6-61
0 0.1 0.2 0.3 0.4 0.5
0
20
40
60
80
100
120
x [m]
hx [W/m2·K]
0 0.1 0.2 0.3 0.4 0.5
-300
-250
-200
-150
-100
-50
0
x [m]
(¶T/¶y)0 [K/m]
6-62
6-87 The ratio of the average convection heat transfer coefficient (h) to the local convection heat transfer coefficient (hx) is to
be determined from a given relationship for hx.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Analysis From the given relationship for hx = Cx-0.5
5.0
Lx CLh−
==
the entire plate length is
5.05.0
0
5.0
0
2
21 −− ==== CLL
L
C
dxx
L
C
dxh
L
h
LL
x
Taking the ratio of h to hx at x = L, we get
2== −
−
=5.0
5.0
2
CL
CL
h
h
Lx
Discussion For laminar flow and constant properties, it should be noted that ratio of the average Nusselt number over the
entire plate length to the local Nusselt number at the end of the plate is also
2Nu/Nu Lx =
=
.
6-88E Two airfoils with different characteristic lengths are placed in airflow of different free stream velocities at 1 atm and
60°F. The heat flux from the second airfoil is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are
constant. 3 Both airfoils are geometrically similar.
Analysis The relation for Nusselt, Prandtl, and Reynolds numbers
is given as
Pr)(Re,Nu g=
where
hL
cp
VL
6-63
6-89 Two irregularly shaped objects with different characteristic lengths and the same surface temperatures are placed in
atmospheric airflow of different air velocities at 250 K. The average heat flux from the first object is known. The average
convection heat transfer coefficient from the second object is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Both objects are geometrically similar.
Analysis The relation for Nusselt, Prandtl, and Reynolds numbers is given as
Pr)(Re,gNu =
where
k
hL
Nu =
,
k
c
Pr p
=
, and
VL
Re =
Then
/sm 01)m 5.0)(m/s 20(LV
2
11
c
6-64
Momentum and Heat Transfer Analogies
L
Re
6-65
6-93 A metallic airfoil is subjected to air flow. The average friction coefficient is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible.
Properties The properties of air at 25C and 1 atm are (Table A-15)
C W/m335.1
)( 2
−
−
Q
ss
0.000363=
==
C)J/kg m/s)(1007 5)(kg/m 184.1(
)7296.0(C) W/m(1.3352
Pr2
3
3/22
2/3
p
fVc
h
C
6-94 A metallic airfoil is subjected to air flow. The average friction coefficient is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible.
Properties The properties of air at 25C and 1 atm are (Table A-15)
= 1.184 kg/m3, cp =1.007 kJ/kgK, Pr = 0.7296
Analysis First, we determine the rate of heat transfer from
W2083
s) 602(
C)150160(C)J/kg kg)(500 50(
)( 12airfoilp, =
−
=
−
=t
TTmc
Q
Then the average heat transfer coefficient is
C W/m335.1
C)25155)(m 2(1
W2083
)(
)( 2
2=
−
=
−
=⎯→⎯−=
TTA
Q
hTThAQ
ss
ss
where the surface temperature of airfoil is taken as its average temperature, which is (150+160)/2=155C. The average
friction coefficient of the airfoil is determined from the modified Reynolds analogy to be
0.000181=
==
C)J/kg m/s)(1007 10)(kg/m 184.1(
)7296.0(C) W/m(1.3352
Pr2
3
3/22
2/3
p
fVc
h
C
Air
25C
5 m/s
Air
25C
10 m/s
L=3 m
6-66
6-95 The windshield of a car is subjected to parallel winds. The drag force the wind exerts on the windshield is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible.
Properties The properties of air at 0C and 1 atm are (Table A-15)
= 1.292 kg/m3, cp =1.006 kJ/kgK, Pr = 0.7362
Analysis The average heat transfer coefficient is
)(
−=
Q
TThAQ
ss
Air
0C
80 km/h
6-67
6-97 Air is flowing in parallel to a stationary thin flat plate over the top and bottom surfaces. The rate of heat transfer from
the plate is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 The edge effects are negligible.
Properties The properties of air (1 atm) at the film temperature of Tf = (Ts + T∞)/2 = 20°C are given in Table A-15:
= 1.204
kg/m3, cp = 1007 J/kg∙K,
= 1.516 10−5 m2/s, and Pr = 0.7309.
Analysis The flow is over the top and bottom surfaces of the plate, hence the total surface area is
2
6-68
6-98 Air is flowing in parallel to a stationary thin flat plate over the top surface. The rate of heat transfer from the plate is to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Properties The properties of air (1 atm) at the 100°C are given in Table A-15:
= 0.9458 kg/m3, cp = 1009 J/kg∙K,
= 2.306
10−5 m2/s, and Pr = 0.7111.
