10–17
10–32
Solution A frustum shaped body is rotating at a constant
0.1 Pas = 0.1 Ns/m2 at 20C and 0.0078 Pas at 80C.
Analysis The velocity gradient anywhere in the oil of
film thickness h is V/h where V =
r is the tangential velocity.
Then the wall shear stress anywhere on the surface of the
dA
h
r
dAdF w
dA
h
r
rdFd
2
T
dAr
hA
2
T
dAr
h
W
A
2
2
sh T
h
h
h
h
r
r
r324
0
0
0
Bottom surface: A relation for the bottom surface is obtained by replacing D by d,
h
d
W32
42
bottomsh,
Side surface: The differential area for the side surface can be expressed as
rdzdA
2
. From geometric considerations, the
dDd
D = 12 cm
L = 12 cm
10–33
Solution We are to determine the torque required to rotate the outer cylinder of two
concentric cylinders, with the outer cylinder rotating and the inner cylinder stationary.
Assumptions 1 The fluid is incompressible and Newtonian. 2 End effects (top and bottom)
h
Inner cylinder
y
u
10–19
10–34
Solution A large plate is pulled at a constant speed over a fixed plate. The space between the plates is filled with
engine oil. The shear stress developed on the upper plate and its direction are to be determined for parabolic and linear
velocity profile cases.
𝑑𝑦 =𝜇𝑑
𝑑𝑦 56.568 𝑦 = 37.712𝜇 𝑦3/2 𝑦=0
or
= 0.421 N/m2
Since dynamic viscosity of oil is 0.8374 Pa ∙s (see Table A7). If we assume a linear profile we will have
PROPRIETARY MATERIAL. © 2017 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
10–35C
Solution We are to define and discuss surface tension.
10–36C
Solution We are to determine whether the level of liquid in a tube will rise or fall due to the capillary effect.
10–37C
Solution We are to define and discuss the capillary effect.
10–38C
Solution We are to analyze the pressure difference between inside and outside of a soap bubble.
10–39C
Solution We are to compare the capillary rise in small and large diameter tubes.
10–21
10–40
Solution An air bubble in a liquid is considered. The pressure difference between the inside and outside the bubble is
to be determined.
10–43
Solution The diameter of a soap bubble is given. The gage pressure inside the bubble is to be determined.
Analysis The pressure difference between the inside and the outside of a
bubble is given by
R
PPP s
i
4
0bubble
In the open atmosphere P0 = Patm, and thus
bubble
P
is equivalent to the gage
pressure. Substituting,
2
, bubble
4(0.025 N/m) 100 N/m 100 Pa
0.00200/2 m
i gage
D = 5.00 cm:
2
, bubble
4(0.025 N/m) 4 N/m 4 Pa
0.0500/2 m
i gage
PP
Discussion Note that the gage pressure in a soap bubble is inversely proportional to the radius (or diameter). Therefore,
the excess pressure is larger in smaller bubbles.
10–44E
Solution A slender glass tube is inserted into kerosene. The capillary rise of kerosene in the tube is to be determined.
0.0280.06852 = 0.00192 lbf/ft. The density of kerosene at 68F is
= 51.2 lbm/ft3.
The contact angle of kerosene with the glass surface is given to be 26.
Analysis Substituting the numerical values, the capillary rise is determined to be
2
32
2 0 00192 lbf/ft cos26
2 cos 32 2 lbm ft/s
1 lbf
51 2 lbm/ft 32 2 ft/s 0 015 12 ft
0 0539 ft
s..
hgR . . . /
.
0.650 in
Discussion The capillary rise in this case more than half of an inch, and thus it is clearly noticeable.
Soap
bubble
P
10–45
Solution The force acting on the movable wire of a liquid film suspended on a U-shaped wire frame is measured. The
N/m 0.15 )m 2(0.08
N 024.0
2b
F
s
10–46
Solution A capillary tube is immersed vertically in water. The height of water rise in the tube is to be determined.
10–47
Solution A capillary tube is immersed vertically in water. The maximum capillary rise and tube diameter for the
maximum rise case are to be determined.
Assumptions 1 There are no impurities in water, and no contamination on the surfaces of the tube. 2 Water is open to the
Liquid
film
b
F
10–48
Solution A steel ball floats on water due to the surface tension effect. The maximum diameter of the ball is to be
determined, and the calculations are to be repeated for aluminum.
