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You may have wondered where the velocity profile

for laminar flow between two parallel plates came from in

homework problem 4.40. It was given as

u = [4 u_max y (h - y)]/h^2

where u is the velocity in the x direction, y = 0 starts at the

bottom plate, and y = h at the top plate. The transverse

velocity components v and w are zero, and the flow is

steady. This velocity profile comes from the Navier-Stokes

equation for x momentum

d,t u + u d,x u + v d,y u + w d,z u = - (d,x p)/rho + g_x +

nu del^2 u

where d,t etc. indicate partial differentiation.

The first term on the left is zero by the steady state

assumption. The second is zero because u does not change

with x. The third is zero because v is zero. The fourth is

zero because w = 0.

The streamwise pressure gradient is constant and

negative, so d,x p = - K. Gravity has no horizontal term, so

g_x = 0. The last term is nu [ d,x^2 u + d,y^2 u + d,z^2 u ]

= nu [d^2 u / dy^2 ], since u depends only on y. Thus

d^2 u / dy^2 = K/mu

is the ordinary differential equation that must be integrated.

The first integral is du/dy = - Ky/mu + C, where C is a

constant. The second integral gives u = - Ky^2/mu + Cy +

D, where D is a constant. By the no slip condition u = 0 at y

= 0 and y = h. From the first of these boundary conditions,

D = 0. By the second C = Kh/2mu . Thus;

u = [K/2mu] y [h - y] .

Now u = u_max at y = h/2, so [K/2mu] = - 4 u_max / h^2.

Substituting for [K/2mu] gives

u = [4 u_max y (h - y)]/h^2

which was the given parabolic velocity profile of the

homework problem.

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Where did the expression eps = nu (du/dy)^2 come

from that was substituted into the energy equation in the

solution of 4.40 given in the last lecture?

Solution: The viscous dissipation rate per unit mass

eps is given by [tau dot dot e ]/ rho, where the viscous stress

tensor tau = 2 mu e . The rate of strain tensor e has

components eij = [d,xj ui + d,xi ui ]/2, but only e12 = e21 =

du/dy / 2 are not zero. Thus eps = 2 mu e^2 = 2 mu

[(du/dy)^2 /2 + (du/dy)^2 /2 ] = nu (du/dy)^2 , as expected.

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Problems 4.56, 65, 69, 86.

4.56. The velocity potential phi = Kxy, K = constant gives

a set of hyperbolas symmetric about the x and y axes. The

orthogonal streamlines represent a stagnation point flow

against a plate tilted at an angle of 45 degrees up from the x

axis about the origin.

4.65. We calculated the z vorticity for this flow last time and

found it was zero. We also calculated the velocity potential

phi to be K theta.

4.69. Find the stream function and plot some streamlines

for the combination of a counterclockwise line vortex K at

(x,y) = (a,0) and an equal line vortex placed at (-a,0).

Solution: Because the equations of motion are linear,

we can add the individual stream functions for the two

vortices. The first is psi_1 = - K ln[(x-a)^2 + y^2]^1/2 ,

and the second is psi_2 = - K ln[(x+a)^2 + y^2]^1/2 .

Plotting this gives a "cat's-eyes" pattern.

4.86. SAE 10 oil has a parabolic velocity profile in laminar

flow between the plates, with average velocity = 2 u_max / 3

as shown in example 4.11, p 238 of the text. Thus v_ave =

K h^2 / 3 nu , where K is the pressure gradient and h = 4

mm is half the plate separation. For the oil rho = 870

kg/m^3, and mu = 0.104 kg/m s. From the manometer

reading the pressure difference for a meter distance

downstream is 7463 Pa. This is K. Solve for v_ave =

0.383 m/s. The flow rate Q = v_ave A = 0.383 x 0.008 =

0.00306 m^3/s m.

*****

1. Irrotational (ideal) flows.

* Superposition of ideal flows.

**uniform flow

**source

**counterclockwise vortex

2. Viscous flows.

3. Dimensional analysis.

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Superposition of ideal flows.

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Recall that the assumptions for the existence of ideal

flows are that 1. the flow is steady, 2. the fluid is constant

density, 3. the flow is two-dimensional, and 4. the flow is

irrotational. Assumptions 1-3 establish that a stream function

psi will exist that satisfies the continuity equation del dot v = 0.

Assumption 4 guarantees that the velocity can be expressed

as the gradient of a velocity potential phi. Because the

density is constant and the flow is steady, phi must satisfy

LaPlace's equation (substitute d,x phi and d,y phi in del dot v = 0).

del^2 phi = 0.

Because the flow is irrotational, the streamfunction psi also

satisfies LaPlace's equation (substitute d,y psi and - d,x psi

in w_z = d,x v - d,y u).

del^2 psi = 0.

Because LaPlace's equation is linear, any number of

solutions can be added to produce another solution.

If psi_1 and psi_2 are solutions of del^2 psi = 0, then

psi_3 = psi_1 + psi_2 is a solution, by subsitution. This is

the principle of superposition. Quite sophisticated ideal

flows can be constructed by combinations of the stream

functions and potential functions of simple ideal flows.

