Fluid Mechanics for MAP/Differential Analysis of Fluid Flow

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Differential relations for a fluid particle

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We are interested in the distribution of field properties at each point in space. Therefore, we analyze an infinitesimal region of a flow by applying the RTT to an infinitesimal control volume, or , to a infinitesimal fluid system.

Conservation of mass

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The differential control volume dV and the mass flux through its surfaces


The conservation of mass according to RTT


 


or in tensor form


 


The differential volume is selected to be so small that density   can be accepted to be uniform within this volume. Thus the first integral in 1 is:


 


The flux term (second integral term) in the equation of conservation of mass can be analyzed in groups:


 


Let's look to the surfaces perpendicular to  


 


 


Similarly, the flux integrals through surfaces perpendicular to   and   are


 


 


Hence the flux integral reads;


 


The conservation of mass equation becomes:


 


Droping the  , we reach to the final form of the conservation of mass:


 


This equation is also called continuity equation. It can be written in vector form as:


 


For a steady flow, continuity equation becomes:


 

.


For incompresible flow, i.e.  :


 

Example

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For a two dimensional, steady and incompressible flow in   plane given by:


 


Find how many possible   can exist.

For incompressible steady flow:

 


in two diemnsions

 


Thus,

 


This is an expression for the rate of change of   velocity while keeping

  constant. Therefore the integral of this equation reads


 


Thus, any function   is allowable.

Example

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Compressible and unsteady flow inside a piston

Consider one-dimensional flow in the piston. The piston suddenly moves with the velocity   . Assume uniform   in the piston and a linear change of velocity   such that   at the bottom ( ) and   on the piston ( ), i.e.


 


Obtain a function for the density as a function of time.


The conservation of mass equation is:


 


For one-dimensional flow and uniform  , this equaiton simplifies to


 


 


 



 


 


 


 


The same problem can be solved by using the integral approach with a deforming control volume.

The differential equation of linear momentum

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The differential control volume dV and the flux of   (momentum per unit volume in i-direction) through the surfaces perpendicular to   axis


The integral equation for the momentum conservation is


 


For the first integral we assume   and   are uniform within dV, and dV is so small that:


 


Analyze the flux of the   momentum terms through the faces perpendicular to the axis:


 


First consider the flux of   (momentum per unit volume in i-direction) through the surfaces perpendicular to   axis:


 


 


Similarly, the momentum flux through the surfaces in other directions read

 

,


 

.


Rearranging the equation for   we obtain:


 


We can simplify further:

 


 


Hence

 


Let's look to the forces on the exposed on the diffrential control volume:


 


Here, only gravitational force is considered as a body force. Thus,


 


 
Differential surface forces

Surface forces are the stresses acting on the control surfaces.   can be resolved into three components.   is normal to dA.   are tangent to dA:


 


 


  is a normal stress whereas   is a shear stress. The shear stresses are also designated by  .


 
Stresses on the surface of differential control volume

Thus, the surface forces are due to stresses on the surfaces of the control surface.


 
 
The positive stress directions









We define the positive direction for the stress as the positive coordinate direction on the surfaces (e.g. on ABCD) for which the outwards normal is in the positive coordinate direction . If the outward normal represents the negative direction (A'B'C'D'), then the stresses are considered positive if directed in the negative coordinate directions.


The stresses on the surface   are the sum of pressure plus the viscous stresses which arise from motion with velocity gradients:


 


  has a minus sign since the force due to pressure acts opposite to the surface normal.


 
Stresses on the surface of the differential control volume in the x1 direction






Let us look to the differential surface force in the   direction:


 


Noting that   and   (4),


 


Thus in tensor form the differential surface forces in  'th direction can be written as


 


Note that   is a symmetric tensor, i.e.


 


Hence, the diffential surface forces reads:


 


Inserting   and   into (2),


 

and canceling   we obtain


 


Expanding the substatial derivative at the left hand side,


 


We obtain the the most general form of momentum equation which is valid for any fluid (Newtonian, Non-newtonian, Compressible, etc.). It is non-linear due to the   term at the LHS. Efect of Newtonian and Non-newtonian properties appears in the formulation of the viscous stresses  .   will introduce also non-linearity when the fluid is non-Newtonian.


It should be noted that these formulations are based on stress conception which was thought to exist in fluids in motion. However it is known that   can be expressed as momentum transfer per unit area and time. Thus it can be considered as molecular momentum transport term. Derivations based on this concept requires a molecular approach (which is lengthy). The students should be aware that   causes momentum transport when there is a gradient of velocity.

