The Hong Kong University of Science & Technology
Department of Civil & Environmental Engineering
CIVL2510
Practice exercise
Problem 1
A thin plate is separated from two fixed plates by very viscous liquids µ1 and µ2 , respectively, as in Fig. 1.
The plate spacings h1 and h2 are unequal, as shown. The contact area is A between the center plate and
each fluid.
(a) Assuming a linear velocity distribution in each fluid, derive the force F required to pull the plate at
velocity V .
(b) Is there a necessary relation between the two viscosities, µ1 and µ2 ?
Figure 1: Problem 1
Solution
(a) Assuming a linear velocity distribution on each side of the plate, we obtain
µ1 V µ2 V
+
A Ans. (a)
F = τ1 A + τ2 A =
h1
h2
The formula is of course valid only for laminar (nonturbulent) steady viscous flow.
(b) Since the center plate separates the two fluids, they may have separate, unrelated shear stresses, and
there is no necessary relation between the two viscosities.
Problem 2
An approximation for the boundary-layer shape in Figs. 2 (a) and Figs. 2 (b) is the formula
πy u(y) ≈ U sin
,
0≤y≤δ
2δ
1
where U is the stream velocity far from the wall and δ is the boundary layer thickness, as in Fig. 2 (b). If
the fluid is helium at 20◦ C and 1 atm, and if U = 10.8 m/s and δ = 3 cm, use the formula to
(a) Estimate the wall shear stress τw in Pa.
(b) Find the position in the boundary layer where τ is one-half of τw .
(a)
(b)
Figure 2: Problem 2
Solution
For helium, take R = 2077 m2 /(s2 -K) and µ = 1.97E-5 kg/m-s.
(a) Then the wall shear stress is calculated as
τw = µ
π
∂u
πy πµU
= µ U cos
=
∂y y=0
2δ
2δ y=0
2δ
Numerical values:
τw =
π(1.97E-5 kg/m-s)(10.8 m/s)
= 0.11 Pa
2(0.003 m)
Ans. (a)
A very small shear stress, but it has a profound effect on the flow pattern.
(b) The variation of shear stress across the boundary layer is a cosine wave, τ = µ(du/dy):
τ (y) =
πy πy τ
πµU
w
cos
= τw cos
=
2δ
2δ
2δ
2
when
πy
π
2δ
= , or : y =
2δ
3
3
Ans. (b)
Problem 3
The system in Fig. 3 is used to calculate the pressure p1 in the tank by measuring the 15-cm height of
liquid in the 1-mm-diameter tube. The fluid is at 60◦ C. Calculate the true fluid height in the tube and the
percentage error due to capillarity if the fluid is (a) water or (b) mercury.
2
Figure 3: Problem 3
Solution
This is a somewhat more realistic variation of Ex. 1.9. You can check Ex.1.9 in the textbook. Use values
from that example for contact angle θ:
3
(a) Water at 60◦ C: γ ≈ 9640 N/m , θ ≈ 0◦ :
h=
or:
4Υ cos θ
4(0.0662 N/m) cos(0◦ )
=
= 0.0275 m,
3
γD
(9640 N/m )(0.001 m)
∆htrue = 15.0 − 2.75 cm ≈ 12.25 cm (+22% error)
Ans. (a)
3
(b) Mercury at 60◦ C: γ ≈ 132200 N/m , θ ≈ 130◦ :
h=
or:
4(0.47 N/m) cos 130◦
4Υ cos θ
=
= −0.0091 m,
3
γD
(132200 N/m )(0.001 m)
∆htrue = 15.0 + 0.91 ≈ 15.91 cm (−6%error)
Ans. (b)
Problem 4
Derive an expression for the capillary height change h for a fluid of surface tension Υ and contact angle θ
between two vertical parallel plates a distance W apart, as in Fig. P1.70. What will h be for water at 20◦ C
if W = 0.5 mm?
Figure 4: Problem 4
3
Solution
With b the width of the plates into the paper, the capillary forces on each wall together balance the weight
of water held above the reservoir free surface:
ρgW hb = 2(Y b cos θ),
h≈
or:
2Y cos θ
ρgW
Ans.
3
For water at 20◦ C, Y ≈ 0.0728 N/m, ρg ≈ 9790 N/m , and θ ≈ 0◦ . Thus, for W = 0.5 mm,
h=
2(0.0728 N/m) cos 0◦
3
(9790 N/m )(0.0005 m)
≈ 0.030 m ≈ 30 mm
Ans.
Problem 5
Early mountaineers boiled water to estimate their altitude. If they reach the top and find that water boils
at 84◦ C, approximately how high is the mountain?
