White).

Conservation of momentum (non-inertial frames).

Example (earth).

Mechanical energy (the Bernoulli equation).

2. Examples.

3. Homework hints

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

Conservation of momentum.

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

Last time we derived the Reynolds transport

theorem relating the rate of change of a conserved fluid

property B for a system control volume to that for the same

initial control volume, but moving in an arbitrary fashion

(or remaining fixed). This is useful because most

conservation laws refer to systems, but many control

volumes needed in practical problems are not for systems.

We saw that the conservation of momentum for a control

volume is dP/dt|_CVsystem = dP/dt|_CVgeneral + int [ rho

v v_r dot dA] gives the sum of the forces on the left hand

side from Newton's law. For fixed control volumes in

steady state the sum of the forces is just int [ rho v v dot

dA] . Moving the area integral to the left hand side of the

equation changes its interpretation from that of an outward

momentum flux integral ( rho v is momentum per unit

volume and v dot dA is the local volume flux out of the

CV) to that of an integrated inertial force due to the inertial

stress tensor - rho v v. For example, the forces required to

keep a rocket stationary on a test stand are equal and

opposite to the inertial forces acting on the nozzle surface

area - rho v v dot A. A rocket in outer space with no

forces moves by causing exhaust gases to move rapidly in

the opposite direction, so that the net momentum of the

rocket system does not change.

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

Conservation of momentum in accelerating coordinate

frames.

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

Strictly speaking, Newton's law does not work in

coordinate systems fixed to the earth because earth-bound

coordinate systems are accelerating due to the earth's

rotation with respect to inertial coordinates. The rate of

change of momentum is equal to the sum of the forces on a

system (Newton's law) only in an "inertial" coordinate

system moving at constant velocity without rotation with

respect to the universe. The difference between

accelerations in the two systems can be considered

(fictitious) forces in the accelerating coordinate system,

just as gravitational force can be simulated by centripetal

acceleration in a space station. The time derivative of

vector B_I in an inertial coordinate system,

dB_I/dt|_I = dB_I/dt|_R + Omega cross B_I

in a rotating coordinate system, where Omega is the

angular velocity of the rotating coordinate system and the

origins of the inertial and rotating coordinate systems are

the same. Thus,

dr/dt|_I = dr/dt|_R + Omega cross r ,

where r is the same position vector in both the rotating and

inertial coordinate systems at the initial time, or

v_I = v_R + Omega cross r,

where

v_I = dr/dt|_I

and

v_R = dr/dt|_R.

The acceleration of the position r in the inertial frame

a_I = dv_I/dt|_I = dv_I/dt|_R + Omega cross v_I.

Substituting the expression

v_I = v_R + Omega cross r

gives

a_I = a_R + dOmega/dt cross r + 2 Omega cross

v_R + Omega cross ( Omega cross r ). Newton's law

states that a_I = sum of forces. Thus,

a_R = sum of forces - [dOmega/dt cross r + 2

Omega cross v_R + Omega cross ( Omega cross r )],

where all the accelerations in the brackets are "fictitious

forces" when placed on the left hand side of the

conservation of momentum equation. Such forces are

required to compensate for the coordinate system

accelerations.

The most famous "fictitious force" is the Coriolis

force - 2 Omega cross v_R, which accounts for the

direction of wind motions about high and low pressure

points in the northern and southern hemispheres of the

earth's atmosphere. Horizontal motions in the north are

deflected to the right by Coriolis force, and to the left in the

south, because the earth spins from west to east.

We saw examples in hydrostatics of the role of

accelerations. Without accelerations the forces per unit

mass in a fluid with v = 0 were - (grad p)/ rho + g = 0

where g is gravitational force per unit mass and - g is the

gravitational acceleration. With acceleration a, then - (grad

p)/ rho + g = a. Clearly we could consider the acceleration

a as a force per unit mass -a if we moved it to the left hand

side of this equation. Note that numerous accelerations are

neglected in applying Newton's law to laboratory (earth-

bound) coordinate systems...the rotation of the earth on its

axis, about the centers of gravity with the moon and the

sun and the center of the galaxy, etc. ... because they are

small compared to the forces of interest.

