Emittance R Emittance R Config. Factor R

MEF 312 – EXAMPLE PROBLEMS
1458. What surface area must be provided by the filament of a 100-W evacuated light globe
where T =2 482 oC and Ɛ = 0.38 for the filament? Assume the ambient temperature to be 25.6 oC.
Given: q12 = 100 W

T1 = 2 428 oC + 273 = 2 701 K
Ɛ1 = 0.38
T2 = 25.6 oC + 273 = 298.6 K
Required:
A=?
Solution:
Case 2: F12 = 1
ω1 ω2
emittance R config. factor R emittance R
1−ε 1 1 1−ε 2
A1ε 1
A 1 F 12 A1 ε 2
1−ε 1 1 1−ε 2
∴ R t= + +
A 1 ε 1 A1 F 12 A2 ε 2
1−ε 2
Since A2 is large, then the term ≈ 0. Hence,
A2 ε 2
1−ε 1 1
Rt = +
A1 ε 1 A 1( 1)
1−ε 1 + ε 1
¿
A 1 ε1
1
Rt =
A 1 ε1
And so,
σ ( T 14 −T 24 )
q 12=
1
A1 ε 1
q 12=A 1 ε 1 σ ( T 14 −T 24 )
100 W
A1=
W
(
( 0.38 ) 5.67 x 10−8 2
m −K 4 )
[ ( 2 701 )4−( 298.6 )4 ] K 4
( 100 )2 cm2
A1=8.722 x 10−5 m2 x
1 m2
A1=0.872cm 2
3./p. 77 - Brown and Marco

Determine the net radiant heat exchange between two parallel oxidized iron plates when the
distance between them is 60 cm and each plate measures 3 m by 3 m. The surface of one is at
95 oC and of the other is at 38 oC.
Given: two parallel oxidized iron plates

Ɛ1 = Ɛ2 = 0.736
A = 3 x 3 m2
T1 = 95 oC + 273 = 368 K
T2 = 38 oC + 273 =311 K
d = 60 cm (distance between plates)
Required: q12 = ?
2
Solution:
Case 7: F12 in Fig 19/12
W
R1 = L/D = 3/0.6 = 5
R2 = W/D = 3/0.5 = 5
F12 = 0.68
1−ε 1 1 1−ε 2
∴ R t= + +
A 1 ε 1 A1 F 12 A2 ε 2
1−0.736 1 1−0.736
¿ + +
( 3 m ) ( 0.736 ) ( 3 m ) ( 0.68 ) ( 3 m )2 ( 0.736 )
2 2
Rt =0.243 11/m 2
σ ( T 14 −T 24 )
q 12=
Rt
W
5.67 x 10−8 2
4
[ ( 368 )4−( 311 )4 ] K 4
= m −K
0.243 11/m2
q 12=2095.5 W
7./p. 77 – Brown and Marco

Determine the net radiant heat exchange between the closed ends of a hollow cylinder 30 cm in
diameter and 10 cm long. The ends are of Cr-Ni alloy with one end heated electrically to a
temperature of 1 040 oC and the other end maintained at 52 oC. Neglect the effect of the side wall
of the cylinder.
Given: closed ends of a hollow cylinder made of Cr-Ni alloy

D = 30 cm
h = 10 cm
T1 = 1 040 oC + 273 = 1 313 K
T2 = 52 oC + 273 = 325 K
Ɛ1 = 0.76
Ɛ2 = 0.64
r2 2
Required:
q12 = ?
r1 1
Solution:
Case 6: F12 from Fig. 19/11

R1 = r1/L = 15/10 = 1.5
R2 = r2/L = 15/10 = 1.5 and 1/R1 = 0.67
F12 = 0.51
1−ε 1 1 1−ε 2
∴ R t= + +
A 1 ε 1 A1 F 12 A2 ε 2
1−0.76 1 1−0.64
¿ + +
π π π
( 0.30 m )2 ( 0.76 ) ( 0.30 m)2 ( 0.51 ) ( 0.30 m )2 ( 0.64 )
4 4 4
Rt =40.164 68/m 2
σ ( T 14 −T 24 )
q 12=
Rt
W
5.67 x 10−8 2 4
[ ( 1 313 )4 −( 325 )4 ] K 4
m −K
¿
40.164 68/m2
q 12=4 180 W

