Exchange of Thermal Radiation Between Surfaces Separated by Transparent Medium Rajendra Karwa Chapter First Online: 19 June 2020

Abstract Radiation heat exchange between two black surfaces depends on the shape and size of the bodies, relative position and distance between them while the radiation exchange between gray bodies also depends on their emissivity. For calculation of the radiation exchange between two black surfaces, basic integral equation of the shape factor has been derived which represents the fraction of the total radiation emitted by a surface intercepted by other surfaces. Salient features of the shape factor have been presented and numbers of illustrative examples on calculation of the shape factor for different systems have been given.

Radiation exchange between gray bodies is a complex process. Electrical analogy-based method is given to solve such problems wherein the actual system is reduced to an equivalent electric network. Concept of radiation shield, which reduces the radiation heat transfer from a body by placing a thin opaque partition between the surfaces, has been explained in Sect. 11.9 . In the end of the chapter, Newton’s law of cooling is presented.

Keywords Shape or geometrical or view factor Shape factor characteristics Radiosity Irradiation Electrical analogy Radiation network Radiation shield Newton’s law of cooling This is a preview of subscription content,

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Review Questions 11.1 Define shape or geometrical or view factor in reference to the radiation heat exchange. Derive a general relation of the factor for radiation energy exchange between two blackbodies.

11.2 Prove the following relation for two surfaces 1 and 2 exchanging radiation energy

$$A_{1} F_{12} = A_{2} F_{21},$$

where

A and

F are area and shape factors, respectively.

11.3 Discuss salient features of the radiation shape factor.

11.4 Define irradiation and radiosity. Find the net rate at which the radiation energy leaves a gray surface in terms of radiosity and emissivity.

11.5 Using the definition of radiosity and irradiation develop an expression for the radiant energy exchange between two gray bodies.

11.6 Prove that the radiation heat exchange between two long concentric cylinders is given by

$$q_{12} = A_{1}{\sigma }\left( {T_{1}^{4} {-}T_{2}^{4} } \right)/[1/{{\varepsilon_{1} }} + \, \left( {A_{1} /A_{2} } \right)(1/{{\varepsilon_{2} }} {-} \, 1)],$$

where

ε _{1} and

ε _{2} are the emissivities of the inner and the outer cylinders, respectively. Assume that there is no radiation heat loss from the cylinder ends.

11.7 For a system consisting of two diffuse gray surfaces 1 and 2 at different temperatures T _{1} and T _{2} , respectively, connected by a single refractory surface, draw the radiation network (electrical analog) and write the equation of q _{12} .

11.8 A system consists of three diffuse gray surfaces which see each other and nothing else. Draw the radiation network (electrical analog). Show all the resistances on the network. Also write the nodal equations and discuss how will you proceed to find heat transfer from each surface?

11.9 Show that a thin black screen (radiation shield) introduced between two black surfaces reduces the radiation heat transfer by half. Develop general relation of reduction in heat transfer for n shields.

11.10 Discuss various methods of reducing the temperature measurement error from thermometer and thermocouples.

11.11 State Newton’s law of cooling and show that the temperature of a body falls exponentially when the body is rejecting heat by convection and radiation with a small temperature difference.

11.12 Define radiation heat transfer coefficient.

11.13 How would you determine specific heat of a solid using Newton’s law of cooling?

Problems 11.1. Hot flat plate of small area

∆A _{1} moves a distance

d from location A directly below sensor S of small area

∆A _{2} at a vertical distance

H = 0.8 m as shown in Fig.

11.80 . Determine the distance

d at which the sensor signal will be 50% of that directly below the sensor.