*******

Example: How large are "fictitious forces" due to the

earth's rotation?

Coriolis force = - 2 Omega cross v. For horizontal

velocities, this force is largest near the poles. Omega is 2

pi /24 x 3600 = 7x10^-5 radians/sec. A velocity of about

7x10^4 m/s is required to produce 1 g of Coriolis

acceleration at the north pole (artillery shell trajectories at

high latitudes are indeed affected significantly). A racing

car at 100 m/s is deflected by only 0.0014 g by Coriolis

forces at the north or south pole, and not at all at the

equator. The centrifugal force Omega cross ( Omega cross

r ) due to the earth's rotation is twice this at the equator,

with a value of 0.003 g (r is about 6 megameters).

*******

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

Conservation of mechanical energy for a stream tube

(Bernoulli's equation).

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

Consider the momentum equation for a Newtonian

fluid with constant physical properties (including density

and viscosity) in the form:

part v/ part t = - grad B + v cross omega + div ( tau

/ rho )

where tau is the viscous stress tensor 2 mu e , B is the

Bernoulli group, and e is the rate of strain tensor. We can

derive the mechanical energy equation; that is, an equation

for the rate of change of the kinetic energy per unit mass

v^2/2, by taking the dot product of the equation with the

velocity v. The left hand side becomes

part (v^2 / 2) / part t

which is what we want.

The first term on the right becomes

- div (v B) + B div v = - div (v B)

because div v = 0 from the continuity equation with rho =

constant.

The second term on the right becomes

v dot ( v cross omega ) = 0

because v cross omega is perpendicular to v.

The third term on the right becomes

v dot div ( tau / rho ) = div [ v dot tau / rho ] - [ tau

/ rho ] dot dot [ grad v ]

where grad v is the velocity gradient tensor. The term [ tau

/ rho ] dot dot [ grad v ] is the viscous dissipation rate

epsilon per unit mass, and we can show that eps = 2 nu

e^2, so eps is always positive. Combining all terms gives

part (v^2 / 2) / part t = - div (v B) + div [ v dot tau

/ rho ] - eps,

which is the mechanical energy equation we need.

The famous "Bernoulli equation" can be found by

applying this equation to the steady flow of a constant

property fluid in a stream tube. A stream tube is formed

from a cylinder of adjacent stream lines. Take the entrance

to the tube to be point 1 and the exit as point 2. The kinetic

energy KE in the tube is the integral of rho v^2 / 2 over the

control volume. The derivative of KE with time is found

from Leibnitz' Rule and is equal to zero by our steady state

assumption, so

d KE /dt = int [ rho part (v^2 / 2) / part t dV] + int

[ rho v^2 / 2 v_S dot dA] = 0

where the surface integral is zero for the fixed control

volume because v_S = 0.

Now we can substitute the kinetic energy equation

for part (v^2 / 2) / part t in the volume integral, and convert

two of the resulting integrals to surface integrals by the

divergence theorem

- int [ rho vB dot dA ] + int [ v dot tau dot dA ] -

int [ eps dV] = 0

where the first integral is the flow of B into the control

volume, the second is rate of viscous work being done on

the control volume per unit time, and the third is minus the

rate that mechanical energy is being lost within the control

volume per unit time due to internal viscous friction.

Recall B = p/ rho + v^2 / 2 + gz has the units of energy per

unit mass (it is specific enthalpy minus specific internal

energy).

The first term is just (B_1 - B_2)(dm/dt), where

dm/dt is the mass flow rate through the stream tube. The

second term is -dW_s/dt, the rate of "shaft work" being

done on the surroundings per unit time. The third term is

-dW_l/dt, the "lost work" per unit time. Dividing by dm/dt

gives the Bernoulli equation

B_1 = B_2 + gh_s + gh_f

where the shaft work and lost work terms are expressed as

head losses h_s and h_f , respectively. Along a steady

stream tube, h_s is positive if a turbine is encountered and

positive shaft work is extracted, and h_s is negative if a

pump is encountered. The frictional head loss h_f is

always positive. If no work is done or extracted, and there

are no losses, then the Bernoulli group B is constant and

"where the speed is greatest the pressure is least".

