Document M7BMkDzdYkd4m5wOq1qpVRQM

52 CHAPTER 5 1959 Guide Clan Table 3 .... Radiation Factors or Enussrvities, * For the determination of factor Fg in Equation 3 Fraction of Blade-Body Radiation At 50-100 F At 1000F' ALiOtfJhity for Solar Radiation 1 A small hole in a large box, sphere, furnace, or enclosure...................... 0.97 to 0.99 0.97 to 0.99 2 A Black Don-metallic surfaces such as asphalt, carbon, slate, paint, paper. 0.90 to 0.98 0.90 to 0.98 3 Red brick and tile, concrete and stone, rusty steel and iron, dark paints (red, brown, green, etc.)...................... .................................................... 0.85 to 0.95 0.75 to 0.90 4 Yellow and buff brick and stone, firebrick, fire clay................................. 0.85 to 0.95 0.70 to 0.85 5 White or light-cream brick, tile, paint or paper, plaster, whitewash... 0.85 to 0.95 0.60 to 0.75 6 Window glass................................................................................. ................. 0.90 to 0.95 7 Bright aluminum paint; gilt or bronze paint.............................................. 0.40 to 0.60. '8 Dull brass,'copper, or aluminum; galvanized steel; polished iron......... 0.20 to 0.30 0.30, to 0.50 9 Polished brass, copper, monel metal............................................................. 0.02 to 0.05 0.05 to 0.15 id Highly polished aluminum, tin plate, nickel, chromium................ ......... 0.02 to 0.04 0.05.to 0.10 * EmiMTitM of other "iii mey be found in Reference 4. * ReQecte about 8 percent. 0.97 to 0.99 0.85 to 0.98 0.65 to 0.80 0.50 to 0.70 0.30 to 0.50 Transparent* 0.30 to 0.50 0.40'to 0.65 0.30 to 0.50 0.10 to 0.40 - The temperature drop, At, through an individual resistance may then be calculated from the relation: At = Rq,, (15) where R is the resistance in question. The problem is now reduced to one of evaluating the in dividual resistances of the system. This entails suitable manipulation of the rate Equations i, 2, and 5 to produce expressions of the form: dicate the magnitudes of the thermal conductivities, k, to be employed in the expressions of Table 6, after dividing k by 12. The solution applicable to the problem depicted in Fig. 8, for tiie calculation of Ri and Rz,' is-Case 2 in Table 6. Thus for a I-ft length of 2 in. nominal size pipe (I. D. = 2.067 in., 0. D. =* 2.375 in.) insulated with 1 in. of material hav ing a conductivity of 0.025: " - Y U88 . lo- , 033 R* ------- ----r " 8.5 X 10- (hr) (F deg) per Btu. 2x X 26 X 1 where q'is the heat transfer rate, and At is the potential drop or temperature difference through the resistance RTable 6 lists such solutions for six different conduction sys tems. Table 4 in Chapter 9 and Table 1 of this chapter in 1 188 R " 2, X 0.025 X I - 3 9 <hr) P" ^ The convection resistances to heat transfer from the pipe Toap F Dog -30 -20 -10 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 * FrnmftT- P Table A .... Heat Transmission by Ratiiation for Black-Body Conditions* Exproaed in flfu per (tquarw fof) (hear) 0 -1 -2 -3 -4 --5 --6 -7 -8 -9 59.3 65.2 71.4 78.0 ' 68.7 64.7 70.8 77.4 58.2 64.1 70.1 76.7 57.7 63.5 69.5 76.0 57.2 62.9 68.9 75.4 56.7 62.3 68.3 74.7 56.2 . 61.7 67.7 74.0 , 55.7 ` 61.1 67.1 73.4 . 55.2 60.5 66.4 72.7 54.7 59.9 65.8 72.1 0 + 1 +2 +3 +4 +5 +6 +7 +8 +9 78.0 85.0 92.4 100 109 118 127 137 148 159 170 183 196 211 78.7 85.7 93.3 101 110 119 128 138 149 160 171.'' 184. 