Heat Transfer–

Practical Lecture 13 (Solved Problems)

120. A long, hemicylindrical (1 m radius) shaped furnace used to heat treat sheet metal products iscomprised of three zones. The heating zone (1) is constructed from a ceramic plate of emissivity0.85 and is operated at 1600 K by gas burners. The load zone (2) consists of sheet metal prod-ucts, assumed to be black surfaces maintained at 500 K. The refractory zone (3) is fabricatedfrom insulating bricks having an emissivity of 0.6. Assume steady-state conditions, diffuse, graysurfaces, and negligible convection.

(a) What is the heat rate per unit length of the furnace that must be supplied by the gas burnersfor the prescribed conditions?

Solution:

The heat transfer that must be supplied by the gas burners corresponds to the net radiativeheat transfer rate from Surface 1 – heat transfer by convection and conduction is notrelevant. There are two equivalent procedures to obtain the corresponding net radiativeheat transfer rate per unit furnace length: (a) the radiation network approach; and (b) thedirect approach.

The direct approach will be herein considered. The objective of this approach is to obtainfor each surface the corresponding radiosity Ji. For each surface whose radiosity is unknownEquation (1) (Equation (2)) should be considered if the temperature (net radiative heattransfer rate) is known. The denominator of the LHS of Equation (11) corresponds to thesurface radiative resistance and the denominator of the RHS of the same equation is thespace (or geometric resistance)

Ebi − Ji(1− εi) / (εiAi)

=N∑j=1

Ji − Jj(AiFij)

−1 (1)

qi =N∑j=1

Ji − Jj(AiFij)

−1 (2)

The current enclosure for radiative heat exchange analysis is comprised by 3 (N) indepen-dent opaque, diffusive and gray surfaces, which one with a uniform temperature, irradiation

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and radiosity. In the present problem, since Surface 3 is a reradiating surface – surfaceinsulated at one side and exchanging heat by radiation exclusively by the opposite side –the net radiative heat transfer rate is zero, i.e., q3 = 0. Since Surface 2 is a black surface(ε2 = 1), J2 = Eb2 – Surface 2 has a negligible surface radiative resistance. Therefore,only two equations developed for Surface 1 and 3 have to be considered to obtain thecorresponding radiosities (J1 and J3).

Since Surface 1 (Surface 3) has a known temperature (net radiative heat transfer rate)Equation (3) (Equation (4)) is applied.

q1 ≡Eb1 − J1

(1− ε1) / (ε1A1)=

J1 − J2(A1F12)

−1 +J1 − J3

(A1F13)−1 (3)

0 = q3 ≡J3 − J1

(A3F31)−1 +

J3 − J2(A3F32)

−1 (4)

Since F31 = F32 (due to the furnace geometry – symmetry rule), Equation (4) can bewritten as shown in Equation (5).

J3 =J1 + J2

2(5)

Substituting Equation (5) in Equation (3) and considering F12 = 0 and F13 = 1 andJ2 = Eb2, Equation (6) is obtained.

J1 =Eb2/2 + (ε1Eb,1) / (1− ε1)

1/2 + ε1/ (1− ε1)(6)

Replacing in Equation (6) Eb1 and Eb2 by σT 41 (= 5.67 × 10−8 × 16004 W m−2) and σT 4

2

(= 5.67 × 10−8 × 5004 W m−2), respectively and ε1 by 0.85, J1 is calculated and thenconsidering Equation (5) J3 is obtained – see the corresponding values in Equations (7)and (8) .

J1 ≈ 341747.604 W m−2 (7)

J3 ≈ 172645.678 W m−2 (8)

Finally, the heat transfer rate per unit length of the furnace that must be supplied by thegas burner – equal to the net radiative heat transfer rate from Surface 1 – is calculated inEquation (9).

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q′1 =q1L≡ ε1A1 (Eb1 − J1)

L (1− ε1)⇔ q′1 =

ε1 (W1L) (Eb1 − J1)L (1− ε1)

⇔ q′1 =ε1W1 (Eb1 − J1)

(1− ε1)⇔

⇔ q′1 =0.85× 1× [(5.67× 10−8 × 16004)− 341747.604]

(1− 0.85)⇔

⇔ q′1 ≈ 169101.924 W m−1(169 kW m−1

)(9)

(Note that only the calculation of J1 – through Equation (6) – would be enough to answerthis question. However, J3 was also evaluated because this value will be required in thefollowing question.)

In alternative to the direct method, the radiation network approach can also be applied toobtain the net radiative heat transfer from Surface 1 as follows. The resistance networkrepresentation is presented in the following figure.

