0 =+ Analysis a When the stack is exposed to 10 m/s winds, the heat transfer will be by forced convection... The rates of heat transfer from the vessel by natural convection, conduction,
Trang 1Review Problems
9-99 A cold cylinder is placed horizontally in hot air The rates of heat transfer from the stack with and
without wind cases are to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The local
atmospheric pressure is 1 atm
Properties The properties of air at 1 atm and the film temperature of (Ts+T∞)/2 = (40+10)/2 = 25°C are (Table A-15)
2 5
K003356.0K)27325(
11
1
C W/m
02551
0
=+
Analysis (a) When the stack is exposed to 10 m/s winds, the heat transfer will be by forced convection We
have flow of air over a cylinder and the heat transfer rate is determined as follows:
020,64/sm10562.1
m)m/s)(0.110
(Re
7296.0()020,64(027.0PrRe027
0
C W/m87.45)8.179(m1.0
C W/m
02551.0
2 5
3 -1
2 2
3
10953.2)7296.0()
/sm10562.1(
)m1.0)(
K1040)(
K003356.0)(
m/s81.9(Pr)(
)10953.2(387.06.0Pr
/559.01
387.06
.0
2
27 / 8 16 / 9
6 / 1 6 2
27 / 8 16 / 9
6 / 1
Nu
m1.0
C W/m
02551
Q&
Trang 29-100 A spherical vessel is completely submerged in a large water-filled tank The rates of heat transfer
from the vessel by natural convection, conduction, and forced convection are to be determined
Assumptions 1 Steady operating conditions exist 2 The surface temperature is constant
Properties The properties of water at the film temperature of (Ts+T∞)/2 = (30+20)/2 = 25°C are (Table A-9)
1 - 3
2 7 3 3
K10247
skg/m10891
0
C W/m
607
0
kg/m997
Analysis (a) Heat transfer in this case will be by natural convection
The characteristic length in this case is L c = D = 0.3 m Then,
9 2
2 7
3 -1
3 2
2
3
10029.5)14.6()
/sm10937.8(
)m3.0)(
K2030)(
K10247.0)(
m/s81.9(Pr)(
)10029.5(589.02Pr
/469.01
589.0
16 / 9
4 / 1 9 9
/ 4 16 / 9
4 / 1
=+
×+
=+
2
2
m2827.0m)3.0(
C W/m0.293)8.144(m3.0
C W/m
607.0
(b) When buoyancy force is neglected, there will be no convection currents (since β = 0) and the heat
transfer will be by conduction Then Rayleigh number becomes zero (Ra = 0) The Nusselt number in this case is
Then
C W/m047.4)2(m3.0
C W/m
607
(c) In this case, the heat transfer from the vessel is by forced convection The properties of water at the free
stream temperature of 20°C are (Table A-9)
01.7Pr
kg/m.s10
798.0
/sm10004.1/
kg/m.s10
002.1
C W/m
0.598k
kg/m998
3 C
30
@
,
2 6 - 3 3
ρ
Trang 3The Reynolds number is
760,59/sm10004.1
m)m/s)(0.3(0.2
10002.1)01.7()760,59(06.0)760,59(4.02
PrRe06.0Re4.02
4 / 1
3
3 4
0 3 / 2 5
0
4 / 1 4 0 3 / 2 5
0
hD
Nu
μμ
The heat transfer coefficient is
C W/m3.875)1.439(m3.0
C W/m
598
forced hA T T
Trang 49-101 A vertical cylindrical vessel looses heat to the surrounding air The rates of heat transfer from the
vessel with and without wind cases are to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The local
atmospheric pressure is 1 atm
Properties The properties of air at 1 atm and the film
temperature of (Ts+T∞)/2 = (60+0)/2 = 30°C are (Table A-15)
2 5
K003300.0K)27330(
11
1
C W/m
02588
0
=+
Analysis (a) When there is no wind heat transfer is by natural
convection The characteristic length in this case is the height
