Document 70OONOn7kyrmpjrOv67pB02q8

56 No. 2. CHAPTER 5 1960 Guide Table 6.... Solutions for Some Steady-State Thermal Conduction Problems*- b Sysfwn Expression* for Mm Redstone* R Entorinj into the Equation: q = td/R (Btu per hour) Rat wall or curved wall if currafcu re is small (wall thickness lee* than 0.1 of inside diaa eter). at y SurfoCRWt&A v k *-l-- '-k Radial flow through a right circular cylinder. Lon* qrbbw oftengUvM r^j'T.j log. ^ R ~ 2rkN (See footnote e). 3. The buried cylinder. . (* IwjqWrifj ottengtt^N . 4. Radial flow in a hollow sphere. \ log.(_^s) " 2t4AT 2t4AT a For - g 3, satisfactory approximation is: , 2a log, -- R r - 2*ktf 'a COSO1 - T 2*kN 11 4rk 5. The straight fin or rt>d heated at one end. 0 l <1/ Conduction erass-MCtion e.A 1INI IN 6. Finned surface of area HB. Surface vtt, HB R = ,------- ^------r (see footnotes d and e). h.p tanh mL For ml > 23, tanh mL 1 m -- VKp/kA A " conduction cross-section area. p -- perimeter of cross-section A. h, ** unit conductance to the surroundings from the fin surface. k = thermal conductivity fin material. At -- wall temperature--ambient temperature. r i* + *) h, tanh ml ~f- HB vi-vi At defined as in Case 5 above. * TtwdimeMWMto be employed n length <rf dbaeaaioop, L, r feet; nnlta of 4 " Btu per (boor) (square foot) (Fahrenbeit defree far an* fool t*irhim); unite of A. Btu per (hour) (equera foot) (Fahrenheit tfc*ree); unite of are*, A - equere (set. b The thermal conductivity, i, in these solutions should be taken at the averse* material temperature. * Lo|( s * tAB Iocm s. . * This rrTuuMiuii can also bo employed as an epprorimstina for tapered fin* or of annular fin* by employing average magnitudes of A and p. * tanb b tbs hyperbolic tangent. satis D /l , Heat Transfer 57 convection between the water and the pipe wall. The equa tion for this case is the following: h. - 13.9(1,)- "^ 07) where Um * 5 fps 13.9 X 6.9 X 3.62 494 Btu per (hr) (sq ft) (F deg). 0.703 This heat transfer rate is through the inner surface of the pipe and it is, therefore, this area that determines the re sistance fti. A = rD -- 0.542 sq ft per unit length of pipe, and therefore 11 ` KA ` 494 X 0342 fit = 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: (18) RADIATION BETWEEN ADJACENT RCCTANCt.es IN PERFCKPICUIAR FtANE* 1. RATIO G.CNCTM OR unique sioe of --that RECTANGLE ON WHOSE AREA THE MEAT TRAMFQt EQUATION IS RASED) (LENGTH OF COMMON ioO-Vxin saEtcm 2-RATIO QXNSTM OF UNIQUE SIDE OF OTHER RECTANCLQ + (JiNGTM CT COMMON sto$-zA in shetcm Pig. 5 . Geometrical Factor F for Direct Radiation Be tween Adjacent Rectangles in Perpendicular Planes* 20 F * 0.364 ft > P, = one atmosphere therefore. / 20 y he = 0.271 \0364/ - 0.737 Btu per (hr) (sq ft) (F deg) Using the surface area of the insulation, the value of the resistance per unit length is determined. 1.14 sq ft Fig. 6.... Geometrical Factor F for Direct Radiation Between Opposed Parallel Rectangle and Discs of Equal Size* IL C. Hottel, Radiant bst transmission, UfeefenKol gagiaama. July 1830. pp. 700 to 701). h*A 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.- (tip)*- - 0.211(77)- (19)