6-21
6-46 For air flowing over a flat plate at 5 m/s, the effect of plate location on the wall shear stress is to be determined.
Assumptions 1 Isothermal condition exists between the flat plate and fluid flow. 2 Properties are constant. 3 Edge effects are
negligible.
Properties The properties of air at 20°C are ν = 1.516 10−5 m2/s and ρ = 1.204 kg/m3 (Table A-15).
Analysis The friction coefficient and the wall shear stress can be determined with
5.0
−
Vx
2
2
V
The problem is solved using EES, and the solution is given below:
“GIVEN”
V=5 [m/s]
“PROPERTIES”
nu=1.516e-5 [m^2/s]
rho=1.204 [kg/m^3]
“ANALYSIS”
C_f=0.664*(V*x/nu)^(-0.5)
tau_w=C_f*rho*V^2/2
x [m] τw [N/m2]
0.01 0.1740
0.02 0.1230
0.03 0.1005
0.04 0.0870
0.06 0.07104
0.08 0.06152
0.10 0.05503
0.15 0.04493
0.20 0.03891
0.30 0.03177
0.40 0.02751
0.50 0.02461
0.60 0.02246
0.80 0.01945
1.00 0.01740
Discussion The wall shear stress decreases with increasing x. This is because as the air flows along the plate, the velocity
gradient decreases with increasing velocity boundary layer thickness.
0 0.2 0.4 0.6 0.8 1
0
0.04
0.08
0.12
0.16
0.2
x [m]
tw [N/m2]
6-22
6-47 A flat plate is positioned inside a wind tunnel. The minimum length of the plate necessary for the Reynolds number to
reach 1 105 is to be determined. The type of flow regime at 0.2 m from the leading edge is to be determined.
Assumptions 1 Isothermal condition exists between the flat plate and fluid flow. 2 Properties are constant.
Properties The kinematic viscosity for air at 20°C is
= 1.516 10−5 m2/s (Table A-15).
Analysis The Reynolds number is given as
c
VL
m/s 60
cr .
V
6-48 For air flowing over a flat plate, the plate length to achieve a Reynolds number of 1 108 at the end of the plate and the
distance from the leading edge of the plate at which transition would occur are to be determined.
Assumptions 1 Isothermal condition exists between the flat plate and fluid flow. 2 Properties are constant. 3 Edge effects are
negligible.
Properties The kinematic viscosity of air at 25°C is ν = 1.562 10−5 m2/s (Table A-15).
Analysis (a) The Reynolds number is given as
c
VL
=Re
For the Reynolds number to reach 1 108, we need a plate length of
m 39.1=
==
−
m/s 40
)101)(/sm 10562.1(Re 825
V
Lc
(b) The distance from the leading edge for the transition to take place with a critical Reynolds number of 5 105 is
−
)105)(/sm 10562.1(Re 525
cr
6-25
Convection Equations and Similarity Solutions
6-51C For steady, laminar, two-dimensional, incompressible flow with constant properties and a Prandtl number of unity
v
u
6-54C In a boundary layer during steady two-dimensional flow, the velocity component in the flow direction is much larger
than that in the normal direction, and thus u >> v, and
xv /
and
yv /
are negligible. Also, u varies greatly with y in the
6-56C For steady two-dimensional flow over an isothermal flat plate in the x-direction, the boundary conditions for the
velocity components u and v, and the temperature T at the plate surface and at the edge of the boundary layer are expressed as
follows:
6-57C An independent variable that makes it possible to transforming a set of partial differential equations into a single
u, T
T
6-27
6-62 A shaft rotating in a bearing is considered. The power required to rotate the shaft is to be determined for different fluids
in the gap.
Assumptions 1 Steady operating conditions exist. 2 The fluid has constant properties. 3 Body forces such as gravity are
negligible.
Properties The properties of air, water, and oil at 40C are (Tables A-15, A-9, A-13)
Air:
= 1.91810-5 Ns/m2
Water:
= 0.65310-3 Ns/m2
Oil:
= 0.2177 Ns/m2
Analysis A shaft rotating in a bearing can be
approximated as parallel flow between two large plates
with one plate moving and the other stationary.
Therefore, we solve this problem considering such a
25 cm
)m 2(0.0005
2
5 cm
5600 rpm
6-28
6-63 Parallel flow of oil between two plates is considered. The velocity and temperature distributions, the maximum
temperature, and the heat flux are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body
forces such as gravity are negligible. 4 The plates are large so that there is no variation in z direction.