[Ans. The sensor signal is proportional to the radiation leaving plate (area ∆A _{1} ) and intercepted by the sensor of area ∆A _{2} . Hence, at location A , \(q_{A} = I_{n1} \frac{{\Delta A_{1} \Delta A_{2} \cos 0\cos 0}}{{H^{2} }} = I_{n1} \frac{{\Delta A_{1} \Delta A_{2} }}{{H^{2} }}\) and at location B, \(q_{B} = I_{n1} \frac{{\Delta A_{1} \Delta A_{2} \cos \theta \cos \theta }}{{s^{2} }} = I_{n1} \frac{{\Delta A_{1} \Delta A_{2} \cos^{2} \theta }}{{s^{2} }}\) . Ratio of the q _{B} and q _{A} is \(\frac{{q_{B} }}{{q_{A} }} = \frac{{\cos^{2} \theta }}{{s^{2} }} \times H^{2}\) \(= \frac{{H^{2} /s^{2} }}{{s^{2} }} \times H^{2} = \frac{{H^{4} }}{{s^{4} }} = \frac{{H^{4} }}{{(H^{2} + d^{2} )^{2} }}\) . Substituting \(\frac{{q_{B} }}{{q_{A} }} = 0.5\) and H = 0.8, we get d = 0.514 m.]

11.2. Determine sh6ape or configuration factor

F _{12} for a hemispherical shell and a flat surface forming an enclosure, Fig.

11.81 .

[Ans. From reciprocity relation, F _{12} = (A _{2} /A _{1} )F _{21} ; F _{21} + F _{22} = 1, but F _{22} = 0, so F _{21} = 1 hence F _{12} = (A _{2} /A _{1} ) = πr ^{2} /(2πr ^{2} ) = 0.5]

11.3. Two diffuse-black surfaces, a small flat disc 1 of area

dA _{1} and a large disc 2 of radius

R are parallel to each other and directly opposed. A line joining their centers is normal to both the surfaces. The larger disc is located at a height

H from the smaller one. Determine the shape factor

F _{12} .

11.4. A rolled steel sheet (

ε _{1} = 0.4) at 600 K is lying in a large space, which can be regarded at an average temperature of 300 K. Estimate the heat loss per m

^{2} of the plate area from one side of the plate by radiation only.

11.5. The Sun can be regarded as nearly a spherical radiation source of diameter

D _{s} of approximately 1.4 × 10

^{9} m and is at a distance

s = 1.5 × 10

^{11} m from the Earth. If the solar flux outside the Earth’s atmosphere, i.e. the solar constant is 1350 W/m

^{2} , determine the emissive power of the Sun and its surface temperature.

[Ans. From Example 11.9 , \(q_{12} = \sigma T_{s}^{4} \frac{{D_{s}^{2} }}{{4s^{2} }}\) or \(E = \sigma T_{s}^{4} = q_{12} \times \frac{{4s^{2} }}{{D_{s}^{2} }} =\) \(1350 \times \frac{{4 \times (1.5 \times 10^{11} )^{2} }}{{(1.4 \times 10^{9} )^{2} }}\) \(= 6.2 \times 10^{7}\) W/m^{2} ; surface temperature of the Sun, \(T_{s} = \left( {\frac{E}{\sigma }} \right)^{1/4} =\) \(\left( {\frac{{6.2 \times 10^{7} }}{{5.67 \times 10^{ - 8} }}} \right)^{1/4} = 5750\) K.]

11.6. Prove that the interchange factor for two concentric spheres is

$$\left[ {\frac{1}{{\varepsilon_{1} }} + \left( {\frac{{R_{1} }}{{R_{2} }}} \right)^{2} \left( {\frac{1}{{\varepsilon_{2} }} - 1} \right)} \right]^{ - 1}$$

where

ε _{1} and

ε _{2} are the total emissivities of the two spheres. The outside radius of the inner sphere is

R _{1} and inside radius of the outer sphere is

R _{2} .

11.7. Two concentric spheres, 210 mm and 300 mm in diameter, with the space between them evacuated are to be used to store liquid air at –153°C in a room at 27°C. The surfaces of the spheres are flushed with aluminium (

ε = 0.03). Find the rate of evaporation of liquid air if its latent heat of vaporization

L is 200 kJ/kg. Assume that the outer sphere temperature is equal to the room temperature.

11.8. Two parallel discs of 600 mm diameter each are spaced 300 mm apart. The temperatures of the discs are maintained at 800 K and 500 K with emissivities of 0.2 and 0.4, respectively. The discs are located in a very large space whose walls are maintained at 310 K. Determine the rate of heat loss by radiation from the inside surface of each disc.

[

Ans. L =

R = 0.3 m. From Table

11.1 ,

F _{12} = 1 + (

L ^{2} /2

R ^{2} )[1 – (1 + 4

R ^{2} /

L ^{2} )

^{1/2} ] = 0.382.

