Separation Process Principles- 2n - Seader & Henley - Solutions Manual

0478 00110 00368 lb waterlb dry air exercise 1822

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Unformatted text preview: sychrometric chart and equations. Given: From Example 18.1, air at 131oF with a humidity of 0.03 lb water/lb dry air, but with a total pressure for this exercise of 1.5 atm. Find: All of the psychrometric values listed below, using a psychrometric chart, steam tables, and Table 18.3. Analysis: (a) Molal humidity. From (18-6), Molal humidity = Hm = 0.03(28.97/18.02) = 0.048 mol water/mol dry air (b) Saturation humidity From (18-7), using the vapor pressure of water at 131oF = 0.155 atm, Saturation humidity, Hs = 18.02 ( 0.155 ) = 0.072 lb water/lb dry air 28.97 (1.5 − 0.155 ) (c) Relative humidity Relative humidity, HR = pH 2 O s pH 2 O 0.048 (1.5) 1.048 = = 0.443 or 44.3% 0.155 (d) Percentage humidity From (18-9) using the saturation humidity computed in part (b), HP = 0.03 × 100% = 41.7% 0.072 (e) Humid volume From (18-10), humid volume us given by, vH = 0.730 (131 + 460 ) RT 1 H 1 0.03 + = + = 10.4 ft3/lb dry air P M air M H2 O 1.5 28.97 18.02 (f) and (g) The small pressure change has a negligible effect. Therefore, cs = 0.254 Btu/lb dry air-oF And H = 57.4 Btu/lb dry air Exercise 18.18 Subject: Humidity conditions for a mixture of n-hexane and nitrogen. Given: Gas at 70oF and 1.1 atm with a relative humidity for hexane of 25%. Find: Humidity conditions given below. Analysis: At 70oF, from Fig. 2.4, the vapor pressure of n-hexane is 2.4 psia or 0.163 atm. (a) Partial pressure of hexane. From (18-8), p = psHR = 0.163(0.25) = 0.041 atm (b) Humidity From (18-5), H = M nC6 pnC6 ( M N2 P − pnC6 ) = 86.17 0.041 = 0.119 lb nC6/lb dry N2 28.02 (1.1 − 0.041) (c) Percentage humidity Need saturation humidity, Hs, from (18-7), Hs = M nC6 ( s p nC 6 M N2 P − p nC s 6 ) = 86.17 0.163 = 0.535 lb nC6/lb dry N2 28.02 (1.1 − 0.163) From (18-9), HP =(0.119/0.535) x 100% = 22.2% (d) Mole fraction of n-hexane ynC6 = pnC6 P = 0.041 = 0.037 1.1 Exercise 18.19 Subject: Humidity conditions for a mixture of toluene and air. Given: Gas at 180oF and 1 atm with a relative humidity for toluene of 15%. Find: Humidity conditions given below. Analysis: Use Fig. 18.20 for the air-toluene system at 1 atm. Using Fig. 18.20: H = 0.21 lb toluene/lb dry air Ts = 112oF To obtain the wet-bulb temperature, which is not equal to the adiabatic saturation temperature for the toluene-air system, a large extrapolation is needed, giving Tw = approximately 118oF. 2 From Table 18.5, psychrometric ratio = N Le/ 3 = 1.908 Exercise 18.20 Subject: Wet-solid temperature profile in a continuous, adiabatic, direct-heat dryer Given: Entering air at To = 180oF and 1 atm, with a relative humidity of 15%. Air flows cocurrently to the water-wet solid, which maintains the wet-bulb temperature of the entering air (Ts = Tw). The air exits the dryer at 120oF. The temperature of the air in the dryer, Tg, experiences an exponential decrease with distance through the dryer according to the relation, Tg = Ts + (To − Ts ) exp ( −0.1377 z ) (1) where, z = distance through the dryer in feet. Assumptions: Isobaric and adiabatic. Find: Calculate and plot the var...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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