Properties The properties of oil at the average temperature of (40+15)/2 = 27.5C are (Table A-13):
k = 0.145 W/mK and
= 0.605 kg/ms = 0.605 Ns/m2
Analysis (a) We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two
plates, and thus v = 0. Then the continuity equation reduces to
v
u
u
6-29
C62.0=
−
++
−
=
−++
−
==
2
222
2
22
1
12
max
m) 0007.0(
m) (0.0004155
m 0007.0
m 0.0004155
)C W/m2(0.145
m/s) 8)(s/mN 605.0(
C51m) 0004155.0(
m 0007.0
C)1540(
2
)0004155.0( L
y
L
y
k
V
Ty
L
TT
TT
(c) Heat flux at the plates is determined from the definition of heat flux,
( )
24 W/m103.28 −=
−
−
−=
−
−
−=−−
−
−=−=
=
m/sN 1
W1
)m 2(0.0007
m/s) 8)(s/mN 605.0(
m 0007.0
C)1540(
)C W/m.145.0(
2
01
2
22
2
12
2
12
0
0L
V
L
TT
k
kL
V
k
L
TT
k
dy
dT
kq
y
( )
24 W/m102.25 =
+
−
−=
+
−
−=−−
−
−=−=
=
m/sN 1
W1
)m 2(0.0007
m/s) 8)(s/mN 605.0(
m 0007.0
C)1540(
)C W/m.145.0(
2
21
2
22
2
12
2
12
L
V
L
TT
k
kL
V
k
L
TT
k
dy
dT
kq
Ly
L
Discussion A temperature rise of about 35C confirms our suspicion that viscous dissipation is very significant. Calculations
are done using oil properties at 27.5C, but the oil temperature turned out to be much higher. Therefore, knowing the strong
dependence of viscosity on temperature, calculations should be repeated using properties at the average temperature of about
45C to improve accuracy.
6-31
−
−
−++
−
==
222
2
22
1
12
max
m) (0.0002968
m 0.0002968
m/s) 8)(s/mN 605.0(
C)1540(
2
)0002166.0( L
y
L
y
k
V
Ty
L
TT
TT
6-32
6-65 The flow of fluid between two large parallel plates is considered. The relations for the maximum temperature of fluid,
the location where it occurs, and heat flux at the upper plate are to be obtained.
Assumptions 1 Steady operating conditions exist. 2 The fluid has constant properties. 3 Body forces such as gravity are
negligible.
Analysis We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two
plates, and thus v = 0. Then the continuity equation reduces to
Continuity:
0=
+
y
v
x
u
⎯→
0=
x
u
⎯→ u = u(y)
V
6-33
Therefore, maximum temperature will occur at the lower plate surface, and it s value is
k
V
TTT 2
)0(
2
0max
+==
.The heat flux at the upper plate is
L
V
L
kL
V
k
dy
dT
kq
Ly
L
2
2
2
==−=
=
6-34
6-66 The flow of fluid between two large parallel plates is considered. Using the results of Problem 6-45, a relation for the
volumetric heat generation rate is to be obtained using the conduction problem, and the result is to be verified.
Assumptions 1 Steady operating conditions exist. 2 The fluid has constant
properties. 3 Body forces such as gravity are negligible.
Analysis The energy equation in Prob. 6-55 was determined to be
2
2
2
V
dy
Td
The steady one-dimensional heat conduction equation with constant
heat generation is
0
gen
2
2
=+ k
e
dy
Td
(2)
Comparing the two equation above, the volumetric heat generation rate is determined to be
2
gen
=L
V
e
Integrating Eq. (2) twice gives
43
2
gen
3
gen
2
)( CyCy
k
e
yT
Cy
k
e
dy
dT
++−=
+−=
Applying the two boundary conditions give
B.C. 1: y=0
00 3
0
=⎯→⎯=−
=
C
dy
dT
k
y
B.C. 2: y=L
2
gen
040 2
)( L
k
e
TCTLT
+=⎯→⎯=
Substituting, the temperature distribution becomes
−+= 2
2
2
gen
01
2
)(
L
y
k
Le
TyT
Maximum temperature occurs at y = 0, and it value is
k
Le
TTT 2
)0(
2
gen
0max
+==
which is equivalent to the result
k
V
TTT 2
)0(
2
0max
+==
obtained in Prob. 6-55.
V
Fluid
L
6-36
The maximum temperature is
C53.3=
+=+=
−+==
m/sN 1
W1
)C W/m8(0.17
m/s) 425.9)(s/mN 05.0(
C50
8
)2/(
2/
2
)2/(
22
2
0
2
2
2
0max
k
V
T
L
L
L
L
k
V
TLTT
(b) The rates of heat transfer are
( )
−=−−=−=
=
2
01
2
22
0
0L
V
A
kL
V
kA
dy
dT
kAQ
y
6-37
6-68 The oil in a journal bearing is considered. The velocity and temperature distributions, the maximum temperature, the
rate of heat transfer, and the mechanical power wasted in oil are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body
forces such as gravity are negligible.
Properties The properties of oil at 50C are given to be
k = 0.17 W/mK and
= 0.05 Ns/m2
Analysis (a) Oil flow in journal bearing can be
approximated as parallel flow between two large plates
with one plate moving and the other stationary. We take the
x-axis to be the flow direction, and y to be the normal
direction. This is parallel flow between two plates, and
thus v = 0. Then the continuity equation reduces to
v
u
u
12 m/s
6 cm
3000 rpm
20 cm
6-38
2
yV
dT
6-40
15 cm
6-70 The oil in a journal bearing is considered. The bearing is cooled externally by a liquid. The surface temperature of the
shaft, the rate of heat transfer to the coolant, and the mechanical power wasted are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body
forces such as gravity are negligible.
Properties The properties of oil are given to be k = 0.14 W/mK and
= 0.03 Ns/m2. The thermal conductivity of bearing is
given to be k = 70 W/mK.
Analysis (a) Oil flow in a journal bearing can be
approximated as parallel flow between two large plates
with one plate moving and the other stationary. We take
the x-axis to be the flow direction, and y to be the normal
4 cm
5200 rpm