Question 13.5: A cryogenic fluid flows through a long tube of 20-mm diamete...
A cryogenic fluid flows through a long tube of 20-mm diameter, the outer surface of which is diffuse and gray with ε_{1} = 0.02 and T_{1} = 77 K. This tube is concentric with a larger tube of 50-mm diameter, the inner surface of which is diffuse and gray with ε_{2} = 0.05 and T_{2} = 300 K. The space between the surfaces is evacuated. Calculate the heat gain by the cryogenic fluid per unit length of tubes. If a thin radiation shield of 35-mm diameter and ε_{3} = 0.02 (both sides) is inserted midway between the inner and outer surfaces, calculate the change (percentage) in heat gain per unit length of the tubes.
Known: Concentric tube arrangement with diffuse, gray surfaces of different emissivities and temperatures.
Find:
1. Heat gain by the cryogenic fluid passing through the inner tube.
2. Percentage change in heat gain with radiation shield inserted midway between inner and outer tubes.
Schematic:
Assumptions:
1. Surfaces are diffuse and gray and characterized by uniform irradiation and radiosity.
2. Space between tubes is evacuated.
3. Conduction resistance for radiation shield is negligible.
4. Concentric tubes form a two-surface enclosure (end effects are negligible).
Analysis:
1. The network representation of the system without the shield is shown in Figure 13.11, and the heat rate may be obtained from Equation 13.25, where
q_{12} = \frac{σA_{1}(T_{1}^{4} – T_{2}^{4})}{\frac{1}{ε_{1}} + \frac{1 – ε_{2}}{ε_{2}}\left(\frac{r_{1}}{r_{2}}\right)} (13.25)
q = \frac{σ(\pi D_{1}L)(T_{1}^{4} – T_{2}^{4})}{\frac{1}{ε_{1}} + \frac{1 – ε_{2}}{ε_{2}}\left(\frac{D_{1}}{D_{2}}\right)}
Hence
q' = \frac{q}{L} = \frac{5.67 × 10^{-8} W/m^{2} · K^{4} (\pi × 0.02 m)[(77 K)^{4} – (300 K)^{4}]}{\frac{1}{0.02} + \frac{1 – 0.05}{0.05}\left(\frac{0.02 m}{0.05 m}\right)}\\ q' = -0.50 W/m
2. The network representation of the system with the shield is shown in Figure 13.12, and the heat rate is now
q = \frac{E_{b1} – E_{b2}}{R_{tot}} = \frac{σ(T_{1}^{4} – T_{2}^{4})}{R_{tot}}
where
R_{tot} = \frac{1 – ε_{1}}{ε_{1}(\pi D_{1}L)} + \frac{1}{(\pi D_{1}L)F_{13}} + 2\left[\frac{1 – ε_{3}}{ε_{3}(\pi D_{3}L)}\right] + \frac{1}{(\pi D_{3}L)F_{32}} + \frac{1 – ε_{2}}{ε_{2}(\pi D_{2}L)}
or
R_{tot} = \frac{1}{L}\left\{\frac{1 – 0.02}{0.02(\pi × 0.02 m)} + \frac{1}{(\pi × 0.02 m)1} + 2\left[\frac{1 – 0.02}{0.02(\pi × 0.035 m)}\right] + \frac{1}{(\pi × 0.035 m)1} + \frac{1 – 0.05}{0.05(\pi × 0.05 m)}\right\}\\ R_{tot} = \frac{1}{L}(779.9 + 15.9 + 891.3 + 9.1 + 121.0) = \frac{1817}{L}\left(\frac{1}{m^{2}}\right)
Hence
q' = \frac{q}{L} = \frac{5.67 × 10^{-8} W/m^{2} · K^{4} [(77 K)^{4} – (300 K)^{4}]}{1817 (1/m)} = -0.25 W/m
The percentage change in the heat gain is then
\frac{q'_{w} – q'_{wo}}{q'_{wo}} × 100 = \frac{(-0.25 W/m) – (-0.50 W/m)}{-0.50 W/m} × 100 = -50\%
Comment: Because the geometries are concentric and the specified emissivities and prescribed surface temperatures are spatially uniform, each surface is characterized by uniform irradiation and radiosity distributions. Hence the calculated heat transfer rates would not change if the cylindrical surfaces were to be subdivided into smaller radiative surfaces.
