Solutions Manual
for
Fundamentals of Thermal Fluid Sciences
5th Edition
Yunus A. Çengel, John M. Cimbala, Robert H. Turner
McGraw-Hill, 2017
Chapter 10
INTRODUCTION AND PROPERTIES OF
FLUIDS
PROPRIETARY AND CONFIDENTIAL
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10-2
No-Slip Condition and Classification of Fluid Flows
of pressure (such as a liquid) is commonly referred to as an “incompressible fluid,” although it is more proper to refer to
incompressible flow. The flow of compressible fluid (such as air) does not necessarily need to be treated as compressible
since the density of a compressible fluid may still remain nearly constant during flow – especially flow at low speeds.
are in error by less than a couple percent.
Analysis External flow is the flow of an unbounded fluid over a surface such as a plate, a wire, or a pipe. The flow
their classification.
Analysis A fluid in direct contact with a solid surface sticks to the surface and there is no slip. This is known as
the no-slip condition, and it is due to the viscosity of the fluid.
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.
10-5C
Solution We are to define forced flow and discuss the difference between forced and natural flow. We are also to
discuss whether wind-driven flows are forced or natural.
fan. In natural flow, any fluid motion is caused by natural means such as the buoyancy effect that manifests itself as the rise
of the warmer fluid and the fall of the cooler fluid. The flow caused by winds is natural flow for the earth, but it is
forced flow for bodies subjected to the winds since for the body it makes no difference whether the air motion is caused
by a fan or by the winds.
10-6C
Analysis The region of flow (usually near a wall) in which the velocity gradients are significant and frictional
effects are important is called the boundary layer. When a fluid stream encounters a solid surface that is at rest, the fluid
outside the boundary layer is typically irrotational (individual fluid particles move, but do not rotate).
Vapor Pressure and Cavitation
10-7C
Solution We are to define and discuss cavitation.
drops below the vapor pressure. The vapor bubbles collapse as they are swept away from the low pressure regions,
10-8C
The higher the pressure, the higher the saturation or boiling temperature.
saturation pressure at a given pressure are equivalent.
10-4
10-9C
boiling (or saturation) temperature of a pure substance depends on pressure and increases with it.
stop until the temperature has time to reach its new (higher) boiling temperature. A pressure cooker uses this principle.
Analysis The vapor pressure Pv of a pure substance is defined as the pressure exerted by a vapor in phase
equilibrium with its liquid at a given temperature. In general, the pressure of a vapor or gas, whether it exists alone or in
humidity is 100%, the partial pressure and the vapor pressure are equal.
10–11E
Properties The vapor pressure of water at 70F is 0.3632 psia.
The minimum pressure in the pump is 0.1 psia, which is less than the vapor pressure. Therefore, there is danger of
higher fluid temperatures.
Properties The vapor pressure of water at 20C is 2.339 kPa.
Discussion Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater
at higher fluid temperatures.
10–13
Properties The vapor pressure of water at 30C is 4.246 kPa.
Analysis To avoid cavitation, the pressure anywhere in the flow should not be allowed to drop below the vapor (or
Therefore, the pressure should be maintained above 4.246 kPa everywhere in flow.
Discussion Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater
at higher fluid temperatures.
10–14
Solution The minimum pressure in a pump is given. It is to be determined if there is a danger of cavitation.
Analysis To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation)
pressure at the given temperature, which is
sat 20 C 2 339 kPa
v@
P P .
The minimum pressure in the pump is 2 kPa, which is less than the vapor pressure. Therefore, a there is danger of
cavitation in the pump.
Discussion Note that the vapor pressure increases with increasing temperature, and thus there is a greater danger of
cavitation at higher fluid temperatures.
Viscosity
10–15C
internal frictional force that develops between different layers of fluids as they are forced to move relative to each other.
Viscosity is caused by the cohesive forces between the molecules in liquids, and by the molecular collisions in gases. In
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.
Analysis (a) For liquids, the kinematic viscosity decreases with temperature. (b) For gases, the kinematic
Discussion You can easily verify this by looking at the appendices.
Analysis (a) The dynamic viscosity of liquids decreases with temperature. (b) The dynamic viscosity of gases
increases with temperature.
Discussion A good way to remember this is that a car engine is much harder to start in the winter because the oil in the
engine has a higher viscosity at low temperatures.
10–19C
Solution We are to compare the settling speed of balls dropped in water and oil; namely, we are to determine which
given. The viscosity of the fluid is to be determined.
to be
2
s/ftlbf 102.72
4
1–3232 ft) 5)(s 60/250(ft) 12/3(4
ft) 12ft)(0.035/lbf 2.1(
4
LnR
T
Discussion This is the viscosity value at temperature that existed during
the experiment. Viscosity is a strong function of temperature, and the values can be significantly different at different
temperatures.
l = 0.035 in
fluid
10–21
surface are to be determined.
Assumptions 1 The inclined surface is plane (perfectly flat, although tilted). 2 The friction coefficient and the oil film
thickness are uniform. 3 The weight of the oil layer is negligible.
Properties The absolute viscosity of oil is given to be
= 0.012 Pas = 0.012 Ns/m2.
block is given. Then the force balance gives
:0
x
F
020sin20cos 11 N
fFFF
(1)
:0
y
F
020sin20cos
1 WFF f
N
(2)
Friction force:
1N
ffFF
(3)
Substituting Eq. (3) into Eq. (2) and solving for FN1 gives
N 150
W
applied on the bottom surface of the block due to the oil. Because
50 cm
at the bottom and the lower surface of the block at the top. Then
the shear force is expressed as
22
-4
0 8 m/s
0 012 N s/m 0 5 0 2 m 4 10 m
2 4 N
shear w s
s
FA
V
Ah
.
