****

Notes:

The first exam is after the break today.

It will be open book, closed notes, except for one page

(both sides) of handwritten material that you may want to

prepare (no Xeroxes of anything please). The exam will

cover Chapters 1, 2 and 3 of White, plus the lectures.

*****

1. Integral Relations for Control Volumes (Chapter 3 of

White).

The rate-of-strain and rotation tensors.

The deformation rate and angular velocity of a fluid

particle.

Relationship of vorticity to angular velocity.

2. Differential relations for a fluid particle (Chapter 4 of

White).

The conservation of vorticity equation.

3. Midterm review.

============

The velocity near a fluid particle.

============

In the derivation of the momentum and energy

equations of fluid systems and control volumes within

fluids we have frequently encountered the velocity gradient

tensor and the related rate-of-strain tensor. In order to

understand the physical significance of these quantities it is

useful to consider the velocity field in the neighborhood of

a fluid particle. By a fluid particle we mean a small system

control volume that is spherical at time t. What happens to

the system control volume (fluid particle) as a function of

time? What about the fluid in its vicinity? How big can

the system be and still be considered a fluid particle?

Clearly there is a point in the system control

volume where the velocity is zero, corresponding to the

center of mass. We choose this point as the origin of the

coordinate system. The velocity dv at a distance dx away

from the origin is given by the chain rule

dv_i = ( partial v_i / partial x_j ) dx_j ,

or

dvi = vi,j dxj

introducing the "comma" notation for partial

differentiation. Decompose the velocity dv into

components parallel and perpendicular to dx; that is, dv||

and dv#, respectively. Decompose the velocity gradient

tensor into symmetric and antisymmetric components e and

Omega

vi,j = (vi,j + vj,i)/2 + (vi,j - vj,i)/2 = eij + Oij

where

eij = (vi,j + vj,i)/2 = eji

and

Oij = (vi,j - vj,i)/2 = - Oji

so

dv = (e + O) dot dx = dv|| + dv# .

Now take the dot product of dx with dv , making these

substitutions for dv

dx dot dv = dx dot dv|| + dx dot dv# = dx dot (e +

O) dot dx

but

dx dot dv# = 0

because the vectors are perpendicular, and

dx dot O dot dx = 0

because O is antisymmetric (expand out, reverse dummy

indices, use definition of antisymmetric tensor to show that

dx dot O dot dx = - dx dot O dot dx ; therefore zero).

Hence,

dx dot dv|| = dx dot e dot dx

so

dv|| = e dot dx.

But since

dv = dv|| + dv# = e dot dx + O dot dx

we see that

dv# = O dot dx

by subtracting the expression for dv|| from dv.

However, the perpendicular velocity dv# is related

to the angular velocity W by the expression

dv# = W cross dx = O dot dx.

where the i componet of dv# is

dv#i = eijk Wj dxk

using the alternating tensor eijk representation of the cross

product C = AxB

Ci = eijk Aj Bk : eijk = 1 for ijk = 123, 231, 312 ;

eijk = -1 for ijk = 132, 321, 213 ; eijk = 0 for ijk = 111,

112, 113, 221, 222, 223, 331, 332 and 333.

This notation is identical to using the determinant definition

of the cross product components. An important identity

we need is

eijk eklm = dil djm - dim djl

where d is the identity tensor with components

dij = dji = 1 for ij = 11, 22, 33 and dij = 0 for ij =

12, 13, 23.

Now, compute the quantity w cross dx for comparison

with W cross dx = O dot dx, where w is the vorticity

vector defined as del cross v. Thus,

(w cross dx)i = eijk wj dxk = eijk ejlm vm,l dxk =

ekij ejlm vm,l dxk = (dkl dim - dkm dil) vm,l dxk = (vi,k -

vi,k) dxk = 2 Oij dxj = 2 (W cross dx)i .

But if A cross B = C and D cross B = C it follows that A =

D. Therefore,

w = 2 W ;

that is, the vorticity vector is twice the angular velocity

vector of any fluid particle. This is an important physical

interpretation of the vorticity field.

