Properties The emissivities of surfaces are given to be ε1 = 0.5 and ε2 = 0.7.
Analysis The net rate of radiation heat transfer between the two spheres is
( )
[ ] ( ) [ ( ) ( ) ]
W
=2641
⎟⎠
⎜ ⎞
⎝
− ⎛ +
−
⋅
= ×
⎟⎟
⎠
⎞
⎜⎜
⎝
− ⎛ +
= −
−
2
4 4
4 2 8 2
22 12
2 2 1
24 14 12 1
m 3 . 0
m 15 . 0 7 . 0
7 . 0 1 5 . 0
1
K 500 K
800 K W/m 10 67 . 5 m) 3 . 0 ( 1 1
π ε
ε ε
σ
r r T T Q& A
ε= 0.35
D1 = 0.3 m T1 = 800 K ε1= 0.5 D2 = 0.6 m
T2 = 500 K ε2 = 0.7
Tsurr =30°C T∞ = 30°C
Radiation heat transfer rate from the outer sphere to the surrounding surfaces are
W 1214
] ) K 273 30 ( ) K 500 )[(
K W/m 10 67 . 5 ](
m) 6 . 0 ( )[
1 )(
35 . 0 (
) (
4 4
4 2 8 2
4 24
2
=
+
−
⋅
×
=
−
=
π −
σ
ε surr
rad FA T T
Q&
The convection heat transfer rate at the outer surface of the cylinder is determined from requirement that heat transferred from the inner sphere to the outer sphere must be equal to the heat transfer from the outer surface of the outer sphere to the environment by convection and radiation. That is,
W 1427 1214
12− =2641− =
= rad
conv Q Q
Q& & &
Then the convection heat transfer coefficient becomes
( )
[ ]
C W/m 6.40 2⋅°
=
=
−
= ∞
h h
T T hA Qconv
K) 303 -K (500 m) 6 . 0 ( W
1427 2
2 2 .
π
&
13-47 A spherical tank filled with liquid nitrogen is kept in an evacuated cubic enclosure. The net rate of radiation heat transfer to the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be ε1 = 0.1 and ε2 = 0.8.
Analysis We take the sphere to be surface 1 and the surrounding cubic enclosure to be surface 2. Noting that , for this two-surface enclosure, the net rate of radiation heat transfer to liquid nitrogen can be determined from
12 =1 F
( )
[ ] ( ) [ ( ) ( ) ]
W
=228
⎥⎥
⎦
⎤
⎢⎢
⎣
− ⎡ +
−
⋅
− ×
=
⎟⎟⎠
⎜⎜ ⎞
⎝
− ⎛ +
− −
=
−
=
−
2 2
4 4
4 2 8 2
2 1 2
2 1
4 2 4 1 12 1
21
m) 6(3
m) 2 ( 8 . 0
8 . 0 1 1 . 0
1
K 240 K
100 K W/m 10 67 . 5 m) 2 (
1 1
π π
ε ε ε
σ
A A T T Q A
Q& &
Liquid N2
Vacuum Cube, a =3 m T2 = 240 K ε2 = 0.8
D1 = 2 m T1 = 100 K ε1= 0.1
13-48 A spherical tank filled with liquid nitrogen is kept in an evacuated spherical enclosure. The net rate of radiation heat transfer to the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be ε1 = 0.1 and ε2 = 0.8.
