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100 CHAPTER 5 1952 Guide Following the-electrical analogy, when there is a thermal current-flowing through several resistances in series, the resistances are additive: fiT = Ri + R + + -l" tt, (7) Similarly, conductance is the reciprocal of resistance, and for heat-flow through several resistances in parallel, the conductances are additive: 11 1 R--,--1-R---i-- 1-R-z---h (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 Table 6. Solutions.for Some Steady-State'Thermal Conduction. Problems8- b No. Systsm Expressions for the resbtaace.lt entering Into the conation: ! 61/R (Btu per hour) l. Flat wall or curved wall if curvature is small (wall thickness less than 0.1 of inside dia meter). o . Jl - 7 kA Surfsce are^A Radial flow through a right circular cylinder, [at ' . UngcyGndtr oMertei^N r.JWl ur94 R - 2wkN (See footnote *), R~ / txtN ~ tSkS Fo* y 3, a satisfactory approximation b: tog.-- co*b-> --' 2TkH " Fig. 8. Heat Transfer Conditions in an Insulated Cold Water Line steady-state' conductionproblems.. 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 the 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, <?rc, Btu per hour, may be thought of as flowing through the parallel _ resistances Rr and Rc, associated with the insulation surface radiation and convection transfer. Then the flow is through the resistance offered to thermal conduction by the insulation, R3, through the pipe wall resistance, It - r# 4vft The straight fin or rod heated at one end. i * Conduction cross-section 'Area. A Finned surface of area BB. * " A.rtanhmL footnote, d nd .). For ml > 2.3, tanh 1 m -- -y/kiP/kA A -- conduction crosa-occtlon area. P -- perimeter of cross section A. As " unit conductance to the surroundings from the fin surface. k -- thermal conductivity fin material. 6S -- wall temperature--ambleattemperature (* 4 O R m A. (-J-taohM-3)bB yn - _ J?-*! U 61 defined as In Case $ above. Btu di^ensiona to be employed in these solutions are: length of dimension p, L, t = feet: units of k fom)nLffi!uS Wuare foot) (Fahrenheit degree for one foot thickness); units of h, Btu per (hour) (square ^ iranrenheit degree); units of area. A = square feet. e thermal conductivity, k, in these solutions should be taken at the average material temperature. ^ Ioge * = 2.303 logic s. Dlovi^3 eiPression can also be employed as an approximation for tapered fins or of annular fins by emaverage magnitudes of A and p. * tauh is the hyperbolic tangent.