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PROBLEM 13.88 

KNOWN:  Cylindrical furnace of diameter D = 90 mm and overall length L = 180 mm. Heating

elements maintain the refractory liming (ε  = 0.8) of section (1), L1 = 135 mm, at T1 = 800°C. The bottom (2) and upper (3) sections are refractory lined, but are insulated. Furnace operates in a

 pacecraft environment.s F 

IND:  Power required to maintain the furnace operating conditions with the surroundings at 23°C.

SCHEMATIC: 

ASSUMPTIONS:  (1) All surfaces are diffuse gray, (2) Uniform radiosity over the sections 1, 2, and

, and (3) Negligible convection effects.3 

ANALYSIS:  By defining the furnace opening as the hypothetical area A4, the furnace can be

represented as a four-surface enclosure as illustrated above. The power required to maintain A1 at T1 

is q 1, the net radiation leaving A1. To obtain q 1 following the methodology of Section 13.2.2, we

must determine the radiosity at all surfaces by simultaneously solving the radiation energy balance

equations for each surface which will be of the form, Eqs. 13.14 or 13.15.

( )

 N j j bi i

1i i i i i j 1

J JE Jq 

1 / A 1/ A Fε ε =

−−= =

−  ∑

 j

  (1,2)

Since ε4 = 1, J4 = E b4, so we only need to perform three energy balances, for A1, A2, and A3,respectively

A1:( )

 b1 1 1 31 2 1 4

1 1 1 1 12 1 13 1 14

E J J JJ J J J

1 / A 1/ A F 1/ A F 1/ A Fε ε 

− −− −= + +

−  (3)

A2:2 32 1 2 4

2 21 2 23 2 24

J JJ J J J0

1/ A F 1/ A F 1/ A F

−−= + +

  −  (4)

A3:3 1 3 2 3 4

3 31 3 32 3 34

J J J J J J0

1/ A F 1/ A F 1/ A F

− − −= + +   (5)

 Note that q 2 = q 3 = 0 since the surfaces are insulated (adiabatic). Recognize that in the above equation

set, there are three equations and three unknowns: J1, J2, and J3. From knowledge of J1, q 1 can be

determined using Eq. (1). Next we need to evaluate the view factors. There are N2 = 4

2 = 16 view

factors and N(N – 1)/2 = 6 must be independently evaluated, while the remaining can be determined

 by the summation rule and appropriate reciprocity relations. The six independently determined Fij are:

 By inspection:  (1) F22 = 0 (2) F44 = 0

Coaxial parallel disks: From Fig. 13.5 or Table 13.5,

Continued …..

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PROBLEM 13.88 (Cont.)

( )1/ 2

2224 4 2F 0.5 S S 4 r / r  = − −

⎧ ⎫⎡ ⎤⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭

(3)  ( )1/ 2

2224F 0.5 18 18 4 1 0.05573= − − =

⎧ ⎫⎡ ⎤⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭2 24

2 2 4 42 22

1 R 1 0.250S 1 1 18.00 R r / L 45 /180 0.250 R r / L 0.250

R 0.250

+ += + = + = = = = = =  

 Enclosure 1-2-2′  : from the summation rule for A2,

(4) 

F21 = 1 – F22’ = 1 – 0.09167 = 0.9083

where F22′ can be evaluated from the coaxial parallel disk relation, Table 13.5. For these surfaces, R 2 

= r 2/L1 = 45/135 = 0.333, R 2′ = r 2/L1 = 0.333, and S = 11.00. From the summation rule for A1,

(5) F11 = 1 – F12 – F12′ = 1 – 0.1514 – 0.1514 = 0.6972

and by symmetry F12 = F12′ and using reciprocity

( )( )[ ]12 2 21 1F A F / A 0.090m 2 / 4 0.9083 / 0.090m 0.135m 0.1514π π = = × × × =

  Enclosure  2′   -3-4: from the summation rule for A4,

(6) F43 = 1 – F42′ - F44 = 1 – 0.3820 – 0 = 0.6180

where F44 = 0 and using the coaxial parallel disk relation from Table 13.5, with R 4 = r 4/L2 = 45/45 =

1, R 2′ = r 2/L2 = 1, and S = 3.

The View Factors: Using summation rules and appropriate reciprocity relations, the remaining 10

view factors can be evaluated. Written in matrix form, the Fij are

0.6972* 0.1514 0.09704 0.05438

0.9083* 0* 0.03597 0.05573*

0.2911 0.01798 0.3819 0.3090

0.3262 0.05573 0.6180* 0*The Fij shown with an asterisk were independently determined.

From knowledge of the relevant view factors, the energy balances, Eqs. (3, 4, 5), can be solved

simultaneously to obtain the radiosities,

2 21 2 3J 73,084 W / m J 67, 723W / m J 36, 609 W / m= = =

2

The net heat rate leaving A1 can be evaluated using Eq. (1) written as

( )

( )

( )

2 b1 1

1 21 1 1

75,159 73, 084 W / mE Jq 317 W

1 / A 1 0.8 / 0.8 0.03817 mε ε 

−−= = =

−   − ×