Document 8VJ1NOMkrVwpb1OVmYx8O6M1o

100 CHAPTER 5 Similarly, conductance is the reciprocal of resistance, and for heat flow ^ through several resistances in parallel, the conductances are additive: S Ct ~ Rr ~ R, + R,+ R3 + ' ' ' + (8)' Practical Heat Transfer Problems ||J The use of these relations for resistance and conductance makes pos-j sible the solution of many practical heat transfer problems. As discussed ' in Chapters 9, 28 and 36, the practical analyses of heat transfer in building J|I walls, in fin-tube coils and in pipe coverings, are usually computed by this&J method. The same resistance analysis may be applied to complicated S' Heat Transfer System 101 T ihermal Conduction ***< S.ePeSS-r the '* W. into g = At/R (Btu per hour) Surface area, A Radial fl0w trough a right circular cyUndeii r At R --i2o-r-gk-N.5? (See footnote c). The buried cylinder. r Ik a U At: tp-1 -1 /jY'V tong cylinder <rf. lengthy N ^dial flow in a hollow sphere log. ^ + yWr^ coeh-i+ r?) S = MN " Mclf For ' & 3, satisfactory approximation is: , 2o , , o loge -- cosh 1 - ft =--------- - T 2rkN 2rkN 1 r0 Fig. 8. Heat Transfeb Conditions in an Insulated Cold Water Line The straight fin or rod heated at one end j steady-state conduction problems. Table 6 gives the resistances in si common cases of steady-state conduction. I A complete analysis by the resistance method is well illustrated b; considering the heat transfer from the air outside to the cold water insidi of an insulated pipe. The temperature gradients and the nature of thi resistance analysis are indicated by the two sketches of Fig. 8. 0. k Conduction cross-section (ambient " hap tanh mL* v(see footnotes d and a). For ml > 2.3, tanh m L 1 m --*y/hap/kA A -- conduction cross-section area . p -- perimeter of cross-section A. ha = unit conductance to the surroundings from the fin surface. k =* thermal conductivity fin material. At ~ wall temperature--ambient temperature Since air is sensibly transparent to radiation, there will be some heal transfer by both radiation and convection to the outer insulation surface. Finned surface of area HB. The mechanisms act in parallel on the air side. The total transfer byradiation and convection then passes through the insulating layer and thr* pipe wall by thermal conduction, and thence by convection and radiatioi = / 2 (s + g) \ hi (tanh wl + ij HB into main cold water streams. (Radiation is not significant on the water- J side as liquids are sensibly opaque to radiation, although water transmil energy in the visible region). The contact resistance between the insula 1tion and the pipe wall is presumed to be equal to zero. A AeP A /2 ha m ~ y kA y At defined as in Case 5 above. Referring to Fig. 8, the heat transferred for a given length N of pipe?- 5rc, Btu per hour, may be thought of as flowing through the parallel resistances Rr and Rc, associated with the insulation surface radiation an<h.| convection transfer. Then the flow is through the resistance offered t<' thermal conduction by the insulation, Rz, through the pipe wall resistance! Btu 3^*6 dmiensions to be employed in these solutions are: length of dimension p. L,r = feet; units of t = (hour) (square foot) (Fahrenheit degree for one foot thickness); units of h, Btu per (hour) (square j' Fahrenheit degree); units of area, A -- square feet. t ^he thermal conductivity, fc, in these solutions should be taken at the average material temperature. d Loge z * 2.303 logwi. nlov;?^*8 can also be employed as an approximation for tapered fins or of annular fins by em- ;,n* average magnitudes of A and p. knh is the hyperbolic tangent.