Many fluid mechanics books have been written devoted to

this process. Chapter 8 of White shows some of the most

useful applications and reviews some of the elementary

flows. Some are:

=*=*=*=*=*

** Uniform stream iU; psi = Uy, phi = Ux

=*=*=*=*=*

Check the given values of psi and phi by differentiation

to give U = d,x phi = d,y psi = u; 0 = d,y phi = - d,x psi = v.

=*=*=*=*=*

** Line source or sink; psi = m @ , phi = m ln r

=*=*=*=*=*

Consider a vertical source of length b emitting

Q vol/time. The velocity in the r direction v_r is Q/2 pi r b

= m/r, where m = Q/2 pi b, and v_@ = 0. From the

definitions of velocity components v_r and v_@ in terms

of derivatives of psi and phi, we can derive the given

expressions for phi and psi. Set v_r = m/r = d,@ psi / r.

Integrating gives psi = m@ + f(r). v_@ = -d,r psi = 0. Thus

psi = g(@) only, so f(r) is a constant.

In terms of the velocity potential phi, v_r = d,r phi =

m/r, so phi = m ln r + f(@). Substitute in v_@ = d,@ phi / r ,

giving 0 = d,@ f(@) = df/d@, an ordinary derivative, so f = const.

=*=*=*=*=*

** Line vortex; psi = -K ln r , phi = K @ .

=*=*=*=*=*

We proved this using u = -Kx/(x^2 + y^2)

and v = Ky/(x^2 + y^2). We can also prove it using

the Stokes theorem that the area integral of curl v equals

the line integral of v dot dl around the area in the counter-

clockwise direction. The line integral is defined as the

"circulation" gamma, so gamma / area equals the area

averaged vorticity. Note that all the vorticity is concentrated

at the origin, since the vortex flow is irrotational everywhere

else. The velocity v_@ = d,@ phi / r = K/r. Therefore

gamma = 2 pi r v_@ = 2 pi K

giving the physical interpretation of K as the circulation

of the vortex over 2 pi, just as the physical interpretation

of m for a source is Q/ 2 pi b.

In terms of psi, v_@ = - d,r psi . Thus,

d,r psi = - K/r , so psi = -K ln r + f(@). Substitute this

in v_r = d,@ psi / r to give df/d@ = 0, so f is a constant.

You might want to see some of these flows by observing

particle motions, streamlines and potential lines for various

combinations in the "Fluid Mechanics" computer program

provided with your text. You can change the parameters as

you like, and observe the effects.

The combination of a sink plus a vortex at the origin

thus has psi = -m @ - K ln r , and phi = -m ln r + K @ . The

flow is a good representation of the external flow induced by

a tornado.

Flow past a cylinder is represented by psi = U sin @

[r - R^2 / r ] as shown in Fig. E 4.8 p 222 of the text. This

is a uniform flow plus a doublet. A doublet is formed by

allowing two vortex lines of opposite sign to merge. If we

add a vortex we get lift, and this potential flow begins to

approximate the influence of a wing.

A uniform flow plus a source is described by psi =

Ur sin @ + m @ . This gives the Rankine half body flow of

Fig. 4.15, p 233. Adding a sink further down stream would

close the body. Merging the source and sink produces a

doublet. Lots of such combinations have been studied, as

you can see by looking ahead at Chapter 8.

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Viscous flows.

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Previously we discussed the laminar flow between

two fixed parallel plates separated by 2h with a pressure

gradient, and found a parabolic velocity profile. What

happens if the plates move? What happens if the sign of the

pressure gradient changes, or is zero?

The flow between long concentric cylinders with

various rotation rates is called a Couette flow, and can be

quite complex as the relative velocity of rotating cylinders

increases to for so-called Taylor vortices. The appropriate

coordinate system is cylindrical, and the conservation

equations are given in Appendix E, p 725. The steady

Couette flow with rotating cylinders is described by the theta

momentum equation

del^2 v_@ = v_@ / r^2 = r^-1 d,r [ r d,r v_@ ] .

This is solved by assuming the velocity has the form v_@ =

r^m and substituting, giving m = +,- 1. The solution is of

the form v_@ = C_1 r + C_2 r^-1. The constants are

determined by the boundary conditions at r = r_inner and

r_outer.

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Dimensional analysis and similarity (Chapter 5)

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The basis of dimensional analysis is the principle of

dimensional homogeneity for all physical equations. Many

problems of engineering interest are too complex to be

solved analytically, so experiments must be done to test

regularities in the relationships between possible

combinations of dimensional quantities. The Buckingham Pi

theorem is used to organize dimensional variables into the

minimum number of dimensionless ("Pi") groups. The theorem

states that if we have n relevant dimensional parameters that form

a dimensional matrix of rank r, then the number of dimensionless

"Pi" groups is only n-r. Because r is usually equal to the number

of fundamental dimensions (eg, 3 for MLt) this considerably reduces

the experimental effort needed to characterize the problem.