Document ykDavR3xvXeOE0pw0n5Z9QxGE

110 CHAPTERS :1949.Guide. six different conduction systems. Table 2 in Chapter 6 and Table 1 ofthis chapter indicate the magnitudes of the thermal conductivities, k, to be employed in the expressions of Table 5, after dividing k by 12. The solution applicable to the problem depicted in Fig. 4,- for the cal' culation of Rt and Rz, is case 2 in Table 5. Thus for a 1 fl length of 2 in. nominal size pipe (I. D. = 2.067 in., O. D. = 2.375 in.) insulated with 1 in. of material having a conductivity of 0.025: , 1.188 l0g*033 . Rt *= 2-*-----X----2-8---X--- 1- = 8.6 X 10_< hr Fahrenheit degree per Btu. ; , 2.188 los* fill R, = 2--x---X---0.02~ 5 X 1 = 3.9 hr Fahrenheit degree 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, Fig. 4. Heat Transfer Conditions in an Insulated Cold Water Line Re, are dependent on the flow conditions prevailing at these surfaces, and on the thermal properties of the fluids. The unit conductances for thermal convection, he, Btu per (hour) (square foot) (Fahrenheit degree), have been determined by test for many flow systems. These data may be employed to predict the conductances for simitar flow systems. Table 2 summarizes some empirical equations expressing such test results. For the problem under consideration -(Fig. 4) case'3 of Table 2 is ap plicable for the calculation of the cold water side convection resistance RiCorresponding to the water velocity of 5 fps, the mass velocity is : G = 5 (ft per sec) X 62.4 (lb percuft) X 3600 (sec per hr) = 11.2 X 105 lb per (hour) (square foot). -The inside.diameter of the pipe D is 2.067/12 = 0.1725 ft. The average water, film temperature will be estimated as 36 F (mixed mean fluid temperature of 34 F). Then case 3, Table 2 yields: /II O y TArtO. A. = 0.00480(1 + 0.36) --= 660 Btu per (hr) (sq ft) (F deg). The transfer area on which this conductance is based is the inside tube Fundamentals of Heat Transfer - ; ; area. Associated with 1 ft length of pipe there" are : 2.067 V x -- X 1 => 0.542 sq ft. Thus the resistance for 1 ft of tube length is: Rl = herDX 1 = 650 X* 0.542 = 2"8 X 10~, tr Fahrenhei` dereB Btu. Case 9, Table 2 is applicable for calculating the free thermal convection resistance, Re, existing between the surrounding air and the insulation. The air temperature is given as 120 F. As an approximation a 20 deg temperature difference between the air and the pipe surface will be as sumed. D = 4.375/12 = 0.364 ft. Then case 9 yields: he 0.63 Btu per (hour) (square foot) (Fahrenheit degree). (13) This result may not be deemed conservative inasmuch as the expression is for still air. If, however, the air is not still, but flows at approximately 5 mph or 7 fps the mass velocity corresponds to: O = 7 X 0.07 X 3600 = 1770 lb air per (hour) (square foot). A magnitude of k = 0.014 Btu per (hour) (square foot) (Fahrenheit degree per one foot thickness) applied to case 4 yields: .: he = 0.45 + 0.178(1770 )-(/0--.014V) = 0.017 + 2.8 -- 2.8 Btu per (hour) (square foot) (Fahrenheit degree). This conductance is based on 1 sq ft of outside lagging area. . Thus, since there are ir X (4.375/12) = 1.14 sq ft of outside lagging area associated with 1 ft length of pipe: Re =------------ = 0.312 hr Fahrenheit degree per Btu. 2.8X1.14 The radiation resistance, R,, which acts in parallel with the convection resistance, Re, for the transfer of heat to the surface of the insulation, may be calculated. For the purposes of this illustrative problem.lt will be assumed that the insulated pipe is exposed to (sees) surroundings, which exist at 120 F. Then the angle factor, Fa, is unity and for an estimated surface emissivity of 0.9 (see Table 3), F = 0.9. As a first approximation the insulation surface temperature will be estimated , as 20 deg below the surroundings at 120 F. Then the radiation per degree of temperature difference, by Equation 3 (or more conveniently by Table 5) divided by the temperature difference will be: At = = 1.17 Btu per (hour) (square foot) (Fahrenheit degree). 20 > The outside surface area of the insulation associated with-1; ft of pipe length was previously calculated as 1.14 sq ft: Thus : ' ft, = ^ ^ ^ ^ = 0.75 hr Fahrenheit degree per Btu.,