AuthorMircea Orasanu

Ideal fluids and Euler’s equation

To obtain the equation of motion for a fluid we appeal to Newton’s Second Law–the mass of a fluid element times its acceleration is equal to the net force acting on that fluid element. If we take an element of unit volume, then we have

where is the force per unit volume on a fluid element. This force may have several contributions. The first is the `internal'' force which is due to viscous dissipation, which we will ignore for right now. The second set are`

body forces” which act throughout the volume of the fluid, such as the gravitational force. The third force is due to pressure gradients within the fluid. To see how this works, consider a cube of fluid, with dimensions , , and , as shown in Fig. 2.6. Derivation of the force on a fluid element due to pressure gradients.

The force on the top face at position x is , while the force on the bottom face is ; subtracting, we see that the net force in the x-direction is , so the pressure per unit volume is . Repeating for the y and z directions, we find the net force per unit volume . Therefore, if we ignore viscosity and gravity for the moment, we have

Now comes the tricky part. What is the acceleration of the fluid? We want the acceleration of a particular element of the fluid; the coordinates of this fluid element change in time as the fluid flows. In a time interval , the x-coordinate changes by , the y-coordinate by , and the z-coordinate by . The velocity then becomes

To calculate the acceleration, we need to find the rate of change of the velocity:

We see that the acceleration is not simply . The reason for this is that even if , so that the velocity at a given point is not changing, that doesn’t mean that a fluid element is not accelerating. A good example is circular flow in a bucket. If the flow is steady, then at a point in the bucket , even though a fluid element in the bucket is experiencing a centripetal acceleration. The term is nonlinear, and is the source of all of the difficulties in fluid mechanics. Pulling together all of the pieces, we have for our equation of motion

This is known as Euler’s equation . This equation and the equation of continuity are the governing equations of nonviscous fluid flow.

For an incompressible fluid, the Euler equation can be written in a somewhat different form which is often useful for applications. We use the following identity from vector calculus:

[see the PQRG, p. 161, next to last equation]. Next we use , and the fact that the fluid is incompressible so that is constant, to rewrite Eq. (2.20) as

This is still a complicated equation, but it has interesting consequences. If we apply × to both sides of this equation, and remenber that the curl of a gradient is zero, we obtain an equation that tells us how the vorticity moves:

/t + ×(×v) = 0.

In particular, this equation shows that remains zero at all times if it is zero initially. It also it allows us to check if some can stay fixed in time and produce a steady flow.

Next: Bernoulli’s equation Up: Steady flows of incompressible Previous: Steady flows of incompressible

Notes by A. Dorsey, revised by V. Celli, U. of Virginia

Wed Sep 10 01:02:02 EDT 1997 Differential Equations of Mass and Momentum Conservations:

For incompressible flow:

Continuity:

NS equation:

` where`

Solution Procedure:

Step 1: Assumptions

• Used to simplify NS equation (reducing PDE to ODE)

• Common assumptions are:

o Incompressible ® r and m are constant

o Steady state ® (time independent)

o Fully developed ® (space independent)

• Other assumptions may include:

o Neglect gravity ®

o No pressure gradient ®

Step 2: Boundary Conditions

• Needed for solving integration constants in Step 5

• Sketching the problem will help to define boundary conditions

Figure 2.6:

Step 3: Continuity Equation

•

Step 4: Momentum Equation

• x – momentum:

• y – momentum:

Step 5: Find the Velocity Profile

• Integrate the resulting momentum equation(s) from Step 4

• Apply boundary conditions from Step 2 to solve for integration constants

Step 6: Solve for Flow Rate or Wall Shear Stress

• flow rate:

• wall shear stress:

Example:

A conveyor belt, which moves at constant velocity Uc, transports oil from an oil bath to a conduit above. The vacuum pump assists in the process by imposing a negative pressure gradient in the flow direction (i.e., dp/dx < 0). Assume the flow is fully developed in the x-direction. Find the velocity profile and flow rate.. We shall first choose an infinitesimally small control around the point (x, y, z) and then shrink this volume to a point by allowing DV ® 0. Recall from the Newton’s second law of motion:

where the left hand side represents the sum of the external forces and the right hand side represents the inertia forces. Since we have already derived the expression for

as the acceleration at a point, we’ll use this result here.

Let us review some facts about stresses. The stress tensor that you were taught in the strength of materials and represented by the matrix below has 9 components.

