Document V31K61zbmvyKDoOa3N4KNK2Xp

660 CHAPTER 28 1956 Guide r, <= outer radius of insulation, inches. h = thermal conductivity of insulation, Btu per (hour) (square foot). (Fahren- heit degree per inch). h = temperature of inner surface of insulation, Fahrenheit degrees. U -- temperature of outer surface of insulation, Fahrenheit degrees. It is convenient to work from the outer surface of the insulation, since the loss through the covering must be determined from the outer surface loss by means, of surface loss curves such as given in Fig. 4. The curves * 'J -- J-.1-4 T'nsitil'uip Via. 3. After the true heat loss is obtained, the loss per square foot of pipe sur face can be calculated from the relationship: 9i = g-OVn) where qj = Btu per (hour) (square foot outer surface of pipe). The heat loss through two or more thicknesses of insulation applied to a pipe can be calculated by means'of the equation: t\ -- ti <7o = ---------- -- -- ------------- r. log. n- r,, log. Tt + +- ki (2) where ra = outer radius of second'Jayer of insulation, inches, r, outer radius of last layer of insulation, inches. Pipe and Industrial Insulation 661 The method of solving Equation 2, which is the most difficult of the two, is given in Example S. Example 3: Compute the heat loss per linear foot of pipe surface per hour from a 6-in. pipe, insulated with a 3-in. thickness of diatomaceous silica, 1900 F maximum type, and a 2-in. thickness of 85 percent magnesia. The pipe is operating at a tem perature of 1200 F and is exposed to a room temperature of 80 F. Solution: In figuring the heat loss from Equation 2, it is necessary to first make an assumption for the outer surface temperature It and the temperature between the diatomaceous silica and 85 percent magnesia insulation, so that the mean tempera ture of each material can be obtained and the thermal conductivity corresponding to the mean temperature of each material substituted in the formula. First assume an outer surface temperature of 140 F and a temperature of 570 F between the two ma terials corresponding to a mean temperature of (1200 + 570) + 2 or 885 F for the dia- 7. 1,: 2 3Table Pipe Covering Factobs to be Applied to :Figs1 and Type of Pipe Insui+atiox Mean TeifpEBATUBB, Fahrenheit 85% MAGNESIA MOULDED AMOSITE AND BINDER. : LAMINATED,ASBESTOS PAPER (35-40 Per In.) CORRUGATED_AND LAMINATED'ASBESTO; PAPER 4 Ply Per In. 6 Ply Per In. 8 Ply Per In. CALCIUM SlUCATE CELLULAR GLASS: \ DIATOMACEOUS SILICA (22 lb 1 cu ft) DIATOMACEOUS SILICA (25 lb 1 cu ft) . MINERAL WOOL (Rock, Slag or Glass) Low Temp. (Asphalt or Resin Bonded) Low Temp. (Fine liber Resin Bonded) High Temp. (Blanket,.Metal Reinforced) PLASTICS (Foamed) RUBBER (Foamed) WOOL FELT HAIR FELT OR HAIR FELT PLUS JUTE- 40 1.05 -- -L. -- "7J.U5 '.r^` 0.80 0.63 -v0.74 0.66 0.83 0.77 70 1Q0 1.05 1.05 0.89 1.08 1.49 l.!54 1.85 1.37 1.30 1.32 -- 0.97 1.08 1.10 ---- " 0.83 0.89 0.64 0.65 0.78 0.78 0.84 0.72 0.68 0.86 | 0.89 0.78 0.81 200 3Q0 500 1.05 1.05 1.05 0.95 1.01 1.09 1.13 1.07 ,1.24 1.70 1.87 _1.48 1.62 _l.!43 1.52 1.00 1.03 .1.13 1.20 1.29 *-- 1.24 T -- 1.44 _0.98 0.68 0.73 0.90 0.98 1.05 _ * _i_ 1 - I ^1-- -1 700 _ -- _ , _ 1.18 1.39 _ -- 900 -- _ 1.14 1.34 - me conductivities of these two materials'at mean temperatures ( interpolated from Table 6, are 0.796 and 0.467 Btu, respectively. ____, These values are substituted in Equation 2 and a trial calculation made. For a nominal 6-in. steel pipe: r, = 3.312, r, = 6.312, and r, = 8.312. Then, 1200 - 140 8.312 log. 6.312 3.312 8.312 log. 6.312 1060 6.74 + 4.89 = 91.1 Btu. 0.796 0.467 , The temperature drop from the outer surface of the insulation to the surrounding air for a heat loss of 91.1 Btu is found from Fig. 4 to be 55 deg for a 16-in. O.D. cylin drical surface, or 55 + 80 F room temperature = 135 F surface temperature. Since a surface temperature of 140 F was assumed, it is evident that a temperature closer tp 135 F, or, for instance, 136 F should be used for recalculation: 1200 136 9 = 6.74 + 4.89 91.5 Btu. Since the temperature drop through each material is equal to the heat flow times the actual resistance of each material, the temperature drop through the diatomaceous silica is 91.5 X 6.74 = 617 F, Or the temperature between the two insulating materials is (1200 -- 617) = 583 F. Since a temperature of 570 F between the two materials was