********

Example: Consider the case of a tank of water with an exit

port a distance h below the surface. What is the velocity of

the fluid exiting the port?

Apply Bernoulli's equation to the streamline from

the surface to the port. There are no losses or shaft work

terms, so B is constant.

B_1 = B_2

where B_1 = p_a / rho + 0 + 0 and B_2 = p_a / rho + v^2

/ 2 + gz_2. Now z_2 is -h if z_1 = 0. Therefore v =

(2gh)^1/2 (ans).

*********

********

Example: How does a pitot tube work? What is the

pressure at the entrance of an impact tube used to measure

velocity by comparison to the static pressure in the

surroundings?

Consider the streamline that stagnates at the

entrance to the impact tube. Again, B_1 = B_2. The z

values are the same at 2 and 1. Thus the pressure

difference p_2 - p_1 = rho v^2 / 2, where v=0 at 2 and v at

1, and can be detected with a manometer between the

points.

********

********

Homework hints

********

3.5 Plot the given data and show that the turbulent velocity

profile is different from a parabolic profile expected for

laminar flow. Integrate graphically to get Q = 0.006 m^3/s.

3.10 Integrate the enthalpy flux over the velocity profile.

3.38 Use the RTT for mass conservation to show V=V_o/2h.

3.43 Use the RTT for momentum and mass for a CV

enclosing the pipe. The forces in the x direction include

pressure forces and flange forces on the inlet and exit.

These equal int[rho v v dot dA] for the inlet and exit,

which are both in the minus x direction! Summing gives

F_flange = -750N in the x direction and 14 N in the y

direction.

3.102 Apply RTT for mass and momentum for a CV

enclosing the hydraulic jump from h_1 to h_2. Neglect

wall shear, so the only forces are due to hydrostatic pressure.

From the mass balance rho V_1 h_1 b = rho V_2 h_2 b.

From the mass-mom. balances, rho g h_1 h_1 b / 2 -

rho g h_2 h_2 b / 2 = rho V_1 h_1 b [ V_2 - V_1]. Some

algebra gives h_2 / h_1 = -1/2 + (1/2)sqrt[1+8V_1^2 / (gh_1)].

White).

Conservation of energy...the first law of

thermodynamics for fluids.

The total specific energy of a fluid particle.

The general mechanical energy equation of a fluid

particle.

The internal energy of a fluid particle.

2. Examples.

3. Homework hints.

Design project D3.1.

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

Conservation of energy.

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

Above we derived a form of the conservation of

mechanical energy equation for a constant density fluid

partial ke /partial t = - div vB + div (v dot tau / rho )

- eps

where ke = v^2 / 2 is the kinetic energy per unit mass, B is

p/rho + ke + pe, tau is the viscous stress tensor 2 mu e (for

a Newtonian fluid), and eps is the viscous dissipation rate

per unit mass (tau / rho ) dot dot grad v =

eps = 2 nu e dot dot e. This last

expression results from the identity

grad v = e + Omega

where e is the rate of strain tensor with components (

partial v_i / partial x_j + partial v_j / partial x_i )/2 , and

Omega is the rotation tensor ( partial v_i / partial x_j -

partial v_j / partial x_i )/2. Clearly e is symmetric because

e_ij = e_ji and Omega is antisymmetric because Omega_ij

= - Omega_ji. Substituting 2 mu e for tau in eps = (tau /

rho ) dot dot grad v gives

eps = 2 nu e dot dot e

because e dot dot Omega = 0 (this is proved by reversing

the dummy indices and substituting e_ij = e_ji and

Omega_ij = - Omega_ji, so that e dot dot Omega = - e dot

dot Omega). We see that eps must always be positive if nu

is positive (as it must be), since e dot dot e = e^2 is

positive.

We applied this equation to the steady flow of a

constant property fluid in a streamtube and found the

Bernoulli equation

B_1 = B_2 + gh_s + gh_f

where h_s is the head loss due to viscous working on the

surroundings, and h_f is the "lost work" head loss due to

irreversible internal friction.

