Document Qp91J3Q6v4jondEMD7nByvok

76 Chapter-3 1945 Guide -Table-8.--Solutions-for-Some Steady-StateThermal Conduction Problems**1 Expressions for the resistance R entering into' No. System the equation: g = &1/R (Btu per hour) 1. Flat wall or curved wall if curvature ts small (wall thickness less than 0.1 of inside dia meter). R L kA Surface area. A 3. The buried cylinder. s'At.T'- - -I- ~ k t:tp-ts a Long cylinderjTrj of length, N 4. Radial flow in a hollow sphere. R " 2-rkN (See footnote c). 2a log, -;.R for ~r > 3 (See footnote c). cosh-1 --r 2rkN ' ' _1_ 1 R n r0 4*k 5. ' The straight fin or rod heated at one end. Conduction cross-section area. A * " J-ptanhmL (sM footnotes d and e)' For ml > 2.3, tanh m L 1 m = -\/hp/kA tambfeni A = conduction cross-section area. ` P perimeter of cross-section A. h* = unit conductance to the surroundings from the fln surface. k =* thermal conductivity fin material. A/ = wall temperature--ambient temperature 6. Finned surface of area HB. aThe dimensions to be employed in these solutions are: length of dimension p, L, r feet; units of k =* Btu per hou> peT square foot per degree Fahrenheit ior one foot thickness; units of h, Btu per hour per square foot per degree Fahrenheit; units of area, A = square feet. bThe thermal conductivity, k, in these solutions should be taken at the average material temperature (see Table 5). Log* x ** 2.303 logu x. dThis expression can also be employed as an approximation for tapered fins or of annular fins by employ ing average magnitudes of A and p. Tanh is the hyperbolic tangent. Fundamentals of Heat Transfer 77- can be obtained from this relation. Then the heat transfer current for the length of pipe (iV, ft) can be established by-the relation:-------------------- ^_ Sic (Btu per hour) = U (10) For a unit length of the pipe the heat transfer rate is: . (Btu per hour foot) = (n) The temperature drop, At, through an individual resistance may then be calculated from the relation: At = i? Src where R is the resistance in question. The problem- is now reduced to one of evaluating the individual resist ances of the system. This entails suitable integration of the-rate Equa- , tions 1, 2 and 3 to produce expressions of the form: where g is the heat transfer rate, and At is the potential drop or tempera ture difference through the resistance R. Table 8 lists such solutions for six different conduction systems. Table 2 in Chapter 4 and Table 1 of this chapter indicate the magnitudes of the thermal conductivities, k, to be employed in the expressions of Table 8. The solution applicable to the problem depicted in Fig. 4, for the calculation of and R3, is case 2 in Table 8. Thus for a 1 ft length of 2 in. nominal size pipe (I. D. = 2.067 in., O. D. = 2.375 in.) insulated with 1 in. of cork: 1.188 R, log; Fo33 2x X 26 X 1 8.5 X 10"* hr degree Fahrenheit per Btu. R, 3.9 hr degree Fahrenheit per Btu. The convection resistances to heat transfer from the pipe wall to the cold water, Rlt and from the air to the surface of the insulating material, Rc, are dependent on the flow conditions prevailing at these surfaces, and on the thermal properties of the fluids. The unit conductances- for thermal convection, h, Btu per hour per square foot per degree F'ahrenheit, have been determined by test for many flow systems. These data may be employed to predict the conductances for similar flow systems. Table 5 summarizes some empirical equations expressing such test results. For the problem under consideration (Fig. 4) case 3 of Table 5 is applicable for the calculation of the cold water side convection resistance R-i- Corresponding to the water velocity of 5 fps, the mass velocity is: *7 = 5 (ft per sec) X 62.4 (lb per cu ft) X 3600 (sec per hr) = 11.2 X 105 lb per hour per square foot. 1 he inside diameter of the pipe D is 12 U.l, 20 it.