a* From last time, momentum per unit volume and

forces per unit volume.

b* Same, in terms of total stress tensor.

c* Rearranged, to get substantive derivative of

velocity.

d* Assuming constant rho and mu, to give Navier

Stokes equation.

e* Rearranged Navier Stokes, to reveal the Bernoulli

group B and the inertial-vortex force vxw.

2. Hydrostatics exercises.

+++++++++

Conservation of momentum equations.

+++++++++

Different problems in fluid mechanics require

different forms of the conservation of momentum equations,

and the different forms give useful physical interpretations

and insights. From the conservation of momentum principle

applied to a system control volume, we found that

(1) partial rho v / partial t + div ( rho v v ) + grad p - div

[ tau ] - rho g = 0

everywhere in a continuous fluid. Moving the last four

terms to the right hand side

=====

a* partial rho v / partial t = - grad p - div ( rho v v )+

div [ tau ] + rho g .

=====

The terms on the right are pressure, inertial, viscous, and

gravity forces per unit volume, and the term on the left is the

rate of change of momentum per unit volume. In terms of

the total stress tensor T,

=====

b* partial rho v / partial t = div [ T ] + rho g.

=====

where T = - p d - rho v v + tau is the sum of the pressure,

inertial and viscous stress tensors, respectively.

This shows why local velocities change. What about

an individual fluid particle? For this we need the substantive

derivative Dv/Dt. Go back to (1) and rearrange the two left

terms:

partial rho v / partial t = rho partial v / partial t + v

partial rho / partial t

div ( rho v v ) = rho (v dot grad) v + v div ( rho v ).

Adding these together gives:

rho partial v / partial t + v partial rho / partial t +

rho (v dot grad) v + v div ( rho v ) = rho Dv/Dt

from the continuity equation and the definition of Dv/Dt.

Hence

=====

c* rho Dv/Dt = - grad p + div [ tau ] + rho g

=====

which shows that the momentum per unit volume of a fluid

particle changes due to pressure, viscous and gravity forces

per unit volume.

The most popular form of the momentum equations

is the "Navier-Stokes" version, for a Newtonian fluid with

constant density rho and viscosity coefficient mu.

From c* and these assumptions, dividing by rho and

bringing it inside derivatives:

partial v /partial t + (v dot grad) v = - grad (p/ rho) +

g + div [ tau / rho ] .

However, div [ tau / rho ] = div [ 2 mu e / rho ] = nu del^2

v from the continuity equation for constant density div v = 0.

Please prove as an exercise. Thus,

=====

d* partial v /partial t + (v dot grad) v = - grad (p/ rho) +

g + nu del^2 v

=====

which are the Navier-Stokes equations...the most commonly

quoted form of the momentum equations in fluid mechanics.

Let us rearrange the NS equations by using the

vector identity (v dot grad) v = grad (v^2)/2 - vxw, where w

= curl v is the vorticity vector. We will prove this later, and

show that the vorticity is twice the angular velocity of a fluid

particle. Also, write the gravity vector as the gradient of the

gravitational potential -gz. Then

=====

e* partial v /partial t = - grad B + vxw + nu del^2 v

=====

where B is the Bernoulli group B = p/ rho + (v^2)/2 + gz,

and vxw and nu del^2 v are the inertial vortex force and

viscous force per unit mass, respectively.

This is a particularly useful form of the momentum

equations. For steady, irrotational, non-viscous flows

(inviscid) we see from e* that B is a constant. This is the

basis for statements like "where the speed is greatest the

pressure is least" that you may have heard in high school to

explain why airplanes fly. Later, we will use this equation

to derive the conservation of mechanical energy equation.

The non-linear term vxw is the cause of turbulence.

Perturbed vortex sheets induce such forces in the direction of

the perturbations. These amplify the perturbations to form

the eddies of turbulence. Note that viscosity does not cause

turbulence as stated on page 1 of the textbook, it

damps it out. For turbulence to exist, forces due to vxw

must be larger than forces due to nu del^2 v. The ratio

(vxw/ nu del^2 v) is the Reynolds number Re, which must

be larger than some critical value for turbulence to exist.

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

Hydrostatics exercises.

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

------------------------

In problem 2.18, 1 meter of SAE 30 oil is over 2

meters of water, which is over 3 meters of fluid x, which is

over 1/2 meter of mercury in a tank. The pressure at the top

of the tank is 101.33 kPa, and the bottom pressure is 242

kPa. What is the specific gravity of fluid X?