of the vessel, L c = L=3m.Then,
11 2
2 5
3 -1
2 2
3
10477.1)7282.0()
/sm10608.1(
)m3)(
K060)(
K003300.0)(
m/s81.9(Pr)(
/ 1 11
4
/
1
35 thusand 1.0
<
156.0)7282.0/10477.1(
)3(3535
Gr
L D Gr
.528)
10477.1(1.01
C W/m
02588
(b) When the vessel is exposed to 20 km/h winds, the heat transfer will be by forced convection We have
flow of air over a cylinder and the heat transfer rate is determined as follows:
5 2
/sm10608.1
m)m/s)(1.03600
/10020(
7282 0 ( ) 10 455 3 ( 027 0 Pr Re 027
0
Nu = 0 805 1 / 3 = × 5 0 805 1 / 3 = (from Table 7-1)
C W/m07.18)1.698(m0.1
C W/m
02588.0
Trang 59-102 A solid sphere is completely submerged in a large pool of oil The rates of heat transfer from the
sphere by natural convection, conduction, and forced convection are to be determined
Assumptions 1 Steady operating conditions exist 2 The surface temperature is constant
Properties The properties of oil are given in problem statement
Analysis (a) For conduction heat transfer, β = 0
and Ra = 0 Then the Nusselt number is
Nu = 2
Then
m5.0
C W/m
22
W/m22.0
C)J/kgs)(1880kg/m
010.0(
kg/m010.0
)kg/mm)(888m/s)(0.5(1.5
010.0)45.85()600,66(06.0)600,66(4.02
PrRe06.0Re4.02
4 / 1 4
0 3
/ 2 5
0
4 / 1 4 0 3 / 2 5
0
+
=
s k
hD
Nu
μμ
The heat transfer coefficient is
C W/m4.662)1506(m5.0
C W/m
22
(c) Heat transfer in this case will be by natural convection Note that the fluid properties in this case are to
be evaluated at the film temperature of 40ºC The characteristic length in this case is L c = D = 0.5 m Then,
10 2
2
3 -1
2 2
3
10522.3)5.65()
/sm876/007.0(
)m5.0)(
K2060)(
K00070.0)(
m/s81.9(Pr)
C)J/kgs)(1965kg/m
007.0(
)10522.3(589.02Pr
/469.01
589.0
16 / 9
4 / 1 10 9
/ 4 16 / 9
4 / 1
=+
×+
=+
+
Nu
C W/m2.105)4.250(m5.0
C W/m
21
Discussion The heat transfer from the sphere by forced convection is much greater than that by natural convection and the heat transfer by natural convection is much greater than that by conduction
Trang 69-103E A small cylindrical resistor mounted on the lower part of a vertical circuit board The approximate
surface temperature of the resistor is to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The local atmospheric pressure is 1 atm 4 Radiation effects are negligible 5 Heat transfer through the connecting
wires is negligible
Properties The properties of air at 1 atm and the
anticipated film temperature of (Ts+T∞)/2 =
(220+120)/2 = 170°F are (Table A-15E)
Q&
Resistor 0.1 W
D = 0.2 in
Air
T∞ = 120°F
1 -
2 3
R001587.0R)460170(
11
0
FBtu/h.ft
01692
0
=+
properties and h We will check the accuracy of this guess later and repeat the calculations if necessary
The characteristic length in this case is the diameter of resistor, L c = D=0.2in Then,
8.343)7161.0()
/sft10222.0(
)ft12/2.0)(
R120220)(
R001587.0)(
ft/s2.32(Pr)(
2 2 3
3 -1
2 2
)8.343(387.06
.0Pr
/559.01
387.06
.0
2
27 / 8 16 / 9
6 / 1 2
27 / 8 16 / 9
6 / 1
Nu
2 2
2
2
ft00175.04/ft)12/2.0(2)ft12/3.0)(
ft12/2.0(4/2
F.Btu/h.ft138.2)105.2(ft12/2.0
FBtu/h.ft
01692.0
=+
=+
π
A
Nu D
°
=+
F.Btu/h.ft138.2(
Btu/h )412.31.0(F
120)
s s
s
Q T T T
T hA
&
which is sufficiently close to the assumed temperature for the evaluation of properties Therefore, there is
no need to repeat calculations
