Document Xz7pDGNGqa9kzrLQBRKvZ0O3y

668 CHAPTER 37 i ~. 1943 Guide through the tower counter to the direction of water flow at the rate of 30,000 lb of.dry air per hour. ' - The cross-section of the tower.is to be 8 ft x 8 ft and the packing is to be of the type producing a rate coefficient as indicated in curve No. 2 of Fig. 6.- Eor this type of packing the average cross-sectional area for air flow will be 36 sq ft.:< . . - -r ' Solution: " \ ' `` , ' Initial air enthalpy = 25.1 Btu per pound of dry air. ,.. Final air enthalpy : (hi - h,) = ^.(1, .- 1.) (A,-251) = ?^Wi(110-80) hi = 61.1 Btu per pound dry air. . A numerical integration (Table 4) is employed to determine the Number of Transfer Units (NTIT) required. Temperature increments of 2 F are used between successive determinations of the quantity p-i-j-. The energy balance indicates that the enthalpy increments corresponding to these temperature increments are: Ah = Af = X 2 = 2.4 Btu per pound of dry air. The result of the integration is that the Number of Transfer Units required (NTU) = 1.72. Unit ^s mass velocity, -yG- = 3--0g0g0--0 = 830 lb per (hour) (square foot average cross-sectional air flow area). From Curve 2, Fig. 6, Ka = 138 Btu. Fig. 6. . Unit Conductances fob Various Types of Packing Construction Spray Apparatus 669 The tower volume required is: V ^ $- (NTU) = X 1.72 = 217 X 1.72 = 374 cu ft. Ka loo . Height of the packed section is: 374 8X8 5.9 ft. The graphical solution for the Number of Transfer Units required for the desired performance is plotted in Fig. 7. ^ is plotted as a function of h,'and the area , under the curve from the, initial enthalpy of the air, 25.1 Btu per pound, to the final enthalpy of the air, 61.1 Btu per pound is 1.72, the Num.ber of Transfer Units ,required. The effect of the rapid decrease in potential due to the cooling of the water is indicated by comparison of area Ai. A t, and At of Fig. 7. Each represents the Number of Transfer Units required to achieve a water temperature reduction of about 10 F. Ai = 0.471 NTU (110 to 100 F) A, = 0.595 NTU (100 to 90 F) A, = 0.657 NTU ( 90 to 80 F) The use of the logarithmic mean driving potential is illustrated by applying Equation 7 to Example 1: NTU (Ka) V = hi -U G Afam Atom (h\ - hi) - (h\ - h,) = 3Q.7 - 18.4 = ^ h'-hi .30.7 lp8e - . l0Ee lU A- - h. NTU = 61.1 - 25.1 24 = 1.5 The Number of Transfer Units required as determined by use of the logarithmic mean driving potential equals 1.5 which compares favorably with the correct magnitude, 1.72. Table 4. Numerical Integration for the Number of Transfer Units, Water Temperature Interval F Dbg. Mean Water Temperature F Deg 80-82 82-84 84-86 86^88 88-90 81 83 85 87 89 90^92 92-94 94-96 96-98. . 98-100 91 . 93 95 . 97 99 100-102 102-104 104-106 106-108 108-110 101 103 105 107 109 Mean Air Enthalpy, />& Btu per Lb Dry Air Saturated Air hnEnthalpy, Btu per Lb Dry Air . 26.3 28.7 31.1 33.5 35.9 44.6 46.9 49.2 51.7 54.4 . 38.3 40.7 43.1 45.5 47.9 . 57.1 60.0 63.0 66.2 69.6 50.3 52.7 55.1 57.5 59.9 73.2 77.0 80.9 85.1 89.5 Enthalpy Potential h" - h* Ah h" - ha 18.3 18.2 18.1 18.2 18.5 : : 0.131 0.132 0.133 0.132 0.130 18.8 . 19.3 19.9 20.7 21.7 : 0.128 0.1240.121 0:116 0.111 22.9 24.3 25.8 27.6 29.6 0.105 0.099 0.093 0.087 0.081 ../ -1.723 ;