E _{b1} =

σT _{1} ^{4} = 23224,

E _{b2} =

σT _{2} ^{4} = 3543.75,

E _{b3} =

σT _{3} ^{4} = 523.63. The network is shown in Fig.

11.82 .

A _{1} = A _{2} = 0.2828 m^{2} , (1 – ε _{1} )/A _{1} ε _{1} = 14.147, (1 – ε _{2} )/A _{2} ε _{2} = 5.305, F _{13} = F _{23} = 1 – F _{12} = 0.618, 1/A _{1} F _{12} = 9.258, 1/A _{1} F _{13} = 1/A _{2} F _{23} = 5.723.

The nodal equations are (E _{b1} – J _{1} )/14.147 + (E _{b3} – J _{1} )/5.723 + (J _{2} – J _{1} )/9.258 = 0, (E _{b2} – J _{2} )/5.305 + (E _{b3} – J _{2} )/5.723 + (J _{1} – J _{2} )/9.258 = 0; The equations give J _{1} = 5802.76, J _{2} = 2941.66. Hence, q _{1} = (E _{b1} – J _{1} )/14.147 = 1231.44 W, q _{2} = (E _{b2} – J _{2} )/5.305 = 113.49 W, q _{3} = [(J _{1} – E _{b3} ) + (J _{2} – E _{b3} )]/5.723 = 1344.95 W = q _{1} + q _{2} .]

11.9. Following the procedure used for the estimate of

F _{12} in Example

11.36 , determine the heat exchange from surface 2 to 1.

[Ans. Changing the nomenclature of the surfaces, we have A _{1} = 3.2 × 4 m^{2} , A _{2} = 5 × 4 m^{2} , ε _{1} = 0.7, ε _{2} = 0.8, T _{1} = 350 K, T _{2} = 300 K. For Z /X = 5/4 = 1.25, Y /X = 3.2/4 = 0.8, F _{12} ≈ 0.24; R _{t} = [(1 – ε _{1} )/A _{1} ε _{1} + 1/A _{1} F _{12} + (1 – ε _{2} )/A _{2} ε _{2} ] = 0.3715; q _{12} = σ (T _{1} ^{4} – T _{2} ^{4} )/R _{t} = 1054.1 W, which is nearly the same as found earlier.]

11.10. Two parallel rectangular plates 1 and 2, 3 m × 2 m in size and 2 m apart, are joined on their long edge by a third plate

3 as shown in Fig.

11.83 . Determine shape factors

F _{12} and

F _{13.} [Ans. . Surface 1 to 2 : X = 3 m, Y = 2 m and L = 2 m, for X /L = 1.5 and Y /L = 1.0, F _{12} = 0.255; Surface 1 to 3 : F _{13} = 0.225 for Z /X = 0.66 and Y /X = 0.66.]

11.11. Determine the radiation energy impinging upon a 2 m × 2 m wall 1.5 m away from a 50 mm diameter spherical body at 1000 K as shown in Fig.

11.84 . Assume the wall to be a blackbody.

[Ans. The sphere is very small compared to the wall; hence, it can be treated as an infinitesimal disc of area π /4 × 50^{2} = 1963.5 mm^{2} . Considering 1/4th of the wall a , F _{12a} = 0.09 from Eq. (11.18 ) for L _{1} = L _{2} = 1 m and L = 1.5. For the total wall area, F _{12} = 0.09 × 4 = 0.36; q _{12} = A _{1} F _{12} σ (T _{1} )^{4} = 40.08 W.]

11.12. Two parallel square plates, each 5 m

^{2} , are separated by 3 mm distance. One of the plates has a temperature of 2000 K and surface emissivity of 0.7, while the other plate surface has a temperature of 1000 K and a surface emissivity of 0.8. Find the net energy exchange by radiation between the plates.

If a thin polished metal sheet of surface emissivity of 0.15 on both sides is now located centrally between the two plates, find out the altered net heat transfer. The convection and edge effects may be neglected.

