1. Viscous flow in ducts...laminar

2. Viscous flow in ducts...turbulent.

The Moody chart of the Colebrook equation.

Definition of turbulence.

3. Examples.

4. Homework hints.

5. Warning label.

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

Viscous flow in ducts...turbulent.

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

In the last lecture we found for laminar flow in a

circular tube of diameter D the Darcy friction factor f =

64/Re_D, where f = 8 (tau_wall / rho V^2 ). The friction

factors f is a form of Euler number (Eu = "drag"

stress/inertial stress), just like the "drag coefficient" C_D =

(drag force/streamwise projected area)/(rho V^2 /2) defined

in 5.27, p273, and the "cavitation number" of Table 5.2,

p271. Think of the friction factor as the wall stress of the

pipe made dimensionless by the available inertial stress rho

V^2. Recall that the inertial stress tensor is rho v v , and that

the inertial force of fluid at the pipe entrance is rho v v dot A,

where A is the pipe entrance area vector. It is generally a

good idea to have a simple physical model like this in mind

when using dimensional groups and dimensional analysis to

avoid trouble. Beware of "blind" dimensional analysis,

which is juggling of dimensional quantities "taken out of the

air" (without any physical model) in order to find a

dimensionless group to use in a problem.

For turbulent flows, the friction factor f does not

behave the same way with changing Reynolds numbers as it

does for laminar flows. By injecting a thin dye stream the

center of flowing water in a glass tube, Osborne Reynolds

found (in 1893) the flow becomes "sinuous" for values of

VD/nu greater than about 2100. As shown in Fig. 6.12b,

p316, the friction factor erratically increases sharply for

2100 < Re_D < 4000, the transition zone, and then

decreases as

f = 0.316 Re_D^-1/4 , 4000 < Re_D < 10^5

for smooth tubes, and settles out to constant values for tubes

with various roughness ratios e_w/D. For example, for

e_w/D = 0.1% the constant value of f is 0.02. The various

e_w/D curves shown were determined by Nikuradse by

gluing sand grains of known diameter to the walls of pipes

and then measuring the flow rates at different pressure

drops. Recall that tau_w = D delta p / 4 L from a force

balance, and that V can be measured by dividing the volume

flow rate through the pipe by the cross sectional area.

A variety of interpolation formulae have been used to

simulate the empirical results of pipe flow measurements for

various Reynolds numbers and roughness ratios. The

accepted design formula is due to Colebrook

f^-1/2 = -2.0 log [ (e_w/D)/3.7 + (2.51 / Re_D

f^1/2)]

which was plotted by Moody in 1944 and appears as the

"Moody Chart" of Fig. 6.13, p318 in the text. These days

of course such charts can be replaced by entering the formula

in a good scientific calculator with a "solve" capability.

*********

Example: Solve the Colebrook equation for Reynolds

number Re_D, when f is 0.03 and e_w/D = 0.001. Ans.:

From the solver program in a HP100LX palmtop, Re_D =

14,101. This agrees with the Moody chart. You can see

this result computed by the AMES 170 computer at the break

or after class, and try other values of f, e/D, and Re.

*********

========

Definition of turbulence.

========

The dramatic departure of friction factors from the

laminar relationship f = 64 / Re_D occurs because viscous

forces can no longer suppress the non-linear inertial vortex

forces v cross w that arise in the pipe. If the flow is to the

right, the vortex lines are clockwise rings looking

downstream. As the velocity increases, the inertial vortex

force will increase, and be in the outward radial direction,

balanced by an inward radial pressure gradient for laminar

flow.

boundary layer cartoon

This flow becomes increasingly unstable. Any perturbation

from the equilibrium laminar flow will induce an additional v cross w

force in the direction of the perturbation that must be damped

out or balanced by opposite induced viscous forces. When a

critical value of the Reynolds number is exceeded, the

perturbation will grow, induce a nearby perturbation of the

opposite sign that will grow in the opposite direction, and

the two perturbations will pair to form an eddy. Eddies tend

to pair with eddies of the same size to form eddy pairs, and

pairs of eddies pair with pairs of eddies in a cascade to larger

and larger eddy-like flow structures until the pipe is full of

eddies of all sizes up to the size of the pipe. Other turbulent

flows such as boundary layers, jets, and wakes grow from

small scale eddies to larger scale eddies in the same way.

