from where did he get the values of r1,r2,r3

ma T.2 1 12 Steam) 1 T3 EXAMPLE 3-8 Heat Loss through an Insulated Steam Pipe Steam at Toi = 320C flows in a cast iron pipe (k = 80 W/mK) whose inner and outer diameters are D = 5 cm and D2 = 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation with k = 0.05 W/m.K. Heat is lost to the surroundings at Tooz = 5C by natural convection and radiation, with a combined heat transfer coefficient of h2 = 18 W/m2.K. Taking the heat Insulation transfer coefficient inside the pipe to be h = 60 W/m2.K, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation. Properties The thermal conductivities are given to be k = 80 W/m.K for cast 13 iron and k = 0.05 W/m-K for glass wool insulation. Analysis The thermal resistance network for this problem involves four resis- T1 tances in series and is given in Fig. 3-29. Taking L = 1 m, the areas of the hi surfaces exposed to convection are determined to be A1 = 20 L = 277(0.025 m)(1 m) = 0.157 m T2 Tz Towwww A3 = 27773Z = 277(0.0575 m) (1 m) = 0.361 m - www www.T.2 R; R R2 R Then the individual thermal resistances become 1 R = Rcony, 1 = 0.106C/W (60 W/mK) (0.157 m) The temperature drops across the pipe and the insulation are determinec In(2/7) In(2.75/2.5) R = Rpipe = 0.0002C/W from Eq. 3-17 to be 2k, 27(80 W/m-K)(1 m) AT In(rzfr) In(5.75/2.75) pipe QR cipe = (121 W)(0.0002C/W) = 0.02C R2 = Rinsulation = 2.35C/W 2rkl 27(0.05 W/m.K)(1 m) = OR = (121 W)(2.35C/W) = 284C 1 R = Rconv, 2 = 0.154C/W MA; (18 W/mK)(0.361 m) Noting that all resistances are in series, the total resistance is determined to be Rotal = R + R1 + R2 + R = 0.106 + 0.0002 + 2.35 + 0.154 = 2.61C/W Then the steady rate of heat loss from the steam becomes Tool - 7.2 (320 - 5)C = 121 W (per m pipe length) Rotal 2.61C/W 1 AT-alabon insulation 1