Trang 79-104 An ice chest filled with ice at 0°C is exposed to ambient air The time it will take for the ice in the
chest to melt completely is to be determined for natural and forced convection cases
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Heat transfer from the base of the ice chest is disregarded 4 Radiation effects are negligible 5 Heat transfer coefficient is the same for all surfaces considered 6 The local atmospheric pressure is 1 atm
Properties The properties of air at 1 atm and the
anticipated film temperature of (Ts+T∞)/2 =
(15+20)/2 = 17.5°C are (Table A-15)
Ice chest, 0°C
3 cm
Q&
Air, T∞= 20°C
30 cm1
-2 5
K003442.0K)2735.17(
11
1
C W/m
02495
0
=+
properties and h We will check the accuracy of this guess later and repeat the calculations if necessary The characteristic length for the side surfaces is the height of the chest, L c = L = 0.3 m Then,
7 2
2 5
3 -1
2 2
3
10495.1)7316.0()
/sm10493.1(
)m3.0)(
K1520)(
K003442.0)(
m/s81.9(Pr)(
.0
492.01
)10495.1(387.0825.0Pr
492.01
Ra387.0825
.0
2
27 / 8 16 / 9
6 / 1 7 2
27 / 8 16 / 9
6 / 1
×+
+
=
Nu
C W/m923.2)15.35(m3.0
C W/m
02495
2m64.0)m4.0)(
m4.0()m4.0)(
m3.0(
C W/m923.2(
1)
m64.0)(
C W/m
033.0(
m03.0
C)020(1
2 2
2
, ,
−
=+
−
s s
i s o
conv
wall
i s
hA kA L
T T R
(2.923
W23.10C
20
)
s s
s
Q T T T T hA
&
which is almost identical to the assumed value of 15°C used in the evaluation of properties and h
Therefore, there is no need to repeat the calculations
Trang 8The rate at which the ice will melt is
kg/s10066.3kJ/kg7.333
kJ/s1023
⎯→
⎯Δ
kg/s10066.3
(T
C
s+T∞)/2 = (19+20)/2 = 19.5°C are (Table A-15)
1 -
2 5
K00342.0K)2735.19(
11
1
C W/m
0251
0
=+
)m4.0(m/s)3600/100050(
5×
7.362)
7310.0()700,367(664.0PrRe664
C W/m
0251
C W/m76.22(
1)
m64.0)(
C W/m
033.0(
m03.0
C)020(1
2 2
2
, ,
−
=+
−
s s
i s o
conv
wall
i s
hA kA L
T T R
(22.76
W43.13C
20
)
s s
s
Q T T T T hA
&
which is almost identical to the assumed value of 19°C used in the evaluation of properties and h
Therefore, there is no need to repeat the calculations Then the rate at which the ice will melt becomes
kg/s10025.4kJ/kg7.333
kJ/s1043
⎯→
⎯Δ
10025.4
5
m
m t t
m
m
&
&
Trang 99-105 An electronic box is cooled internally by a fan blowing air into the enclosure The fraction of the
heat lost from the outer surfaces of the electronic box is to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Heat transfer from the base surface is disregarded 4 The pressure of air inside the enclosure is 1 atm
Properties The properties of air at 1 atm and the film temperature of
(Ts+T∞)/2 = (32+15)/2 = 28.5°C are (Table A-15)
1 -
2 5
K003317.0K)2735.28(
11
1
C W/m
02577
0
=+
Analysis Heat loss from the horizontal top surface:
The characteristic length in this case is 0.125m
)]
m5.0()m5.0[(
2
)m5.0
2 5
3 -1
2 2
3
10275.1)7286.0()
/sm10594.1(
)m125.0)(
K2532)(
K003317.0)(
m/s81.9(Pr)(
2
m25.0)m5.0(
C.W/m741.3)15.18(m125.0
C W/m
02577.0
Nu L
k h
and Q&top =hA top(T s−T∞)=(3.741 W/m2.°C)(0.25m2)(32−25)°C=6.55 W
Heat loss from vertical side surfaces:
The characteristic length in this case is the height of the box L c = L =0.15 m Then,
6 2
2 5
3 -1
2 2
3
10204.2)7286.0()
/sm10594.1(
)m15.0)(
K2532)(
K003317.0)(
m/s81.9(Pr)
