Document NGQzaZOe1ED5vXrMbzxewxk8E

106 CHAPTER 5 1951 Guide'; Following the electrical analogy, when there is a thermal current flowing through several resistances in series, the resistances are additive: . Rr = Ri + Ri + R. 4- + R pj Similarly, conductance is the reciprocal of resistance, and for heat flow through several resistances in parallel, the conductances are additive: CT Ri + Rs Rr Practical Heat Transfer Problems The use of these relations for. resistance and conductance makes po&. 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 107 m 6 Solutions fob Some Steady-State Thebmal Conduction Problems*.Table Fro. 8. Heat Tbansfeb Conditions in an Insulated Cold Wateb 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 the water i 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, qm Btu per hour, may be thought of as flowing through the parallel resistances Rr and Re, associated with the insulation surface radiation and .* convection transfer. Then the flow is through the resistance offered to .thermal conduction by the insulation, Rt, through the pipe wall resistance, \ * The dimensions to be employed in these solutions are: length of dimension p, L,t fw:tt;; tunits of k -- Btu per (hour) (square foot) (Fa...h..r.e..n...h..e..i.t...d..e..g..r.e..e....f.o..r...o.n..e...f..o..o..t...t.t..ie...t.i.M....i,i),;,u__n_it_s__o_f_i_( _B_t_u^p_err((hboour) (square foot) (--F--a-h--r-e--n--h--e-i-tddeeggrreeee));;uunniittssooffaarreeaa,.AA ==ssqauuaarreeffeeeett.. b The thermal conductivity, t, in these solutions' should be taken at the average material temperature. 0 Log.* == 2.303 logu *. d" TihhiSsexpression can nloo employed as an approximation for tapered fins or of annular fins by em ployyiinnggaavvera--g-e-'-ji-n--a.g--n-i-t-u-des of A and p. * tanh is tHJivnwIinlii lan*nt