Question

Enironment: 1, +298.15 K. po = 1 atm, N, 75.67%, O2 20.35%, H20(g) 3.12%, CO2 0.03% and others 0.83%, by volume. sample 16.12 Determine the chemical exergy of (a) carbon, (b) hydrogen, d methane, (d) carbon monoxide, (e) liquid methanol, (f) nitrogen, (g) oxygen, carbon dioxide, and (i) water in kJ/kg, in respect of the environment in which the pas phase obeys the ideal gas model.

Please answer all questions.

Answer 1

Date: Page no 1 801" The envitonment consist of an Idial gas mixture having the following: Envixonment To=298.15 K, PoElatm Gas Phase : component yeo. N2 75-67 OD 20:35 H₂O(g) 3.12 CO2 0-03 Other 0.83 @ Całbon : C+02 GOL ech = (gc + goz - Jcoz) (To, Po) + Ř To en Yoz y cost Then, with gå data from table A-25; ēCH - (0 + 0-(-394,380)) + (8-314)(298.15) en 0-2035 L0.0003 ēCH 3410,541 KJ/kmol with m= 12:01 from table A-te OCH = 420, 541 34,183 kJ/kg Ams 12.04

Date: Page no: (6) Hydrogen: H2+ 10. H2O OCH = ( 703 + 190,-71420) [10, P.) + Toen (462) (4 40). Then, with d data from table A-25 @CH = CO + 4.CO) -(-228,590)) + (8-314) (298-15) In[0.2015) ēCH. 235,212 KJ/K mol 0.0312 with m= 2.016 kg/kmol, есн 235,212 2.046 CO2 + 2H₂O = 116,673 kJ/kg Ans- (6) Methane : CH4 + 202.- ēCH = [ Jcha +2902 - jcoz - 25420 7 + Ř To en eye? Lycoplus Then with Ji data from table A-25, ēCH [-50,790 +200)-(-394,380)-2-228,590) +(8:314) (298.15) en (0-2035) CO-0003)(0-03127 ēCH = 830,174 kJ/kmoe with m= 16.04 kg/kmol есн - 830,174 16.04 - 51,756 kJ/kg Ans:

Date: Paga no: (3 (d) Carbon monoxide :- GO2 2 Co + 102 ŞCH - (Jco + Joz -7.00,) (T., Pe) + R Toln[ (Yer y42 Ycee with de data from table A-25 2CH = [6133,150) + 4.0)-(-894380)] +(8-81920298:15) en| 0.2035) ēCH = 275,364 kJ/kmol 0.0003 with M = 28.01 Kg/kmol 1.5 275,364 9831 kJ/kg Ans 28.01 (2) Liquid Methanol :- CH3OH + 1 5 02 CO2 + 2H2O ad OCH = (J CH2OH + 1.5 go, - 5 co2 -2ğ H2O) + Ř To en Yo (YH20² spēCH =(-166,290 +1:5(0) – (-394,380) – 2-228,590)) +8-314 (298.15) en (0.2035) ēCH = 7L6, 647 kJ/kmol LC0-0003) (0-0312)

Date: Page no A Nitrogen :- R Toln (1) =(8-314)6126:18) ech = 691.1 kö/kmol (6=7557) with m=28.01 kg/kmol CA E691 L 28.0d 24 7 kJ/kg Ans (9) oxygen ēCH R Toln ( 1 ) =(8-314)(298.15) en (ozone You KJ /kmol 2 OCH 3947 with m = 32.00 kg/kmol ICH = 3947 32 123 3 kJ/kg Ans (6) carbon dioxide :- e CH = Ř To en 2. = (8:31962 (8:314)(2 98-15) en (0.600) Yoo CO2 @CH = 20,108 KJ/kmol with m= 44.01 kg/kmol OCH = 20, 208 456.9 kJ/kg 44.01 (I) water :- Ans • OCH (ģH20 cl) - JH2O(g)) (To, Po) + Ř To en I 0-031 ECH = [(-237,180) -(-228,590)] +(8-314) (298.15) en ēCH = 4•9 kJ/kmol This is a negligible amount that would be zero where the table values for gf not rounded.