Tutorial 4 Solutions

# Tutorial 4 Solutions - CHBE 141 fit/yaw 6’ Swap{0214 2003...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: CHBE 141 . fit/yaw 6’ Swap/{0214 2003/: 1/0; Q1: 20 CH4 (:02 20 cs2 , \$02 10 502 “33"” H20 50 H20 N2 02 N2 02 Also, we know that the NZ/O2 ratio in the air feed stream is 79/21. We could solve this problem either by choosing elements (C, S, H, O, N) or compounds as components. We’ll solve it here using compounds, which means that we need to write the balanced chemical equations. There are only two (SO2 and H20 are already oxidized): CH4 + 202 ~—> €02 +2H20 (R1) C82 + 302 » €02 +2so2 (R2) From steady-state material balance equations for methane and carbon disulﬁde, we ﬁnd the extents of reaction: helium = 20 - \$1 = 0 51 = 20 ﬁCSZput = 20 ‘é2 i: 0 5'2 = 20 Now we use these numbers to solve the remaining material balances: ﬁcozput = 31 + 52 = 40 ﬁsozput =10 + 252 = 50 ﬁmom, = so + 251 = 90 ﬁszut = ﬁNZjn ﬁozput = h02,in " 251 ' 352 = {102,1}; " 100 We have 2 equations involving 4 unknowns. We use the stream composition specifications to provide the last 2 equations: ﬁNZjn = 12 ﬁoz’m 21 ﬁsozpu: = 0 02 z 50 ﬁsozpuz + ﬁcozput + ﬁszut + ﬁozput 50 + 40 + ﬁszut + ﬁozput (Note: mole fraction SO2 in outlet is on a dry basis, so water is not included in the denominator.) We ﬁnd the solution by substitution or by using an equation solver: i102 in = 527 ﬁozpu: = 427 ﬁszut = ﬁN2,in =1983 The % excess oxygen is calculated from the amount of oxygen fed that is in excess of stoiChiometric requirements, considering both reactions: r'z -— I+3' _ wxwm=wxiom=42m 251+3§2 100 or, rounding up just to be safe, minimum oxygen excess required is 430%. The composition of the combustion gases is calculated from the solution of the material balances, knoWing that the total outlet ﬂow (at the chosen basis ) is 40+50+90+1983+427 = 2590 gmol/min: 77 mol% N, 16.6 mol% 02, 1.93 mol% \$02, 3.5 mol% H20, 0.77 mol% C02. ’ ‘ ' ' ' 715M 7LF0+ ﬂm "(Mw‘ MM = 300 -— g g 3100:300§1§3 yufm/ -g'v : gufjgf/yw 52mg 200—; - ‘ (a 9:412 ...
View Full Document

## This note was uploaded on 04/05/2010 for the course CHEBE 241 taught by Professor Ali during the Winter '08 term at UBC.

### Page1 / 4

Tutorial 4 Solutions - CHBE 141 fit/yaw 6’ Swap{0214 2003...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online