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FAQnewred.gif (906 bytes)          

Aeronautics - Fluid Dynamics - Level 3

Flow Equations

Equations Describing Fluid Flow

The flow of most fluids may be analyzed mathematically by the use of two equations. The first, often referred to as the Continuity Equation, requires that the mass of fluid entering a fixed control volume either leaves that volume or accumulates within it. It is thus a "mass balance" requirement posed in mathematical form, and is a scalar equation. The other governing equation is the Momentum Equation, or Navier-Stokes Equation, and may be thought of as a "momentum balance." As will be seen later, the Navier-Stokes equations are the fluid dynamic equivalent of Newton’s second law, force equals mass times acceleration. The Navier-Stokes equations are vector equations, meaning that there is a separate equation for each of the coordinate directions (usually three).

 

There are many methods to derive these equations. One of the simplest, a control volume approach, is used here to demonstrate the origin of each term. These equations may be used to analyze the flow of most common fluids in internal (pipes) or external (wings) flow situations. Mathematically speaking, these equations are extremely difficult to solve in their raw form. The Navier-Stokes equations are second order, non-homogenous, non-linear partial differential equations that require at least two boundary conditions for solution. Most solutions that exist are for highly simplified flow situations where certain terms in the equations have been eliminated through some rational process.

 

Derivation of the Continuity Equation

Let’s start with a small, fixed volume of fluid somewhere in the middle of a flow stream. This elemental volume has sides of lengths Dx, Dy and Dz (see Figure 1).

 

block.gif (1895 bytes)

Figure 1. Illustration of the elemental volume used to derive the equations.

 

These lengths are short enough so that changes in all fluid properties across the volume may be well approximated with linear functions. On the other hand, these dimensions must me large enough so that the fluid may be considered as a continuum (i.e., much larger than the molecular scale). The mass of fluid in this elemental volume depends on the amount of fluid entering and leaving through the faces. The difference between these two is the rate of mass that accumulates in the volume. The rate of mass entering a face is the product of the density, the fluid velocity and the face area. For example, on the side facing the reader, the density (r) is multiplied by the velocity in the x direction (u) and the area of the face Dy Dz. Thus, the mass flux entering the volume through this face is

Eq042.gif (1098 bytes)

The mass leaving the volume on the opposite side of the volume is again the product of density, velocity and area, but the density and velocity may have changed as the fluid passed through the volume. We will express these changes as small quantities (since our volume is small enough), i.e., r+ Dr and u + Du. The mass flux leaving that face is thus

Eq043.gif (497 bytes)

The negative sign tells us that the mass is leaving, rather than entering, the control volume. Performing the same analysis on the mass entering the volume through the other faces of the volume gives us

Eq044.gif (190 bytes)
Eq045.gif (203 bytes)

where v is the y direction velocity and w is the z direction velocity. Similarly, the mass fluxes leaving the volume on the opposite faces are

Eq046.gif (360 bytes)
Eq047.gif (379 bytes)

All of these added together must equal the mass of fluid accumulating in the volume, Eq046a.gif (187 bytes),

Eq048.gif (388 bytes)

Putting all of these together, we have

Eq049.gif (1478 bytes)

Multiplying out the quantities in parentheses results in the cancellation of some terms and the appearance of higher-order terms such as Dr Du Dy Dz. Since the quantities preceded by D are very small, products of these quantities will be extremely small, depending on the number of D terms included in the product. The terms with four of these will be much smaller than the terms with only three D terms. Thus, all higher order terms are neglected. This leaves

Eq054.gif (1851 bytes)

which, when divided by Eq046a.gif (187 bytes) and rearranged, yields

Eq052.gif (641 bytes)

The application of basic calculus (taking the limit as Dt tends to 0) allows us to write this equation as

Eq050.gif (638 bytes)

where the symbol Eq055a.gif (57 bytes)/Eq055a.gif (57 bytes)t, for example, is a "partial derivative" with respect to time.  Partial derivatives are used when the function (velocity or density in this case) depends on several variables (3 position or spatial variables and time, in this case).

The Continuity Equation may be simplified for some common flow situations as follows. If the fluid may be treated as incompressible (as is the case with water or in low velocity air flows), the density will be constant. The Continuity Equation then becomes

Eq055.gif (1458 bytes)

In the case when the flow is steady (all time derivatives are zero), then

Eq050a.gif (551 bytes)

Note that in this equation, the density and velocities are still functions of the spatial coordinates x,y and z.