.0
492.01
)10204.2(387.0825.0Pr
492.01
Ra387.0825
.0
2
27 / 8 16 / 9
6 / 1 7 2
27 / 8 16 / 9
6 / 1
×+
m15.0(4
C.W/m530.3)55.20(m15.0
C W/m
02577.0
k h
and Q&side =hA side(T s−T∞)=(3.530 W/m2.°C)(0.3m2)(32−25)°C=7.41 W
The radiation heat loss is
W95.17])K27325()K27332)[(
.K W/m1067.5)(
m3.025.0)(
75.0(
)(
4 4
4 2 8 2
4 4
=+
−+
×+
+
W200
W)95.1741.755
6
(
f
Trang 109-106 A spherical tank made of stainless steel is used to store iced water The rate of heat transfer to the
iced water and the amount of ice that melts during a 24-h period are to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Thermal resistance of the tank is negligible 4 The local atmospheric pressure is 1 atm
Properties The properties of air at 1 atm and the film temperature of
(Ts+T∞)/2 = (0+20)/2 = 10°C are (Table A-15)
1.5 cmIced water
Di = 6 m 0°C
Q&
T s = 0°C
T∞ = 20°C
1 -
2 5
K003534.0K)27310(
11
1
C W/m
02439
0
=+
2 5
3 -1
2 2
3
10485.5)7336.0()
/sm10426.1(
)m03.6)(
K020)(
K003534.0)(
m/s81.9(Pr)
)10485.5(589.02Pr
/469.01
589.0
16 / 9
4 / 1 11 9
/ 4 16 / 9
4 / 1
=+
×+
=+
+
Nu
2 2
2
2
m2.114m)03.6(
C W/m596.1)5.394(m03.6
C W/m
02439.0
A
Nu D
k
h
and
W3646C)020)(
m2.114)(
C W/m596.1()
≅
=+
=
=+
−+
W759,11])K2730()K27320)[(
.K W/m1067.5)(
m2.114)(
1(
)(
4 4
4 2 8 2
4 4
total
surr s s rad
Q
T T A Q
&
& ε σ
(b) The total amount of heat transfer during a 24-hour period is
kJ/day10
331.1)s/h3600h/day24)(
kJ/s4.15
=Δ
=Q t
Q &
Then the amount of ice that melts during this period becomes
kg 3988
kJ10331
if
Q m mh
Trang 119-107 A double-pane window consisting of two layers of glass separated by an air space is considered The
rate of heat transfer through the window and the temperature of its inner surface are to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Radiation effects are negligible 4 The pressure of air inside the enclosure is 1 atm
Properties We expect the average temperature of the air gap to be roughly the average of the indoor and
outdoor temperatures, and evaluate The properties of air at 1 atm and the average temperature of
(T∞1+T∞2)/2 = (20 +0)/2 = 10°C are (Table A-15)
1 -
2 5
K003534.0K)27310(
11
1
C W/m
02439
0
=+
Analysis We “guess” the temperature difference across the air gap to be
15°C = 15 K for use in the Ra relation The characteristic length in this
case is the air gap thickness, L c = L = 0.03 m Then,
4 2
2 5
3 -1
2 2
3 2
)/sm10426.1(
)m03.0)(
K15)(
K003534.0)(
m/s81.9(Pr)
03.0
m2.1)7336.0()10065.5(42.0Pr
42
0
3 0 012
0 4
/ 1 4 3
0 012 0 4 /
Nu
C W/m688.1)076.2(m03.0
C W/m
02439
=+
++
−
=
++
+
−
=+
++
−
)4.2)(
25(
1)4.2)(
688.1(
1)
4.2)(
78.0(
)003.0(2)4.2)(
10
(
1
020
11
2
, ,
, ,
,
, ,
s o s air s glass s i
i s o
conv air conv glasses cond i conv
o i
A h A h A k
t A h
T T R
R R
R
T T Q&
Check: The temperature drop across the air gap is determined from
C0.16)mC)(2.4 W/m688.1(
W65
→Δ
=
s
Q T T hA
&
which is very close to the assumed value of 15°C used in the evaluation of the Ra number
Trang 129-108 An electric resistance space heater filled with oil is placed against a wall The power rating of the
heater and the time it will take for the heater to reach steady operation when it is first turned on are to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Heat transfer from the back, bottom, and top surfaces are disregarded 4 The local atmospheric pressure is 1 atm
Properties The properties of air at 1 atm and the film temperature of
-2 5
K003096.0K)27350(
11
1
C W/m
02735
0
=+
Analysis Heat transfer from the top and bottom surfaces are
said to be negligible, and thus the heat transfer area in this case
consists of the three exposed side surfaces The characteristic
length is the height of the box, L c = L = 0.5 m Then,
8 2
2 5
3 -1
2 2
3
10244.4)7228.0()
/sm10798.1(
)m5.0)(
K2575)(
K 003096.0)(
m/s81.9(Pr)(
.0
492.01
)10244.4(387.0825.0Pr
492.01
Ra387.0825
.0
2
27 / 8 16 / 9
6 / 1 8 2
27 / 8 16 / 9
6 / 1
×+
m15.0(2)m8.0)(
m5.0
(
C.W/m179.5)68.94(m5.0
C W/m
02735.0
=+
.K W/m1067.5)(
m55.0)(
8.0(
)(
4 4
4 2 8 2
4 4
=+
−+
Q& ε σ
Then the total rate of heat transfer, thus the power rating of the heater becomes
W 311.5
=+
J10514.4C)2575)(
CJ/kg
2006)(
kg45()
kJ10514
Q
Q t t
Trang 139-109 A horizontal skylight made of a single layer of glass on the roof of a house is considered The rate of
heat loss through the skylight is to be determined for two cases
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The local
atmospheric pressure is 1 atm
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T∞)/2 = (-4-10)/2 = -7°C are (Table A-15)
ε = 0.9
t = 0.5 cm
1 -
2 5
K003759.0K)2737(
11
1
C W/m
02311
0
=+
Analysis We assume radiation heat transfer inside the house to be negligible We start the calculations by
“guessing” the glass temperature to be -4°C for the evaluation of the properties and h We will check the accuracy of this guess later and repeat the calculations if necessary The characteristic length in this case is
m)2.5+m2(1
m)m)(2.51
3 -1
2 2
3 2
)/sm10278.1(
)m357.0)(
K )10(4)[
K003759.0)(
m/s81.9(Pr)(
m1
(
C.W/m467.3)56.53(m357.0
C W/m
02311.0
A
Nu L
3
K])27330()2734)][(
27330()2734)[(
.K W/m1067.5(9
0
))(
(
2
3 2 2
4 2 8
2 2
=
+
−++
−+
−++
−
×
=
++
=
−
sky s sky s
Then the combined convection and radiation heat transfer coefficient outside becomes
2 ,combined = o + rad =3.467+3.433=6.90W/m
h
Again we take the glass temperature to be -4°C for the evaluation of the properties and h for the inner
surface of the skylight The properties of air at 1 atm and the film temperature of T f = (-4+20)/2 = 8°C are (Table A-15)
1 -
2 5
K003559.0K)2738(
11
1
C W/m
02424
0
=+
2 2
3 2
)/sm10408.1(
)m357.0)(
K )4(20)[
K003559.0)(
m/s81.9(Pr)(
Nu
Trang 14C W/m
02424
1C
W/m
78.0
m005.0C W/m998.11
C)]
10(20)[
m5.2(1
1
)(
2 2
2
glass
, ,
,
, ,
−
=
++
−
h k
t h
T T A
R R
R
T T Q
glass i
out room s
o combined glas
cond i conv
o i s skylight
&
Using the same heat transfer coefficients for simplicity, the rate of heat loss through the roof in the case of R-5.34 construction is determined to be
W5.12C W/m90.6
1C/W
.m34.5C W/m998.11
C)]
10(20)[
m5.2(1
1
)(
2
2 2
2 ,
,
, ,
=
°+
°+
−
=
++
−
h
R h
T T A
R R R
T T Q
glass i
out room s
o combined cond
i conv
o i s roof
&
Therefore, a house loses 115/12.5 ≅ 9 times more heat through the skylights than it does through an insulated wall of the same size
Using Newton’s law of cooling, the glass temperature corresponding to a heat transfer rate of 115
W is calculated to be –3.3°C, which is sufficiently close to the assumed value of -4°C Therefore, there is
no need to repeat the calculations
Trang 159-110 A solar collector consists of a horizontal copper tube enclosed in a concentric thin glass tube Water
is heated in the tube, and the annular space between the copper and glass tube is filled with air The rate of heat loss from the collector by natural convection is to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Radiation effects are negligible 3 The pressure of air in the enclosure is 1 atm
Properties The properties of air at 1 atm and the average temperature
of (Ti+To)/2 = (60+32)/2 = 46°C are (Table A-15)
Do =9 cm
Di =5 cm, Ti = 60°C Air space
Glass cover
To = 32°C
1 -
2 5
K003135.0K)27346(
11
1
C W/m
02706
0
=+
Analysis The characteristic length in this case is the
distance between the two cylinders
cm22
cm )59(
c
D D
L
and,
100,16)7238.0()
/sm10760.1(
)m02.0)(
K3260)(
K003135.0)(
m/s81.9(Pr)(
2 2 5
3 -1
2 2
m05.0
m09.0ln)
(
ln
5 7/5 - 7/5
3
-4
5 5 / 3 5 / 3 3
c
i o
D D
L
D
D F
[(0.1303)(16,100)] 0.05811 W/m.C7238
.0861.0
7238.0)C W/m
02706.0(386.0
)Ra(Pr861.0
Pr386
.0
4 / 1 4
/ 1
4 / 1 cyl
4 / 1 eff
k
Then the heat loss from the collector per meter length of the tube becomes
W 17.4
m09.0ln
)C W/m
05811.0(2)(ln
o i
i o
T T D D k Q&
Trang 169-111 A solar collector consists of a horizontal tube enclosed in a concentric thin glass tube is considered
The pump circulating the water fails The temperature of the aluminum tube when equilibrium is
established is to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal
gas with constant properties 3 The local atmospheric pressure is 1
atm
Properties The properties of air at 1 atm and the
anticipated film temperature of (Ts+T∞)/2 = (33+30)/2 =
31.5°C are (Table A-15)
2 5
K003284.0K)2735.31(
11
1
C W/m
02599
0
=+
Analysis This problem involves heat transfer from the aluminum tube to the glass cover, and from the outer
surface of the glass cover to the surrounding ambient air When steady operation is reached, these two heat transfers will be equal to the rate of heat gain That is,
length)meter (per
W 20
Q& & &
Now we assume the surface temperature of the glass cover to be 33°C We will check this assumption later
on, and repeat calculations with a better assumption, if necessary
The characteristic length for the outer diameter of the glass cover L c = D o =0.07 m Then,
700,91)7278.0()
/sm10622.1(
)m07.0)(
K3033)(
K003284.0)(
m/s81.9(Pr)(
2 2 5
3 -1
2 2
)700,91(387.06
.0Pr
/559.01
387.06
.0
2
27 / 8 16 / 9
6 / 1 2
27 / 8 16 / 9
6 / 1
Nu
2m2199.0m)1)(
m07.0
C W/m
02599
m2199.0)(
C W/m832.2()
s s
2 2
)K27320()K273(
.K W/m1067.5)(
m2199.0)(
1(
C)30)(
m2199.0)(
C W/m832.2(
×+
rad conv total
T T
Q Q
Q& & &
Its solution is T glass =33.34°C, which is sufficiently close to the assumed value of 33°C
Trang 17Now we will calculate heat transfer through the air layer between aluminum tube and glass cover We will assume the aluminum tube temperature to be 45°C and evaluate properties at the average temperature of
(Ti+To)/2 = (45+33.34)/2 = 39.17°C are (Table A-15)
1 -
2 5
K003203.0K)27317.39(
11
1
C W/m
02656
0
=+
/sm10694.1(
)m01.0)(
K34.3345)(
K003203.0)(
m/s81.9(Pr)(
2 2 5
3 -1
2 2
3 2
m05.0
m07.0ln)
(
ln
5 3/5 - 3/5
3
-4
5 5 / 3 5 / 3 3
c
i o
D D
L
D
D F
[(0.08085)(926.5)] 0.02480 W/m.C7257
.0861.0
7257.0)C W/m
02656.0(386.0
)Ra(Pr861.0
Pr386
.0
4 / 1 4
/ 1
4 / 1 cyl
4 / 1 eff
k
The heat transfer expression is
C)34.33(
m05.0
m07.0ln
)C W/m
02480.0(2)(ln
2
2 1
T T
T D D
4 4
2 2
)K27334.33()K273(
.K W/m1067.5)(
m1571.0)(
1(
)(
m1571.0m)1(m)05.0(
+
−+
i s
T
T T A Q
L D A
σε
ππ
&
The expression for the total rate of heat transfer is
4 2 8
2)(5.67 10 W/m K ( 273K) (33.34 273K)m
1571.0)(
1(
C)34.33(
m05.0
m07.0ln
)C W/m
02480.0(2 W
20
+
−+
×+
T
T
Q Q Q
π
&
&
&
Its solution is T tube =46.3°C,
which is sufficiently close to the assumed value of 45°C Therefore, there is no need to repeat the
calculations
Trang 189-112E The components of an electronic device located in a horizontal duct of rectangular cross section is
cooled by forced air The heat transfer from the outer surfaces of the duct by natural convection and the average temperature of the duct are to be determined
Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The local atmospheric pressure is 1 atm 4 Radiation effects are negligible 5 The thermal resistance of the duct is
22 cfm
100°F
Air 80°F
Properties The properties of air at 1 atm and the
anticipated film temperature of (Ts+T∞)/2 =
(120+80)/2 = 100°F are (Table A-15E)
1 -
2 3
R001786.0R)460100/(
0
FBtu/h.ft
01529
0
=+
Analysis (a) Using air properties at the average temperature of (85+100))/2 = 92.5°F and 1 atm for the
forced air, the mass flow rate of air and the heat transfer rate by forced convection are determined to be
m& =ρV&=(0.07186lbm/ft3)(22ft3/min)=1.581lbm/min
Q&forced =m&c p(T out −T in)=(1.581×60lbm/h)(0.2405Btu/lbm.°F)(100−85)°F=342.1Btu/h
Noting that radiation heat transfer is negligible, the rest of the 180 W heat generated must be dissipated by natural convection,
Btu/h 272
Q& & &
(b) We start the calculations by “guessing” the surface temperature to be 120°F for the evaluation of the properties and h We will check the accuracy of this guess later and repeat the calculations if necessary
Horizontal top surface: The characteristic length is 0.2222ft
ft)6/12+ft2(4
ft)ft)(6/124
2 2
3
10599.5)726.0()
/sft101809.0(
)ft2222.0(R)80120)(
R001786.0)(
ft/s2.32(Pr)(
c top
A A
Nu L
k h
ft4(
F.Btu/h.ft016.1)77.14(ft2222.0
FBtu/h.ft
01529.0
Horizontal bottom surface: The Nusselt number for this geometry and orientation can be determined
from
386.7)10599.5(27.027
FBtu/h.ft
01529
c bottom
Vertical side surfaces: The characteristic length in this case is the height of the duct, L c = L = 6 in Then,
6 2
2 3
3 -1
2 2
3
10379.6)726.0()
/sft101809.0(
)ft5.0(R)80120)(
R001786.0)(
ft/s2.32(Pr)(
Trang 19.0
492.01
)10379.6(387.0825.0Pr
492.01
Ra387.0825
.0
2
27 / 8 16 / 9
6 / 1 6 2
27 / 8 16 / 9
6 / 1
×+
ft4(2
F.Btu/h.ft843.0)57.27(ft
5.0
FBtu/h.ft
01529.0
k h
Then the total heat loss from the duct can be expressed as
)](
)()
()
=++
Q&total &top &bottom &side top bottom side s
Substituting and solving for the surface temperature,
F)80](
FBtu/h
)4843.025082.02016.1[(