10.9. HEAT EFFECTS AND TEMPERATURE VARIATIONS IN ABSORPTION
10.9A. Heat Effects in Absorption
The temperature in an absorption tower can vary from the top to the bottom of the tower when the inlet gas contains a relatively high concentration of solute. Solute absorption by the liquid yields a heat of solution which raises the liquid temperature. Also, if evaporation of the solvent to the gas phase occurs, the liquid temperature is cooled. The curvature of the equilibrium line depends upon the absorption of the solute, the heat-transfer rate between phases, and the evaporation or condensation of the liquid solvent.
Calculated temperature profiles may also have a maximum in the tower. This can occur when the entering gas temperature is considerably below the exit solvent temperature and/or the volatile solvent evaporates. Solvent evaporation cools the liquid near the tower bottom and a temperature maximum occurs. Detailed rigorous heat and mass-balance equations are needed for mass transfer of the solute, mass transfer of the solvent, heat transfer between the gas and liquid, and enthalpy balances. Equations and examples are given by Sherwood et al. (S1), Stockar and Wilke (S7), and Treybal (T2). Note that in actual towers, intercoolers are often used to keep the towers isothermal.
10.9B. Simplified Design Method
A simplified adiabatic design method is sometimes used when the entering gas contains solvent vapor and/or solvent evaporation occurs. Heat of solution of the solute is also included in the design. Then, using an overall enthalpy balance, the exit liquid temperature is calculated. This rise in liquid temperature can then be used to adjust the equilibrium-line curvature. First, an overall enthalpy balance is made which includes sensible heat for inlet and outlet gas temperatures, heat of solution of the solute, heat of vaporization of the solvent, and sensible heat for the inlet and the unknown outlet liquid temperature. This outlet temperature is then obtained.
The equilibrium line for the top entering liquid temperature is plotted, with the given slope at the dilute concentration. Next, an equilibrium line at this higher outlet liquid temperature is plotted. The outlet concentration x1 is then located on this equilibrium line. Then, assuming a linear liquid temperature profile, the temperature of point x, which is halfway between the top and bottom concentrations, is obtained. This x value is then located on the equilibrium line for this new temperature. The final curved equilibrium line is plotted with the slope at the dilute end, through the x at the halfway point, and finally through the x1 point. This method is illustrated in Example 10.9-1.
EXAMPLE 10.9-1. Heat Balance for Nonisothermal Absorption TowerThe gas feed at 20°C and 1 atm to a packed absorption tower contains 6.0 mol % of NH3 (on a dry basis) in air. The inlet gas is saturated with water vapor. The absorbing pure liquid water enters at 25°C. The outlet gas contains 0.5% of NH3 on a dry basis; it is assumed to be saturated with water vapor and leaves at 25°C. For a feed of 100 g mol dry gas, 190 g mol of pure water are to be used for absorption. The heat of solution for 1.0 g mol NH3 gas absorbed in water is ΔH = −8310 cal/g mol (S1). Equilibrium data (P1) are as follows for this system, where y = mx:
Solution: For the inlet gas, where y1 = 0.06, the moles NH3 = 0.06(100) = 6.0 and moles air = 0.94(100) = 94.0. For the outlet gas, where y2 = 0.005, the moles air = 94.0 and moles NH3 = 94.0 [0.005/(1 − 0.005)] = 0.472. The total is 94.472 moles. The moles NH3 absorbed = 6.0 − 0.472 = 5.528. The outlet x1 = (5.528)/(190 + 5.528) = 0.0283. From Appendix A.2-9, the latent heat of water at 25°C is (2442.3 J/g)(1/4.184 J/cal)(18.02 g/g mol) = 10 519 cal/g mol. The vapor pressure of water at 20°C = 2.346 kPa and at 25°C, 3.169 kPa. The moles H2O in the inlet gas at 20°C = (100.0)(2.346)/(101.325 − 2.346) = 2.370 moles vapor. In the outlet gas moles H2O vapor = (94.472)(3.169)/(101.325 − 3.169) = 3.050 moles vapor. The moles of H2O vaporized = 3.050 − 2.370 = 0.680. An enthalpy balance is made for ammonia gas and air in and liquid water at 25°C. For air, from Appendix A.3-3, cp = (1.0048 J/g · K)(1/4.184 J/cal) (28.972g/g mol) = 6.957 cal/g mol · K. For water vapor, from the Appendix, Fig. A.3-1, cp = 8.0 cal/g mol · K, and for NH3 gas, cp = 8.58 cal/g mol · K. The sensible heat for the entering gas at 20°C is as follows. For air, the sensible heat q = 94(6.957)(20 − 25) = −3270 cal. For water vapor, q = (2.370)(8.0)(20 − 25) = −95 cal. For NH3 gas, q = (6.00)(8.58)(20 − 25) = −257 cal. The total sensible heat is −3270 − 95 − 257 = −3622 cal. The latent heat of water vapor entering q = 2.370(10 519) = 24 930 cal. The sensible heat of the entering liquid at 25°C is 0. For the exit gas at 25°C, the sensible heat = 0. The latent heat of the water vapor = 3.050(10 519) = 32 083 cal. The heat of solution of NH3 absorption = 5.528(−8310) = −45 938 cal. From Appendix A.2-11, assuming dilute liquid water, cp = (1.000 cal/g · K)(18.02 g/g mol) = 18.02 cal/g mol · K. The total g moles outlet liquid = 190 + 5.528 − 0.68 = 194.85. The sensible heat q of the outlet liquid = (194.85)(18.02)(T1 − 25) = 3511(T1 − 25) cal, where T1 is the unknown outlet temperature in °C. Equating the heat in to the heat out,
Solving, (T1 − 25) = 10.02°C, which is the temperature increase of the outlet liquid. The outlet T1 = 35.02°C or 35°C. In Fig. 10.9-1, a straight operating line is plotted since the solution is dilute. The equilibrium line for 25°C is also shown with a slope m = 1.005 at the top of the tower. At the tower bottom, at 35°C, the exit x1 = 0.0283. The equilibrium line at 35°C is shown. The point x1 is located on this 35°C line and on the curved equilibrium line for the tower. The assumed linear temperature profile is also shown. Selecting a value of x = (0 + 0.0283)/2 = 0.0142 with a temperature of (35 + 25)/2 or 30°C, the point x1 = 0.0412 is plotted on the 30°C equilibrium line. The final curved equilibrium line is plotted with a slope at the origin of 1.005 at 25°C, through the point x = 0.0142 on the 30°C equilibrium line, and then through the point x1 = 0.0283 on the 35°C equilibrium line. Since the equilibrium line is curved, the number of overall transfer units NOy can be obtained by numerical or graphical integration as before using Eq. (10.6-46). Figure 10.9-1. Operating line and curved equilibrium line for Example 10.9-1. The calculated linear temperature profile is also shown.
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