In stratified flows with no solid boundaries such as

in the ocean or atmosphere, turbulence is constrained in the

vertical direction by buoyancy forces, and in the horizontal

direction by Coriolis forces. Most of the fluid volume is not

turbulent at all. Only rarely does a patch of turbulence

appear, and then it is usually rapidly damped and converted

to internal wave motions by the stable stratification.

Horizontal shears between large current systems can result in

horizontal turbulence (or "2-D" turbulence) that grows to

such large scales that the Coriolis forces of the earth's

rotation can constrain further turbulent growth.

Astrophysical turbulence, and turbulence in fusion reactors,

is often constrained by electro-magnetic forces.

This competition between inertial-vortex and

damping viscous, buoyancy, Coriolis, and other forces leads

to the following definition of turbulence:

motion where the inertial-vortex forces of the eddies are

larger than the viscous, buoyancy, or other forces that arise

to damp out the turbulent eddies.********

Thus, for a flow to be turbulent by this definition eddy

motions must exist, and have Reynolds numbers above

known critical values. If the flow is stably stratified, the

Froude number of the eddies must also be above some

critical value known from theory or experiment. Irrotational

flows cannot be turbulent because their inertial vortex forces

are zero. In pipe, boundary layer, jet, wake, and shear layer

turbulence, energy is extracted from the larger scale

irrotational flow by inducing a non-turbulent cascade from

large scale motions to small as the irrotational fluid is sucked

into the interstices of the turbulent eddies and converted to

turbulence by viscous forces.

turbulence cascade cartoon

Think of the energy transport in such turbulent flows

by drawing a stack of clouds of smaller and smaller sizes,

connected by arrows showing the energy fluxes. The large

blue cloud on top is irrotational, and cascades to smaller and

smaller scales by ideal, irrotational, flow (v cross w forces

zero). This is shown by the arrows going to smaller blues to

the smallest red turbulent cloud. The smallest turbulent clouds

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

where nu is the kinematic viscosity and eps is the viscous

dissipation per unit mass. Smaller eddies can't form because

their Reynolds numbers would be too small. All the small

red turbulent clouds are connected by short red arrows going

both directions, indicating that the mechanism of turbulent

energy transport is local (by eddy pairing) and that there is

continual feedback to smaller scales. The NET flux of

energy from large scales to small is eps watts/kg, but the

local fluxes are much larger. This keeps high Reynolds

number turbulence in a state of "statistical equilibrium",

characterized only by the scale of the velocity fluctuation and

the average dissipation rate eps per unit mass. Since most of

the dissipation rate occurs at the small scales, and since all of

the energy comes from larger scales, all of the intermediate

scales are also characterized by eps.

**********

Example: Based on this idea of a constant flux of energy

from large scales to small and the assumption that there must

be a universal critical Reynolds number limiting the size of

the smallest scale eddies in such an equilibrium flow,

Kolmogorov (1941) proposed two universal similarity

hypotheses for turbulence. The first hypothesis is that the

probability laws for velocity differences for separation

vectors y should depend only on the viscosity nu and the

average dissipation rate per unit mass eps. Find the length

and time scales L_K and T_K of the smallest eddies of

turbulence by dimensional analysis.

Solution: Use the power product method. L_K =

f(eps,nu) [=] eps^a nu^b [=] L [=] (L^2 T^-3)^a (L^2 T^-

1)^b . Solving gives a = -1/4 and b = 3/4. Thus L_K =

(nu^3 / eps)^1/4. We saw in an earlier example that the

actual viscous length scale is larger than L_K by a factor of

Re_crit^1/2. The universal critical Reynolds number is

about 25-100, so the factor is 5 to 10. The Kolmogorov

time T_K = f(eps,nu) = eps^a nu^b [=] T [=] eps^a nu^b

[=] (L^2 T^-3)^a (L^2 T^-1)^b . Solving gives a = -1/2 and

b = 1/2, so T_K = (nu / eps)^1/2. Note that both the length

and time scales decrease with increasing eps, and increase

with increasing nu. This is physically reasonable.

**********

**********

Example: Kolmogorov's second universal similarity

hypothesis is that for separations much larger than L_K, the

velocity differences in turbulence should be independent of

the viscosity nu. Estimate the velocity difference between

two points a distance of 50 cm apart in a room with a fan that

keeps the dissipation rate constant at 10 cm^2 s^-3.

