A paint baking oven consists of a long, triangular duct in which a heated surface is maintained at 1200 K and another surface is insulated. Painted panels, which are maintained at 500 K, occupy the third surface. The triangle is of width W = 1 m on a side, and the heated and insulated surfaces have

Question

A paint baking oven consists of a long, triangular duct in which a heated surface is maintained at 1200 K and another surface is insulated. Painted panels, which are maintained at 500 K, occupy the third surface. The triangle is of width W = 1 m on a side, and the heated and insulated surfaces have an emissivity of 0.8. The emissivity of the panels is 0.4. During steady-state operation, at what rate must energy be supplied to the heated side per unit length of the duct to maintain its temperature at 1200 K? What is the temperature of the insulated surface?

Accepted Answer

Known: Surface properties of a long triangular duct that is insulated on one side and heated and cooled on the other sides.

Find:
1. Rate at which heat must be supplied per unit length of duct.
2. Temperature of the insulated surface.

Schematic:

Assumptions:
1. Steady-state conditions exist.
2. All surfaces are opaque, diffuse, gray, and of uniform radiosity and irradiation.
3. Convection effects are negligible.
4. Surface R is reradiating.
5. End effects are negligible.

Analysis:
1. The system may be modeled as a three-surface enclosure with one surface reradiating. The rate at which energy must be supplied to the heated surface may then be obtained from Equation 13.30:

[latex]q_{1} = -q_{2} = \frac{E_{b1} - E_{b2}}{\frac{1 - ε_{1}}{ε_{1}A_{1}} + \frac{1}{A_{1}F_{12} + [(1/A_{1}F_{1R}) + (1/A_{2}F_{2R})]^{-1}} + \frac{1 - ε_{2}}{ε_{2}A_{2}}}[/latex] (13.30)

[latex]q_{1} = \frac{E_{b1} - E_{b2}}{\frac{1 - ε_{1}}{ε_{1}A_{1}} + \frac{1}{A_{1}F_{12} + [(1/A_{1}F_{1R}) + (1/A_{2}F_{2R})]^{-1}} + \frac{1 - ε_{2}}{ε_{2}A_{2}}}[/latex]

From symmetry, [latex]F_{12} = F_{1R} = F_{2R} = 0.5[/latex]. Also, [latex]A_{1} = A_{2} = W · L[/latex], where L is the duct length. Hence

[latex]q'_{1} = \frac{q_{1}}{L} = \frac{5.67 × 10^{-8} W/m^{2} · K^{4} (1200^{4} - 500^{4}) K^{4}}{\frac{1 - 0.8}{0.8 × 1 m} + \frac{1}{1 m × 0.5 + (2 + 2)^{-1} m} + \frac{1 - 0.4}{0.4 × 1 m}}[/latex]

or

[latex]q'_{1} = 37 kW/m = -q'_{2}[/latex]

2. The temperature of the insulated surface may be obtained from the requirement that [latex]J_{R} = E_{bR}[/latex], where [latex]J_{R}[/latex] may be obtained from Equation 13.31. However, to use this expression [latex]J_{1}[/latex] and [latex]J_{2}[/latex] must be known. Applying the surface energy balance, Equation 13.19, to surfaces 1 and 2, it follows that

[latex]\frac{J_{1} - J_{R}}{(1/A_{1}F_{1R})} - \frac{J_{R} - J_{2}}{(1/A_{2}F_{2R})} = 0[/latex] (13.31)

[latex]q_{i} = \frac{E_{bi} - J_{i}}{(1 - ε_{i})/ε_{i}A_{i}}[/latex] (13.19)

[latex]J_{1} = E_{b1} - \frac{1 - ε_{1}}{ε_{1}W} q'_{1} = 5.67 × 10^{-8} W/m^{2} · K^{4} (1200 K)^{4} - \frac{1 - 0.8}{0.8 × 1 m} × 37,000 W/m = 108,323 W/m^{2}[/latex]

[latex]J_{2} = E_{b2} - \frac{1 - ε_{2}}{ε_{2}W} q'_{2} = 5.67 × 10^{-8} W/m^{2} · K^{4} (500 K)^{4} - \frac{1 - 0.4}{0.4 × 1 m} (-37,000 W/m) = 59,043 W/m^{2}[/latex]

From the energy balance for the reradiating surface, Equation 13.31, it follows that