Solution
From Table A-5 at 84◦ C, vapor pressure pv ≈ 55.4 kPa. We may use this value to interpolate for pressure
in the standard altitude, Table A-6, to estimate
z ≈ 4800 m
Ans.
Problem 6
Oil, with a vapor pressure of 20 kPa, is delivered through a pipeline by equally spaced pumps, each of which
increases the oil pressure by 1.3 MPa. Friction losses in the pipe are 150 Pa per meter of pipe. What is the
maximum possible pump spacing to avoid cavitation of the oil?
Solution
The absolute maximum length L occurs when the pump inlet pressure is slightly greater than 20 kPa. The
pump increases this by 1.3 MPa and friction drops the pressure over a distance L until it again reaches 20
kPa. In other words, quite simply,
1.3 MPa = 1,300,000 Pa = (150 Pa/m)L,
or
Lmax ≈ 8660 m
Ans.
It makes more sense to have the pump inlet at 1 atm, not 20 kPa, dropping L to about 8 km.
Problem 7
Use
dx
dy
dz
dr
=
=
=
u
v
w
V
to find and sketch the streamlines of the following flow field:
u = Kx;
v = −Ky;
where K is a constant.
4
w = 0,
Solution
dy
dz
dx
=−
=
Kx
Ky
0
Cancel the K’s and integrate:
Z
Z
dx
dy
=−
,
x
y
(indicates flow in the x − y plane)
or : ln(x) = − ln(y) + ln(C),
or : xy = C
Ans.
The streamlines are hyperbolas, depending upon the value of C. When C = 0, the streamlines are the x
and y axes. The complete pattern looks like this:
Figure 5: Problem 7
Problem 8
A velocity field is given by u = V cos θ, v = V sin θ, and w = 0, where V and θ are constants. Derive a
formula for the streamlines of this flow.
Solution
Equation may be used to find the streamlines:
dx
dy
dx
dy
=
=
=
,
u
v
V cos θ
V sin θ
Solution:
or:
dy
= tan θ
dx
y = (tan θ) x + constant
Ans.
The streamlines are straight parallel lines which make an angle θ with the x axis. In other words, this
velocity field represents a uniform stream V moving upward at angle θ.
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Problem 9
Sensor A reads 1.5 kPa (gage). All fluids are at 20◦ C. Determine the elevations Z in meters of the liquid
levels in the open piezometer tubes B and C.
Figure 6: Problem 9
Solution
(B) Let piezometer tube B be an arbitrary distance H above the gasoline-glycerin interface. The spe3
3
3
cific weights are γair ≈ 12.0 N/m , γgasoline = 6670 N/m , and γglycerin = 12360 N/m . Then apply the
hydrostatic formula from point A to point B:
2
3
Solve for
ZB = 2.73 m
1500 N/m + (12.0 N/m )(2.0 m) + 6670(1.5 − H) − 6670(ZB − H − 1.0) = pB = 0 (gage)
(23 cm above the gasoline-air interface)
Ans. (b)
(C): Let piezometer tube C be an arbitrary distance Y above the bottom. Then
1500 + 12.0(2.0) + 6670(1.5) + 12360(1.0 − Y ) − 12360(ZC − Y ) = pC = 0 (gage)
Solve for
ZC = 1.93 m
(93 cm above the gasoline-glycerin interface)
Ans. (c)
Problem 10
The tank contains water and immiscible oil at 20◦ C. What is h in centimeters if the density of the oil is
3
898 kg/m ?
6
Figure 7: Problem 10
Solution
3
For water take the density = 998 kg/m . Apply the hydrostatic relation from the oil surface to the water
surface, skipping the 8-cm part:
patm + (898)(g)(h + 0.12)
− (998)(g)(0.06 + 0.12) = patm ,
Solve for
h ≈ 0.08 m ≈ 8.0 cm
Ans.
Problem 11
Determine ∆p between points A and B. All fluids are at 20◦ C.
Figure 8: Problem 11
Solution
Take the specific weights to be
Benzene:
8640 N/m
3
Mercury:
133100 N/m
Kerosene:
7885 N/m
3
Water:
9790 N/m
7
3
3
3
and γair will be small, probably around 12 N/m . Work your way around from A to B:
pA + (8640)(0.20 m) − (133100)(0.08) − (7885)(0.32) + (9790)(0.26) − (12)(0.09)
= pB ,
or, after cleaning up,
pA − pB ≈ 8900 Pa
Ans.
Problem 12
2
As measured by NASA’s Viking landers, the atmosphere of Mars, where g = 3.71 m/s , is almost entirely
carbon dioxide, and the surface pressure averages 700 Pa. The temperature is cold and drops off exponentially: T ≈ To e−Cz , where C ≈ 1.3E-5 m−1 and To ≈ 250 K. For example, at 20,000 m altitude, T ≈ 193 K.