Solution: By the Kolmogorov second hypothesis the

velocity difference V = f(eps, y) = const. eps^a y^b [=] L/T

[=] (L^2 T^-3)^a (L^2 T^-1)^b . Solving gives a = 1/3 and

b = 1/3, so V = const. (eps y)^1/3. We can estimate the

constant by evaluating the velocity difference at the universal

critical Reynolds number R, where y = R^1/2 L_K. Thus R

= const. (eps R^1/2 L_K)^1/3 R^1/2 L_K / nu = const.

R^2/3, so the constant is R^1/3, or between 2.9 and 4.6 for

R between 25 and 100. Taking a value of const. = 3.8 give

V = 3.8 (10x50)^1/3 = 30 cm/s (ans.).

***********

***********

Example: Consider problem 6.18. Water at 20 deg C flows

in a 9 cm diameter pipe under fully developed conditions.

The centerline velocity is 10 m/s. Compute a. Q, b. V, c.

tau_wall, and d. delta p for a 100 m long pipe.

Solution: For water at 20 deg c, take rho = 998

kg/m^3 and mu = 0.001 kg/m s. Check Re = rho V D / mu

= 998x10x0.09/0.001 = 900,000 so the flow will be

turbulent. We can estimate the centerline velocity from the

law of the wall; that is

u_cl / u* = (1/kappa) ln (Re x u* / nu ) + B

where u* = (tau_wall / rho )^1/2 , kappa is von Karman's

constant 0.42, and B = 5. Solve to get u* = 0.350 m/s

(ans). Therefore tau_wall = rho u*^2 = 998x(0.35)^2 =

122 Pa (ans. c.). At such a high Re value the velocity

profile is nearly flat, so we estimate V = 0.85 u_cl = 0.85 x

10 = 8.5 m/s (ans. b.). The volume flowrate Q = VA =

0.054 m^3/s (ans. a.). The pressure drop comes from D

delta p / 4 L = tau_wall. delta p = 122 x 100 / 0.09 =

542,000 Pa (ans. d.).

***********

===========

Homework hints and example problems.

===========

Example 6.19. Assume laminar flow and use the pressure

drop formula 6.44 to get mu = 0.292. Check the turbulence

assumption by assuming a density of oil = 900 kg/m^3 and

calculating the Reynolds number to be 16 (assumption

correct!). You can't estimate the density from the data of a

laminar flow because laminar flows are independent of

density.

6.17. Show that the log layer follows from Theodore

von Karman's assumption that the wall shear stress and boundary

layer shear stress of a turbulent boundary layer can be derived

by assuming a mixing length eddy viscosity

nu_T = rho y^2 d,y u = u(y)y (neglecting the constant).

The constant stress boundary layer has stress tau = nu_T d,y u.

Substitute the proposed eddy viscosity and integrate to get the

logarithmic form.

tau = rho nu_T d,y u = const. = rho y^2 d,y u ^2, so

d,y u = u*/y ; u = u* ln y + const.

6.24. A 6 cm diameter pipe contains glycerin at 20

deg. C flowing at a rate of 6 m^3/h. Verify that the flow is

laminar. The pressure at point A is 2.1 atm, and at point B

at a higher elevation of 12 m the pressure is 3.7 atm. Is the

pressure from A to B or from B to A? What is the indicated

head loss for these pressures?

Solution: For glycerin at the given temperature the

density is 1260 kg/m^3 and the dynamical viscosity mu is

1.49 kg/m s. The Reynolds number Re = 30, so the flow is

laminar. The Bernoulli equation is

B_A = B_B + gh_f

where B_A = 2.1x101350/1260 + V^2 /2 + z_A, and B_B =

3.7x101350/1260 + V^2 /2 + z_A. Substituting into the

Bernoulli equation we find the velocity terms cancel. The

only way for h_f, the frictional head loss, to be positive is

for the flow to be from B to A. The head loss is then 25.1 m

(ans.).

6.44. The Reynolds number for the mercury in the

glass tube of 7 mm diameter, 4 m long, with velocity 3 m/s, is

182,000 ( rho = 13550 kg/m^3 , mu = 0.00156 kg/m^3 from

the tables in the book). Consider the roughness ratio for

glass to be zero. Find the Darcy friction factor f from the

Moody chart to be 0.016. The head loss h_f = f (L/d) (V^2 /2g).

The pressure drop is rho g h_f to give 555 kPa.