A corridor 3 m wide and 12 m long is heated by a baseboard radiator 30 cm high extending the
full width of the 3 m end. The surface of the radiator is covered with flat black lacquer, and the
floor is of marble. Determine the net radiant heat exchange between the radiator and the floor
when the temperature of the radiator surface is 93 oC and the average temperature of the floor is
27 oC.
Given: baseboard radiator installed in a corridor

L x W = 12 m x 3 m (for the corridor)
H = 30 cm (height of radiator)
T1 = 93 oC + 273 = 366 K
T2 =27 oC + 273 =300 K
1
Ɛ1 = 0.98
Ɛ2 = 0.931
Required:
2
q12 =?
Solution:
Case 8: F12 from Fig. 19/13

Readjust the dimensions in the given with that of the nomenclature in the figure.
L = H = 0.30 m
D=W=3m
W = L = 12 m
R1 = L/D = 0.30/3 = 0.10
R2 = W/D = 12/3 = 4
F12 = 0.44
1−ε 1 1 1−ε 2
∴ R t= + +
A 1 ε 1 A1 F 12 A2 ε 2
1−0.98 1 1−0.931
Rt = + +
( 0.30 m) ( 3 m )( 0.98 ) ( 0.9 m ) ( 0.44 ) ( 3 m )( 12 m )( 0.931 )
2
Rt =2.549 99/m 2
σ ( T 14 −T 24 )
q 12=
Rt
W
5.67 x 10−8 2 4
[ ( 366 ) 4−( 300 )4 ] K 4
m −K
q 12=
2.549 99/m2
q 12=218.9W

A 30 x 60 cm open pan containing water at 70 oC is placed in the exact center of a 4.8 x 6 m
room having a ceiling temperature of 24 oC. If the distance from the water surface to the ceiling
is 3 m and the ceiling is painted with an oil paint, what is the net radiant heat exchange between
the water surface and the ceiling?
Given: 30 x 60 cm open pan containing water

T1 = 70 oC + 273 = 343 K (water)
4.8 m x 6 m room
H=3m
T2 = 24 oC + 273 = 297 K (ceiling)
Ɛ1 = 0.96
Ɛ2 = 0.92
Required:
q12 = ?
Solution:
Case 5: F12 from fig. 19/10

L2 =3 m
L1 =2.4 m
D =3 m
D/L1 = 3/2.4 = 1.25

D/L2 = 3/3 = 1
FA = 0.12
F12 = 4(0.12) = 0.48
1−ε 1 1 1−ε 2
∴ R t= + +
A 1 ε 1 A1 F 12 A2 ε 2
1−0.96 1 1−0.92
Rt = + +
( 0.30 m ) ( 0.60 m )( 0.96 ) ( 0.30 m )( 0.60 m ) ( 0.48 ) ( 4.8 m ) ( 6 m )( 0.92 )
Rt =11.808 57 /m 2
σ ( T 14 −T 24 )
q 12=
Rt
W
5.67 x 10−8 2 4
[ ( 343 )4 −( 297 )4 ] K 4
m −K
q 12=
11.808 57 /m2
q 12=29W

Emittance R Emittance R Config. Factor R

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MEF 312 – EXAMPLE PROBLEMS

Given: q12 = 100 W

3./p. 77 - Brown and Marco

Given: two parallel oxidized iron plates

7./p. 77 – Brown and Marco

Given: closed ends of a hollow cylinder made of Cr-Ni alloy

Case 6: F12 from Fig. 19/11

16./p. 78 – Brown and Marco

Given: baseboard radiator installed in a corridor

Case 8: F12 from Fig. 19/13

10./p. 77 – Brown and Marco

Given: 30 x 60 cm open pan containing water

Case 5: F12 from fig. 19/10

D/L1 = 3/2.4 = 1.25

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