Lecture (part 1):
*Viscous flow in ducts, Ch. 6.
*Laminar flow in a circular tube

Viscous flow in ducts.
Many practical problems in engineering require the
transfer of fluids through ducts. Viscous stresses at the
walls of the ducts must be balanced by pressure forces
supplied by pumps, where the pump power is the pressure
drop across the pump times the volume flow rate. The flows
may be laminar or turbulent, or a combination.

Example: Buckingham Pi analysis of the wall stress for
flow in a duct. The stress tau depends on the density rho ,
the fluid velocity V, the dynamical viscosity mu , the
diameter of the duct d, and the roughness of walls e_w .
The dimensional matrix is (tau rho V mu d e_w ; M L t ) = (1
1 0 1 0 0 ; -1 -3 1 -1 1 1 ; -2 0 -1 -1 0 0 ). The rank of the
6x3 matrix is 3, so we expect 3 Pi groups. Choose the core
group to be rho , V, and d. All the dimensions are different
for these variable, they include M, L and t, and these
variables are not as important as tau , mu , and e_w. The
first Pi group comes by making tau dimensionless with rho ,
V, and d: Pi_1 = tau / rho V^2 which is called the "friction
factor" of the duct. The "Darcy friction factor" f is defined
as 8 Pi_1. The second Pi group comes from mu, giving
Pi_2 = mu / rho V d = 1 / Re_d, where Re_d is the Reynolds
number based on the duct diameter d. The third Pi group is
formed from e_w, giving Pi_3 = e_w / d, or the "roughness
ratio". The Moody chart of Fig. 6.13, p318 of the text is a
summary of the theoretical and experimental values of f
versus Re_d and e_w/d.

The pressure drop required from a pump depends on
the wall shear stress, the length of the duct, and its diameter.
For steady state flow through a circular duct of diameter D
and length L the wall stress tau times the tangential area pi D
L equals the pressure drop delta p = (p_1 - p_2) times the
cross sectional area pi D^2 / 4. Solving gives
tau = D delta p / 4L.
Substituting in f = 8 tau / rho V^2
f = 8 tau / rho V^2 = 8 (D delta p / 4L) / rho V^2 =
(D/L) (delta p / rho) / (V^2 /2) .
However, from Bernoulli's equation for a constant cross
sectional area level duct
delta p / rho = g h_f
where h_f is the frictional head loss. Common engineering
practice is to express pump problems in terms of head
losses, so
h_f = f (L/D) (V^2 / 2g)
is the formula for calculating the head loss from the friction
factor, mean duct velocity V, and the L/D ratio. The
expression is dimensionally consistent and reasonable with
respect to all terms. Minor head losses from valves and
fittings in a piping system are expressed as
h_m = K (V^2 / 2g)
where the coefficient K can also be given as
K = f (L_eq / D)
in terms of an equivalent duct length L_eq. The total head
loss for a piping system is the sum of the major and minor
head losses
h_total = h_f + sum h_m = (V^2 / 2g) [fL/D + sum
K] .

Laminar flow in a circular tube.
What is the velocity profile for laminar flow in a
circular duct? What is the friction factor f in terms of the
Reynolds number? We can answer the first question by
considering the forces on a cylinder of radius r in the tube of
length L. Let dp/dx = -K, where K is a positive constant.
Neglect gravity and variations of mu and rho of the
Newtonian viscous fluid, and assume steady flow. The sum
of the forces in the streamwise, or x, direction are
tau dot A_3 + (-p d dot A)_1 + (-p d dot A)_2 = 0
where 1 is the entrance, 2 is the exit, and 3 is the side area of
the element. This equation becomes
[( mu du/dr 2 pi r L) + (p_1 - p_2)(pi r^2)] i = 0 = [(
mu du/dr 2 pi r L) + KL(pi r^2)] i
where i is the unit vector in the x direction and (p_1 - p_2) =
KL. This gives the differential equation
du/dr = - Kr/2mu
with boundary condition u = 0 at r = R = D/2, where D is the
diameter of the cylinder and R is the radius. Integrating with
respect to r gives
u = - Kr^2 / 4 mu + C
where C = KR^2 / 4 mu from the boundary condition. Thus
u = (KR^2 / 4 mu) [1 - (r/R)^2]
is the parabolic velocity profile of viscous flow in a cylinder,
and the maximum velocity v_max = (KR^2 / 4 mu).
What is the average velocity in terms of v_max? We
take the area average
v_ave = V = (1/A) integral (0,R) [u(r)] dA(r) = (1/pi
R^2) integral (0,R) [u(r)] 2 pi r dr = v_max 2 integral (0,1)
[eta (1-eta^2] d eta = v_max / 2
so the average is just half of the maximum velocity. Now
the friction factor f is defined as 8 |tau| / rho V^2 . We can
find tau at the wall from mu du/dr at r=R. Since u = 2V[1-
(r/R)^2], du/dr = - 4 V/R = - 8 V/D, and |tau| = 8 mu V/D.
Thus, the friction factor is
f = 8 |tau| / rho V^2 = 64 mu / rho VD = 64 / Re_D
which is the laminar portion of the Moody chart.

Homework hints.
Problems due: 6.2,7,12,17. Open for questions.

6.1. Look up values of density and viscosity for air so you
can calculate the velocity needed to cause the boundary layer
to become turbulent at Re_D = 2.5 E5. The dimples on the
golf ball are intended to "trip" the boundary layer so it will
go turbulent at lower Re_D values. The turbulence causes
the separation point to move from the front to the back of the
golf ball, thus reducing drag. Turbulent boundary layers
have large "eddy viscosities" that tend to prevent separation.
You may have noticed rows of vortex generators on the top
surface of wings near the leading edge. These are also used to
prevent boundary layer separation, which could cause the
wing to "stall" under conditions of high lift (e.g., in landing).

6.9. Pressure values are given as a function of distance x along
the horizontal smooth pipe with diameter 5 cm, for velocity 10 m/s
of fluid of density 950 kg/m^3. Find the average wall shear stress
and the wall shear stress in the "fully developed" region of the
pipe. A force balance for the pipe, or any pipe section, gives the
relationship tau = D delta p / 4 L . The pressure drop per meter of
pipe delta p / L decreases to 13000 pascals/m in the fully developed
region, giving tau of 163 Pa, less than the overall average of 217.

6.13. Velocity in ft/s are given as a function of distance from the
wall y in inches for air flow at 75 deg F and 1 atm pressure near
a smooth wall in a wind tunnel turbulent boundary layer. To estimate
the shear stress at the wall you fit the data to the "log-layer", eq.
6.21 (most scientific calculators have a "solve" function that is
designed for just this sort of calculation). You will find that u*
values are nearly constant at 3.57 ft/s (rho = 0.0023 slug/ft^3,
mu = 3.8E-7 slug/ft-s). Solve for the shear stress tau = rho u*^2
from the definition of u*. For part b, assume the log-layer extends
to y = 0.22 inches, so u* is still 3.57 ft/s. Use 6.21 to compute u
of 70 ft/s.

6.21. Water is to be siphoned through a tube 1 m long and 2 mm
diameter to a height H below the surface of the tank. What is H for
the flow to become turbulent (take Re_transition to be 2000). Use
Bernoulli's equation with hf = 32 mu L V/ rho g d^2 from laminar
flow theory, so hf = H - V^2/2g . Solve the quadratic for the positive
root, giving V = 0.59 m/s. The Re number at H=50 cm is 1180, so the
flow is laminar there. Find where Re = 2000.

Exam 2, Good luck!