(a) Find an analytic formula for the variation of pressure with altitude. (b) Find the altitude where pressure
on Mars has dropped to 1 pascal.
Solution
(a) The analytic formula is found by integrating Eq. (2.17) of the text:
Z
Z
g z dz
g
g z dz
p
=−
=−
(eCz − 1)
=−
ln
po
R 0 T
R 0 To e−Cz
RTo C
g
(eCz − 1)
or, finally, p = po exp −
RTo C
Ans.(a)
(b) From Table A.4 for CO2 , R = 189 m2 /(s2 -K). Substitute p = 1 Pa to find the altitude:
p = 1 Pa = po exp −
or:
ln
1
700
"
#
2
g
3.71 m/s
Cz
(1.3E-5)z
(e − 1) = (700 Pa) exp −
{e
− 1}
RTo C
(189)(250)(1.3E-5)
= −6.55 = −6.04{e(1.3E-5)z − 1},
Solve for
z ≈ 56, 500 m
Ans.(b)
Problem 13
Gate AB is a quarter-circle 10 ft wide and hinged at B. Find the force F just sufficient to keep the gate
from opening. The gate is uniform and weighs 3000 lbf.
Figure 9: Problem 13
8
Solution
The horizontal force is computed as if AB were vertical:
FH = γhCG Avert = (62.4)(4 ft)(8 × 10 ft2 )
= 19968 lbf
acting 5.33 ft below A
The vertical force equals the weight of the missing piece of water above the gate, as shown below.
FV = (62.4)(8)(8 × 10) − (62.4)(π/4)(8)2 (10)
= 39936 − 31366 = 8570 lbf
The line of action x for this 8570-lbf force is found by summing moments from above:
ΣMB (of FV ) = 8570x = 39936(4.0) − 31366(4.605),
or
x = 1.787 ft
Finally, there is the 3000-lbf gate weight W, whose centroid is 2R/π = 5.093 ft from force F, or 8.0 − 5.093 =
2.907 ft from point B. Then we may sum moments about hinge B to find the force F, using the freebody of
the gate as sketched at the top-right of this page:
ΣMB (clockwise) = 0 = F (8.0) + (3000)(2.907) − (8570)(1.787) − (19968)(2.667),
or
F = 59840
8.0 = 7480 lbf
9
Ans.
Problem 14
Panel AB is a parabola with its maximum at point A. It is 150 cm wide into the paper. Neglect atmospheric
pressure. Find (a) the vertical and (b) horizontal water forces on the panel.
Figure 10: Problem 14
Solution
(b) The horizontal force is calculated from the vertical projection of the panel (from point A down to the
bottom). This is a rectangle, 75 cm by 150 cm, and its centroid is 37.5 cm below A, or (25 + 37.5) = 62.5 cm
below the surface. Thus
N
FH = pCG,H Aprojected = 9790 3 (0.625 m) [0.75 m(1.50 m)] = 6880 N Ans.(b)
m
(a) The vertical force is the weight of water above the panel. This is in two parts (1) the weight of the
rectangular portion above the line AC; and (2) the little curvy piece above the parabola and below line AC.
Recall from Ex. 2.8 that the area under a parabola is two-thirds of the enclosed rectangle, so that little
curvy piece is one-third of the rectangle. Thus, finally,
10
FV = (9790)(0.25)(0.4)(1.5) + (9790)
= 1469 N + 1469 N ≈ 2940 N
1
(0.75)(0.4)(1.5) =
3
Ans.(a)
Problem 15
The bottle of champagne ( SG = 0.96 ) in Fig. is under pressure as shown by the mercury manometer
reading. Compute the net force on the 2 -in-radius hemispherical end cap at the bottom of the bottle.
Solution
First, from the manometer, compute the gage pressure at section AA in the champagne 6 inches above
the bottom:
pAA + (0.96 × 62.4)
2
12 ft
4
− (13.56 × 62.4) 12
ft = patmosphere = 0 (gage)
or: PAA = 272lbf/ft2 (gage)
Then the force on the bottom end cap is vertical only (due to symmetry) and equals the force at section
AA plus the weight of the champagne below AA:
F = FV = pAA ( Area )AA + W6-in cylinder − W2-in hemisphere
π
= (272) (4/12)2 + (0.96 × 62.4)π(2/12)2 (6/12) − (0.96 × 62.4)(2π/3)(2/12)3
4
= 23.74 + 2.61 − 0.58 ≈ 25.8 lbf Ans
Problem 16
The hemispherical dome in Fig below weighs 30 kN and is filled with water and attached to the floor by six
equally-spaced bolts. What is the force in each bolt required to hold the dome down?
