Document Nee9vJ4am2ddKk9EKoKzJjYjw

584 1 Water Trap. $ 84 85 86 87 88 89 90 95 100 105 110 CHAPTER 40 Toblo 3------ Sequence of Mechanical Integration Tower Performance 2 Enthalpy of FtIn h' 3 Enthalpy of Air h. 4 Enthalpy Diff. . h"--h9 6 A Ih'-h.) 7 2_*_ k'-h. 48.22 49.43 50.66 51.93 53.23 54.56 55.93 63.32 71.73 81.34 92.34 38.61 39.81 41.01 42.21 43.41 44.61 45.81 51.81 57.81 63.81 69.81 9.61 9.62 9.65 9.72 9.82 9.95 10.12 11.51 13.92 17.53 22.53 0.1042 0.1041 0.1038 0.1030 0.1020 0.1006 0.0988 010868 0.0718 0.0570 0.0445 . 0.1042 0.1039 0.1034 0.1025 0.1013 0.1002 0.4640 0.3965 0.3220 0.2537 0.1042 0.2081 0.3115 0.4140 0.5153 0.6155 1.0795 1.4760 1.7980 2.0517 1959 Guide 8 Cooling Bono* 1 2 3 4 5 6 11 16 21 26 h. Air enters the tower at wet-bulb temperature (t having an enthalpy of hi, which is plotted vertically below tif. Since the heat removed from the water equals the heat added to tile air, the enthalpy increase per pound of dry air equals the temperature change of the water multiplied by the liquid/gas (L/G) ratio. The enthalpy of the air, therefore, increases along a straight line having a slope equalling L/G and terminating at hi which is vertically below h'. The driving force at any section of the tower is h* -- , the vertical distance be tween the two operating curves. Example 1: It is desired to cool 300 gpm (150,000 lb of water Eer hr) from 110 to 84 P with 3050 cfm (125,000 lb of dry air per r) with an entering wet-bulb temperature of 75 F. Calculate the (NTU) showing the successive steps in tabular form. Solution; The sequence of steps in the mechanical integra tion is shown in Table 3. Water temperature is entered in Column 1 in even, but not necessarily equal, increments. One-degree increments are used from 84 to 90 F where the potential difference is small, and the effect of-the reciprocal is large. Little accuracy is sacrificed by using 5-degree increments from 90 to 110 F where the potential difference is larger and its reciprocal smaller. Column 2 shows the enthalpy of saturated air corresponding to the tempera tures in Column 1. Air enters the tower at a wet-bulb tempera ture of 75 F, having an enthalpy of 38.61 Btu per lb of dry air. This value is entered at the top of Column 3. The enthalpy of t.h. e a.ir .incr_eases .by M.. =* LdO (j TMTh. e Lf /G ratti.o .is11.4-n550r,U,0n0^000, = 1 .2 so bh equals 1.2 Btu from 84 to 90 F and 6 Btu from 90 to 110 F. These increments are added to the initial enthalpy to obtain the successive values in Column 3. The enthalpy potential dif ference in Column 4 is the difference between Columns 2 and 3. This driving force appears in the denominator of Equation 2 so the reciprocal of Column 4 is entered in Column 5. The (NTU) within each increment equals the temperature change multiplied by the average of the entering and leaving values from Column 5. The products are entered in ColumD 6. The summation of Column 6 in Column 7 gives the (NTU) for the cooling ranges in Column 8. Hence, the mechanical integration reduces the given conditions to the numerical value of 2.0517 Transfer Untie. Example t: The cooling tower designed to handle the condi tions in Example 1 has a nil height of 8 ft and a plan area of 96 sq ft. Calculate the unit-volume and overall coefficients and explain their significance. Solution: The (NTU) of 2.0517 as determined in Example 1 signifies the tower must cool the water 2.0517 deg per Btu of mean driving force. This cooling is accomplished in 8 ft of fill height, so the unit-volume coefficient is = 0.2565. This means that each cubic foot of tower must cool the water 0.2565 degrees per Btu of mean driving force. The water loading is = 1563 lb per (sq ft) (hr). The overall coefficient is 0.2565 X 1563 = 400.91. This signifies that the tower must trans fer 400.91 per cu ft (hr) (Btu of mean driving force). Column 7 of Table 3 shows how the (NTU) increases with cooling range. A plot of Columns 7 vs. 8 can be used to de termine the (NTU) for intermediate cooling ranges. This curve can also be used to determine temperature distribution be cause equal increments of transfer units correspond to equal increments of tower height. The linear relationship between (NTU) and height provides the hsis of the procedure used to analyse crossflow cooling towers.11 Since the wader temperature varies both horizontally and vertically in a cross-flow tower a double integration is re quired. This is accomplished by using equal increments of transfer units, representing equal increments of distance, and calculating the corresponding temperatures. Counter-flow integration progresses from the bottom of the tower upward. The cross-flow integration starts at the top of the air inlet, and progresses downward and inward. Evaporative Apparatus for Heat Rejection Required and Available Coefficients Any set of performance conditions can be accomplished by an infinite number of cooling tower designs. Calculating the (NTU) serves to evaluate the