6.78. Use Bernoulli's equation for the streamline

starting at the top of the open reservoir, ending at the top

of the closed reservoir, where the pressure is known.

Calculate h_f for the pipe connecting the tanks using

Bernoulli's equation and the known pressure and heights,

to give h_f = 5.43 m. Find the velocity from the

Moody chart by iteration. Assume turbulent flow and guess

a value of f, using the roughness ratio for commercial steel,

where e_w = 0.046 mm. Compute an estimate of V from

h_f = f (L/d) (V^2 /2g). Find the Reynolds number and

a better guess for f from the Moody chart for the commercial

steel wall roughness. Keep it up until f converges to 0.0205.

6.85. The pump with the h_p versus Q performance

curve shown in Fig. P6.80 delivers 0.7 m^3/s of methanol

through 95 m of cast iron pipe. What is the pipe diameter?

Look up the density and viscosity of methanol. Look up the

roughness length for cast iron (0.26 mm). From Bernoulli's

equation the head loss h_f = h_p for a constant diameter,

horizontal pipe (note: h_p = - h_shaft). Express the equation

h_f = f (L/d) (V^2 /2g) in terms of Q, to get

h_f = f (L/d) (V^2 /2g) = 8fLQ^2/pi^2 g d^5.

Find the constants a and b in the "parabolic" pump performance

curve h_p = a + bQ^2 (h_p = 80 -20Q^2). Substitute and solve

for d = 0.559 f^1/5. Guess f = 0.02 and compute d. From

Re = 4 rho Q/ pi mu d find Re = 4.61x10^6. From the roughness

ratio and the Moody chart find a better estimate of f. Iterate to

d = 0.255 m.

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

WARNING: Many people think turbulence is any flow that

wiggles, and that the turbulent cascade is from from large

scales to small. They are wrong. A poem, due to L. F.

Richardson, is partly responsible for this misleading

folklore: "Big whorls have little whorls, that feed on their

velocity, and smaller whorls have smaller whorls, and so on

to viscosity (in the molecular sense)". The poem (which is a

rip-off of one about fleas by another author) should be

corrected to read: "Little whorls at viscous scales, form and

pair with more of, whorls that grow by vortex forces, Slava

Kolmogorov!". (Slava is a Russian word meaning "Glory".

You can hear it sung by the chorus to the Tsar in the opera

"Boris Gudonov").

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

1. Turbulent boundary layers...

-laminar sublayer thickness

-constant stress

-the "logarithmic-overlap layer"

-The "law of the wall".

2. Bernoulli obstruction theory.

3. Homework hints and examples.

Turbulent boundary layers.

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

Consider the flow at the entrance to a duct from a

stagnant reservoir. The fluid will be irrotational everywhere

because it has not been in contact with a wall before entering

the duct. We know from the vorticity equation

D,t w = w dot e + nu del^2 w

that the only way the vorticity w can change if the fluid is

irrotational is by viscous diffusion at the wall due to the term

nu del^2 w (the term w dot e = 0 if w = 0). This is

equivalent to the more intuitive fact that the only way for a

fluid particle in the flow to be heated is for it to touch the

(hot) wall from the temperature conservation equation

D,t T = alpha del^2 T

where alpha is the thermal diffusivity and D,t is the

substantive derivative. The temperature equation is in

exactly the same form as the vorticity equation for

irrotational flow. Vorticity must diffuse from the wall of the

duct into the laminar boundary layer of the fluid entering the

duct to make the entering fluid rotational, just as temperature

must diffuse into the temperature boundary layer in order to

heat the fluid.

At first the boundary layer of the fluid entering the

duct will be laminar no matter how fast it enters. This is

because it will be very thin, and no matter how big the

entrance velocity V may be, the Reynolds number Vd/nu will

be less than Re_crit. if the boundary layer thickness d

approaches zero. How fast will the boundary layer

thickness grow? We can find this by dimensional analysis

(in view of the vorticity equation). The thickness d will

depend on the time t of diffusion and the diffusivity of the

vorticity nu. Thus d (t,nu) = const. t^a nu^b = const. (nu t

)^1/2 , just as the thickness of the thermal boundary layer

will be d_T = const. (t alpha )^1/2. However, the time t is

equal to the distance from the entrance x divided by the

velocity V. Therefore

d = (x nu / V)^1/2 = Re_x ^-1/2

where Re_x = Vx/nu is the Reynolds number based on the

mean velocity and the distance x from the entrance of the

duct. As shown in Fig. 6.9, p306 of the text, the laminar

boundary layer will grow until d is greater than about 5 or 10

times the quantity nu / u*, where u* is the "friction velocity"

(tau/rho)^1/2 , and then it will become turbulent. As you

might expect, nu / u* is equal to the Kolmogorov scale L_K

= (nu^3 / eps )^1/4 , because the Kolmogorov scale is the

size of the largest possible viscous force dominated flow

according to the first universal Kolmogorov hypothesis.