The net radiative heat transfer rate per unit length is computed through Equation (10)considering the difference between the emissive powers of Surfaces 1 and 2 and the surfaceand space resistances in between the nodes that represent such emissive powers in theradiation network representation. (Note that the equivalent resistance is developed as acombination of series and parallel resistance schemes.)

q′1 =q1L≡ 1

L

Eb1 − Eb2

1−ε1ε1A1

+ 1

A1F12+[(A1F13)−1+(A3F32)

−1]−1

⇔

⇔ q′1 =σ (T 4

1 − T 42 )

1−ε1ε1W1

+ 1

W1F12+[(W1F13)−1+(W2F23)

−1]−1

⇔

⇔ q′1 =5.67× 10−8 × (16004 − 5004)1−0.850.85×1 + 1

1×0+[(1×1)−1+(1×1)−1]−1

⇔

⇔ q′1 = 169101.927 W m−1

(10)

(b) What is the temperature of the insulating brick surface for the prescribed conditions?

Solution:

Since Surface 3 is a reradiating surface, q3 = 0 and, consequently, G3 = J3 = Eb3. There-fore, the equilibrium temperature of Surface 3 is calculated in Equation (11) taking intoconsideration Equation (8).

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J3 = Eb3 ⇔ J3 = σT 43 ⇔ T3 =

4

√J3σ⇔ T3 = 4

√172645.678

5.67× 10−8⇔

⇔ T3 ≈ 1320.971 K

(11)

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122. A row of regularly spaced, cylindrical heating elements (1) is used to cure a surface coating thatis applied to a large panel (2) positioned below the elements. A second large panel (3), whosetop surface is well insulated, is positioned above the elements. The elements are blackbodiesmaintained at T1 = 600 K, while the panel has an emissivity of 0.5 and is maintained at T2 =400 K. The cavity is filled with a nonparticipating gas and convection heat transfer occurs atsurfaces 1 and 2, with 10 W m−2 K−1 e 2 W m−2 K−1, respectively. (Convection at the insulatedpanel (3) may be neglected.)

(a) Evaluate the mean gas temperature Tm.

Solution:

The mean gas temperature can be evaluated based on an overall energy balance applied toa gas representative control volume – see the dashed lines in the figure which represent thecontrol volume boundaries. (The control volume length is equal to the pitch of the heatingelements (s = 50 mm).)

Assuming steady-state conditions (Est = 0), no internal generation of thermal energy – nogas-phase reactions and nonparticipating gas – (Eg = 0), and heat transfer rates to/fromthe control volume given by the Newton’s law of cooling, the mean gas temperature canbe evaluated as follows – see Equation (12).

Ein − Eout + ��Eg = ��Est ⇔ h1A1 (T1 − Tm)− h2A2 (Tm − T2)⇔

⇔ h1πDL (T1 − Tm)− h2sL (Tm − T2)⇔ Tm =h1πDT1 + h2sT2

h1πD + h2s⇔

⇔ Tm =10× π × 0.025× 600 + 2× 0.050× 400

10× π × 0.025 + 2× 0.050⇔

⇔ Tm ≈ 577.411 K

(12)

(b) What is the rate per unit axial length at which electrical energy must be supplied to eachelement to maintain its prescribed temperature?

Solution:

The required electrical power per unit length is calculated as the sum of the convectiveheat transfer rate per unit length and the net radiative heat transfer rate per unit lengthonce all the electrical energy is converted into thermal energy – see Equation (13).

q′1,elec = q′1,total ≡ q′1,conv + q′1,rad (13)

The convective contribution to the total heat transfer rate from each heating element perunit axial length is calculated in Equation (14) considering the gas temperature calculated

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previously (see Equation (12)).

q′1,conv = h1πD (T1 − Tm)⇔ q′1,conv = 10× π × 0.025× (600− 577.411)⇔⇔ q′1,conv ≈ 17.741 W m−1

(14)

The radiative contribution is herein calculated considering the radiation network approach.The radiative heat exchange between the three surfaces that comprise the enclosure canbe represented through the network illustrated in the following figure. The surfaces are as-sumed as opaque, diffusive, and gray with uniform temperature, irradiation, and radiosity.Since the surface of the heating elements (Surface 1) is black the corresponding radiativeresistance is negligible and J1 = Eb1. Surface 2 has a finite surface radiative resistance(since ε2 6= 1) and consequentely, J2 6= Eb2 – note the presence of this resistance in the net-work representation ((1− ε2) / (ε2A2)). Surface 3 is a reradiating surface since this surfaceis insulated at the outer side and has a negligible convection heat transfer rate with thegas – consequently, the corresponding net radiative heat transfer rate is zero and J3 = Eb3.

The net radiative heat transfer rate from Surface 1 can be calculated according to Equation(15). Note that the denominator of RHS of Equation (15) corresponds to the summationof all (surface radiative and geometric) resistances between the nodes considered at thenumerator – nodes for which the blackbody emissive power are equal to Eb1 and Eb2.

q1,rad ≡ q′1,radL =Eb1 − Eb2

1

A1F12+[(A1F13)−1+(A2F23)

−1]−1 + 1−ε2

ε2A2

(15)

The view factors F12, F13, and F23 must be evaluated. Equations (16) and (17) can beapplied due to the reciprocity relations. In Equation (17), F31 is equal to F21 (symmetryrule).

