate, the first two terms cancel out and we obtain
 
Dividing Equation (11.19) by volume of the control volume (LWΔx), we have
 
 
As Δx→0 , The Equation (11.20)simplified to
 
 
The Newton’s law of viscosity (here, shear stress is defined as momentum flux) is given by
 
 
By integrating the Equation (11.25), we have
 
or
 
 
The above equation requires two boundary conditions for determining c1 and c2.
 
Boundary conditions
 
1 At x=0 the liquid surface is in contact with air where the shear stresses at both gas liquid phases should be equal. Thus,
 
 
Since both may be assumed Newtonian fluids, we have
 
 

where ρg is the density and µg is the viscosity of air. Thus

 
 
Since, µg and ρg is much smaller than µ and ρ, and Equation (11.30) may be approximately written as
 
 
Substituting above boundary condition in Equation (11.26), we obtain
 
 
 2. At x=δ no slip boundary condition may be applied, i.e.,
 
at
 
Thus, from Equation (11.27), we get
 
 
or
 
 
Finally the velocity profile is obtained as
 
 
or