Analysis The flow is over the top surface of the metal foil, hence the surface area is
2
6-70
6-100 Air at 1 atm is flowing over a flat plate. The friction coefficient and wall shear stress at a location 2 m from the leading
edge are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Properties The properties of air (1 atm) at 20°C are given in Table A-15:
= 1.204 kg/m3, cp = 1007 J/kg∙K,
= 1.516 10−5
m2/s, and Pr = 0.7309.
Analysis At the location x = 2 m from the leading edge, the Reynolds number is
6
25 10235.9
/sm 10516.1
)m 2)(m/s 70(
Vx
Applying the modified Reynolds analogy,
3/23/2
,Pr
PrRe
Nu
PrSt
2x
x
x
xf
C==
Substituting the given correlation for Nusselt number, we get
2.03/2
3/18.0
,Re03.0Pr
PrRe
PrRe03.0
2
−
== x
x
x
xf
C
or
2.0
,Re06.0 −
=xxf
C
The friction coefficient at x = 2 m is
0.002427=== −− 2.062.0
,)10235.9(06.0Re06.0 xxf
C
The wall shear stress is
2
N/m 7.16=== 2
)m/s 70)(kg/m 204.1(
)002427.0(
2
232
,
V
Cxfs
Discussion At x = 2 m from the leading edge, the flow is turbulent. Since the Reynolds number at that location is greater than
5 105.
6-101 Metal plates are subject to parallel air flow cooling. The average convection heat transfer coefficient for the plates are
to be determined from a given average friction coefficient over each plate.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Properties The properties of air at 20°C and 1 atm are ν = 1.516 10−5 m2/s, ρ = 1.204 kg/m3, cp = 1007J/kgK, and Pr =
0.7309 (Table A-15).
Analysis The Reynolds number for the plate is
55
)m 1)(m/s 5(
VL
6-71
6-102 A flat plate is subjected to air flow parallel to its surface. The effect of air velocity on the average convection
heat transfer coefficient for the plate is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Properties The properties of air at 20°C and 1 atm are ν = 1.516 10−5 m2/s, ρ = 1.204 kg/m3, cp = 1007 J/kgK, and Pr =
0.7309 (Table A-15).
Analysis The air velocity for ReL < 5 × 105 is
VL
)105/s)(m 10516.1(
Re 525
−
6-72
0 4 8 12 16 20
0
10
20
30
40
50
60
70
V [m/s]
h [W/m2·K]
Turbulent flow
Laminar flow
6-73
Special Topic: Microscale Heat Transfer
6-103 It is to be shown that the rate of heat transfer is inversely proportional to the size of an object.
Analysis Consider a cylinder of radius r and length l. The surface area of this cylinder is
)(2 rlrA +=
and its volume is
lrA 2
=
. Therefore, the area per unit volume is
rl
rl )(2 +
which, for a long tube l <<r, becomes
r
A2
=
V
. Similarly, it can be
shown that the surface area to volume ratio is
r
3
for a sphere of radius r, and
r
6
for a cube of side r.
Note that as r becomes smaller, the surface to volume ratio increases. Specifically, this means that while the surface
area is about the same order of that of the volume of macroscale (meter, centimeter scale) objects, but the surface becomes
million or more times the volume as the size of the object goes to micrometer scale or below. Since, convective heat transfer
is proportional to A(T – T), heat flow increases as A increases.
6-104 For specified wall and fluid temperatures, the heat flux at the wall of a microchannel is to be determined.
Assumptions Steady operating conditions exist.
Properties The properties for both cases are given.
Analysis: (a) The gas and wall temperatures are Tg =100C = 373 K, Tw = 50C = 323 K. Then,
ww
T
T
wg y
T
y
T
TT
−
=
+
−
=− )5.0(
667.2
667.12
1
12
Pr1
2
2
K/m 80
625.0
323373
625.0 =
−
=
−
=
wg
w
TT
y
T
Therefore, the wall heat flux is
2
W/m12==
−K/m) K)(80W/m15.0(
w
y
T
k
(b) Repeating the same calculations for a different set of properties,
ww
T
T
wg y
T
y
T
TT
+
−
=
+
−
=− )5(
12
22
8.0
8.02
Pr1
2
2
K/m 5
10
323373
10 =
−
=
−
=
wg
w
TT
y
T
2
W/m0.5==
−K/m) K)(5W/m1.0(
w
y
T
k
r
r
r
A/
V
=2/r
A/
V
=3/r
A/
V
=6/r