Document km8w4zNw1B3X94KOynk5mNJ1V

584 CHAPTER 28 1949: Guide transferred per square foot of outer surface of the insulation is given by the equation: k(t, - tt) 9o = . r, U log. n (1) Pipelnsiilation 585 : After'the true, heat loss is obtained, the loss per square foot of pipe surface can be calculated from the relationship: 9* = 9(r/n) where qi = Btu per (hour) (square foot outer surface of pipe). Fig. 2. Heat Loss Through 1J In. Thick 85 per cent Magnesia Type Covering where 9o =* Btu per (hour) (square foot of outer surface of insulation). r, = outer radius of pipe or inner radius of insulation, inches. r =' outer radius of insulation, inches. k = thermal conductivity of insulation, Btu per (hour) (square foot) (Fahrenheit degree per inch). ti = temperature of inner surface of insulation, Fahrenheit degrees. tt .= 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. - . Fig. 3. Heat Loss Through 2 In. Thick 85 per cent . Magnesia Ttpe Covering The heat loss through two or more thicknesses of insulation applied to a pipe can be calculated by means of the equation: ., tl -- tt 9 = r. 1log, -- r, 1log, - Tl Tl + --:------+* At As (2) where rt -- outer radius of second layer of insulation, inches. r, = outer radius of last layer of insulation, inches. - .; r ' ' The method of solving Equation 2, which is the most' difficult of the two, is given in Example 3. Example S. Compute the heat loss per linear foot of pi{)e surface per hour from a 6-in. pipe, insulated with a 3-in. thickness of diatomaceous silica, and a 2-in. thickness of 85 per cent magnesia. The pipe is operating at a temperature of 1200 F <and is exposed to a room temperature of 80 F. . I*