A1F12 = A2F21 (16)

A1F13 = A3F31 ⇔ A1F13 = A3F21 (17)

The view factor F21 can be calculated with Equation (18) that was developed for the aninfinite plane and row of cylinders – see the figure below and find this equation in Table13.1 of the textbook “Fundamentals of Heat and Mass Transfer”, Sixth Edition (SeventhEdition).

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Fij = 1−

[1−

(D

s

)2]1/2

+D

stan−1

[(s2 −D2

D2

)1/2]

(18)

Therefore, the view factor F21 is calculated in Equation (19).

F21 = 1−

[1−

(D

s

)2]1/2

+D

stan−1

[(s2 −D2

D2

)1/2]⇔

⇔ F21 = 1−

[1−

(25

50

)2]1/2

+25

50tan−1

[(502 − 252

252

)1/2]⇔

⇔ F21 ≈ 0.658

(19)

The view factor F23 can be computed based on the summation rule applied for Surface 2– see Equation (20). Note that F22 is equal to zero because Surface 2 is a planar surface.

F21 + F22 + F23 = 1⇔ F23 = 1− F21 (20)

The following equation (Equation (21)) is obtained by substituting Equations (16), (17)and (20) in Equation (15).

q′1,rad =(Eb1 − Eb2) /L

1

A1F12+[(A1F13)−1+(A2F23)

−1]−1 + 1−ε2

ε2A2

⇔

⇔ q′1,rad =(Eb1 − Eb2) /L

1

A2F21+[(A3F21)−1+(A2(1−F21))

−1]−1 + 1−ε2

ε2A2

⇔

⇔ q′1,rad =Eb1 − Eb2

1

sF21+{(sF21)

−1+[s(1−F21)]−1}−1 + 1−ε2

ε2s

⇔

⇔ q′1,rad =sσ (T 4

1 − T 42 )

1

F21+[(F21)−1+(1−F21)

−1]−1 + 1−ε2

ε2

(21)

Replacing the parameters of Equation (21) by the corresponding values, the net radiativeheat transfer rate per unit length of Surface 1 (heating element) is obtained – see Equation

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(22).

q′1,rad =sσ (T 4

1 − T 42 )

1

F21+[(F21)−1+(1−F21)

−1]−1 + 1−ε2

ε2

⇔

⇔ q′1,rad =0.050× 5.67× 10−8 × (6004 − 4004)

1

0.658+[(0.658)−1+(1−0.658)−1]−1 + 1−0.5

0.5

⇔

⇔ q′1,rad ≈ 138.263 W m−1

(22)

Finally, replacing the results obtained in Equations (14) and (22) in Equation (13), thetotal heat transfer rate per unit axial length (equal to the total rate of electrical energyper unit axial length required) is computed – see Equation (23).

q′1,total = q′1,conv + q′1,rad ⇔ q′1,total = 17.741 + 138.263⇔ q′1,total = 156.004 W m−1 (23)

(c) What is the heat flux to the coated panel (2)?

Solution:

The total heat flux to the coated panel (Surface 2) has two contributions – due to convectionand radiation heat transfer modes – as shown in Equation (24).

q′′2,total = q′′2,conv + q′′2,rad (24)

The convective contribution is computed in Equation (25) taking into account the meangas temperature calculated in Equation (12).

q′′2,conv = h2 (Tm − T2)⇔ q′′2,conv = 2× (577.411− 400)⇔ q′′2,conv = 354.822 W m−2 (25)

The radiative contribution is calculated as follows. Since Surface 3 is a reradiating surface(q3 = 0), the net radiative heat transfer rate to Surface 2 per unit length (−q′2,rad) is equalto the net radiative heat transfer rate from Surface 1 per unit length (q′1,rad). The netradiative heat transfer rate to Surface 2 (coated panel) per unit length is calculated inEquation (26).

q′1,rad = −q′2,rad ⇔ q′2,rad = 138.263 W m−1 (26)

The net radiative heat flux to Surface 2 is evaluated in Equation (27) taking into accountthe net radiative heat transfer rate to Surface 2 per unit length (Equation (26)) and thepitch of the heating elements (s).

q′′2,rad =q′2,rads⇔ q′′2,rad =

138.263

0.050⇔ q′′2,rad = 2765.260 W m−2 (27)

Finally, the heat flux to the coated panel is calculated by replacing the values obtained in

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Equation (25) and (27) in Equation (24) – see Equation (28).

q′′2,total = q′′2,conv + q′′2,rad ⇔ q′′2,total = 354.822 + 2765.26⇔

⇔ q′′2,total = 3120.082 W m−2(28)

Since the hydrothermal conditions are time-independent (steady-state problem) the heatflux to the coated surface must be absorbed by the coating (for instance to promote speciesvaporization (latent heat) or converted into chemical bonding energy by curing reactions).

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