Document 5BxpxLx1Kq8RDmvrDaZGeVZV

580 CHAPTER 27 1950 Guide' transferred ' per square foot of outer surface of the insulation'is givenby the equation :' Hh -- tO go - r, log. - (1) Pipe Insulation 581 After the-true heat loss is obtained, the loss per square foot of pipe surface can be calculated from the relationship: q> - go(r/ri) where q-, -- Btu per (hour) (square foot outer surface of pipe). Fig. 2. Heat Loss Through 11 In. Thick 85 per cent Magnesia Type Covering \ where go = Btu per (hour) (square foot of outer surface of insulation)^ ft = outer radius of pipe or inner radius of insulation, inches, n "= outer radius of insulation, inches. '- k == thermal conductivity of insulation, Btu per (hour) (square foot) (Fahrenheit degree per inch). e;'li =;temperature of inner surface of insulation, Fahrenheit degrees, 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'byhiehns of siirface loss curves such as given in Fig. 4. ` ' ' Fig. 3..Heat Loss Through.2 In. Thick 85 per cent Magnesia Type - Covering j ' The heat loss through two or more thicknesses of insulation applied to a pipe can be calculated by means of the equation: g = r, liog. --r* r. liog. -r> ri , r, + + kt (2) where - n = outer radius of second layer of insulation, inches. r, = outer radius of last layer of insulation, inches. 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 pipe surface per hour from a 6-in. pipe, insulated with a 3-in. thickness of diatomaceous silica, and a 2-in. thickness of- 85 peri cent magnesia.' The pipe is operating at a temperature of 1200 F and is exposed to a room temperature of 80 F.