***

Solution:

***

Integrate the hydrostatic equation dp/dz = - rho g

from the bottom to the top of the tank. Thus, int [dp] =

p_bottom - p_top = sum gamma_i delta z_i , where gamma

= rho g for the various fluids. Table A.3 has the densities at

20 C that are needed. Substituting:

242000 Pa = 101330 + (8720)(1.0) + (9790)(2.0) +

gamma_X (3.0) + (133100)(0.5),

to give gamma_X = 15273 N/m^3. The specific gravity

rho_X / rho_water = gamma_X / gamma_water =

15273/9790 = 1.56 (ans.).

------------------------

------------------------

In problem 2.146 a helium balloon floats in a tank of

water. If the tank accelerates to the right at 5 m/s^2, which

way will the balloon string lean?

=======

Solution:

=======

The equations of motion for the accelerating system

are:

rho a = - grad p + rho g

so grad p = rho (g - a) in the water as well as in the helium

of the balloon. Because g is down and a is to the right, then

g - a is down to the left, which is the direction of the

pressure gradients in both the fluids. Because the density in

the balloon is so much less than that of water, hydrostatic

forces will develop forcing the balloon up and right, that

must be balanced by string forces down and to the left. You

can see this happening in a water vortex. Any bubbles move

toward the vortex center, in the direction of the acceleration.

A bubble in a bottle of liquid in space will also move in the

direction of any acceleration of the bottle. It is as though

gravity were turned on in the opposite direction.

The angle of the tether is tan^-1 (a/g) = tan^-1

(5/9.81) = 27 degrees (ans), independent of the size of the

balloon or the density difference between fluids.

White).

Derivation of the Reynolds' Transport Theorem.

Application to the conservation of mass.

Conservation of momentum.

Angular momentum.

2. Control-volume exercises.

3. Homework hints.

+++++++++

Integral Relations for Control Volumes.

+++++++++

Suppose we have some conserved quantity B, with

concentration beta per unit mass. Presumably we have a

physical law describing how B varies with time for a

system (a specified quantity of matter. For example, if B

is the mass of a system, then beta is 1 and the time

variation of B is zero. Often we want to know how B

varies with time for leaky volumes that contain different

systems at various times. The connection between the time

variation of B for a system and B for a different (leaky)

control volume is called the Reynolds Transport Theorem.

*****

System control volume

*****

B is the integral of rho beta over the control volume

of interest. The time rate of change of B is given by

Leibnitz' rule for differentiating integrals:

(1) dB/dt|syst = int [ rho beta dV] + int [ rho beta v dot

dA]

where the surface velocity v_S is replaced by the fluid

velocity v because the surface of a system control volume

is not "leaky".

*****

Arbitrarily moving control volume

*****

Now consider the time rate of change of B for an

arbitrary moving control volume that happens to coincide

exactly with the system control volume above at time t.

The surface of this control volume is generally not moving

with the fluid velocity v, but with velocity v_S. The time

rate of change of B for this arbitrarily moving control

volume is:

(2) dB/dt|CV = int [ rho beta dV] + int [ rho beta v_S

dot dA] .

*****

Reynolds transport theorem

*****

Subtracting (2) from (1) gives the Reynolds Transport

Theorem:

(3) dB/dt|syst = dB/dt|CV + int [ rho beta v_r dot dA]

where the two volume integrals cancel because they are

integrals with identical integrands over the same volume,

and v_r = (v - v_S) is the relative velocity of the fluid of

the system with respect to the arbitrarily moving control

volume.

Notice that if the control volume is fixed, the

relative velocity equals the fluid velocity. Also, if the

control volume surface moves with the fluid, the area

integral in (3) vanishes showing that dB/dt|syst =

dB/dt|CV, as expected.

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

Conservation of mass.

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

As an illustration, consider the integral

conservation of mass for a fixed control volume, where

fluid enters and leaves through a number of ports of

different sizes. Assume the fluid has constant density

everywhere, and all the velocities are constant in time, and

perpendicular to and constant over the port areas. The total

mass in both the system and fixed control volumes is a

constant M, where M is the integral of rho over the volume

(beta = 1). Thus, the Reynolds Transport Theorem of (3)

becomes 0 = 0 + int [ rho v dot dA], since v_r = v - 0.

Since rho was assumed constant everywhere, the area

integral reduces to the sum of the mass flows out minus

the sum of the mass flows into the control volume equals

zero.

If the densities of the port fluids had been different,

the negative of the area integral would be dB/dt|CV -

dB/dt|syst = dM/dt|CV - dM/dt|syst = dM/dt|CV - 0, which

is the rate of accumulation of mass in the fixed control

volume.