[Ans. The gap is very small; hence, equation of infinite parallel planes can be used. q _{12} = f _{12} A _{1} σ (T _{1} ^{4} – T _{2} ^{4} ), where f _{12} = [1/ε _{1} + 1/ε _{2} – 1]^{−1} without shield. f _{12} = [(1/ε _{1} + 1/ε _{3} – 1) + (1/ε _{2} + 1/ε _{3} – 1)]^{−1} with shield. Substitution of the values gives (q _{12} )_{without shield} = 2.533 × 10^{6} W, (q _{12} )_{with shield} = 303.5 × 10^{3} W.]

11.13. Which curve of Fig.

11.85 represents correctly the variation of the radiation shape factor for parallel black planes of finite size?

11.14. A thermocouple probe of emissivity 0.85 gives the temperature reading of 600 K of a gas flowing through a 0.9 m diameter duct having wall temperature of 500 K. Find out the error in the temperature measurement. The convection coefficient between the gas and thermocouple is 125 kJ/(m

^{2} h °C).

[Ans. The thermocouple bead area is very small compared to the pipe surface area, hence f _{12} = ε _{1} . The heat balance equation gives [hA _{1} (T _{g} – T _{c} ) = A _{1} ε _{1} σ (T _{c} ^{4} – T _{s} ^{4} )], where T _{c} is the thermocouple reading, T _{g} the true gas temperature, and T _{s} the pipe wall temperature. Substitution of the values gives, error = (T _{g} – T _{c} ) = 93.14°C.]

11.15. A mercury-in-glass thermometer, hanged in a room, indicates a temperature of 25°C. The walls of the room are at a temperature of 40°C. Calculate the true temperature of the room air if the thermometer bulb emissivity is 0.9 and the heat transfer coefficient from the room air to the thermometer bulb is 8 W/(m

^{2} K).

11.16. Air at atmospheric pressure flows over a thermocouple bead with a velocity

U of 2.5 m/s. The temperature of the pipe wall is 650 K and the thermocouple indicates a temperature of 850 K. The diameter of the thermocouple bead

d is 1.0 mm and emissivity

ε is 0.3. Determine the true temperature of the air.

[Ans. Thermophysical properties of air at 850 K (for trial): k = 0.0603 W/(m K), μ = 3.756 × 10^{−5} kg/(m s), ρ = 0.4149 kg/m^{3} , Pr = 0.692; Re = ρUd /μ = 27.55; Nu = 4.28 from Eq. ( 8.55 ) neglecting (μ _{∞} /μ _{w} ); h = Nuk /d = 258 W/(m^{2} K); For T _{bead} = 850 K and T _{wall} = 650 K, equation h (T _{a} – T _{bead} ) = εσ (T _{bead} ^{4} – T _{wall} ^{4} ) gives T _{a} = 872.6 K; Iteration may be carried out using thermophysical properties at mean air temperature T _{m} = (872.6 + 850)/2 ≈ 861 K.]

11.17. Figure

11.86 shows a cylindrical cavity whose surface can be assumed to be gray (

ε = 0.6). Find the rate of emission from the cavity to the surrounding at 293 K if the cavity surface temperature is 900°C.

Fig. 11.86 A cylindrical cavity (Problem 11.17)

[Ans. Assume cavity to be covered with a black surface 2 at 293 K as shown in the figure. ε _{1} = 0.6, T _{1} = 1173 K, R = 0.1 m; H = 0.1 m; A _{1} = πDH + (π /4)D ^{2} = 0.09425 m^{2} ; A _{2} = π R ^{2} = 0.031416 m^{2} ; F _{12} = A _{2} /A _{1} = 0.3333; R _{t} = (1 – ε _{1} )/(A _{1} ε _{1} ) + 1/(A _{1} F _{12} ) = 38.9; q _{12} = σ (T _{1} ^{4} – T _{2} ^{4} )/R _{t} = 2748.72 W; Note (1 – ε _{2} )/(A _{2} ε _{2} ) = 0]

11.18. Determine view factor F _{12} for the following configurations:

(i) A sphere lying on an infinite horizontal plane, Fig.

11.87 a

(ii) Long inclined planes where plate end edge indicated by B in Fig. 11.87 b is 100 mm above the longitudinal axis of plane 1 passing through center C in the figure.

[Ans. (i) Put a parallel infinite plane 3 as shown in figure. Then F _{11} + F _{12} + F _{13} = 1; for spherical surface F _{11} = 0. Since F _{12} = F _{13} (by symmetry), F _{12} = F _{13} = 0.5. From reciprocity relation F _{21} = (A _{1} /A _{2} ) F _{12} . Since A _{2} = ∞, F _{21} = 0.

