Document XOOX9QoK8mMMM391MgJRDrN0d

54 CHAPTER 5 1959 Guide Table 6 .... Solutions for Some Steady-State Thermal Conduction Problems*- SytJem Expratttotu for Ifte fig&tonc* It Entering into the Equation; q - Al/R (Btu per hour) Flat wall or curved wall if curvature is small (wall thick ness less than 0.1 of inside diameter). XJ Radial flow through a right circular cylinder. The buried cylinder. Radial flow in a hollow sphere. log, R ' 2rkN (See footnote c). log. cosh"1 <^/1 + 2xiN 2vkN For - ^ 3, satisfactory approximation is: r 2a a log. -- cosh 1 - 2rkN 2r*JV ^ 1^ Ti r. The straight fin or rod heated at one end. Finned surface of area HB. ty Surface area. HB - (see footnotes d and e). h.p tanh mL For ml > 2.3, tanh mL ns 1 m -- y/Kp/kA A = conduction cross-section area, p * perimeter of cross-section A. h, = unit conductance to the surroundings from the f surface. i = thermal conductivity fin material. At -- wall temperature--ambient temperature- ( + 4) tanh ml + * ) HB m - yJhtPA - Ay/3khsl At defined as in Case 5 above. * The dimension* to be employed Lb thenr nlii*i--11 are: length of dimension p, L, t -- feet; units ofi a Btu per (hour) (square foot) (Fahienhcat decree for nu foot thick****); unita of h, Btu per (boor) (square foot) (Fahrenheit decree); unita ot area, A -- aquare feet. b The thermal conductivity, i. is these solutions ahould be Ultra at the awac* niaSerial temperature. s Lac. * " 1.303 loci* z. S This expression can Im be employed aa an approximation for tapered fine or of fina by employing average magnitudes of A and j>. * tanh b the hyperbolic tangent. Heat Transfer 55 convection between the water and the pipe wall. The equa tion for this case is the following: 4,-13*)'"^ (17) where u. -- 5 fps 2.067 D 12 34+36 35 F 2 h. - 13,9 X^7^X 3~ - 494 Btu per (hr) (sq ft) (F deg). This heat transfer rate is through the inner surface of the pipe and it is, therefore, this area that determines the re sistance R, . A = *D -- 0.542 sq ft per unit length of pipe, and therefore A.A 494 X 0.542 Ri = 3.73 X 10"* (hr) (F deg) per Btu. Case 11 of Table 2 fits the conditions of the problem if only free convection heating of the pipe is assumed. The equation in this case is as follows: Fig. 5 .... Geometrical Factor F for Direct Radiation Be tween Adjacent Rectangles in Perpendicular Planes* where At = 20F D - 0.364 ft P = P, = one atmosphere therefore, / 20 Y ht = 0.271 I ---- ) \0364/ he = 0.737 Btu per (hr) (sq ft) (F deg) Using the surface area of (he insulation, the value of the resistance per unit length is determined. A=v x 1 " 114 8(1 ft Fig. 6 .... Geometrical Factor F for Direct Radiation Between Opposed Parallel Rectangle and Discs of Equal Size* H. C. Hottel, Radiant beat tranamisnoo, (Mechanical Bnfineerin#, July 1930. pp. 700 to 709). * M 0.737 X 1.14 R, = 1.19 (hr) (F deg) per Btu. This result may not be deemed conservative inasmuch as the expression is for still air. If, however, the air is not still, but flows at approximately 5 mph or 7 fps, the heat transfer equation for forced convection would apply. This equation is Case 5 of Table 2. ft*TM.) - 0.21 (19)