Assumptions 1 The water is pure, and its
temperature is constant. 2 The ball is dropped on
ss DF
and
6/
3
DggmgW
V
When the ball floats, the net force acting on the ball in the vertical direction is zero. Therefore, setting
WFs
and solving
6
10–25
Review Problems
10–50
Solution A relation is to be derived for the capillary rise of a liquid between two large parallel plates a distance t
apart inserted into a liquid vertically. The contact angle is given to be
.
cos2)( ssurface whtwgFW
Canceling w and solving for h gives the capillary rise to be
h
Air
Liquid
10–51
Solution A journal bearing is lubricated with oil whose viscosity is
R
At start up at 20C:
mN 4.34 m 0008.0
m) 55.0)(s 60/1500(m) 04.0(4
)skg/m 1.0(
4-13232
LnR
T
During steady operation at 80C:
mN 0.347 m 0008.0
m) 55.0)(s 60/1500(m) 04.0(4
)skg/m 008.0(
4-13232
LnR
T
Discussion Note that the torque needed to overcome friction reduces considerably due to the decrease in the viscosity
of oil at higher temperature.
10–52
Solution A U-tube with a large diameter arm contains water. The difference between the water levels of the two arms
Analysis Any difference in water levels between the two arms is due to surface tension effects and thus capillary rise.
Noting that capillary rise in a tube is inversely proportional to tube diameter there will be no capillary rise in the arm with a
large diameter. Then the water level difference between the two arms is simply the capillary rise in the smaller diameter
h
10–53E
Solution The minimum pressure on the suction side of a water pump is given. The maximum water temperature to
avoid the danger of cavitation is to be determined.
10–54
Solution The variation of the dynamic viscosity of water with absolute temperature is given. Using tabular data, a
273.15
1.78710-3
278.15
1.51910-3
283.15
1.30710-3
293.15
1.00210-3
303.15
7.97510-4
313.15
6.52910-4
333.15
4.66510-4
353.15
3.54710-4
373.15
2.82810-4
Analysis Using EES, (1) Define a trivial
function “a=mu+T” in the equation window, (2)
= 0.000001475*EXP(1926.5/T) [used initial guess of a0=1.810–6 and a1=1800 in mu=a0*exp(a1/T)]
At T = 323.15 K, the polynomial and exponential curve fits give
10–55
fluid is Newtonian.
Analysis The velocity profile is given by
2
max
4)( hyhyuyu
where h is the distance between the two plates, y is the vertical distance
umax
h
max
4)( hyhyuyu
10–29
10–56
Solution Two immiscible Newtonian liquids flow steadily between two large parallel plates under the influence of an
applied pressure gradient. The lower plate is fixed while the upper one is pulled with a constant velocity. The velocity
profiles for each flow are given. The values of constants are to be determined. An expression for the viscosity ratio is to be
developed. The forces and their directions exerted by liquids on both plates are to be determined.
Assumptions 1 The flow between the plates is one-dimensional. 2 The fluids are Newtonian.
(a) The velocity profiles should satisfy the conditions 𝑉1 ℎ =10, 𝑉2 −ℎ = 0 and 𝑉1 0 =𝑉2 0 . It is clear that
𝑉1 0 = 6 m/s.
𝑉1 ℎ =10: 10 = 6 + 𝑎×0.5 −3× 0.5 2→𝑎= 9.5
𝑉2 0 = 6 = 𝑏+𝑐×0−9 0 2→𝑏= 6
𝑑𝑦 𝑦=−ℎ 𝐴= 10−3N∙s/m2
0.79 × 7.5 −18𝑦 𝑦=−0.5
16.5 × 4 m2 = 0.0835 Nto the right
Liquid 2
10–30
10–57
Solution A cylindrical shaft rotates inside an oil bearing at a specified
speed. The power required to overcome friction is to be determined.
PROPRIETARY MATERIAL. © 2017 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
10–59 ….. 10–62
Solution Students’ essays and designs should be unique and will differ from each other.
key to explaining this phenomenon. If we think of surface tension like a skin on top of the water, somewhat like a
concentration, in fact, we would expect that the razor blade or paper clip could not float at all.
Discussion If the razor blade or paper clip is fully submerged (breaking through the surface tension), it sinks.