Analysis The net rate of radiation heat transfer to liquid nitrogen can be determined from
( )
[ ] ( ) [ ( ) ( ) ]
W
=227
⎥⎥
⎦
⎤
⎢⎢
⎣
− ⎡ +
−
⋅
= ×
⎟⎟
⎠
⎞
⎜⎜
⎝
− ⎛ +
= −
−
2 2
4 4 4
2 8 2
22 12
2 2 1
24 14 12 1
m) (1.5
m) 1 ( 8 . 0
8 . 0 1 1 . 0
1
K 100 K
240 K W/m 10 67 . 5 m) 2 (
1 1
π ε
ε ε
σ
r r T T Q& A
Liquid N2
Vacuum
D1 = 2 m T1 = 100 K ε1 = 0.1 D2 = 3 m
T2 = 240 K ε2 = 0.8
13-49 Prob. 13-47 is reconsidered. The effects of the side length and the emissivity of the cubic enclosure, and the emissivity of the spherical tank on the net rate of radiation heat transfer are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
D=2 [m]
a=3 [m]
T_1=100 [K]
T_2=240 [K]
epsilon_1=0.1 epsilon_2=0.8
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant"
"ANALYSIS"
"Consider the sphere to be surface 1, the surrounding cubic enclosure to be surface 2"
Q_dot_12=(A_1*sigma*(T_1^4-T_2^4))/(1/epsilon_1+(1-epsilon_2)/epsilon_2*(A_1/A_2)) Q_dot_21=-Q_dot_12
A_1=pi*D^2 A_2=6*a^2
a
[m] Q& 21 [W]
2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
226.3 227.4 227.9 228.3 228.5 228.7 228.8 228.9 228.9 229 229 229.1 229.1 229.1 229.1 229.1 229.1 229.2 229.2 229.2 229.2
2 4 6 8 10 12
226 226.5 227 227.5 228 228.5 229 229.5
a [m]
Q21 [W]
ε1
Q& 21
[W]
0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9
227.9 340.9 453.3 565 676 786.4 896.2 1005 1114 1222 1329 1436 1542 1648 1753 1857 1961
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
200 400 600 800 1000 1200 1400 1600 1800 2000
ε1
Q21 [W]
ε2 Q& 21
[W]
0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9
189.6 202.6 209.7 214.3 217.5 219.8 221.5 222.9 224.1 225 225.8 226.4 227 227.5 227.9 228.3 228.7
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
185 190 195 200 205 210 215 220 225 230
ε2
Q21 [W]
13-50 A circular grill is considered. The bottom of the grill is covered with hot coal bricks, while the wire mesh on top of the grill is covered with steaks. The initial rate of radiation heat transfer from coal bricks to the steaks is to be determined for two cases.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Coal bricks, T1 = 950 K, ε1 = 1 0.20 m
Steaks, T2 = 278 K, ε2 = 1 Properties The emissivities are ε = 1 for all surfaces since they are black or
reradiating.
Analysis We consider the coal bricks to be surface 1, the steaks to be surface 2 and the side surfaces to be surface 3. First we determine the view factor between the bricks and the steaks (Table 13-1),
75
(It can also be determined from Fig. 13-7).
Then the initial rate of radiation heat transfer from the coal bricks to the stakes becomes
W
When the side opening is closed with aluminum foil, the entire heat lost by the coal bricks must be gained by the stakes since there will be no heat transfer through a reradiating surface. The grill can be considered to be three-surface enclosure. Then the rate of heat loss from the coal bricks can be determined from
1
13-51 The base and the dome of a hemispherical furnace are maintained at uniform temperatures. The net rate of radiation heat transfer from the dome to the base surface is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not
considered.
T1 = 400 K ε1 = 0.7 T2 = 1000 K ε2 = 1
D = 5 m Analysis The view factor is first determined from
rule) (summation
1 1
surface) (flat 0
12 12
11 11
=
→
= +
= F F
F F
Noting that the dome is black, net rate of radiation heat transfer from dome to the base surface can be determined from
kW
=759
×
=
−
⋅
×
−
=
−
−
=
−
=
−
W 10 594 . 7
] ) K 1000 ( ) K 400 )[(
K W/m 10 67 . 5 )(
1 ](
/4 ) m 5 ( )[
7 . 0 (
) (
5
4 4
4 2 8 2
24 14 12 1 12 21
π
σ
εAF T T
Q Q& &
The positive sign indicates that the net heat transfer is from the dome to the base surface, as expected.
13-52 Two perpendicular rectangular surfaces with a common edge are maintained at specified temperatures. The net rate of radiation heat transfers between the two surfaces and between the horizontal surface and the surroundings are to be
determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Properties The emissivities of the horizontal rectangle and the surroundings are ε = 0.75 and ε = 0.85, respectively.
Analysis We consider the horizontal rectangle to be surface 1, the vertical rectangle to be surface 2 and the surroundings to be surface 3. This system can be considered to be a three-surface enclosure. The view factor from surface 1 to surface 2 is determined from
27 The surface areas are
2 2
Note that the surface area of the surroundings is determined assuming that surroundings forms flat surfaces at all openings to form an enclosure. Then other view factors are determined to be
18
We now apply Eq. 13-35b to each surface to determine the radiosities.
Surface 1:
Solving the above equations, we find
3 2 2 2
1=2937 W/m2, J =13,614 W/m , J =1501 W/m J
Then the net rate of radiation heat transfers between the two surfaces and between the horizontal surface and the surroundings are determined to be
W
13-53 Two long parallel cylinders are maintained at specified temperatures. The rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings are to be determined.