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

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:

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}}}              (13.30)

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}}}

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

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}}

or

q'_{1} = 37  kW/m = -q'_{2}

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

\frac{J_{1}  –  J_{R}}{(1/A_{1}F_{1R})}  –  \frac{J_{R}  –  J_{2}}{(1/A_{2}F_{2R})} = 0              (13.31)

q_{i} = \frac{E_{bi}  –  J_{i}}{(1  –  ε_{i})/ε_{i}A_{i}}              (13.19)

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}

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}

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

\frac{108,323  –  J_{R}}{\frac{1}{W × L × 0.5}}  –  \frac{J_{R}  –  59,043}{\frac{1}{W × L × 0.5}} = 0

Hence

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

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

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 ε_{R}.

3. This problem may also be solved using the direct approach. The solution involves first determining the three unknown radiosities J_{1}, J_{2}, and J_{R}. 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

\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}}              (13.21)

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

\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}}

\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}}

0 = \frac{J_{R}  –  J_{1}}{(A_{R}F_{R1})^{-1}} + \frac{J_{R}  –  J_{2}}{(A_{R}F_{R2})^{-1}}

Canceling the area A_{1}, the first equation reduces to

\frac{117,573  –  J_{1}}{0.25} = \frac{J_{1}  –  J_{2}}{2} + \frac{J_{1}  –  J_{R}}{2}

or

10J_{1}  –  J_{2}  –  J_{R} = 940,584                (1)

Similarly, for surface 2,

\frac{3544 –  J_{2}}{1.50} = \frac{J_{2}  –  J_{1}}{2} + \frac{J_{2}  –  J_{R}}{2}

or

-J_{1} + 3.33J_{2}  –  J_{R} = 4725                (2)

and for the reradiating surface

0 = \frac{J_{R}  –  J_{1}}{2} + \frac{J_{R}  –  J_{2}}{2}

or

-J_{1}  –  J_{2} + 2J_{R} = 0                 (3)

Solving Equations 1, 2, and 3 simultaneously yields

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

Recognizing that J_{R} = σT^{4}_{R}, it follows that

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