Note that “s” represeThe general nonhomogeneous differential equation is given by

(1)

and the homogeneous equation is

(2)

(3)

Now attempt to convert the equation from

(4)

to one with constant coefficients

(5)

by using the standard transformation for linear second-order ordinary differential equations. Comparing (3) and (5), the functions and are

(6)

(7)

Let and define

RGY

Author Horia Orasanu

` Review of Classical Mechanics`

Coordinates and Trajectories

Two of the primary goals of classical physics is to describe the location of an object in space and how the object’s location will change with time.

A. Coordinates

`A set of numbers that uniquely describes mathematically the spatial location of an object is called the object's coordinates.`

```
```Some important points about coordinates:

a) An object has an actual physical location independent of our describing it mathematically. Thus, no particular coordinate system is required by the physical world. We choose the system to simplify mathematical calculations.

b) Between any two coordinate systems there are a unique set of equations (transformation equations) that allows us to take measurements made in one system to predict measurements made in the other system.

c) Since there is no preferred coordinate system, the true mathematical expressions of the laws of the physical world should be unchanged in form under coordinate transformations (i.e. Invariant under coordinate changes).

`d) The origin for any coordinate system is arbitrary. Since the value for a coordinate depends on the origin chosen, you should specify your origin in every problem.`

Before Class Exercise:

In the x-y coordinate system, the position of the ball shown below is

Write the position vector for the ball in the u-v orthogonal coordinate system.

Write the two transformation equations needed to convert spatial measurements in the x-y frame into measurements in the u-v frame.

Trajectories

`The set of values that an object's coordinates take over an interval of time is called a trajectory.`

```
```The central problem in classical mechanics is to find the equations of motion and then solve these equations to find the object's trajectory (i.e. how the bodies position changes with time).

We will see later that the very concept of a trajectory is impossible in quantum mechanics (Chapter 2).

This leads us to the following questions (answered in Chapter 3):

1. What quantities should be used to describe a quantum system?

2. What do we use instead of trajectories to mathematically describe the time evolution of a quantum system?

3. What replaces the equations of motion in solving quantum systems?

`4. What is the connection that links the results of quantum mechanics to those of classical physics?`

C. Constraints

`In many physics problems, the values of an object coordinates are limited in some manner. An example is a particle moving in the x-y plane. In this case, the particle's z-coordinate is limited to some constant values. Such a condition is known as a constraint condition. `

```
```

`Constraints will reduce a system's degrees of freedom and the number of independent coordinates needed to specify the system.`

Holonomic and Non-Holonomic Constraint

`If the constraint condition can be written as an equation with the right hand side equal to zero then the constraint is said to be holonomic otherwise the constraint is said to be non-holonomic.`

EXAMPLE: A particle moves in the x-y plane with z = 3. The constraint condition can be written as so it is a holonomic constraint.

Generalized Coordinates

`It is always possible to write the location of an object in terms of cartessian, cylindrical, or polar coordinates. In more complicated problems, it is often more convenient to use a different set of coordinates due to the form of the constraint equations and forces.`

```
```Consider a lure attached to a fishing line as it unwraps from the spool as shown below:

You can describe the location of the lure by using either the cylindrical coordinates (r,) or the length of the line, l, and the angle the line makes with the spool, . You could also describe the motion in terms of r and or in terms of l and . A wise choice can greatly simplify the math.

`Independent coordinates that serve to uniquely determine the orientation and location of a system in physical space are called generalized or canonical or good coordinates.`

E. Generalized Velocity

`A generalized velocity is the time derivative of a generalized coordinate.`

Before Class Exercise:

The location of a particle in cylindrical coordinates is given by .

Show that the velocity of the particle is .

Hint: Write and in terms of and .

Author Mircea Orasanu

So it is easy to deduce that, the inverse of this problem is to consider the orthogonal straight tangents as fixed and the ellipse given as mobile, returning well to the problem of Professor Valdivieso. Clearly the locus, solving this problem, it will be a portion of the arc of a circle in ortante positive.

However, Professor Valdivieso, even accepting that this was a good solution, required to me follow the long road, namely the analytic geometric deduction directly as initialy requested. Result: I was not able to find such a solution.

The years passed and my attempts were in vain to find a solution following the method that I had being requested. Several colleague whom I consulted , could not resolve it.

So, already retired and close to fulfilling 85 years, I went back on this issue. The Google scholar was not help to find what I was are looking for. Finally, using the Central Library at the Catholic University and recalling those old fragile , but extraordinary in its content French books, I found what I was looking for in:

y=mx + n, is the equation of a secant to the ellipse, with m= as secant and n = ordinate at the intersection of the line with the axis y.

