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 DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full 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 steadystate 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
 Winter '08
 ali

Click to edit the document details