*********

Example: Show that nu / u* = L_K. Soln.: The viscous

dissipation rate eps = 2 nu e^2, where e is the rate of strain

tensor with components eij = (ui,j + uj,i)/2. Near the wall

eij = [e11 e12 e13 ; e21 e22 e23 ; e31 e32 e33] = [0

(du/dy)/2 0 ; (du/dy)/2 0 0 ; 0 0 0] . Therefore e^2 = eijeij =

e1je1j + e2je2j + e3je3j = e11e11 + e12e12 + e13e13 +

e21e21 + e22e22 + e23e23 + e31e31 + e32e32 + e33e33 =

[(du/dy)/2]^2 + [(du/dy)/2]^2 = (du/dy)^2 /2 . Thus eps

near the wall is 2 nu (du/dy)^2 /2 = nu (du/dy)^2 , and

du/dy = (eps/nu)^1/2 . What is the "friction length" nu / u*

?

nu / u* = nu / (tau/rho)^1/2 = nu / (mu du/dy

/rho)^1/2 = nu / (nu (eps/nu)^1/2 )^1/2 = nu^3/4 / eps^1/4

= L_K QED

Note: According to Fig. 6.9 the dimensionless velocity u^+

= u/u* is equal to the dimensionless length y^+ = yu*/nu in

the laminar boundary layer. This is because du/dy = tau /

mu = u*^2 / nu in this layer, which can be integrated to give

u = u*^2 / nu y + const., where const. = 0 because u = 0 at

y = 0. Thus u/u* = y nu / u*, as advertised. We see that the

critical Reynolds number indicated by the laminar boundary

layer thickness is Re_crit. = ud/ nu = 25 to 100 because y+

and u+ become turbulent between 5 and 10. Previously we

saw that Kolmogorov's turbulence theory says there should

be a universal critical Reynolds number Re_crit. =

(L/L_K)^2 = (y+^2)_crit., so we conclude from Fig. 6.9

that the universal critical Reynolds number is in the range 25

to 100.

**********

Once the boundary layer becomes turbulent, the flow

becomes chaotic. Eddies begin to form at the Kolmogorov

scale. These pair to form larger eddies, and the pairs of

eddy pairs pair with other pairs of eddy pairs, and so forth

as the turbulence diffuses into the surrounding irrotational

flow. Eventually the boundary layers meet near the duct

center and the flow is fully turbulent. What is the mean

velocity profile du(y)/dy? According to Fig. 6.9 (the law of

the wall) the velocity is proportional to the logarithm of the

distance from the wall in the turbulent boundary layer. Why

is this?

In steady state, each layer of the boundary layer must

be subjected to the same stress as the stress at the wall.

Otherwise there would be a net force, and since there is no

force available to offset such a force the layer would

accelerate, contrary to our steady state assumption. This is

the physical basis for assuming that du(y)/dy is a function of

only y, the fluid density rho, and the wall shear stress tau.

By dimensional analysis we find du/dy = const. u*/y .

Integrate to find

u - u(cL_K) = u* ln (y/cL_K) / kappa, or

u / u* = ln (u nu/u*) / kappa + B

where the universal constants are B = 5 and kappa (the von

Karman constant) is 0.42. This logarithmic boundary layer

form has been widely tested and confirmed by experiments

in the laboratory and in the field.

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

Bernoulli obstruction theory.

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

Many commonly used engineering flow meters are

based on the so-called Bernoulli obstruction theory. By

restricting the flow at high Reynolds number in a pipe with

an orifice plate, long radius nozzle, or a venturi nozzle, the

flow speeds up, the pressure falls according to Bernoulli's

equation, and by measuring the pressure drop the flow rate

can be computed from the Bernoulli obstruction theory.

The generalized obstruction is shown in Fig. 6.35.