If one imagines that this secant, remaining parallel to itsef, moves herself, and so that the two points of intersection with the curve are close to each other indefinitely, it is clear that at the limit, the secant will be tangent.

Therefore,for y= mx+n representing a tangent is necessary and sufficient that it have a common solution with the equation:

of the ellipse.

This is equivalent to saying that the equation resulting from the elimination of y equations between the two precedents equations, namely:

must have two equal roots. Expressing this condition results:

where it obtain

Replacing the equation of the line y=mx+n equations obtain two parallel tangents to the given ellipse :

` ( 1 )`

```
```

` ( 2 )`

The perpendicular to the first of these equations is:

` ( 3 )`

These equations and the ellipse are referred to as coordinate axes, to the proper axis of the ellipse.

Let’ see now SECTION B)

Distance from a given point to a given straight line

Often we express the distance from one point to a line in terms of factors in the equation of the line. To do this, we compare two forms of the equation of a line, as follows:

General Equation: Ax + By + C = 0

Normal Form:

The general equation and the normal form represent the same line. Hence

A , coefficient x in general equation, is proportional to coefficient of x in the normal way.

By the same token, B is proportional to sen , and C is proportional

to –p.

Calling k the constant of proportionality, obtain:

Raising the square and then adding obtain:across P, parallel to LK. The distance of this straight from the origen is OS, and the difference between OS and P is d

We get an expression for d, based on the coordinates of P, as follows:

and

Returning to expressions for and p in terms of A,B, and C

(Coefficients of the general equation), we have:

The formula for d,the denominator in each of the expressions is the same. Therefore we can combine terms, as follows:

The formula for d, denominator in each of the expressions is the same. Therefore we can combine terms, as follows:

Now, apply this formula to the equations (1) and (3) to establish the distances from the point matching corresponding coeficients. The result is:

= ,

Removing m among these equations will provide a new equation that expressed a constant relationship between the coordinates of any point of locus requested; this equation is then the locus we are looking for:

In particular, if one wants to see the locus described by the center of the ellipse mobile, is necessary make:

and find:

y= , removing m is done immediately, taking the square of precedents equations and adding immediately. The result is:

equation of a circle,into which is inserted the arch which is the locus sought; radius of that circle is:

That was it. To this day I can not explain that Professor Raul Valdivieso expected that any students could develop this problem, in an examination or in any other circumstance, this problem.

It is not that there is a much simpler solution? . I venture to think that has to be one,because the distinguished professor was very demanding but very thoughtful and fair.

Raúl, Civil Engineer ,Pontificia Universidad Católica de Chile.

Vector Differential Operators

Rectangular (Cartesian) coordinate s

Cylindrical c The most basic introduction to mechanics requires vector concepts in addition to scalar ones. Electric and magnetic vector fields are fundamental concepts for understanding the interactions of charged particles and the behavior of light. Our calculus techniques must be generalized to include differential and integral operations on vector quantities.

Concepts of primary interest:

Vector and scalar fields

Gradient

The Gradient Operators

[VCA.9]

The gradient operators have been written with the differential operators rightmost to emphasize that the derivatives do not operate on the unit vectors or coordinate forms involved in the representation of the gradient operator itself. These operators are occasionally written with their coordinate directions on the right. This variant is just a notation, and it is not intended that their evaluation be altered. As represented above, the differential operators are to act on everything to their right.

Cartesian-Coordinate Cube: Flux evaluation for the faces with normal directions and – .

An expression for the divergence in Cartesian coordinates can be developed using a small coordinate ‘cube’ with edges that are coordinate line elements. The cube has corners at (x,y,z) and at (x+x,y+y,z+z). The flux of out of the volume is:

[VCA.12]

The sides naturally form pairs for each coordinate. One such pair per example provides a model for the complete calculation.

Only terms for two of the six sides are displayed, the sides perpendicular to the x-axis with and – being the respective outward-directed normals. The symbol represents a mean value of y where (y< <y+y). Substituting into the definition,

Defn:

Noting that and using the definition of the partial derivative, this expression reduces to:

` (Cartesian) [VCA.13]`

Note that in the Cartesian representation there are contributions to the divergence of a vector field from any component that varies when its associated coordinate varies. A non-zero contribution results if depends on x ( ). The Cartesian form is very suggestive and leads to the notation: . Both are read as the divergence of where, by the inflection of your voice, the quantity is clearly specified to be a vector field.

` ‘grad’, ‘del’ or , is more formally known as: nabla.`