Linear momentum equation for Newtonian Fluid: "Navier-Stokes Equation"

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For a Newtonian fluid, the viscous stresses are defined as:

 


Note that derivation of this relation is beyond the scaope of this course.

Thus, the momentum equation (6) becomes


 


For a flow with constant viscosity ( ):


 


since,

 


then,

 


For an incompressible flow

 


hence assuming that the viscosity is constant, it can be easily shown that the momentum equation reduces to

 

Euler's equation: Inviscid flow

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When the velocity gradients in the flow is negligible and/or the Reynolds number takes very high values, the viscous stresses can be neglected:

 

Since, the viscous stresses are proportional to viscosity:

 

for flows, where   is neglected, the flow is called frictionless or inviscid, although there is a finite viscosity of the flow. Accordingly, the linear momentum equation reduces to

 

Euler's equation in streamline coordinates

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Differential control volume along streamline coordinates and the forces on it for a inviscid flow
 
Differential control volume along streamline coordinates and the forces on it for a inviscid flow

Euler's equation take a special form along and normal to a streamline with which one can see the dependency between the pressure, velocity and curvature of the streamline.



To obtain Euler's equation in s-direction, apply Newton's second law in s-direction in the absence of viscous forces.


 


Omitting   would deliver


 


Since

 


then the Equler's equation along a streamline reads

 


For a steady flow and by neglecting body forces,


 


it can be seen that decrease in velocity means increase in pressure as is indicated by the Bernoulli equation.

To obtain Euler's equation in n direction, apply Newton's second law in the absence of viscous forces and for a steady flow.


 


 


Since,

 


Then,

 


For a steady flow, the normal acceleration of the fluid is towards the center of curvature of the streamline:


 


Hence,

 


For an unsteady flow,

 


For steady flow neglecting body forces, the Euler's equation normal to the streamline is


 


which indicates that pressure increases in a direction outwards from the center of the curvature of the streamlines. In other words, pressure drops towards the center of curvature, which, consequently creates a potential difference in terms of pressure and forces the fluid to change its direction. For a straight streamline  , there is no pressure variation normal to the streamline.

Bernoulli equation: Integration of Euler's equation along a streamline for a steady flow

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For a steady flow, Euler's equation along a streamline reads,


 


If a fluid particle moves a distance ds, along a streamline, since every variable becomes a function of  :


 


 


 

Integration of the Euler equation between two locations, 1 and 2, along   reads

 

For incompressible flow   and after changing the notation as:   and  , the integration results in

 

or in its most beloved form:

 


In other words along a streamline:

 

Note that due to the assumptions made during the derivation, the following restrictions applies to this equation: The flow should be steady, incompressible, frictionless and the equation is valid only along a streamline.


Different forms of Bernoulli equation

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The common forms of Bernoulli equation are as follows:


Energy form (per unit mass)

 


Pressure form


 


Head form


 

Static, stagnation and dynamic pressures

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How do we measure pressure? When the streamlines are parallel to the wall we can use pressure taps.

If the measured location is far from the wall, static pressure measurements can be made by a static pressure probe.

The stagnation pressure is the value obtained when a flowing fluid is decelerated to zero velocity by a frictionless flow process. The Stagnation pressure can be calculated as follows:


 


 


when  


 


where the last term is the dynamic pressure.


 
Pitot tube used in racing car
 
Figure: a)Jet is impinging to the wall and stagnating at the point of impingment b)Schematics of a of a pressure tap in a channel c)Static pressure probe

If we know the pressure difference  , we can calculate the   velocity.


 


The stagnation pressure is measured in the laboratory using a probe that faces directly upstream flow.

Such a probe is called a stagnation pressure probe or Pitot tube . Thus, using a pressure tap and a Pitot tube one can measure local velocity:


 


 


Thus, measuring   one can determine  .

 
Pitot-static tube for velocity measurement

However, in the absence of a wall with well defined location, the velocity can be measured by a Pitot-static tube. The pressure is measured at B and C; assuming  .

Hence,

 

Unsteady Bernoulli equation

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The Euler's equation along a streamline is:


 


along ds,


 


hence,


 


Integration between two points along a streamline is:


 


For incompressible flow,  , thus the integral reads


 


The unsetady Bernoulli equation involves the integration of the time gradient of the velocity between two points.: