9-81
Natural Convection inside Enclosures
9-90C We would disagree with this recommendation since the air space introduces some thermal resistance to heat transfer.
9-92C The effective thermal conductivity of an enclosure represents the enhancement on heat transfer as result of convection
currents relative to conduction. The ratio of the effective thermal conductivity to the ordinary thermal conductivity yields
Lc
9-82
9-94 A rectangular enclosure consists of two surfaces separated by an air gap, and the ratio of the heat transfer rate for the
horizontal orientation (with hotter surface at the bottom) to that of vertical orientation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Thermal properties are constant.
Properties The properties of air at Tf = (Ts + T∞)/2 = 10°C are k = 0.02439 W/m∙K,
= 1.426 × 10−5 m2/s, Pr = 0.7336 (from
Table A-15). Also, β = 1/Tf = 3.534 × 10−3 K-1.
Analysis The characteristic length for both cases is Lc = 0.1 m. The Rayleigh number is
3-12
3
)7336.0(
)m 1.0(K )1030)(K 003534.0)(m/s 81.9(
Pr
)(
Ra
+
=
−
=−
cs
L
LTTg
9-83
9-95 The absorber plate and the glass cover of a flat-plate solar collector are maintained at specified temperatures. The rate of
heat loss from the absorber plate by natural convection is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Heat loss by radiation is
negligible. 4 The air pressure in the enclusure is 1 atm.
Properties The properties of air at 1 atm and the average temperature of
(T1+T2)/2 = (80+40)/2 = 60C are (Table A-15)
1–
25
K 003003.0
K)27360(
11
7202.0Pr
/sm 10896.1
C W/m.02808.0
=
+
==
=
=
=
−
f
T
k
Analysis For
= 0
, we have horizontal rectangular
enclosure. The characteristic length in this case is the
distance between the two glasses Lc = L = 0.025 m
Then,
4
225
3-12
2
3
21 10689.3)7202.0(
)/sm 10896.1(
)m 025.0)(K 4080)(K 003003.0)(m/s 81.9(
Pr
)( =
−
=
−
=−
LTTg
Ra
223.31
18
)10689.3(
10689.3
1708
144.11
1
18
Ra
Ra
1708
144.11Nu
3/14
4
3/1
=
−
+
−+=
−+
−+=
+
+
+
+
Then
2
m 5.4m) 3(m) 5.1( === WHAs
W652=
−
=
−
=m 025.0
C)4080(
)m 5.4)(223.3)(C W/m.02808.0( 2
21
L
TT
kNuAQ s
For
= 30
, we obtain
074.3
1
18
)30cos()10689.3(
)30cos()10689.3(
)308.1sin(1708
1
)30cos()10689.3(
1708
144.11
1
18
)cosRa(
cosRa
)8.1(sin1708
1
cosRa
1708
144.11Nu
3/1
4
4
6.1
4
3/16.1
=
−
+
−
−+=
−+
−
−+=
+
+
+
+
W621=
−
=
−
=m 025.0
C)4080(
)m 5.4)(074.3)(C W/m.02808.0( 2
21
L
TT
kNuAQ s
For
= 90
, we have vertical rectangular enclosure. The Nusselt number for this geometry and orientation can be determined
from (Ra = 3.689104 – same as that for horizontal case)
69.1
m 025.0
m 5.1
)7202.0()10689.3(42.0Pr42.0
3.0
012.04/14
3.0
012.04/1 =
=
=
−−
L
H
RaNu
W344=
−
=
−
=m 025.0
C)4080(
)m 5.4)(69.1)(C W/m.02808.0( 2
21
L
TT
kNuAQ s
Discussion Caution is advised for the vertical case since the condition H/L < 40 is not satisfied.
Solar
radiation
Insulation
Absorber
Plate, 80C
Glass
Cover,
40C
1.5 m
L = 2.5 cm
9-84
9-96 Two glasses of a double pane window are maintained at specified temperatures. The fraction of heat transferred through
the enclosure by radiation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant
properties. 3 The air pressure in the enclosure is 1 atm.
Properties The properties of air at 1 atm and the average temperature
of (T1+T2)/2 = (280+336)/2 = 308 K = 35C are (Table A-15E)
1–
25
K 003247.0
K 308
11
7268.0Pr
/sm 10655.1
C W/m.02625.0
===
=
=
=
−
f
T
k
Analysis The characteristic length in this case is the distance between the two
glasses, Lc = L = 0.4 m. Then,
8
3-12
3
21 10029.3)7268.0(
)m 4.0)(K 280336)(K 003247.0)(m/s 81.9(
)( =
−
−
c
LTTg
280 K
336 K
L=0.4 m
H = 1.5 m
Q
Air
9-85
9-97 A double pane window with an air gap is considered. The rate of heat transfer through the window by natural
convection the temperature of the outer glass layer are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The air pressure in the
enclosure is 1 atm. 4 Radiation heat transfer is neglected.
Properties For natural convection between the inner surface of the window and the room air, the properties of air at 1 atm and
the film temperature of (Ts+T)/2 = (18+26)/2 = 22C are (Table A-15)
1–
25
K 00339.0
K)27322(
11
7304.0Pr
/sm 10534.1
C W/m.02529.0
=
+
==
=
=
=
−
f
T
k
For natural convection between the two glass sheets separated by an
air gap, the properties of air at 1 atm and the anticipated average
temperature of (T1+T2)/2 = (18+0)/2 = 9C are (Table A-15)
1–
25
K 003546.0
K)2739(
11
,7339.0Pr
/sm 10417.1 C, W/m.02431.0
=
+
===
== −
f
T
k
Analysis We first calculate the natural convection heat transfer between the room
air and the inner surface of the window.
m 5.1== HLc
9
225
3-12
2
3
10787.2)7304.0(
)/sm 10534.1(
)m 5.1(K)1826)(K 00339.0)(m/s 81.9(
Pr
)(
Ra =
−
=
−
=−
HTTg s
5.169
7304.0
492.0
1
)10787.2(387.0
825.0
Pr
492.0
1
Ra387.0
825.0
2
27/8
16/9
6/19
2
27/8
16/9
6/1
=
+
+=
+
+=Nu
2
2
m 2.4)m 8.2)(m 5.1(
C. W/m858.2)5.169(
m 5.1
C W/m.02529.0
===
=
==
WHA
Nu
H
k
h
s
W96.0=−=−= C)1826)(m 2.4)(C. W/m858.2()( 22
conv ss TThAQ
Next, we consider the natural convection between the two glass sheets separated by an air gap.
Lc = L = 2.0 cm
309,18)7339.0(
)/sm 10417.1(
)m 020.0(K)018)(K 003546.0)(m/s 81.9(
Pr
)(
Ra 225
3-12
2
3
21 =
−
=
−
=−
LTTg
333.1
m 020.0
m 5.1
)7339.0()309,18(42.0Pr42.0
3.0
012.04/1
3.0
012.04/1 =
=
=
−−
L
H
RaNu
Under steady operation, the rate of heat transfer between the room air and the inner surface of the window is equal to the heat
transfer through the air gap. Setting these two equal to each other we obtain the temperature of the outer glass sheet
−
−
2
2
21
C)18(
T
TT
18C
T2
L=2.0 cm
Q
Air
Room air
T=26C
9-86
9-98E Two glasses of a double pane window are maintained at specified temperatures. The rate of heat transfer through the
window by natural convection and radiation, and the R-value of insulation are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant
properties. 3 The air pressure in the enclosure is 1 atm.
Properties The properties of air at 1 atm and the average temperature of
(T1+T2)/2 = (65+40)/2 = 52.5F are (Table A-15E)
1–
23
R 001951.0
R )4605.52(
11
7332.0Pr
/sft 101548.0
FBtu/h.ft. 01415.0
=
+
==
=
=
=
−
f
T
k
Analysis (a) The characteristic length in this case is the distance between the two
glasses, Lc = L = 1 in. Then,
)ft 12/1)(R 4065)(R 001951.0)(ft/s 2.32(
)(
3-12
3
21 =
−
−
c
LTTg
40F
65F
L = 1 in
H = 4 ft
Q
Air
9-87
9-99E Prob. 9–98E is reconsidered. The effect of the air gap thickness on the rates of heat transfer by natural
convection and radiation, and the R-value of insulation is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
g=32.2 [ft/s^2]
sigma=0.1714E-8 [Btu/h-ft^2-R^4]
“ANALYSIS”
L_ft=L*Convert(in, ft)
Ra=(g*beta*(T_1-T_2)*L_ft^3)/nu^2*Pr
Ratio=H/L_ft
Nusselt=0.42*Ra^0.25*Pr^0.012*(H/L_ft)^(–0.3)
A=H*W
Q_dot_conv=k*Nusselt*A*(T_1-T_2)/L_ft
Q_dot_rad=epsilon_eff*A*sigma*((T_1+460)^4-(T_2+460)^4)
Q_dot_total=Q_dot_conv+Q_dot_rad
Q_dot_total=k_eff*A*(T_1-T_2)/L_ft
R_value=L_ft/k_eff
L
[in]
conv
Q
[Btu/h]
rad
Q
[Btu/h]
R-value
[h.ft2.F/Btu]
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
159
162.2
164.6
166.4
167.9
169.2
170.4
171.4
172.3
173.1
173.9
174.6
175.2
175.8
176.4
176.9
177.4
177.9
178.4
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
454.3
0.9783
0.9731
0.9695
0.9666
0.9642
0.9622
0.9604
0.9589
0.9575
0.9563
0.9551
0.9541
0.9531
0.9522
0.9513
0.9505
0.9497
0.949
0.9483
9-88
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
155
160
165
170
175
180
0.94
0.945
0.95
0.955
0.96
0.965
0.97
0.975
0.98
L [in]
Qconv [Btu/h]
Rvalue [h-ft2–F/Btu]
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
400
420
440
460
480
500
Qrad [Btu/h]
9-89
9-100 A simple solar collector is built by placing a clear plastic tube around a garden hose. The rate of heat loss from the
water in the hose per meter of its length by natural convection is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Heat loss by radiation is
negligible. 3 The air pressure in the enclosure is 1 atm.
Properties Based on the problem statement, the properties of air at 1 atm and the anticipated average temperature of (Ti+To)/2
= (65+35)/2 = 50C are (Table A–15)
1–
25
K 003096.0
K)27350(
11
,7228.0Pr
/sm 10798.1 C, W/m.02735.0
=
+
===
== −
f
T
k
Analysis We assume the plastic tube temperature to be 35C.
We will check this assumption later, and repeat calculations, if
necessary. The characteristic length in this case is
6.15
−
−
DD
Do =5 cm
Air space
Plastic cover, To
Water
Plastic cover
T = 26C
9-91
T [W]
Q
[W]
4
14.6
6
13.98
8
13.37
10
12.77
12
12.18
14
11.59
16
11.01
18
10.44
20
9.871
22
9.314
24
8.764
26
8.222
28
7.688
30
7.163
32
6.647
34
6.139
36
5.641
38
5.153
40
4.675
0 5 10 15 20 25 30 35 40
4
6.2
8.4
10.6
12.8
15
T¥ [C]
Q [W]
9-92
9-102 The space between the two concentric cylinders is filled with water or air. The rate of heat transfer from the outer
cylinder to the inner cylinder by natural convection is to be determined for both cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The air pressure in the
enclosure is 1 atm. 4 Heat transfer by radiation is negligible.
Properties The properties of water air at the average temperature of (Ti+To)/2 = (54+106)/2 =80C are (Table A-9)
1–3
27
K 10653.0 ,22.2Pr
/sm 10653.3 C, W/m.670.0
−
−
==
==
k
The properties of air at 1 atm and the average temperature of
(Ti+To)/2 = (54+106)/2 = 80C are (Table A-15)
1–
25
K 002833.0
K)27380(
11
,7154.0Pr
/sm 10097.2 C, W/m.02953.0
=
+
===
== −
f
T
k
Analysis (a) The fluid is water:
5565
−
−
DD
Do = 65 cm
Di =55 cm, Ti = 54C
Fluid space
To =106C
L = 125 cm
9-93
9-103 A hot fluid flowing inside a horizontal tube with a known mass flow rate and temperature difference between the
tube inlet and outlet. The tube is enclosed in a concentric cylindrical thin cover. The concentric outer cover temperature is to
be determined whether it is safe from thermal burn hazards.
Assumptions 1 Steady operating conditions exist. 2 Surface temperatures are constant. 3 Air is an ideal gas with constant
properties. 4 Heat loss by radiation is negligible. 5 The air pressure in the enclosure is 1 atm.
Properties The properties of air at the assumed Tavg = 80°C and 1 atm pressure are k = 0.02953 W/m∙K, ν = 2.097 × 10−5
m2/s, Pr = 0.7154 (Table A-15), and β = 1/Tavg = 1/353 K.