**************

Example: How large can the fluid particle be before the

flow is unstable?

Solution: The stability of a flow is determined by

the ratio of the inertial vortex force term v cross w to the

viscous force nu del^2 v, known as the Reynolds number

Re, where Re = u L / nu, u is a characteristic velocity

scale, L is the length scale and nu is the kinematic

viscosity. If Re > Re_crit the flow is unstable and may

become turbulent. In the expression Re = u L / nu

u = e x

L = x

where e = (e:e)^1/2 is the magnitude, or rms value, of the

rate of strain tensor; so,

Re = e x^2 / nu = Re_crit

where x = x_crit, the largest size fluid particle with stable

flow. Thus,

x_crit = (nu/e)^1/2 (Re_crit)^1/2.

The rms rate of strain e is related to the viscous dissipation

rate eps by the equation eps = 2 nu e:e, so e = (eps/2

nu)^1/2. Therefore the critical length x_crit is proportional

to (nu^3/eps)^1/4 = L_K, the Kolmogorov length scale,

first introduced by A. N. Kolmogorov in a classic paper of

turbulence theory in 1941. Velocities on scales smaller

than L_K are stabilized by viscous forces. Those on larger

scales are unstable, and can break up into a growing

cascade of turbulent eddies. A universal critical Reynolds

number Re_crit of about 25-100 has been determined

experimentally.

***************

==============

Conservation of vorticity equation.

==============

Recall the conservation of momentum equation for

a constant density and viscosity fluid in the form

partial v / partial t = - grad B + v cross w + nu

del^2 v

where B is the Bernoulli group. Taking curl = del cross of

this equation gives on the left

(curl v,t )i= eijk d,j vk,t = d,t eijk d,j vk = d,t eijk

vk,j = wi,t = partial w_i / partial t

which is the rate of change of the i component of vorticity.

The first term on the right, - curl grad B, is zero because

the curl of the gradient of any function is equal to zero.

This is proved by reversing the order of partial

differentiation, reversing two dummy indices, changing

the sign of eijk by reversing two indices, and noting that

the only quantity equal to its own negative is zero

(curl grad B)i = eijk d,j d,k B = eijk d,k d,j B =

eikj d,j d,k B = - eijk d,j d,k B = 0

where d,j is the partial derivative with respect to x_j.

The second term on the right curl (v cross w) has i

component

[curl (v cross w)]i = eijk d,j eklm vl wm = [dil djm

- dim djl] [ vl d,j wm + wm d,j vl ] = vi d,j wj + wj d,j vi -

vj d,j wi - wi d,j vj

where the divergence of the vorticity d,j wj is zero in all

cases (proved as an exercise below), and the divergence of

the velocity d,j vj is zero for constant density fluids (from

the continuity equation). Thus,

[curl (v cross w)]i = wj d,j vi - vj d,j wi

for incompressible fluids. The first term on the right can

be further simplified because the velocity gradient tensor

d,j vi = eij + Oij and wj Oij = 0 because the local rotation

cannot rotate itself (xj Oij was shown above to be [w cross

x /2]i, or v#i, which is zero for x in the direction of w).

Hence,

[curl (v cross w)]i = wj eij - vj d,j wi

where eij is the rate of strain tensor and wj eij is the rate of

stretching of vortex lines. The final viscous term is

[curl nu del^2 v ]i = nu del^2 curl vi = nu del^2 wi

.

Combining all terms gives the vorticity equation for

constant density, constant viscosity Newtonian fluids

d,t wi + vj d,j wi = wj eij + nu del^2 wi

or in vector form

********** D w / Dt = w dot e + nu del^2 w ********.

We see that the vorticity of a fluid particle increases with

time due to vortex stretching and decreases due to viscous

diffusion.