The upstream velocity is V1 through a pipe of diameter D.

The nozzle diameter d = beta D. By continuity, assuming

incompressible (constant density) flow, the throat velocity is

Vthroat = V1 (D/d)^2 = V1 (1/beta^2).

From Bernoulli's equation along a streamline from

upstream passing through the obstruction throat, and

eliminating V1, we find

po = p1 + rho V1^2 / 2 = pthroat + rho Vthroat^2 / 2

; 2(p1-pthroat) = Vthroat^2(1-beta^4).

so

Vthroat = [2(p1-pthroat) / rho (1 - beta^4)]^1/2

which is the ideal result of the theory. Actually there are

slight departures due to friction and the fact that in some

cases the highest velocity, and lowest pressure, occurs

downstream of the nozzle throat at the vena contracta. These

effects are lumped into a "discharge coefficient" C_d in the

expression

Vthroat = C_d [2(p1-pthroat) / rho (1 - beta^4)]^1/2

where usually C_d is less than 1. Plots for C_d values at

different Reynolds numbers and beta values are given for the

orifice plate in Fig. 6.37 (C_d = 0.6 to 0.61 for beta 0.2 to

0.8 and Re 10^5 to 10^7), a long radius nozzle and Herschel

venturi in 6.38 (C_d is 0.975 to 0.995 for the same Re range

and all betas), and a venturi nozzle in 6.39 (C_d is 0.92 to

0.98). Even though C_d values are smaller, the orifice

plates are most popular because they are cheap and simple,

and their flow losses are affordable.

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

Homework hints and examples.

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

Example 6.87. A commercial steel annulus 40 ft long, with radii a =

1 inch and b = 1/2 inch, connects two reservoirs which

differ in surface height by 20 ft. Compute the flow rate in

ft^3 / s through the annulus if the fluid is water at 20 deg C.

Solution: For water take rho = 1.94 slug/ft^3 and

mu = 2.09x10^-5 slug/ft s. For commercial steel, take e_w

= 0.00015 ft. Compute the hydraulic diameter of the

annulus. Dh = 4A/P, where A is the wetted area and P is the

wetted perimeter. A = pi (a^2 - b^2) , and P = 2 pi (a+b),

so Dh = 2(a-b) = 1 inch. Hence,

hf = 20 ft = f (L/Dh) V^2 / 2g = f (40 / 1/12 ) (V^2 /

2x32.2) ;

or

fV^2 = 2.683.

Such an equation must be solved by "cut and try" methods;

that is, guess a value for f, find Re and thus V from the

Moody chart, and compute a new f from the equation above.

Keep doing this until the values of f and V converge.

To use the Moody chart we need the roughness ratio.

For Dh = 1 in and e_w = 0.00015 ft we find e_w/D =

0.00015/(1/12) = 0.0018. Guess f = 0.023, so V = 10.8 ft /

s from the equation. The Reynolds number is then

1.94x10.8x(1/12)/2.09x10^-5 = 83550. Use this to find a

better value of f = 0.0249, Vbetter = 10.4 ft/s , etc. This

converges to f = 0.0250, V = 10.37 ft/s. Thus Q = pi (a^2 -

b^2)V = 0.17 ft^3 / s. ans.

Example 6.141. Gasoline at 20 deg C flows at 105 m^3 / h in a 10

cm diameter pipe. We wish to meter the flow with a thin

plate orifice and a differential pressure transducer which

reads best at about 55 kPa. What is the proper beta ratio for

the orifice?

Solution: For the gasoline rho = 680 kg/m^3, and

mu = 2.92x10^-4 kg/m s. The pipe velocity is V1 = Q/A1 =

(105/3600) / (pi/4)(0.1)^2 = 3.71 m/s. Re_D =

680x3.71x0.1/2.92E-4 = 865,000.

From Fig. 6.37, read C_d = 0.61. Therefore the

throat velocity V_throat = 3.71/beta^2 = C_d

(2x55000/680(1-beta^4))^1/2 ; or beta^2/ (1-beta^4)^1/2 =

0.4778. Solve for beta = 0.66. ans. Now check back with

Fig. 6.37 to see this is consistent, so no further iteration is

needed.

