Question 10.6: A cooled right cylindrical tank 4 m in diameter and 4 m long...
A cooled right cylindrical tank 4 m in diameter and 4 m long has a black interior surface and is filled with hot gas at a total pressure of 1 atm. The gas is composed of CO_2 mixed with a transparent gas that has a partial pressure of 0.75 atm. The gas is uniformly mixed at T_g= 1100 K. Compute how much energy must be removed from the tank surface to keep it cool if the tank walls are all at low temperature so that only radiation from the gas is significant.
The geometry is a finite circular cylinder of gas, and the radiation to its walls will be computed. Using Table 10.2,
| TABLE 10.2 | ||||
| Mean Beam Lengths for Radiation from Entire Medium Volume | ||||
| Geometry of Radiating System | Characterizing Dimension |
Mean Beam Length for Optical Thickness κ_λL_e → 0,L_e, |
Mean Beam Length Corrected for Finite Optical Thickness,^a L_e |
C = L_e/L_{e,0} |
| Hemisphere radiating to element at center of base | Radius R | R | R | 1 |
| Sphere radiating to its surface | Diameter D | \frac{2}{3} D | 0.65D | 0.97 |
| Circular cylinder of infinite height radiating to concave bounding surface Circular cylinder of semi-infinite height radiating to: |
Diameter D | D | 0.95D | 0.95 |
| Element at center of base | Diameter D | D | 0.90D | 0.9 |
| Entire base Circular cylinder of height equal to diameter radiating to: |
Diameter D | 0.81D | 0.65D | 0.8 |
| Element at center of base | Diameter D | 0.77D | 0.71D | 0.92 |
| Entire surface Circular cylinder of height equal to two diameters radiating to: |
Diameter D | \frac{2}{3} D | 0.60D | 0.9 |
| Plane end | Diameter D | 0.73D | 0.60D | 0.82 |
| Concave surface | Diameter D | 0.82D | 0.76D | 0.93 |
| Entire surface Circular cylinder of height equal to one-half the diameter radiating to: |
Diameter D | 0.80D | 0.73D | 0.91 |
| Plane end | Diameter D | 0.48D | 0.43D | 0.9 |
| Concave surface | Diameter D | 0.52D | 0.46D | 0.88 |
| Entire surface | Diameter D | 0.50D | 0.45D | 0.9 |
| Cylinder of infinite height and semicircular cross section radiating to element at center of plane rectangular face Infinite slab of medium radiating to: |
Radius R | 1.26R | 0.9 | |
| Element on one face | Slab thickness D | 2D | 1.8D | 0.9 |
| Both bounding planes | Slab thickness D | 2D | 1.8D | 0.9 |
| Cube radiating to a face Rectangular parallelepipeds 1 × 1 × 4 radiating to: |
Edge X | \frac{2}{3} X | 0.6X | 0.9 |
| 1 × 4 face | Shortest edge X | 0.90X | 0.82X | 0.91 |
| 1 × 1 face | Shortest edge X | 0.86X | 0.71X | 0.83 |
| All faces 1 × 2 × 6 radiating to: | Shortest edge X | 0.89X | 0.81X | 0.91 |
| 2 × 6 face | Shortest edge X | 1.18X | ||
| 1 × 6 face | Shortest edge X | 1.24X | ||
| 1 × 2 face | Shortest edge X | 1.18X | ||
| All faces | Shortest edge X | 1.20X | ||
| Medium between infinitely long parallel concentric cylinders |
Radius of outer cylinder R and of inner cylinder r |
2(R–r) | See Anderson and Handvig (1989) |
|
| Medium volume in the space between the outside of the tubes in an infinite tube bundle and radiating to a single tube: |
||||
| Equilateral triangular array: | Tube diameter | |||
| S = 2D | D, and spacing | 3.4(S–D) | 3.0(S–D) | 0.88 |
| S = 3D | between tube | 4.45(S–D) | 3.8(S–D) | 0.85 |
| Square array: | centers, S | |||
| S = 2D | 4.1(S–D) | 3.5(S–D) | 0.85 | |
| a Corrections are those suggested by Hottel (1954), Hottel and Sarofim (1967) or Eckert and Drake (1959). Corrections were chosen to provide maximum L_e where these references disagree. | ||||
the corrected mean beam length is L_e = 0.60D = 2.4 m. The partial pressure of the CO_2 is 0.25 atm, so that P_{CO_2}L_e = 0.25 × 2.4 = 0.6 atm · m. From Equation 9.62 and Table 9.9,
| Table 9.9 Coefficients cij for Equation 9.62 to Calculate Water Vapor and CO_2 Emittance |
|||||
| j | C_{0j} | C_{1j} | C_{2j} | C_{3j} | C_{4j} |
| Water Vapor, T > 400 K, M = 2, N = 2 | |||||
| 0 | −2.2118 | −1.1987 | 0.035596 | ||
| 1 | 0.85667 | 0.93048 | −0.14391 | ||
| 2 | −0.10838 | −0.17156 | 0.045915 | ||
| Carbon Dioxide, T > 400 K, M = 3, N = 4 | |||||
| 0 | −3.9781 | 2.7353 | −1.9822 | 0.31054 | 0.015719 |
| 1 | 1.9326 | −3.5932 | 3.7247 | −1.4535 | 0.20132 |
| 2 | −0.35366 | 0.61766 | −0.84207 | 0.39859 | −0.063356 |
| 3 | −0.080181 | 0.31466 | −0.19973 | 0.046532 | −0.0033086 |
remembering to convert P_{CO_2}L_e into units of bar-cm, \epsilon _{CO_2} (P_{CO_2}L_e ,T_g ) =0.170 , and C_{CO_2} from Equation 9.64
C_{CO_2}=1+(\Lambda _{CO_2}-1)\Xi _{CO_2} (9.64)
is 1.0, since the mixture total pressure is 1. From Equation 10.114,
G =\epsilon _g σ T^4_g (10.114)
the energy to be removed is
Q_i=GA=\epsilon_{CO_2}σ T^4_gA=0.170\times 5.6704\times 10^{-8} (1100)^424\pi =1064 kW