[latex]\frac{108,323 - J_{R}}{\frac{1}{W × L × 0.5}} - \frac{J_{R} - 59,043}{\frac{1}{W × L × 0.5}} = 0[/latex]

Hence

[latex]J_{R} = 83,683 W/m^{2} = E_{bR} = σT_{R}^{4}[/latex]

[latex]T_{R} = \left(\frac{83,683 W/m^{2}}{5.67 × 10^{-8} W/m^{2} · K^{4}}\right)^{1/4} = 1102 K[/latex]

Comments:
1. We would expect the temperature of the reradiating surface to be higher in regions adjacent to surface 1 and lower in regions closer to surface 2. Our intuition corresponds to the fact that the surface irradiation and radiosity distributions are not uniform, calling into question the validity of Assumption 2. The temperature distribution of the reradiating surface could be determined by use of an analytical or numerical approach, as described in Comment 3 of Example 13.3. If each geometric surface were to be subdivided into 10 smaller elements, however, we would need (3 × 10)² = 900 view factors. Precise prediction of radiation heat transfer rates in enclosures whose geometric surfaces are not characterized by uniform radiosity or irradiation distributions involves a trade-off between accuracy and computational effort.

2. The results are independent of the value of [latex]ε_{R}[/latex].

3. This problem may also be solved using the direct approach. The solution involves first determining the three unknown radiosities [latex]J_{1}, J_{2}[/latex], and [latex]J_{R}[/latex]. The governing equations are obtained by writing Equation 13.21 for the two surfaces of known temperature, 1 and 2, and Equation 13.22 for surface R. The three equations are

[latex]\frac{E_{bi} - J_{i}}{(1 - ε_{i})/ε_{i}A_{i}} = \sum\limits_{j=1}^{N} \frac{J_{i} - J_{j}}{(A_{i}F_{ij})^{-1}}[/latex] (13.21)

[latex]q_{i} = \sum\limits_{j=1}^{N} \frac{J_{i} - J_{j}}{(A_{i}F_{ij})^{-1}}[/latex] (13.22)

[latex]\frac{E_{b1} - J_{1}}{(1 - ε_{1})/ε_{1}A_{1}} = \frac{J_{1} - J_{2}}{(A_{1}F_{12})^{-1}} + \frac{J_{1} - J_{R}}{(A_{1}F_{1R})^{-1}}[/latex]

[latex]\frac{E_{b2} - J_{2}}{(1 - ε_{2})/ε_{2}A_{2}} = \frac{J_{2} - J_{1}}{(A_{2}F_{21})^{-1}} + \frac{J_{2} - J_{R}}{(A_{2}F_{2R})^{-1}}[/latex]

[latex]0 = \frac{J_{R} - J_{1}}{(A_{R}F_{R1})^{-1}} + \frac{J_{R} - J_{2}}{(A_{R}F_{R2})^{-1}}[/latex]

Canceling the area [latex]A_{1}[/latex], the first equation reduces to

[latex]\frac{117,573 - J_{1}}{0.25} = \frac{J_{1} - J_{2}}{2} + \frac{J_{1} - J_{R}}{2}[/latex]

or

[latex]10J_{1} - J_{2} - J_{R} = 940,584[/latex] (1)

Similarly, for surface 2,

[latex]\frac{3544 - J_{2}}{1.50} = \frac{J_{2} - J_{1}}{2} + \frac{J_{2} - J_{R}}{2}[/latex]

or

[latex]-J_{1} + 3.33J_{2} - J_{R} = 4725[/latex] (2)

and for the reradiating surface

[latex]0 = \frac{J_{R} - J_{1}}{2} + \frac{J_{R} - J_{2}}{2}[/latex]

or

[latex]-J_{1} - J_{2} + 2J_{R} = 0 [/latex] (3)

Solving Equations 1, 2, and 3 simultaneously yields

[latex]J_{1} = 108,328 W/m^{2}[/latex] [latex]J_{2} = 59,018 W/m^{2}[/latex] and [latex]J_{R} = 83,673 W/m^{2}[/latex]

Recognizing that [latex]J_{R} = σT^{4}_{R}[/latex], it follows that

[latex]T_{R} = \left(\frac{J_{R}}{σ}\right)^{1/4} = \left(\frac{83,673 W/m^{2}}{5.67 × 10^{-8} W/m^{2} · K^{4}}\right)^{1/4} = 1102 K[/latex]

Question 13.6: A paint baking oven consists of a long, triangular duct in w...

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