Document emk4BjNVympajrDmGRjjYz8Ve

100 CHAPTER 5 1957 Guide Following the electrical analogy, when there is a thermal current flowing through several resistances in series, the resistances are additive: Rt = Ri + Rt + 72, * + Rb (7) Similarly, conductance is the reciprocal of resistance, and for heat flow through several resistances in parallel, the conductances are additive: Ct Rt 72, + 72, + 72, + " ' + 72. (8) Practical Heat Transfer Problems The use of these relations for resistance and conductance makes pos sible the solution of many practical heat transfer problems. As discussed in Chapters 9, 27 and 35, the practical analyses of heat transfer in building walls, in fin-tube coils and in pipe coverings, are usually computed by this method. The same resistance analysis may be applied to complicated Heat Transfer 101 No. meter) System ihePe^itbn!r "" resiston" onterin* into _______ *? ~ At/R (Btu per hour) "rv?d W1" d curvature is small than > of inside dia- p Lit *3 ----kj Surface area, A I Radial flow through a right circular cylinder. Long cylinderT^J of lengthy N | Radial flow in a hollow sphere. p __iI_g_en-o 2-rkN (See footnote c). log, ^ + e*h-> twkN - t-wkN For f ^ 3, satisfactory approximation is: loge - cosh~` R = ------- __________ L 2rkN 2wkN Fig. 8. Heat Transfer Conditions in an Insulated Cold Water Line steady-state conduction problems. Table 6 gives the resistances in six common cases of steady-state conduction. A complete analysis by the resistance method is well illustrated by considering the heat transfer from the air outside to the cold water inside of an insulated pipe. The temperature gradients and the nature of the resistance analysis are indicated by the two sketches of Fig. 8. Since air is sensibly transparent to radiation, there will be some heat transfer by both radiation and convection to the outer insulation surface. The mechanisms act in parallel on the air side. The total transfer by radiation and convection then passes through the insulating layer and the pipe wall by thermal conduction, and thence by convection and radiation into main cold water streams. (Radiation is not significant on (he water side as liquids are sensibly opaque to radiation, although water transmits energy in the visible region). The contact resistance between the insula tion and the pipe wall is presumed to be equal to zero. Referring to Fig. 8, the heat transferred for a given length N' of pipe, qn, Btu per hour, may be thought of as flowing through the parallel resistances R, and Rr, associated with the insulation surface radiation and ffl convection transfer. Then the flow is through the resistance offered to' thermal conduction by the insulation, Ri, through the pipe wall resistance, The straight fin or rod hosted at one end. lambieni Finned surface of area HB. n hep tanh mL; (see footnotes d and e). For mi > 2.5, tanh m L - 1 m =\/&aP/M A = conduction cross-section area. p ** perimeter of cross-section A. he =* unit conductance to the surroundings from the fin surface. k = thermal conductivity fin material. At = wall temperature--ambient temperature u = r*-M)1 \ tanh m l +* ) HB m * yk kA ~ yA /% jt& At defined aa in Case & above. * The dimensions to be employed in these solutions are: length of dimension p, L,r -- feet; unite of t 1=3 ftoot) (Fahrenhe(ist qdueagrreeefo);outn) i(tFsaohf raerneah,eiAt d~egrseqeufaorer ofeneet.foot thickness)', units of A, Btu per (hour) (square k,* LTohge* t*hearm2.a30l 3colongdiouxc.tivity, in these solutions should be taken at the average material temperature. This expression can also be employed as an approximation for tapered fins or of annular fins by cmP oymg average magnitudes of A and p. is the hyperbolic tangent.