Document 2j9Yz0ky9RoZvKdxd0odm52J5

HEATING VENTILATING AIR CONDITIONING GUIDE 1943 Table 8. Solutions for Some Steady-State Thermal Conduction Problems*1* System Flat wall or curved wall if curvature is small (wall thickness less than 0.1 of inside dia meter). Expressions for the resistance R entering into the equation: ' q = At/R (Btu per hour) l kA Surface area, A Radial flow through a right circular cylinder. At n^i log -~ 2vkN (See footnote e). The buried cylinder. Long cylinder Trj of length. N iN3a.Radial flow in a hollow sphere. ilog-2^0- R " 2MV ! R for r > 3 . (See footnote c). 1 r 4vk cos.h-,1 -- 2ikN The straight fln or rod heated at one end. Conduction It cross-section area, A tjmbient Finned surface of area HB. R -- rhanP ta--nrh--ml (see footnotes d and e). . For ml > 2.3, tanh mini 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 =3 wall temperature--ambient temp. (* + *) A3 / 2 .\ ha I---- tanh m l + $ J HB -V5-V At defined as in Case 5 above. The dimensions to be employed in these solutions are: length of dimension P,l,r =* feet; units of k + Btu per hour per square foot per degree Fahrenheit for one foot thickness; units of ht 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). cLog x o 2.3C3 logu x. . dThis expression can also be employed as an approximation for tapered fins or of annnlar fins by employ- * ing average magnitudes of A and p. Tanh is the hyperbolic tangent. 82 CHAPTER 3. FUNDAMENTALS OF HEAT TRANSFER where the resultant parallel resistance Rt is.obtained from: J_ =_L+J_ Provided the individual resistances may be evaluated, the total resistance can be obtained from this relation. Then the heat transfer current for the length of pipe (N, ft) can be established by the relation: gre (Btu per hour) = (10) For a unit length of the pipe the heat transfer rate is: (Btu per hour foot) = (H) The temperature drop, At, through an individual resistance may then be calculated from the relation: A: -- R Qrc 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: = 1T . (12) where g is the heat transfer rate, and A t 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, ft, to be employed in the expressions of Table 8. The solution applicable to the problem depicted in Fig. 4, for the calculation of Ri and Ri, is case 2 in Table 8. Thus for a 1 ft length of 2 in. nominal size pipe (L D. = 2.067 in., O. D. = 2.375 in.) insulated with 1 in. of-cork: 1.188 Ri log. = s--.. 1.033 , = 8:5 X 10-4 hr degree Fahrenheit per Btu. Zx X m) X 1 2.188 R, = 2x log. 1.188 X 0.025 X 1 = 3.9 hr degree Fahrenheit per Btu. The convection resistances to heat transfer from the pipe wall to the cold water, Ri, 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, ft, Btu per hour per square foot per degree Fahrenheit, have been determined by test for mapy flow systems. These data may be employed to predict the conductances for similarflow systems. Table 5 summarizes some empirical equations expressing such test results.-- 83