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

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online