 

Derivation of the Momentum (Navier Stokes) Equations

Again we start with a small, fixed volume of fluid somewhere in the middle of a flow stream with sides of lengths Dx, Dy and Dz (see Figure 1). The law of the conservation of momentum states that the rate of change of momentum in the control volume must equal the net momentum flux into the control volume plus any external forces acting on the control volume. We will first deal with the momentum change and flux, then with the external forces. Recall that momentum, the product of mass and velocity, is a vector quantity. This derivation will be based on the momentum in the x direction in Cartesian coordinates. Similar derivations may be demonstrated for the y and z direction. This would make a good exercise to better understand this material. There are thus three different momentum equations that together comprise the Navier-Stokes Equations.

 

Momentum Change and Flux

The time rate of change of momentum within the control volume is

Eq051.gif (344 bytes)

where Eq055a.gif (57 bytes)/Eq055a.gif (57 bytes)t  is the partial derivative operator with respect to time presented above. The flux of momentum in the x direction into each face of the control volume is the product of the mass flux (mf x DA) and the x direction velocity,

Eq056.gif (204 bytes)

where DA is the surface area. For the side facing the reader, the momentum flux is thus

Eq053.gif (207 bytes)

The momentum flux out the opposite side is

Eq058.gif (530 bytes)

The mass flux into the face with normal vector in the negative y direction is as derived for the continuity equation above, or rv. The momentum flux into this face is thus

Eq057.gif (207 bytes)

For the face opposite, the momentum flux out is

Eq059.gif (550 bytes)

Using a similar analysis, it is easily shown that the momentum flux into and out of the faces with normal vectors in the z direction are

Eq060.gif (757 bytes)

Our first expression of the momentum equation comes from adding all of these terms together as expressed in the law of the conservation of momentum

Eq062.gif (2023 bytes)

where SFx is the sum of the external forces on the control volume. The momentum fluxes into the control volume cancel with the first part of the momentum fluxes out of the control volume. Performing this cancellation and moving the momentum fluxes to the left hand side of the equation gives

Eq061.gif (1015 bytes)

Using the product rule, the momentum change and fluxes can be expanded to

Eq063.gif (1401 bytes)

where it is noted that the last four terms in parentheses are the continuity equation times u. Since this must be zero, that leaves

Eq064.gif (860 bytes)

In the same method we can calculate the force in the y and z directions:

Eq064b.gif (844 bytes)

Eq064c.gif (870 bytes)

Derivation of Forces

We now turn our attention to the right hand side of the equation, where the forces on the control volume are represented. There are two types of forces to be included: body forces and surface forces. Body forces act on the entire control volume. The most common body force is that due to gravity. Electromagnetic phenomena may also create body forces, but this is a rather specialized situation. The body force due to gravity is the component of the acceleration due to gravity in the x-direction (gx) times the mass of the fluid control volume (density times volume), or

Surface forces act on only one particular surface of the control volume at a time, and arise due to pressure or viscous stresses. The stress on a surface of the control volume acts in the outward direction, and is given the symbol sij, with two subscripts. The first subscript i indicates the normal direction of the face on which the stress acts, while the second subscript j indicates the direction of the stress. For example, using the cube above, the x axis is the normal direction to the back y-z face of the cube, the y axis is the normal direction to the left x-z face of the cube and the z axis is the normal direction to the top x-y face of the cube.   Also, the -x axis is the normal direction to the front y-z face of the cube, the -y axis is the normal direction to the right x-z face of the cube and the -z axis is the normal direction to the bottom x-y face of the cube. 

The force due to the stress is the product of the stress and the area over which it acts. Thus, on the faces with normals in the x-direction (DyDz), the forces acting in the x-direction due to the direct stresses are

The sum of these two forces is

Similarly, on the faces with normals in the y-direction (DxDz), the forces in the x-direction due to shear stresses sum to

and on the faces with normals in the z-direction (DxDy), the forces in the x-direction due to shear stresses sum to

The sum of all surface forces in the x-direction is thus

The stress sxx includes the pressure p (acting inward, and, hence, has a negative sign) and the normal viscous stress txx. The stresses syx and szx include only viscous shearing stresses tyx and tzx. This gives the force in the x-direction as:


Newtonian/Non-Newtonian Fluids

Most fluids may be classified as Newtonian fluids. A Newtonian fluid is one in which the viscous stress is linearly proportional to the rate of deformation ( ~ du/dy). The constant of proportionality is the viscosity, .  Air would be considered a low viscosity Newtonian fluid, while water would be a medium viscosity Newtonian fluid.  Motor oil and Maple Syrup are high viscosity Newtonian fluids.  Fluids that do not follow the Newtonian behavior law include toothpaste, blood and paints.  For an incompressible Newtonian fluid the viscous stresses are

Some of these terms can be cancelled out using the continuity equation.   The remaining terms, combined with the body force term and put into the equation for the force in the x-direction, give

This gives, as the final expression of the x-momentum equation,

Recall that there are corresponding momentum equations in the y and z directions, namely

and

The derivation of these equations would be a good exercise for the viewer.