(iii) F _{11} + F _{12} + F _{13} = 1. Since F _{12} = F _{13} (by symmetry), F _{12} = F _{13} = 0.5. From reciprocity relation F _{21} = (A _{1} /A _{2} ) F _{12} . Since A _{2} = LW √2, A _{1} = 2WL , F _{21} = (2WL /LW √2) × 0.5 = 0.707.]

11.19. Two infinite parallel plates, one is black (1) and other (2) is gray with emissivity 0.8, are at temperatures 1000 K and 600 K, respectively. Determine the irradiation on plate 1.

11.20. A small object at 500 K and spectral emissivity as given in Fig.

10.11 is suspended in a large furnace (wall temperature 1300 K and total emissivity 0.8). Determine (a) hemispherical total surface emissivity and absorptivity of the object, (b) net radiative flux to the surface.

[Ans . (a) (i) Hemispherical total emissivity: λ _{1} T = 2 × 500 = 1000 for which F _{0-1000} = 0.321 × 10^{−3} ; λ _{2} T = 7 × 500 = 3500 for which F _{0-3500} = 0.382635; Thus, F _{1000-3500} = 0.382635 - 0.321 × 10^{−3} = 0.382314; F _{3500-∞} = 1–0.382635 = 0.617365; \(\varepsilon = \varepsilon_{1} (F_{0 - 1000} ) + \varepsilon_{2} (F_{1000 - 3500} ) + \varepsilon_{3} (F_{3500 - \infty } )\) = 0.1 × 0.321 × 10^{−3} + 0.4 × 0.382314 + 0.2 × 0.617365 = 0.27643; (ii) Absorptivity of the object for the radiation from furnace wall at 1300 K: λ _{1} T = 2 × 1300 = 2600 for which F _{0–2600} = 0.18312; λ _{2} T = 7 × 1300 = 9100 for which F _{0–9100} = 0.892582; Thus, F _{2600-9100} = 0.892582 – 0.18312 = 0.709462; F _{9100–∞} = 1 – 0.892582 = 0.107418; Hemispherical total absorptivity, \(\alpha = \alpha_{1} (F_{0 - 2600} ) + \alpha_{2} (F_{2600 - 9100} ) + \alpha_{3} (F_{9100 - \infty } )\) = 0.1 × 0.18312 + 0.4 × 0.709462 + 0.2 × 0.107418 = 0.32358; (b) Net radiative flux to the surface, \(q^{\prime\prime}\) = G – J = G – ρG – εE _{b} (500) = (1 – ρ )G – εE _{b} (500) = αE _{b} (1300) – εE _{b} (500) = 0.32358 × 5.67 × 10^{−8} × 1300^{4} – 0.27643 × 5.67 × 10^{−8} × 500^{4} = 51421 W/m^{2} .]

11.21. A vertical copper plate [

ρ = 8950 kg/m

^{3} ,

c = 380 J/(kg K),

k = 375 W/(m K)] at an initial uniform temperature of 300°C is suspended in a room where the ambient air and surroundings are at 25°C. Plate measures 0.25 m × 0.25 m in area and is 0.02 m in thickness. The rate of cooling is found to be 0.08 K/s when plate temperature was 275°C. Plate surface emissivity is 0.2. Determine the convection heat transfer coefficient.

[Ans . For air at mean film temperature of ½(275 + 25) = 150°C, from Table A5, ρ = 0.8370 kg/m^{3} , c = 1017.1 J/(kg K), μ = 2.3769 × 10^{−5} N s/m^{2} and k = 0.03522 W/(m K). The plate rejects sensible heat both by convection and radiation. At any instant, the energy balance gives \(- \rho (A_{s} \delta )c\frac{dT}{d\tau } = h(2A_{s} )(T_{s} - T_{\infty } ) + \varepsilon (2A_{s} )\sigma (T_{s}^{4} - T_{sur}^{4} )\) , where δ is thickness and A _{s} is area of plate. Equation gives \(h = \frac{1}{{(T_{s} - T_{\infty } )}}\left[ { - \frac{dT}{d\tau }\frac{\rho \delta c}{2} - \varepsilon \sigma (T_{s}^{4} - T_{sur}^{4} )} \right]\) ; For dT /dτ = - 0.08 K/s, \(h = \frac{1}{(548 - 298)}\left[ {\frac{0.08 \times 8950 \times 0.02 \times 380}{2} - 0.2 \times 5.67 \times 10^{ - 8} \times (548^{4} - 298^{4} )} \right]\) \(= 7.15\) W/(m^{2} K).]

