Document ybgoJMgmM05Gpr01RQ2OBJ0Mn

HEATING VENTILATING AIR CONDITIONING GUIDE 1942 Table 9. Some Low Temperature Radiation Emissivities for Commercial Surfaces3 Surface Emissivity e Metal Surfaces Aluminum: Brass: Copper: Zinc, Galvanized Sheet Iron: 0.04-0.06 0.11-0.19 0.096 0.22 0.6 0.023 0.07 0.8 0.95 0.87 0.94--0.97 0.94 0.228 0.276 Painted Surfaces 0.3-0.7 0.8-0.95 Miscellaneous ~ 0.96 0.93--0.945 0.952 0.80 0.75 0.90--0.97 0.85 0.931 0.90--0.95 0.95 0.924 0.91 0.80--0.93 0.95 0.91 0.90 0.87 0.95 0.89 aThese emissivities should not be employed for solar radiation calculations. For this type of calculation reference should be made to Heat Transmission, by W. H. McAdams and The Calculation of Heat Trans mission, by Margaret Fishenden and Owen A. Saunders. 80 CHAPTER 3. FUNDAMENTALS OF HEAT TRANSFER The resultant resistance of Rc and Rr acting in parallel (see Fig. 4) can now be evaluated as: -J- = -----1- -|j- = q + q yg = 4.51 Btu per hour per degree Fahrenheit. F, -- 0.222 hr degree Fahrenheit per Btu. The individual resistances for a 1 ft length of pipe applying to the illustrative problem depicted in Fig. 4 have now been calculated and are summarized as follows: Ri convection from the pipe wall to the cold water = 2.8 X 10-J hr degree Fahren heit per Btu. R, conduction through the pipe wall = 8.5 X 10~` hr degree Fahrenheit per Btu. F, conduction through the cork insulation = 3.9 hr degree Fahrenheit per Btu. R, parallel convection and radiation from the surroundings = 0.22 hr per degree Fahrenheit per Btu. Then Ft the overall resistance surroundings to cold water = Fi + Rt + F, hr degree Fahrenheit per Btu. R, = 4.1 Note that the controlling resistances are R3 and R4. That is, the neglect of Ri and R2 would not significantly influence the total resistance, J?t. On the basis of this resistance calculation the heat transfer from the surroundings to the cold water may be evaluated as: -3jrjC- = At -- -1--2-0- j---j---3-4- = 021,0B.tu per lhour per feoot. or about 0.175 tons of refrigeration per 100 ft of pipe. Since the calculation is based on.a 1 ft pipe length: 3rC = 21 Btu per hour. The temperature drops through the various resistances are now readily evaluated by Equation 6 as: At air to insulation surface = R< grc = 0.22 X 21 = 4.6 F. At through the insulation = F, <?rc = 3.9 X 21 =82 F. At through the pipe wall = F, qTQ = 8.5 X 10~` X 21 = 0.02 F. At pipe wall to cold water = Fi qrc = 2.8 X 10~3 X 21 = 0.06 F. The solution was obtained .on the assumption that the air temperature and the outside temperature differed by 20 F. In order to obtain a slightly better estimate of the rate of heat transfer the numerical solution should be repeated using the temperatures calculated from the previous listed temperature differences. The foregoing problem serves to illustrate a general method of solving steady-state heat transfer problems. There are many problems which cannot be approximated by steady-state solutions. For instance, the problem of pipe line insulation in transient service; the behavior of auto matically controlled thermoflow circuits; or the periodic absorption of solar energy by roof and wall structures during the day and nocturnal radiation to the cold sky at night. The transient heat transfer problem differs from the steady-state in that energy storage rates need to be 81