 

Derivation of the Energy Equation

In situations where the fluid may be treated as incompressible and temperature differences are small, the continuity and momentum equations are sufficient to specify the velocities and pressure (that is, four equations [continuity+3 momentum] and four unknown quantities [u,v,w and p]).  If the flow is compressible (r is not constant), or if heat flux occurs (temperature not constant), at least one additional equation is required.  In some of these instances, the energy equation may be used.  In the derivation, we use the fact that work is the dot product of velocity and force and the fact that work and energy are related to each other.

The energy equation is, of course, a scalar equation, meaning that it has no particular direction associated with it.  The procedure for deriving the energy equation is similar to those presented for the continuity and momentum equations.   In this case, the change in energy of the fluid within the control volume is equal to the net thermal energy transferred into the control volume plus the rate of work done by external forces.  The energy of the fluid is expressed in this case as the sum of the absolute thermodynamic internal energy per unit mass, e, and the kinetic energy per unit mass, 1/2 V2, (V is the magnitude of the velocity vector).   The change in total energy per unit volume of the fluid in the control volume is

As was found above for the momentum transfer into and out of the control volume, the net transfer of energy per unit volume through the control volume is

This equation is obtained by replacing the momentum term (density times velocity) by the energy term (density times the sum of the internal and kinetic energies).   The net thermal energy transferred into the control volume is determined by the heat flux qi , positive for heat going from within the control volume to the surroundings in the ith-direction (that is, the x-,y- or z-direction).   The total heat per unit volume transferred into the control volume is

The rate of work per unit volume being done by the surface forces is found by multiplying the stress, sij, by the velocity in the j-direction for each i face.  Similar to the procedure above for the stresses in the momentum equation, the net rate of work being done from all sides is

Lastly, the rate of work per unit volume done by the gravity force vector (g = gx i + gy j + gz k ), is

Putting all of these terms together, we have


Fourier's Heat Conduction

We will now use Fourier's Law of Heat Conduction that relates the heat flow in the ith direction, qi, to the rate of change of temperature in the ith direction, namely,

                                                                            qi = -kiA Eq055a.gif (57 bytes)T/Eq055a.gif (57 bytes)xi             i=1,2,3 represents the x,y,z directions

where ki  is the heat conduction coefficient in the ith-direction, A represents the surface area perpendicular to the ith-direction, and T represents the temperature of the flow.  From the Zeroth Law of Thermodynamics, heat flows from a location of higher temperature to that of a lower temperature.  If, for example, this is in the x-direction, then Eq055a.gif (57 bytes)T/Eq055a.gif (57 bytes)x is negative.  But since heat flow is considered positive when flowing from the control volume to the surroundings (meaning, in this case, in the positive x-direction), then for qx to be positive, we need the minus sign as indicated.  Thus the heat flow rate per unit volume terms

become

Eq055a.gif (57 bytes)(kx Eq055a.gif (57 bytes)T/Eq055a.gif (57 bytes)x)/Eq055a.gif (57 bytes)x + Eq055a.gif (57 bytes)(ky Eq055a.gif (57 bytes)T/Eq055a.gif (57 bytes)y)/Eq055a.gif (57 bytes)y + Eq055a.gif (57 bytes)(kz Eq055a.gif (57 bytes)T/Eq055a.gif (57 bytes)z)/Eq055a.gif (57 bytes)z

which, in the case of constant heat conduction coefficient (kx = ky = kz = k), become


The rate of work per unit volume being done by the surface forces is found by multiplying the stress sij by the velocity in the j direction for each i face. Similar to the procedure above for the stresses in the momentum equation, the net rate of work per unit volume being done from all sides is


Using the relationship between surface stresses and velocity gradients for incompressible Newtonian fluids, this becomes


Lastly, the rate of work done per unit volume by gravity is


Putting all of these terms together, we have

This equation demonstrates that, per unit volume, the change in energy of the fluid moving through a control volume is equal to the rate of heat transferred into the control volume plus the rate of work done by surface forces plus the rate of work done by gravity. This expression of the energy equation is valid for most applications. However, some specialized situations may require additional terms representing the contributions of other sources (electromagnetic forces, etc.).

 


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Updated: June 02, 2009