Document 6wbxQdmdVDMojZ5eMb2gLrvE6

Hi i 104 CHAPTER 5 1955 Guidec|| Heat Transfer 105 Case IT of Table 2 fits the conditions of the problem if only free convec-fj tion heating of the pipe is assumed. The equation in this case is as follows:? tures are used with Fig. 7, a value for -g--fr is determined directly. Ca/'e / p\o6 /AtV" W WK = 0.271 (16)| p~p~ = 1.4 Btu per (hr) (F deg) (sq ft). where therefore, At = 20 F D = 0.364 ft p = P0 = one atmosphere / 20 V s fc--0371\o5S/ , hc = 0.737 Btu per (hr) (sq, ft) (F deg). 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) and the radiation resistance, R,, is then the following: R' = 1.33 X 1.14 Using the surface area of the insulation, the value of the resistance per unit* length is determined. - /4.375'\ "'V 12 / 1.14 sq ft Rc KA 0.737 X 1.14 Rc = 1.19 (hr) (F deg) per Btu. This result may not be deemed conservative inasmuch as the expression for still air. If, however, the air is not still, but flows at approximately mph or 7 fps, the heat transfer equation for forced convection would appl; This equation is Case 5 of Table 2. (u p)0-8 = o.2ii(r()---g^- (17)1 Ti = -|_ 460 = 570 Rankine (Fahrenheit absolute) Rr = 0.659 (hr) (F deg) per Btu. The resultant resistance of R,, and Rr acting in parallel (see Fig. 8) can now be evaluated as: i=k+k=i+oi=454 Btu per (hr) (F deg) R, = 0.216 (hr) (F deg) per Btu. The overall resistance, Rt, surroundings to cold water, is the sum of R\ + Ri + R3 + Rt = 4.12 (hr) (F deg) per Btu for 1-ft length of pipe. Note that the controlling resistances are R3 and Rt, and that neglect of both R\ and would not significantly influence the total resistance, Rr. On the basis of this resistance calculation, the heat transfer from the surroundings to the cold water may be evaluated as: (,, - t, Rr 120-34 = 20.8 Btu per (br) (ft) 4.12 and ,, = 7 fps ( 520\ =0076 Ur 0.0694 lb per cu ft D = 0.364 ft 0.211(570) `3(7 X 0.Q694)0-8 heaverage) = (0.364)-` hc(vige> - 2.73 Btu per (hr) (sq ft) (F deg) i Be (Forced Convection) = he A 2,73 X 1.14 or about 0.175 tons of refrigeration per 100 ft of pipe. Since the calculation is based on a 1-ft pipe length, q,, = 20.8 Btu per hr. The temperature drops through the various resistances are now readily evaluated by Equation 14 as: Al -- gR t> -- /,, (air to insulation surface) = qRi = 20.8 X 0.216 = 4.49 F deg -- to (through the insulation) = qRi = 20.8 X 3.9 = 81.2 F deg (through the pipe wall) = qRz = 20.8 X 8.5 X 10~* = 0.018 F deg ~ U (pipe wall to cold water) = qRi - 20.8 X 3.73 X 10- - 0.078 F deg ffe = 0.321 (hr) (F deg) per Btu. . .This solution was obtained on the temperature distribution assumptions LThe radiation resistance, Rr, which acts in parallel with the resistance"' just calculated, can be computed with the aid of Fig. 7. The pipe wall;;,I assumed at 100 F sees the surroundings at 120 F. If these two tempera| j initially made. It is apparent that a better solution could be obtained if the whole problem were reiterated using the temperature distribution just wlculated.