Document 08DaVvmj4kw3oqzpeeMJxwJR

HEATING VENTILATING AIR CONDITIONING GUIDE 1944 For the problem under consideration (Fig. 4) case 3 of Table 5 is applicable for the calculation of the cold water side convection resistance Ri. Corresponding to the water velocity of 5 fps, the mass velocity is: G = 5 (ft per sec) X 62.4 (lb per cu ft) X 3600 (sec per hr) = 11.2 X 105 lb per hour per square foot. The inside diameter of the pipe D is 2-' 0^67 = 0.1725 ft. The average water film temperature will be estimated as 36 F (mixed mean fluid temperature of 34 F). Then case 3, Table 5 yields: h -- 0.00486 (1 + 0.36) ^ {q^2725)<^ ~ = 650 Btu per hour per square foot per de gree Fahrenheit. . The transfer area on which this conductance is based is the inside tube area. Associated with 1ft length of pipe there are: *X X 1 = 0.542 sq ft. Thus the resistance for 1.ft of tube length is: F. = 1 1 AxD XI = 650 X 0.542 ~ 2 8 X 1(H hr degree Fahrenheit per Btu. Case 9, Table 5 is applicable for calculating the free thermal convection resistance, Rc, existing between the surrounding air and the insulation. The air temperature is given as 120 F. As an approximation a 20 F temperature difference between the air and the pipe surface will be assumed. Then case 9 yields: D = = 0364 ft- (20 X00 0,,.3-6,4/) = 0.63 Btu per hour per square foot per degree Fahrenheit. (13) 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 mass velocity coiresponds to: G = 7 X 0.07 X 3600 = 1770 lb air per hour per square foot. A magnitude of k -- 0.014 Btu per hour per square foot per degree Fahren heit for one foot thickness applied to case 4 yields: * =045 (S)+0178 (i77o)oM (S)" = 0.017:.+. 2:8 = 2.8'Btu per hour per square foot per degree Fahrenheit. This condu' ctan4ce37is5 based on 1 sq ft of ou. tsid. e lagging area. Thus, since there are it X = T.14 sq ft of outside lagging area associated with 1 ft length of pipe: . / : 84 CHAPTER 3. FUNDAMENTALS OF HEAT TRANSFER Rc = , -1. = 0.312 hr degree Fahrenheit per Btu. The radiation resistance, Rr, which acts in parallel with the convection resistance, Rc, for the transfer of heat to the surface of the insulation, may be calculated. For the purposes of this illustrative problem it will be assumed that the insulated pipe is exposed to (sees) surroundings, which exist at 120 F. Then the angle factor, Fa, is unity and for an estimated surface emissivity of 0.9 (see Table 6), Fe = 0.9. As a first approximation the insulation surface temperature will be estimated as 20 F lower than -the surroundings at 120 F. Then the radiation per degree of temperature difference, by-Equation 3 (or more conveniently by Table 8). divided by the temperature difference will be: hi = ----3---- 2q70) -- 1.17 Btu per hour per square foot per degree Fahrenheit- The outside surface area of the insulation associated with 1 ft of pipe length was previously calculated as 1.14 sq ft. Thus: Rt = i y 2 14 = hr degree Fahrenheit per Btu. The resultant resistance of Fc and Rr acting in parallel (see Fig. 4) can now be evaluated as: = = oil2 + 075 = 4'54 Btu per hour per d<*ree Fahrenl,eit' . Rt = 0.22 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 10-3 hr degree Fahren heit per Btu., Rt conduction through the pipe wall = 8.5 X JO-4 hr degree Fahrenheit per Btu. Rt conduction'through' the cork insulation = 3.9 hr degree Fahrenheit per Btu. Rt parallel.convection and radiation from the surroundings = 0.22 hr per degree Fahrenheit per Btu. Then: i?T = the overall resistance surroundings to cold water = Ri + Rt + Rt.+,F< = 4.1 hr degree Fahrenheit per Btu. Note that the controlling resistances are R3 and Ra. That is, the neglect of R\ and Ri would not significantly influence the total resistance, 2?x- On the basis of this resistance calculation the heat transfer from the surroundings to the cold water may be evaluated as: -g2r+e = At = -1--2-0- ------3-4- = 21 ,,Btu per ,hour per ,foo.t N Rt 4.1 or about 0.175 tons of refrigeration per 100 ft of pipe. Since the calculation is based on a 1 ft pipe length: grc = 21 Btu per hour.