Lecture (part 1)

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:

*******Turbulence is defined as an eddy-like state of fluid
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").
===============

Lecture (part 2):

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 final exam is on July 31 (101A) and Aug. 2 (103A).
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 (turbulence is an eddy-like state of
fluid motion where the inertial-vortex forces of the eddies are
larger than any of the other forces that tend to damp them
out). Know how to compute forces on areas from stress
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.