6.97. Consider the rectangular wooden test section as a circular

pipe with hydraulic diameter D_h = 4A/P, where A is the cross

sectional area (45x95 cm^2), and P is the "wetted" perimeter

(45+95+45+95) so you can use the Moody chart to find the pressure

drop using delta p = f (L/D_h) ( rho V^2 / 2). The roughness length

for wood is about e_w = 0.2 mm. Look up the density and viscosity

of air at sea level (1.22 kg/m^3 and 1.5x10^-5 kg/m s). Find the

Reynolds number and roughness ratio to give f_Moody = 0.0158.

Compute the pressure drop. The force pA times velocity Q/A is the fluid

power pQ so the power to the fan is pQ/ eta , where eta is the efficiency,

giving 6900W. Note that if the wood roughness length were 0.8 mm

the power would increase by 35%.

6.105. What gage pressure is required to deliver 0.005 m^3 / s of

water to the reservoir at 500 m elevation through 1200 m of 5 cm

diameter cast iron pipe, with two 45 and four 90 degree flanged long

radius elbows, a globe valve, and a sharp exit into the reservoir. We

use the Moody chart to find f = 0.0315 from the velocity V = Q/A, the

Reynolds number DV/nu and the roughness length for cast iron (look

up). Apply the Bernoulli equation for a streamline extending from the

entrance of the pipe to the surface of the reservoir. The gage pressure

p1-patm over rho g is then equal to the 100 m difference in elevations plus

the head losses h_f for the pipe and the fittings, valve and entrance. For

the pipe this is [V^2 / 2 g] times f(L/D). For each fitting this is [V^2 / 2 g]

times K for the fitting (K=0.2 for 45 deg elbows, 0.3 for 90 degree elbows,

8.5 for a globe valve, 1 for the submerged entrance, -1 for the velocity

entering the pipe, bringing it over from B_1), giving a total head of 353 m.

The pressure is then rho g h_ftotal = 3.46 MPa.

6.136 Assume the air is turbulent in the 8 cm pipe, and that the velocity of

the turbulence is given by the "law of the wall" logarithmic profile. We get

the centerline velocity from the pitot tube, which will be somewhat higher

than the mean velocity V = Q/A needed to compute the pipe volume flow (rate)

Q and the (smooth) wall shear stress. The 40 mm water level difference of the

manometer colored water levels gives the pressure difference along the axial

steamline, which is V_centerline^2 times rho / 2. Guess V = 0.85 VCL to

find Re and fsmooth = 0.0175. Use equation 6.59 V=VCL/(1+1.33f^1/2) to

find a better estimate of V (based on the law of the wall). This procedure

converges to V=21.69 m/s. Compute Q=VA and tau = f rho V^2 / 8 = 1.23 Pa.

6.145 Look up the density and viscosity of gasoline (680 kg/m^3 and 2.92x10^-4

kg/m s). Use the "curve-fit" formula for thin orifice plates 6.132 with beta=0 (corner

taps) to give a discharge coefficient C_d = 0.596. Beta is 2x10^-4. The pressure

drop across the orifice is rho g h(t), so Q=C_d A (2 g h(t))^1/2 = 0.000829 h^1/2.

Q is also equal to -Atank d,t h(t), giving a differential equation dh/h^1/2 = -0.001056 dt

that can be integrated between the two levels of 2 and 1.6 meters to give the discharge

time of 4.7 minutes.

*****

The ground rules are the same as for the first two

exams: open book, closed notes, two pages of handwritten

notes. Please bring a blue book or some paper, and a

scientific calculator. Good luck!

*****

========

Final Review:

========

The format of the exam will be similar to the

previous exams. Expect to do some derivations. Be sure

you can derive the Reynolds Transport Theorem, the

vorticity equations, Bernoulli's equation, the mechanical

energy equation, the continuity equation, etc. In your

review, you should be sure you understand all the

homework problems assigned, and go over all the worked

out problems in the text and posted lectures. Read over all

the problems at the end of Chapters 1-6 of the text and think

about how you would solve them. Be sure you can give the

definition of turbulence (

fluid motion where the inertial-vortex forces of the eddies are

larger than any of the other forces that tend to damp them

out

tensors. Know how to do Buckingham Pi and power

product dimensional analyses. Review hydrostatics. Be

able to solve problems for laminar and turbulent flows in

circular and non-circular tubes using the Moody chart, or by

solving the conservation equations for fluid mechanics you

have derived. Know the assumptions and conditions for the

existence of stream functions and potential functions. Know

how the various fluid flow meters described in Chapter 6

work. Please bring questions to class today.