*Viscous flow in ducts, Ch. 6.

*Laminar flow in a circular tube

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Viscous flow in ducts.

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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.

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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.

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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] .

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Laminar flow in a circular tube.

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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.

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Homework hints.

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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.

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