Question

霸子

求解

A
Gas
1000 kg/hr
100% Liquid Solvents
***8.26
圖 P8 26 表示一從言 N. 及 NE: 之氣
體流中回收 NH3 之程序。
往上流經程序的氣體流中含有N,及
NH,但不浴S;而經程序往下流
出的液體流中NE 及液體S,但不言
N,
離開程序之氣體這中 NH3之質量
分率與離之液體B中 NH3質
量率有隙,其腺係為:
Process
1000 kg/hr
10 mol% NH3
90 mol% N2
Gas
P8.26
Liquid
YB
PS.26 中的数婆計算支流4及
B之流幸和花成·若以莫耳關係式
Asia - 21, 來取代之質量關係式,你
能解此題嗎?

A (mol)
mol fr.
NH, X
N, X
1.00
1000 kg S
S 1.00
mass fr.
NH,
Gas
B kg liquid
s
1.00
1000 kg F (mol)
mol fr.
NH 0.10
N 0.90
1.00
Step 5: Basis 1.0 hr
Step 6: The unknowns are A, B, X, *R,,,,,a total of 6.
Steps 7 and 8: Three compound balances can be made, N2, NH3, and S. One relation is
given:
0.2
Two summations exist:
and Emº-1
7
MW
hence, we have an adequate number of balances (unless some are not independent) to
solve the problem.
Step 4: (revisited) Find the moles of F by getting the average molecular weight of F.
Basis: I g mol F
mole fr
mass (g)
NH, 0.10
17.03
1.703
N2 0.90
28.0
25.200
1.00
26.903
1000 kg F 2 kg mol F
26.903 kg F
:=37.171kg mol

In
Out
(B)
NH3 (kg mol): 0.10 (37.171) + 0 (1000) MWs
(A)
17.03
N2 (kg mol): 0.90 (37.171) + 0(1000) MWS = 0 (B)+, (A)
S(kg): 0 (1000)
1000
- (B)S +O(A) MWA
*W+x= 1.00
W. + -1.00
M = 20
N2 and S are tie components. Let XII, AEn, let xxA=n, and B-m,, etc.
Then n + n =A , and m, +m=B
Step 9: A total balance of 1000 + 1000 in kg, but not if in moles,
can be used in lieu of one of the above balances. They give
4
(B)
3.717 -
17.03
+x, (A)
33.453 = x, (A)
1000 W (B)
XH, +x=1.00
6, +1.00
= 2008,
17.03 x
17.03 * + 28 x
专x,
X
17.03 x
28 - 11 x
1000
33.453 x
3.717
x
2(28-11%)
1-X
1.
17 x
2(28-11%)
1000
33.453 x
FO) = 3.717 -
X
2(28-11%)
17 x
1-X
1- 1-
2(28-11X)

質均