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

Conservation of Momentum.

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

For B = P, the total momentum, then beta is the

velocity v (momentum per unit mass) in (3). The left hand

side is the sum of all forces on the system by Newton's

second law of motion. The right hand side depends on the

assumptions given for the arbitrarily moving control

volume. If it is fixed, and the flow rates and density

distributions are constant then the momentum is constant in

the control volume (first term zero). The area integral is int

[ rho v v dot dA], which is minus the sum of all the inertial

forces on the fixed control volume due to the inertial stress

tensor - rho v v dotted into the various areas through

which there is flow from or into the fixed control volume.

Moving this term to the left hand side shows that the rate

of change of the momentum in a fixed control volume

equals the sum of the forces on the system contained, plus

the inertial forces due to flows through the surface.

**********

Example 3.11 (p134 text): The von Karman

momentum integral theory for boundary layers illustrates

the application of integral conservation methods for both

momentum and mass. The control volume chosen about a

segment of the leading edge boundary layer of a plate

parallel to a uniform flow of velocity U. The boundary

layer is delta thick at a distance x = L. The streamlines

intersecting y = delta were at y = h at x = 0. The control

volume is bounded at the top by these streamlines, at the

bottom by the wall fluid, and by surfaces normal to the

flow u = U at x = 0 and u = u(y) at x = L.

Integral conservation of x momentum using (3)

gives the sum of forces on the left side, zero for the time

derivative of momentum on the right, and the surface

integral of rho v v dot dA only over the upstream and

downstream surfaces because v dot dA = 0 for all other

CV surfaces.

The forces in the x direction on the system CV

consist only of -D, where D is the frictional drag on the

plate. Note that the pressure is constant throughout the

boundary layer and in the external flow. A constant

pressure surrounding a control volume exerts no force

(this follows from the divergence theorem...the volume

integral of grad p = 0 if p = constant). Viscous forces are

zero except at the bottom of CV. Gravity acts in the y

direction and is irrelevant. Thus:

-D = 0 + int [ rho v v dot b dy]_x=0 + int [ rho v v

dot b dy]_x=L = - rho b UU h + rho b int [ u u dy]_x=L.

From the integral conservation of mass for the

control volume:

0 = 0 + int [ rho v dot dA]_x=0 + int [ rho v dot

dA}_x=L = - rho U h b + int [ rho u(y) b dy]_x=L

or

rho U h b = int [ rho u(y) b dy ]_x=L,

which is substituted into the momentum equation, to give:

D = rho U^2 b delta int [ u/U{1- u/U}],

which is the von Karman momentum integral equation.

The integral term is a drag coefficient because it is a drag

force D divided by rho U^2 times the projected area of the

boundary layer. Von Karman showed the integral was

2/15 assuming a parabolic velocity profile, which as within

10% of the exact (Blasius) velocity profile value discussed

in Chapter 7 on boundary layer theory.

The parabolic velocity profile is obtained by setting

u(y) = a + by + cy^2, with the boundary conditions that u

= 0 at y = 0 (no slip assumption), u = U at y = delta, and

du/dy = 0 at y = delta, giving u(y) = U(2 eta - eta^2),

where eta = y/delta. Substituting u/U = (2 eta - eta^2) into

the integral gives 2/15.

**********

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

Conservation of angular momentum.

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

If we take the cross product of the integral

momentum equation with the distance r from an origin,

then we get the total angular momentum of the material

about this origin. For a fixed control volume, the left hand

side becomes the sum of the torques of the forces about the

origin. The right hand side is the rate of change of the

integrated r cross v dm values (beta is r cross v), plus the

flux of angular momentum out of the control volume.

2. Homework hints.

2.13. Integrate dp = - rho g dz from bottom to top through

both the water and the gasoline to get two equations in two

unknowns (the pressure difference and the height h of the

third fluid under the gasoline). Eliminate the pressure

difference and solve for h = 1.52 m.

2.28. Use equation 2.27 and the temperature and pressure

ratios to estimate the lapse rate B, giving 0.00899 deg F /

ft. Estimate the altitude z = delta T / B = 5400 ft.

2.44. The manometer only indicates pressure losses in the

pipe due to friction (392 psf).

Example: 2.91. The bolts must resist the weight of the missing

water from a cylinder surrounding the dome and standpipe

(90700 N).

Example: 2.113. The weight of the spar buoy equals the weight of

the displaced water (W_steel = 5.4 lbf).