Now let us apply the first law of thermodynamics

to a system of fluid contained by a system control volume

to derive an equation for the specific energy, derive

another form of the mechanical energy equation without

assuming the density is constant, and subtract the two to

interpret the viscous dissipation rate eps. According to the

first law, the heat flow dQ/dt into the system minus dW/dt

done by the system on the surroundings equals the rate of

change of the total energy dE/dt within the system.

The heat flow is int [ k grad T dot dA] = int [ div (k

grad T) dV], where k is the thermal conductivity of the

fluid and T is the temperature.

The work done on the surroundings per unit time is

the pressure work rate dW_p/dt = int [ p v dot dA] = int [

div pv dV] plus the viscous work rate dW_v/dt = - int [ v

dot tau dot dA] = - int [ div (v dot tau ) dV] .

The rate of change of energy in the system dE/dt =

(d/dt) int [ rho e dV] = int [ (partial rho e / partial t) dV] +

int [ rho e v dot dA] = int [ {(partial rho e / partial t) + div (

rho e v )} dV] , where e is the specific total energy u + ke

+ pe and u is the internal energy per unit mass.

Collecting these terms in the first law of

thermodynamics gives

int [ {(partial rho e / partial t) + div ( rho e v ) - div

(v dot tau) + div (pv) - div (kgradT) } = 0.

Now the system control volume was chosen

arbitrarily, and can be made smaller and smaller about any

internal point, so that the {.} term becomes constant by the

continuum hypothesis. Factoring {.} out of the integral

shows that it must be zero at every point in the fluid (since

the remaining integral is finite), or identically zero,

according to the first law of thermodynamics. Thus

(partial rho e / partial t) + div ( rho e v ) - div (v dot

tau) + div (pv) - div (kgradT) = 0

is the total energy equation for a fluid.

Expanding the first two terms gives (from the

continuity equation)

rho De/Dt - div (v dot tau) + div (pv) - div

(kgradT) = 0

which shows how the total energy per unit volume of a

fluid particle changes with time.

Previously we found the following form for the

momentum conservation of a fluid particle

rho Dv/Dt - div tau + grad p - rho g = 0

where g is the gravitational force - grad (gz), and z is up.

Taking the dot product of the velocity v with this equation

gives

rho D(ke + pe)/Dt - v dot div tau + v dot grad p = 0

which is the mechanical energy equation for a fluid particle

with no assumptions of constant properties. Subtracting

this from the total energy equation for a fluid particle,

using the definition of e, gives

rho Du/Dt - tau dot dot grad v + p div v - div

kgradT = 0

where u is the internal energy. Thus the internal energy of

a fluid particle per unit volume increases with positive rho

eps = tau dot dot grad v, increases due to heating at the rate

div kgradT (note that this is positive for a cold spot,

negative for a hot spot), and increases if -p div v is

positive (corresponding to compression of the fluid particle

from the continuity equation).

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

2. Examples.

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

*********

Example 6.1. Consider the solid fuel rocket of problem

3.35. The nozzle diameter is 18 cm and the exit velocity is

1150 m/s. Take a fixed control volume around the rocket,

perpendicular to the flow out of the nozzle. The Reynolds

transport theorem for the mass of the control volume gives

zero on the left hand side (for the rate of change of the

mass of the system) and int [ partial rho / partial t dV] + int

[ rho v dot dA] on the right hand side, where int [ rho v

dot dA] is the mass flow rate out of the control volume at

the nozzle. The molecular weight is given as 28, the gas

constant R is 8313/28 = 297 m^2/s^2 K for the gas.

Hence the exit gas density is

rho_exit = p/RT = 90,000 Pa / (297)x(750 K) =

0.404 kg/m^3.

The rocket mass increases at rate int [ partial rho / partial t

dV] equal to minus the flow rate of mass out int [ rho v dot

dA]. Both are negative. int [ rho v dot dA] = rho_exit

A_exit V_exit = (0.404)(pi/4)(0.18^2)(1150) = 11.8 kg/s.

Therefore the mass increase of the rocket is - 11.8 kg/s

(ans).

