Document 37p4w3yxDqm3axkyxdyDY7KrO

104 CHAPTER 5 ,1958 Guide y Case 11. of Table 2 fits the conditions of the problem if only free conyec tion heating of the pipe is assumed. The equation in this case is as follows / p\o.s /aiV! 0271U) \d) where therefore, At = 20 F D = 0.364 ft P = FV= one atmosphere = -271^/ 20 jV d' if '4' 4 A, = 0.737 Btu per (hr) (sq ft) (F deg). 'ft? 'f' Using the surface area of the insulation, the value of the resistance per unit ) length is determined. /4.375\ X 1 = 1.14 sq.ft =1 * ~ hcA = 0.737 X 1.14 Rc = 1.19 (hr) (F deg) per Btu. ;'C 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 heat transfer equation for forced convection would apply;? This equation is Case 5 of Table 2. (u.p)- he,TMr,,,) = 0.211(7',)'" D< (17)f Tt _ 100 + 120 400 = 570 Rankine (Fahrenheit absolute) : 7 fps 0076f/5e20)\ = n,0694 lb per cu ft D = 0.364 ft 0.211(570) "(7 X 0.0694)0-8 hc(eveeeee) = (Q 304)<M he(,,,,.- 2.73 Btu per (hr) (sq ft) (F deg) and 1 Re (Forced Convection) =[ he A 2.73 X 1.14 Re 0.321 (hr) (F deg) per Btu. The radiation resistance, Rt, which acts in parallel with the resistance just calculated, can be computed with the aid of Fig. 7. The pipe wallf assumed at 100 F sees the surroundings at 120 F. If these two temperaj ; Heat Transfer. 105 tures are used with Fig. 7, a value for is determined directly. he = 1.4 Btu per (hr) (F deg) (sq ft.). FaFe The angle factor, FA, is unity, and for an estimated surface emissivity of 0.95 (see Table 3), FE = 0.95. Therefore, h, = 1.4 FaFe = 1.4 X 1 X 0.95 hr = 1.33 Btu per (hr) (F deg) (sq ft) r- and the radiation resistance, Rr, is then the following: _ JL 1 ' ~ h,A = 1.33.X 1.14 Re = 0.659 (hr) (F deg) per Btu. The resultant resistance of Rc and Rr acting in parallel (see Fig. 8) can now be evaluated as: k k+k= =ofc+oiBtu per (hr) (F deg) The overall resistance, /St, surroundings to cold water, is the sum of Ri + Rs + Rs + R* = 4.12 (hr) (F deg) per Btu for 1-ft length of pipe. Note that the controlling resistances are /Ss and Rt, and that neglect of both /Si and ZSs would not significantly influence the total resistance, Rt- On the basis of this resistance calculation, the heat transfer from the surroundings to the cold water may be evaluated as: ?? = l--l< = ^ = 20.8 Btu per (hr) (ft) N Rt 4.12 or about 0.175 tons of refrigeration per 100 ft of pipe. Since the calculation is based on a 1-ft pipe length,. qn = 20.8 Btu per hr. The temperature drops through the various resistances are now readily evaluated by Equation 14 as: . . . .. At qR t0 -- 3 (air to insulation surface) == qR* a 20.8 .X 0.216 = 4.49 F deg - t*j (through the insulation) ^ qRt ==* 20.8 X 3.9 = 81.2 F deg .. - t,, (through the pipe wall) = qRt = 20.8 X 8.5 X 10-* = 0.018 F deg - 1, (pipe wall to cold water) = = 20.8 X 3.73 X UT1 - 0.078 F deg . solution was obtained on the temperature distribution assumptions initially made. It is apparent that a better solution could be obtained if the whole problem were reiterated using the temperature distribution just calculated.