=======================

Midterm review

=======================

The first midterm examination covers the material

of the first three chapters of White, plus the lectures. You

should be able to derive the continuity equation by either

the shrinking system control volume method or the

shrinking fixed finite volume element method, and the

momentum and energy equations by the shrinking system

control volume method. Exam questions are often taken

from those given at the end of the chapters, so read over all

the problems and think about how you would solve them

(or just solve them). Show all of your work. You must

include explanations in words of what you are doing as

well as writing down equations. Make it very clear that

you understand all aspects of your problem solutions. Be

sure to write down all units and check all equations for

dimensional consistency and reasonableness.

************

Example: Problem 2.42 describes a two fluid manometer

system. Small pressure differences between reservoirs of

fluid 1 are measured by the height difference h between

levels of fluid 2 in the manometer, where rho_2 is greater

than rho_1 and the levels of fluid 1 in the reservoirs are

equal.

Solution: The vertical momentum equation is dp/dz

= - rho g, or dp = - rho g dz. Integrating this equation

from reservoir B to reservoir A gives

p_A - p_B = rho_1 g (h_1 - h) + rho_2 g h+ rho_2

g h_3 - rho_2 g h_3 - rho_1 g h_1

so

p_A - p_B = (rho_2 - rho_1) g h.

Thus, if (rho_2 - rho_1) is very small, h will be very large

for a given p_A - p_B, so the manometer is very sensitive.

**************

**************

Example: Problem 3.149 has a cylindrical jet of alcohol

striking a vertical plate. A force of 425 N is required to

hold the plate stationary. Assuming there are no losses in

the nozzle, estimate (a) the mass flow rate of alcohol and

(b) the absolute pressure in the 5 cm diameter tube before

the 2 cm nozzle exit. The SG of alcohol is 0.79 and the

density of water is 998 kg/m^3.

Solution: Take a control volume enclosing the

impinging jet and the plate, and consider the conservation

of x momentum for the fixed control volume, where x is in

the direction of the jet. The Reynolds transport theorem

for the control volume is dPx /dt |system = dPx /dt |CV +

int [ rho v v dot dA ]. These terms are

dPx /dt |system = - F , the force on the plate.

dPx /dt |CV = 0 , for this steady state flow.

int [ rho v v dot dA ] = - rho v^2 A_jet

so

F = rho v^2 A_jet .

We know all the terms in this equation except the jet

velocity v. F = 425 N. rho = 0.79x998 kg/m^3. A_jet =

pi x (0.02)^2 / 4, to give v = 41.4 m/s. The mass flow

rate dm/dt = rho A v = 10.3 kg/s (ans. a).

We can find the mass flow in the tube from the

integral conservation of mass, and the pressure from

Bernoulli's equation. Consider a control volume enclosing

the tube and the jet. Reynolds transport theorem for mass

is dM/dt|system = dM/dt|CV + int [ rho v dot dA], where

the terms are

dM/dt|system = 0 by the conservation of mass.

dM/dt|CV = 0 for this steady state flow.

int [ rho v dot dA] = - rho v_1 A_1 + rho v_2 A_2,

where A_1 is the 5 cm diameter cross section and A_2 is 2

cm. Hence v_1 = 41.4 m/s (2/5)^2 = 6.63 m/s. There are

no viscous losses or viscous working terms in the nozzle,

so B_1 = B_2 along all streamlines

B_1 = p_1/rho + v_1^2 / 2 + g z = B_2 = p_2/rho

+ v_2^2 / 2 + g z

where z_1 = z_2 = z, p_2 = p_atm = 101,000 Pa, so p_1

= 760,000 Pa (ans. b).

*********

*********

Exercise: Show that div curl v = 0.

d,j wj = d,j ejkl d,k vl = ejkl d,j d,k vl

but we can reverse the order of the partial derivatives

d,j wj = ejkl d,k d,j vl

and because j and k are dummy indices

d,j wj = ekjl d,j d,k vl = - ejkl d,j d,k vl ,

so that

d,j wj = - d,j wj = 0. QED.