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

8 atm and c 12 atm analysis note that 1000of is way

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Unformatted text preview: on, cracks, splits, and checks. The disadvantage is cost. Exercise 18.15 Subject: Air conditions from a psychrometric chart and equations. Given: Air, for a direct-heat dryer, entering at 250oF and 1 atm with a wet-bulb temperature of 105oF. Find: All of the psychrometric values listed below, using a psychrometric chart, steam tables, and Table 18.3. Analysis: (a) Humidity Humidity, H = 0.0158 lb water/lb dry air (b) Molal humidity Molal humidity = Hm = 0.0158(28.97/18.02) = 0.0254 mol water/mol dry air (c) Percentage humidity % humidity, HP, is not applicable at these conditions because the vapor pressure is greater than the pressure. (d) Relative humidity Relative humidity, HR = pH 2 O s pH 2 O 0.0254 (14.696 ) = 1.0254 = 0.0122 or 1.22% 29.83 which is in agreement with Fig.18.17. (e) Saturation humidity is not applicable because vapor pressure is > than pressure. (f) Humid volume From (18-10), humid volume us given by, vH = 0.730 ( 250 + 460 ) RT 1 H 1 0.0158 + = + = 18.3 ft3/lb dry air P M air M H2 O 1 28.97 18.02 which is in agreement with Fig. 18.17. (g) Humid heat From (18-11), humid heat is given by, cs = 0.24 + 0.45 ( 0.0158 ) = 0.247 Btu/lb dry air - oF (h) Enthalpy From (18-12) with To = 32oF and for water, λo = 1075 Btu/lb H = 0.247(250 – 32) + 1075(0.0158) = 70.8 Btu/lb dry air Exercise 18.15 (continued) (i) Adiabatic-saturation temperature For the air-water system, adiabatic-saturation temperature = wet-bulb temperature = 105oF (j) Mole fraction of water in the air Obtain this from the molal humidity. yH 2 O = 0.0254 = 0.0248 1.0254 Exercise 18.16 Subject: Air conditions from a psychrometric chart and equations. Given: Air, for a direct-heat dryer, entering at 200oF and 1 atm with a relative humidity of 15%. Find: All of the psychrometric values listed below, using a psychrometric chart, steam tables, and Table 18.3. Analysis: (a) Wet-bulb temperature Tw = 127.5oF (b) Adiabatic-saturation temperature For the air-water system, adiabatic-saturation temperature = wet-bulb temperature = 127.5oF (c) Humidity Humidity, H = 0.0835 lb water/lb dry air (d) Percentage humidity From (18-9) using the saturation humidity computed in part (e), HP = 0.0835 × 100% = 3.69% 2.262 (e) Saturation humidity From (18-7), using the vapor pressure of water at 200oF = 11.526 psia, Saturation humidity, Hs = (11.526 ) 18.02 = 2.262 lb water/lb dry air 28.97 (14.696 − 11.526 ) (f) Humid volume From (18-10), humid volume us given by, vH = 0.730 ( 200 + 460 ) RT 1 H 1 0.0835 + = + = 18.9 ft3/lb dry air P M air M H2 O 1 28.97 18.02 which is in agreement with Fig. 18.17. (g) Humid heat From (18-11), humid heat is given by, cs = 0.24 + 0.45 ( 0.0835 ) = 0.278 Btu/lb dry air - oF Exercise 18.16 (continued) (h) Enthalpy From (18-12) with To = 32oF and for water, λo = 1075 Btu/lb H = 0.278(200 – 32) + 1075(0.0835) = 136.5 Btu/lb dry air (i) Partial pressure of water in the air. From (18-8), s pH 2O = 0.15 pH2 O = 0.15(11.526) = 1.73 psia Exercise 18.17 Subject: Air conditions from a p...
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