Analysis With the assumption that Tavg = 80°C, the outer surface temperature is estimated as
C402avg =−= io TTT
The Rayleigh number is
)m 0125.0(K )40120()K 27380)(m/s 81.9(
)(
312
3
−+
−
−
coi LTTg
9-94
9-104 Two surfaces of a spherical enclosure are maintained at specified temperatures. The rate of heat transfer through the
enclosure is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Nitrogen is an ideal gas with constant properties. 3 Radiation heat
transfer is negligible
Properties The properties of nitrogen at Tf = (Ts + T∞)/2 = 150°C are k = 0.03416 W/m∙K,
= 2.851 × 10−5 m2/s, Pr = 0.7025
(from Table A-16). Also, β = 1/Tf = 0.002364 K-1.
Analysis The characteristic length in this case is determined from
510
−
−
DD
9-96
9-106 Prob. 9-105 is reconsidered. The rate of natural convection heat transfer as a function of the hot surface
temperature of the sphere is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
D_1=0.15 [m]
D_2=0.25 [m]
T_1=350 [K]
T_2=275 [K]
“PROPERTIES”
Fluid$=’air’
k=Conductivity(Fluid$, T=T_ave)
Pr=Prandtl(Fluid$, T=T_ave)
rho=Density(Fluid$, T=T_ave, P=101.3)
mu=Viscosity(Fluid$, T=T_ave)
nu=mu/rho
beta=1/T_ave
T_ave=1/2*(T_1+T_2)
g=9.807 [m/s^2]
“ANALYSIS”
L=(D_2-D_1)/2
Ra=(g*beta*(T_1-T_2)*L^3)/nu^2*Pr
F_sph=L/((D_1*D_2)^4*(D_1^(-7/5)+D_2^(-7/5))^5)
k_eff=0.74*k*(Pr/(0.861+Pr))^0.25*(F_sph*Ra)^0.25
Q_dot=k_eff*pi*(D_1*D_2)/L*(T_1-T_2)
T1
[K]
Q
[W]
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
–
–
–
0.8153
3.202
6.032
9.139
12.45
15.92
19.52
23.23
27.04
30.93
34.89
38.92
43.01
47.15
260 280 300 320 340 360 380 400 420 440 460
0
10
20
30
40
50
60
70
T1 [K]
Q [W]
9-97
9-107 Two surfaces of a spherical enclosure are maintained at specified temperatures. The rate of heat transfer through the
enclosure is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3
Radiation heat transfer is not considered.
Properties The properties of air at the average temperature of
Tavg = (Ti + To)/2 = (320 + 280)/2 = 300 K = 27C and 1 atm
pressure are: k = 0.02566 W/m·K, Pr = 0.7290, ν = 1.58 × 10−5
m2/s (Table A-15), and β = 1/Tavg = 1/300 K.
Analysis We have a spherical enclosure filled with air. The
characteristic length in this case is the distance between the two
spheres,
m 05.02/)2.03.0(2/)( =−=−= ioc DDL
The Rayleigh number is
5
225
312
2
3
10775.4
)729.0(
)/sm 1058.1(
)m 05.0(K )280320()K 300)(m/s 81.9(
Pr
)(
Ra
=
−
=
−
=−
−
coi LTTg
The effective thermal conductivity is
0005229.0
]m) 3.0(m) 2.0[(]m) 3.0(m) 2.0[(
m 05.0
)DD()DD(
L
F55/75/74
5
5/7
o
5/7
i
4
oi
c
sph =
+
=
+
=−−−−
K W/m1105.0
)]10775.4)(0005229.0[(
729.0861.0
729.0
K) W/m02566.0(74.0
)Ra(
Pr861.0
Pr
74.0
4/15
4/1
4/1
4/1
eff
=
+
=
+
=sph
Fkk
Then, the rate of heat transfer between spheres becomes
)(
eff oi
c
oi TT
L
DD
kQ −
=
W16.7=−
= K)280320(
m05.0
m3.0m2.0
)1105.0(
Discussion Note that the air in the spherical enclosure acts like a stationary fluid whose thermal conductivity is keff/k =
0.1105/0.02566 = 4.3 times that of air as a result of natural convection currents. Also, radiation heat transfer between spheres
is usually significant, and should be considered in a complete analysis.
9-99
Combined Natural and Forced Convection
9-111C When neither natural nor forced convection is negligible, it is not correct to calculate each separately and to add them
to determine the total convection heat transfer. Instead, the correlation
n
nn /1