11.22. Determine the heat loss from an electric bulb (

T _{s} = 125°C) in a room if the surrounding air temperature is 25°C. The bulb can be approximated to be a sphere of 50 mm diameter. Bulb surface emissivity is 0.88 and the room surface temperature is 27°C.

[Ans. At film temperature t _{fm} = (125 + 25)/2 = 75°C, air properties are ρ = 1.0052 kg/m^{3} , k = 0.0299 W/(m K), μ = 2.0658 × 10^{−5} N s/m^{2} and Pr = 0.697; Heat loss from the bulb is by convection and radiation;

\(q = q_{c} + q_{r} = h_{c} A_{s} (T_{s} - T_{\infty } ) + \varepsilon A_{s} \sigma (T_{s}^{4} - T_{sur}^{4} )\) ; \({\text{Ra}} = \dfrac{{\beta g\left( {T_{s} - T_{\infty } } \right)D^{3} }}{{\nu^{2} }}{ \Pr }\) \(= \dfrac{{1/(75 + 273) \times 9.81 \times \left( {125 - 25} \right) \times 0.05^{3} }}{{(2.0658 \times 10^{ - 5} /1.0052)^{2} }} \times 0.697 = 5.8 \times 10^{5}\) ; From Eq. ( 9.27 ),

\({\text{Nu}}_{\text{m}} = 2 + \frac{{0.589{\text{Ra}}_{\text{d}}^{ 1 / 4} }}{{\left[ {1 + \left( {0.469/\Pr } \right)^{9/16} } \right]^{4/9} }}\) \(= 2 + \dfrac{{0.589 \times ( 5 . 8\times 1 0^{ 5} )^{ 1 / 4} }}{{\left[ {1 + \left( {0.469/0.697} \right)^{9/16} } \right]^{4/9} }} = 14.52\) ; \(h_{c} = {\text{Nu}}_{\text{m}} \dfrac{k}{D}\) \(= 14.52 \times \dfrac{0.0299}{{50 \times 10^{ - 3} }} = 8.68{\text{ W/(m}}^{ 2} {\text{ K)}};\) Hence, \(q = 8.68 \times \pi \times 0.05^{2} \times (125 - 25)\) \(+ 0.88 \times \pi \times 0.05^{2} \times 5.67 \times 10^{ - 8} \times (398^{4} - 300^{4} )\) \(= 13.48{\text{ W}};\) Some heat is lost by conduction to the base.]

11.23. Determine net radiative exchange between the plates of Problem 11.19 per unit area of the plates.

[Ans . Radiosity of surface 2 , J _{2} = E _{2} + ρ _{2} E _{b1} = σε _{2} T _{2} ^{4} + ρ _{2} σT _{1} ^{4} = σε _{2} T _{2} ^{4} +(1 – ε _{2} ) σT _{1} ^{4} = σ [ε _{2} T _{2} ^{4} +(1 – ε _{2} ) T _{1} ^{4} ] = 5.67 × 10^{−8} × [0.8 × 600^{4} + (1 – 0.8) × 1000^{4} ] = 17218 W/m^{2} ; Radiosity of surface 1, which is black, J _{1} = E _{b1} = σT _{1} ^{4} = 5.67 × 10^{−8} × 1000^{4} = 56700 W/m^{2,} q _{12} = J _{1} – J _{2} = 39482 W/m^{2} ; alternatively q _{12} = (E _{b1} − E _{b2} )/(1/ε _{1} − 1/ε _{2} − 1) = ε _{2} (E _{b1} − E _{b2} ) = 0.8 × 5.67 × 10^{−8} × (1000^{4} − 600^{4} ) = 39481.3 W/m^{2} ]

11.24. Determine the shape factor

F _{12} for the configuration shown in Fig.