. . .
.
Replacing the friction force by the shear force in part (a),
Then, our final result is expressed as
12
105 5 57 2
FF ..
0.4 mm
Fshear = wAs
200
F2
V= 0.8 m/s
W = 150 N
FN2
200
F1
W = 150 N
Ff
FN1
y
x
200
200
10–22
Solution The velocity profile of a fluid flowing though a circular pipe is given. The friction drag force exerted on the
pipe by the fluid in the flow direction per unit length of the pipe is to be determined.
Assumptions The viscosity of the fluid is constant.
Analysis The wall shear stress is determined from its definition to be
un
nr
r
d
du
n
n
max
1
10-9
10–23
Solution A thin flat plate is pulled horizontally through an oil layer sandwiched between two plates, one stationary
and the other moving at a constant velocity. The location in oil where the velocity is zero and the force that needs to be
Analysis (a) The velocity profile in each oil layer relative to the fixed wall is as shown in the figure below. The point
of zero velocity is indicated by point A, and its distance from the lower plate is determined from geometric considerations
(the similarity of the two triangles in the lower oil layer) to be
3.0
3
6.2
A
A
y
y
h1=1 mm
h2=2.6 mm
Fixed wall
Moving wall
Vw= 0.3 m/s
F
V = 3 m/s
A
yA
y
10–24
Solution We are to determine the torque required to rotate the inner cylinder of two
are negligible. 3 The gap is very small so that wall curvature effects are negligible. 4 The
gap is so small that the velocity profile in the gap is linear.
Analysis (a) We assume a linear velocity profile between the two walls as sketched – the inner wall is moving at
speed V =
iRi and the outer wall is stationary. The thickness of the gap is h, and we let y be the distance from the outer
wall into the fluid (towards the inner wall). Thus,
and
y du V
uV
h dy h
where
– and
o i i i
h R R V R
Since shear stress
has dimensions of force/area, the clockwise (mathematically negative) tangential force acting along the
and therefore the wall curvature effects are negligible, this approximation should be very good. Another way to think about
this is that when the gap is very small compared to the cylinder radii, a magnified view of the flow in the gap appears
Outer cylinder
Inner cylinder
10–11
10–25
Solution A clutch system is used to transmit torque through an oil film between two identical disks. For specified
rotational speeds, the transmitted torque is to be determined.
Assumptions 1 The thickness of the oil film is uniform. 2 The rotational speeds of the disks remain constant.
Properties The absolute viscosity of oil is given to be
= 0.38 Ns/m2.
21
anywhere in the oil of film thickness h is V /h where
rV )( 21
is the tangential velocity. Then the wall shear stress
h
D
r
h
drr
h
D
r
D
r32
)(
4
)(2)(2
T
4
21
2/
0
4
21
3
2/
0
21
Driving
Driven
10–26
Solution We are to investigate the effect of oil film thickness on the transmitted torque.
Analysis The previous problem is reconsidered. Using EES software, the effect of oil film thickness on the
mu=0.38
n1=1450 “rpm”
D=0.3 “m”
10–13
10–27
Solution The viscosities of carbon dioxide at two temperatures are given. The constants of Sutherland correlation for
carbon dioxide are to be determined and the viscosity of carbon dioxide at a specified temperature is to be predicted and
compared to the value in Table A-23.
Analysis Sutherland correlation is given
where 𝑇 is the absolute temperature. Substituting the given values we have
𝜇1=𝑎 𝑇1
=𝑎 50 +273.15
→1.612 ×10−5=𝑎 323.15
1 + 𝑏/𝑇2
1 + 𝑏
200 +273.15
1 + 𝑏
473.15
which is a nonlinear system of two algebraic equations. Using EES or any other computer code, one finds the following
result:
𝑎= 1.633 ×10−6kg/(m ∙s∙K1/2) 𝑏=265.5 K
The agreement is perfect and within approximately 0.1%.
10–28
Solution The variation of air viscosity for a specified temperature range is to be evaluated using power and
10–29
Solution For flow over a plate, the variation of velocity with distance is given. A relation for the wall shear stress is
to be obtained.
10–30
Solution The velocity profile for laminar one-dimensional flow through a circular pipe is given. A relation for
friction drag force exerted on the pipe and its numerical value for water are to be determined.
where R is the radius of the pipe, r is the radial distance from the center of
the pipe, and umax is the maximum flow velocity, which occurs at the
R
r
0
umax
10–16
10–31
Solution The velocity profile for laminar one-dimensional flow through a circular pipe is given. A relation for
friction drag force exerted on the pipe and its numerical value for water are to be determined.
where R is the radius of the pipe, r is the radial distance from the center of
the pipe, and umax is the maximum flow velocity, which occurs at the
center, r = 0. The shear stress at the pipe surface can be expressed as
R
u
R
r
u
R
r
dr
d
u
dr
du
Rr
Rr
Rr
w
max
2
max
2
2
max
2
2
1
Note that the quantity du/dr is negative in pipe flow, and the negative sign is added to the
w relation for pipes to make
shear stress in the positive (flow) direction a positive quantity. (Or, du/dr = du/dy since y = R – r). Then the friction drag
force exerted by the fluid on the inner surface of the pipe becomes
max
2(2 )
D w s
u
F A RL
R
max
4πμLu
(b) Substituting, we get
max 2
1 N
4 4 (0.0010 kg/m s)(30 m)(7 m/s) 1 kg m/s
D
F Lu
2.64 N
Discussion In the entrance region and during turbulent flow, the velocity gradient is greater near the wall, and thus the
drag force in such cases will be larger.
R
r
0
umax