degree-of-difficulty of the prob lem. The unit-volume coefficient is more specific, but it also applies to an infinite number of towers of a given height. The overall coefficient has the most significance when comparing towers of different designs that are being considered for a given duty. It evaluates the effect of variations in air rate, height, and plan area on the required coefficient. A higher required coefficient indicates that more effective filling is needed. The effectiveness of a filling is determined almost entirely by the density of the splash surface and wetted surface per cubic foot of filled volume. These calculations evaluate the problem, not tire tower. The (mailable coefficient of a specific cooling tower is deter mined from its operating characteristics. The tower must be tested at various conditions and the data from each test re duced to a corresponding coefficient. A cooling tower coeffi cient is not constant, but varies with operating conditions. Its characteristic is established by correlating variations of the coefficient with corresponding operating conditions. The characteristic of a cooling tower closely approximates the relationship expressed by the general equation: Ka = cZX? (7) The cooling tower manufacturer uses this procedure to de termine the characteristic of each cooling tower design. An examination of the steps taken in the solution of Ex ample 1 will show that various air rates could have been speci fied. Increasing the air rate would have reduced the degreeof-difficulty and resulted in a smaller (NTU) for a given set of conditions. Decreasing the air rate would have the opposite effect. Therefore, the required coefficient for a given set of conditions varies with the air rate. A tower will operate at a given set of conditions when the required coefficient equals the available coefficient as determined from Equation 7. DESIGN CONDITIONS The design conditions used in selecting a cooling tower are (1) the circulating water rate, (2) cooling range, (3) entering wet-bulb temperature, and (4) the approach of the coldwater to the wet-bulb temperature. These conditions are se- Rg. 9 .... Relation of Cooling Tower to Process 585 iected by considering the function of the cooling tower in the overall system. The system includes a process having a heat load that is.to be transmitted to the circulating water through a heat exchanger. The cooling tower rejects the heat load to the atmosphere. Certain design conditions of the system affect the design of the cooling tower. They include (i) the heat load, (2) the temperature at which it is to be removed, and (3) the wet-bulb temperature of the atmosphere to which the heat is finally rejected. The cooling tower and beat exchanger are then selected to serve these conditions. The relationship of the cooling tower to the system is shown schematically in Fig. 9, with the process represented by two heat sources op erating in series. Heat Load No definite sequence can be followed in selecting the design conditions because each item affects the choice of the others. The heat load comes closest to being a fixed quantity so can be considered first. The amount of heat rejected by typical processes is shown in Table 4. The heat is rejected at a constant temperature when vapors are condensed, as illustrated by Q* in Fig. 9. A second type of heat load, represented by Qi, consists of reducing the temperature of a fluid. Each type of heat load specific requirements that affect overall design. The tem perature at which the heat is removed from a process may be as important in the overall design as the heat load itself. As this temperature level decreases, the degree-of-difficulty increases at an accelerated rate, calling for larger and more expensive equipment. Higher temperatures usually result in higher operating costs, and loss of capacity in the process. Table 4 .... Heat Rejection of Typical Processes Equipment hr hr bhp-hr Refrigeration Compressors (reciprocating)................................. Refrigeration Compressors (centrifugal)....................................... Refrigeration Absorption System.. Steam Jet Refrigerating System___ Steam Electric Power Plant: 500 kw......................................... 1000 kw......................................... 5000 kw......................................... 7500 kw......................................... 10,000 kw........................ ................ Diesel Engine Jacket & Lube Oil: Four-cycle, Supercharged............ Four-cycle, Non-supercharged... Two-cycle, Crank-case Compressor................................................. Two-cycle, Pump Scavenging (large unit)................................. Two-eyclc, Pump Scavenging (high speed)................................ Natural Gas Engine: Four-cycle....................................... Two-cycle.................... ................. 250 300 550 550 11,210 10,750 8,150 7,700 7,020 2600 3000 2000 2500 2200 45004000