11.88 .

[Ans. \(A_{(13)} F_{(1,3) - (2,4)} = A_{1} F_{12} + A_{1} F_{14} + A_{3} F_{32} + A_{3} F_{34}\) ; From symmetry, \(A_{1} F_{12} = A_{4} F_{43}\) ; From reciprocity relation,\(A_{4} F_{43} = A_{3} F_{34}\) ; Hence, \(A_{3} F_{34} = A_{1} F_{12}\) ; This gives \(A_{(13)} F_{(1,3) - (2,4)} = 2A_{1} F_{12} + A_{1} F_{14} + A_{3} F_{32}\) ; or \(F_{12} = \frac{1}{{2A_{1} }}\left[ {A_{(13)} F_{(1,3) - (2,4)} - A_{1} F_{14} - A_{3} F_{32} } \right]\) ; A _{13} = 9 m^{2} , A _{1} = 3 m^{2} , A _{3} = 6 m^{2} ; \(F_{12} = \frac{1}{2 \times 3}\left[ {9F_{(1,3) - (2,4)} - 3F_{14} - 6F_{32} } \right]\) ; From Fig. 11.4 c for Z /X = 1.0 and Y /X = 1, \(F_{(1,3) - (2,4)} = 0.2\) ; for Z /X = 3.0 and Y /X = 3.0, \(F_{14} = 0.125\) and for Z /X = 0.75 and Y /X = 0.25, \(F_{32} = 0.17\) ; substitution gives \(F_{12} = ({1 \mathord{\left/ {\vphantom {1 {6)}}} \right. \kern-0pt} {6)}}(9 \times 0.2 - 3 \times 0.125 - 6 \times 0.17) = 0.0675.]\)

11.25. Long, inclined black surfaces 1 and 2 in Fig.

11.89 are maintained at 1000 K and 600 K, respectively. Determine temperature of the black insulated surface 3.

[Ans . Since surface 3 is insulated, q _{3} = 0, i.e. q _{13} = q _{32} ; or \(A_{1} F_{13} \sigma (T_{1}^{4} - T_{3}^{4} ) = A_{3} F_{32} \sigma (T_{3}^{4} - T_{2}^{4} )\) ; By symmetry, \(A_{1} = A_{3}\) and \(F_{13} = F_{32}\) ; Hence \(T_{1}^{4} - T_{3}^{4} = T_{3}^{4} - T_{2}^{4} ,\) or \(T_{3} = \left[ {\frac{1}{2}(T_{1}^{4} + T_{2}^{4} )} \right]^{1/4} = \left[ {\frac{1}{2}(1000^{4} + 600^{4} )} \right]^{1/4} = 866.91\,{\text{K}}.\) ]

11.26. A 25 mm diameter pipe is laid horizontal in a room for heating. Condensing steam is flowing through the pipe. If the temperature of the outer surface of the pipe is 105°C and temperature in the room is 20°C, find the required length of the pipe for a heating rate of 1.0 kW. Pipe surface emissivity = 0.8.

[Ans. At t _{fm} = (105 + 20)/2 = 62.5°C, air properties are: ρ = 1.05 kg/m^{3} , μ = 2.0 × 10^{−5} , and k = 0.029 W/(m K), β = 1/T _{m} = 1/(62.5 + 273), Pr = 0.7; Gr \(= \frac{{g\beta \rho^{2} d^{3} \Delta T}}{{\mu^{2} }} = 1.07 \times 10^{5} ,\) \(h = 0.53(Gr_{f} \Pr_{f} )^{0.25} k/d{ = 10} 0 . 1 7 {\text{ W/(m}}^{ 2} {\text{ K)}}\) ; Convection heat transfer, q _{c} = hA∆T = 67.92 W/m, radiation heat transfer q _{r} = εAσ (T _{pipe} ^{4} – T _{room} ^{4} ) = 46.48 W/m. Required length L = q /(q _{c} + q _{r} ) = 8.74 m.]

References Howell JR (1982) A catalog of radiation configuration factors. McGraw-Hill Book Co, New York

Google Scholar Mikheyev M (1968) Fundamental of heat transfer. Mir Publishers, Moscow

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Authors and Affiliations 1. Jodhpur Institute of Engineering and Technology Jodhpur India