197 212 79.4 86.5 94.0 102 111 120 129 139 150 161 173 185 199 214 80.1 87.2 94.8 103 112 121 130 140 151 162 174 187 200 215 80.8 88.0 95.6 104 112 122 131 142 152 163 175 188 201 217' 81.5 88.7 96.4 105 113 123 132 143 153 164 176 189 203 218 82.2 89.4 97.2 105 114 123 133 144 154 166 178 191 204 220 82.9 90.2 98.0 106 115 .124 134 145 155 167 179 192 206 221 83.6 90.9 98.8 107 116 125 135 146 156 168 180 193 207 222 84.3 91.7 99.6 108 117 126 136 147 157 169 182 195 209 224 mill nf mum >132 F to surface at --15 F for effective eroissrvity of 0-85 (103 -- 63.3} 0.94 * J7.7 Blu per (square foot) (hour). Heat Transfer 53 > l| p2 Sydea Two infinite parallel planes. Table '5 ... Net Radiation Solutions Solution -J + - -1 Ramarkt Considering interrefiections. (Refer ence 5) |* 2j 1I1 One radiation shield ^2 between two infinite parallel planes. I-G) 1 . 'm` v*: Considering interrefiections. (Refer. ence 5) 1 I111 |tl2|3l.|.|.|.|ru n radiation shields between two infinite A-rhOO. 'w(!)s 'Considering ence 5) parallel planes. is tfie net radiation exchange without* the shields. interrefiections. (Refer *--s. Two concentric spheres Considering interrefiections and diffuse ( ij\2 \S~^J or two infinitely long cylinders. YY(H'(r' " surfaces. (Reference 5) /C\4*2 Two areas dAi and dA* do, ~ - **Fa(Tx* - TV) Surface diffuse, neglecting intcrreflection. (Reference 5) ^---v | R v ) 12 T", Tube of infinite length parallel to an infinite waI1 2rRN ~ (2) e,t*`r<7'4 " where N is the length of cylinder from which qT is exchanged. Neglecting interrefiections. (Reference 5) {-- r~ j Surfaces are perfect radiators. r Surface element dA and roc- tangle above and parallel to it, with one corner of rectangle contained in normal to dA. See Fig. 4 (Reference 4) fSv. Surfaces are perfect radiators. Adjacent rectangles in perpendieular planes. See Fig. 5 (Reference 4) I | Surfaces are perfect radiators. I I Opposed parallel rectangles | 1 and discs of equalsise. See Fig. 6 (Reference 4) wall to the cold water, Ri, and from the air to the surface of. the insulating material, Rc, are dependent on the flow conditions prevailing at these surfaces, and on the thermal properties of the fluids. These resistances are also directly dependent upon the temperature distribution, and for this reason it is necessary first to guess, on the basis of the prob lem statement, a temperature distribution upon-which to base the initial calculations. Since the values of hc for heat transfer between water and pipe walls are relatively high in this temperature range, it is logical to assume only a small temperature difference between the temperature of the fluid body and the temperature of the pipe wall. For the purpose of an initial guess, this temperature difference will be as sumed to be 2 deg. On the other hand, A* for heat transfer from air to a body is relatively small, and a higher tempera ture difference would be expected between these masses. The value initially assumed here will be 20 deg.Tn summary, the temperature distribution in the system is assumed as follows: *' Fluid temperature = 34 F. Inner pipe wall temperature: = 36 F. Outer insulation surface temperature 9 100 F. Ambient air temperature " 120 F. With these assumptions and the problem statement, it is now possible to calculate values for the convective resist ances. If it is found in the ultimate solution of the problem that the temperature distribution is different from that as sumed, it will then be necessary to repeat the solution pro cedure. If reference now be made to Table 2, it is found that Case 3 of this table is a system similar to that encountered in the