Solution
11
Assuming no leakage, the hydrostatic force required equals the weight of missing water, that is, the water in a 4 − m-diameter cylinder, 6 m high, minus the hemisphere and the small pipe:
Ftotal = W2−m− cylinder − W2−m− hemisphere − W3−cm−pipe
= (9790)π(2)2 (6) − (9790)(2π/3)(2)3 − (9790)(π/4)(0.03)2 (4)
= 738149 − 164033 − 28 = 574088 N
The dome material helps with 30 kN of weight, thus the bolts must supply 574088-30000 or 544088 N .
The force in each of 6 bolts is 544088/6 or Fbolt ≈ 90700 N Ans.
Problem 17
For axial flow through a circular tube, the Reynolds number for transition to turbulence is approximately
2300 , based upon the diameter and average velocity. If d = 5 cm and the fluid is kerosene at 20◦ C, find the
volume flow rate in m3 /h which causes transition.
Solution
Solution: For kerosene at 20◦ C, take ρ = 804 kg/m3 and = 0.00192 kg/m · s. The only unknown in the
transition Reynolds number is the fluid velocity:
Retr ≈ 2300 =
(804)V(0.05)
ρVd
=
, solve for Vtr = 0.11 m/s
µ
0.00192
Then
m3
m3
π
× 3600 ≈ 0.78
Q = VA = (0.11) (0.05)2 = 2.16E − 4
4
s
hr
Ans.
Problem 18
If you disturb a tank of length L and water depth h, the surface will oscillate back and forth at frequency
Ω, assumed here to depend also upon water density ρ and the acceleration of gravity g. (a) Rewrite this as
a dimensionless function. (b) If a tank of water sloshes at 2.0 Hz on earth, how fast would it oscillate on
Mars ( g ≈ 3.7 m/s2 )?
Solution
Ω
T −1
h
{L}
L
{L}
ρ −3
ML
g−2
LT
12
(a) There are five variables and three dimensions {MLT}, hence we expect two pi groups.
The writer thinks Ω should be correlated versus h, so he chooses ( L, ρ, g ) as repeating variables:
Π1 = La ρb g c Ω1
Π2 = La ρb g c h1
Thus
q
Ω Lg
yields Π1
yields Π2
h
=L
h
= f cn L
Note that density drops out, being the only variable containing mass {M}. If the tank sloshes on earth at
2.0 Hz , that sets the value of Π1 , which we use on Mars to get ΩMars at the same h/L.
s
Ωearth
L
gearth
s
= 2.0s−1
Solve for ΩMars ≈ 1.23 Hz
√
L
= 0.639m−1/2 L = ΩMars
2
9.81m/s
s
L
gMars
s
= ΩMars
L
3.7m/s2
Ans.(b)
Problem 19
Under laminar conditions, the volume flow Q through a small triangular-section pore of side length b and
length L is a function of viscosity µ, pressure drop per unit length ∆p/L, and b. Using the pi theorem,
rewrite this relation in dimensionless form. How does the volume flow change if the pore size b is doubled?
Solution:
Establish the variables and their dimensions:
µ, b)
3 Q = fcn(∆p/L,
2
2
L /T
M/L T {M/LT}{L}
Then n = 4 and j = 3, hence we expect n − j = 4 − 3 = 1 Pi group, found as follows:
a
1
Π1 = (∆p/L)a (µ)b ( b)c Q1 = M/L2 T2 {M/LT}b { L}c L3 /T = M0 L0 T0
L : −2a − b + c + 3 = 0; T : −2a − b − 1 = 0,
solve a = −1, b = +1, c = −4
Qµ
Π1 = (∆p/L)b
Ans.
4 = constant
M : a + b = 0;
Clearly, if b is doubled, the flow rate Q increases by a factor of 24 = 16. Ans.
Problem 20
In studying sand transport by ocean waves, A. Shields in 1936 postulated that the bottom shear stress τ
required to move particles depends upon gravity g, particle size d and density ρp , and water density ρ and
viscosity µ. Rewrite this in terms of dimensionless groups (which led to the Shields Diagram in 1936).
Solution:
There are six variables ( τ, g, d, ρp , ρ, ) and three dimensions ( M, L, T ), hence we expect n − j = 6 − 3 = 3Pi
groups. The author used (ρ, g, d) as repeating variables:
13
τ
= f cn
ρgd
ρg 1/2 d3/2 ρp
,
µ
ρ
Ans.
The shear parameter used by Shields himself was based on net weight: τ / [(ρp − ρ) gd].
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