********

********

Example 6.2. Consider a uniform flow past a cylinder

with a V shaped wake, as in problem 3.44. This problem

is similar to the von Karman integral momentum theory for

a boundary layer. Take a control volume of width b about

the cylinder, with streamline surfaces separated by +,- L at

the downstream position, and +,- H upstream. The

velocity is uniform with value U upstream, and u =

(U/2)(1 + y/L) for positive y and (U/2)(1 - y/L) for

negative y in the wake, and U outside, downstream.

Apply the conservation of mass to this fixed

control volume, using the Reynolds transport theorem.

The rate of change of mass for the system is zero, and for

the control volume also zero for this steady flow problem.

This leaves the mass flux integral int [ rho v dot dA] = 0.

Only the upstream and downstream surfaces have nonzero

integrals. Upstream we have - 2 int [ rho U b dy] from y

= 0 to H, giving - 2 rho U b L. Downstream we have + 2

int [ rho (U/2)(1 + y/L) b dy] from y = 0 to L, giving H =

3L/4.

Now apply the conservation of x momentum. The

rate of change of momentum for the system is the sum of

the forces, equal to the negative of the drag force on the

cylinder - F_drag. This equals the rate of change of

momentum in the control volume, plus the surface integral

of the momentum flux int [ rho v v dot dA] which is

nonzero only for the upstream and downstream surfaces.

Upstream we have - 2 int [ rho U^2 b dy] from y = 0 to H.

Downstream we have + 2 int [ rho {(U/2)(1 + y/L)}^2 b

dy. Combining these and H = 3L/4 gives - (3/2) rho U^2

L b upstream and (7/6) rho U^2 L b downstream. Hence

F_drag = (1/3) rho U^2 L b, giving a drag coefficient C_D

= F_drag/rho U^2 L b of 1/3 (ans).

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

3. Homework Hints.

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

3.110. Apply the RTT for conservation of angular momentum.

The sum of the torques equals the integral of r X rho v v dot dA

for a control volume surrounding the sprinkler and perpendicular

to the two nozzles.

3.127. Apply the RTT for conservation of energy to the river.

The heat capacity of water is about 4280 J/kg K. The waste heat

of 55MW equals the mass flow rate of the river times the heat

capacity times the temperature difference, giving 23.2 deg C for

the downstream temperature.

3.142. Apply the Bernoulli equation to a steamline passing from

the lower reservoir surface, through the pump, to the upper reservoir

surface at 50 ft higher elevation. The frictional head

h_f = 27 V^2 / 2g from 3.141, and the shaft work head is minus

the pump head 300 - 50Q. Q = VA, so substitution gives a

quadratic equation in Q. The solution is 2.57 ft^3 / s. The pump

power is rho g Q h_p / eta to give 67 hp.

3.160. The air cushion vehicle is lifted by the pressure difference

between that inside and out, times the skirt area, so from the known

weight the inside pressure must be 1768 above atmospheric. From

Bernoulli's equation across the skirt we get the skirt velocity 30.6

m/s. The power is Q delta p, or 54000 W.

========

Design project D3.1.

========

The problem is to select a pump with minimum cost

with given performance characteristics. It must deliver no

less than 1.0 ft^3 / s of water and rotate no slower than 10

rev/s. The cost is assumed to be proportional to the power

input to the pump. The relevant dimensional parameters of

the pump performance are the pump work rate per mass flow

rate gh_p, the volume flow rate Q, the water density rho, the

shaft rotation rate n, the impeller diameter D, the power to

the water rho Q g h_p , and the power input to the pump.

The relevant dimensionless parameters for the pump are

functions of the dimensionless volume flow rate zeta =

Q/nD^3 . The dimensionless pump head is phi = gh_p / n^2

D^2 = 6.04 - 161 zeta . The efficiency eta = 70 zeta - 91500

zeta^3. The dimensionless power to the water pi_W = P_W

/ (rho n^3 D^5) = zeta phi . The dimensionless power input

P_I is pi_I / eta which should be minimized as a function of

zeta . Since Q and n are fixed, the pump impeller diameter to

give minimum power can be determined.

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