Document n9dvBLODYJmbVamM27g9mDg5w

HEATING VENTILATING AIR CONDITIONING GUIDE 1943 heat transfer performance on logarithmic coordinates, and the factor B should be regarded as a simple constant of proportionality. HEAT-FLOW RESISTANCE In most of the steady-state heat transfer problems encountered in air conditioning applications, more than one of the heat transfer mechanisms is effective, and the thermal current flows through several resistances in series or in parallel. In using the resistance concept the calculations in volved are analogous to the application of Ohm's Law in electricity, viz., Table 7. Heat Transmission by Radiation for Black-Body Conditions3 Expressed in Btu per square fool per hour Temp. Deg F -30 -20 -10 0 0 59.3 65.2 71.4 78.0 -1 58.7 64.7 70.8 77.4 -2 58.2 64.1 70.1 76.7 -3 57.7 63.5 69.5 76.0 -4 57.2 62.9 68.9 ' 75.4 -5 56.7 62.3 68.3 74.7 0 +1 +2 +3 +4 0 78.0 78.7 79.4 80.1 80.8 10 85.0 85.7 86.5 97.2 88.0 20 92.4 93.3 94.0 94.8 95.6 30 100 101 102 103 104 40 109 110 111 112 112 50 118 119 120 121 122 60 127 128 129 130 131 70 137 138 139 140 142 80 148 149 150 151 152 90 159 160 161 162 163: 100 170 171 173 174 175 110 183 184 185 187 188 120 196 197 199 200 201 130 211 212 214 215 217 +5 81.5 88.7 96.4 105 113 123 132 143 153 164 176 189 203 218 -6 56.2 61.7 67.7 74.0 +6 82.2 89.4 97.2 105 114 123 133 144 154 166 178 191 204 220 -7 55.7 61.1 67.1 73.4 -8 55.2 60.5 ` 66.4 72.7 +7 82.9 90.2 98.0 106 115 124 134 145 155 167 179 192 206 221 +8 83.6 90.9 98.8 107 116 .125 135 146 156 168 180 193 207 222 -9 54.7 59.9 65.8 72.1 +9 84.3 91.7 99.6 108 117 126 136 147 157 169 182 195 209 224 *Example; Radiation from walls of room at 32 F to surface at -- 25 F for effective emissivity of 0.95 = (102 -- 62.3) 0.95 = 37.7 Btu per square foot per hour. the heat flow or thermal current, is directly proportional to the thermal potential or temperature difference, and inversely proportional to the thermal resistance: <?rc -- (6) Following the electrical analogy, when there is a thermal current flowing through several resistances in series, the resistances are additive: Hi = + Hi + +........... -f-7?n (7) Similarly, conductance is the reciprocal of resistance, and for heat flow through two resistances in parallel, the conductances are additive: 1 Cr- -- R-t _l_ Ri + i + i + 80 (8) CHAPTER 3. FUNDAMENTALS OF HEAT TRANSFER Practical Heat Transfer Problems The use of these simple relations.for resistance and conductance simpli fies many practical heat transfer problems. As discussed in Chapters 4, 26 and 43, the practical analyses of heat transfer in building walls, in fin-tube coils and in pipe coverings, are usually computed by this method. The same resistance analysis may be applied to complicated steadystate conduction problems. Table 8 indicates the solutions in six common cases of steady-state conduction. A complete analysis by the resistance method is best illustrated by considering the heat transfer from the air outside to the cold water inside of an insulated pipe. The temperature gradients and the nature of the resistance analysis are indicated by the two sketches of Fig. 4. . Since air is sensibly transparent to radiation, there will be some heat transfer by both radiation and convection to the outer insulation surface. The mechanisms act in parallel on the air side. The total current by Fig. 4. Heat Transfer Conditions in the Insulated Cold Water Line radiation and convection then passes through the insulating layer and the pipe wall by thermal conduction, and thence by convection into main cold water streams. Radiation is not significant on the water side as liquids are sensibly opaque to radiation, although water transmits energy in the visible region. The contact resistance .between the insulation and the pipe wall is presumed to be equal to zero. Referring to Fig. 4, the thermal current for a given length N of pipe, grc Btu pier hour, may be thought of as flowing through the parallel resistances Rr and.Re, associated with the insulation surface radiation and convection transfer. Then the flow is through the resistance offered to thermal conduction by the insulation, R3, through the pipe wall resistance, R2, and into the water stream through the convection resistance, Ri. Note the analogy to the direct current electrical circuit problem. A temperature (potential) drop is required to overcome these resistances to the flow of thermal current. The total resistance to heat transfer, Rr, hour degrees Fahrenheit per Btu, is the summation of the individual resistances: . Rt = Ri R, R, R* \ (9) 81