Gas Ab
Gas Ab
Gas Ab
two hundred
pound per moles per hour of a gaseous mixture containing 5 mol HCl per 2
mole of air enters the bottom of the tower and 15 000 lb of pure water enters
the top of the tower per hour. The aqueous HCl solution leaving the bottom of
the tower contains 1 mol of HCl per every 7 mol of H2O. assuming no water
vaporizes in tower, determine the mol of HCl per mol of all in the exit gas
stream.
Soln:
L2=15 000 lb/hr
Y2=?
X2 = X2 = 0
X1 = 1/7
40
1.51
Determine the no. of stages required to twice the minimum water rate.
Solution:
No. of stages: N =
y 2 y 2
y 1 y 2
y 1 y 2
log
1
y 1 y
log
x2 = 0
y2 = 0.0005
T= 90F
550 R
P = 1 atm
14.7 psi
Y1 = 0. 03
Q = 5000 ft/hr
V1 =
5000 ft 3
(14.7 psi)
hr
ft 3
10.73 psi
(550 R)
lbmol R
V1 = 12.45 lbmol/hr
V = V1 (1 - y1) = 12.45 (1- 0.03)
= 12.08 lbmol/hr
V (y1 y2) = L (x1 x2)
Since x2 =0
At 90F = 32.2 C
X=
2. .16 lb NH 3
100 lb H 2O
18 lb H 2 O/lbmolH 2O
17 lb NH 3/lbmolNH 3 )
0.03
12.08 ( 10.03
0.0005
10.0005 ) = L (0.0229 - 0)
L = 16.05 lbmol/hr
L act. = 2L = 2(16.05)lbmol/hr
= 32.10 lbmol/hr
Y1* = 0.185 x1
Y2* = 0.185 x2 = 0
0.03
12.08 ( 10.03
0.0005
x1
)
=
32.10
(
10.0005
1x 1
- 0)
X1 = 0.0113
Y1* = 0.185 (0.0113) = 2.09x 10-3
0.0032.09 x 103
)
0.00050
0.030.0005
log (
)
2.09 x 103
log (
N==
N = 1.52 2
stages
%recovery=95%
P= 1 atm
y2 = ?
Y1 = 0.02
Y* = 2.53x
N=?
% volume = % by mole (gases)
Basis: V1 = 1000lbmol
V = 1000 (1 0.02) = 980 lbmol
Minimum reqd (from the problem)
Y1 = 2.53 xi
Xi =
y 1 0.02
=
2.53 2.53
= 7.9x 10
-3
0.02
Y2 =
N=
y2
1 y 2
; Y2 = 1.02x10-3
y 2 y 2
y 1 y 2
y 1 y 2
log
1
y 1 y
log
7.9 x 103
(min)
= 2386. 11
x1
0.02
980[ 10.02
0 )
x1
1.20x10-3] = 2863. 34 (
1x 1 )
X1 = 6.59x10-3
Y* = 253x
Y1* = 2.53x1 = 2.53 (6.59x10-3) = 0.0167
Y2 = 2.53 x2 = 2.53 (0)= 0
0.020.0167
)
1.02 x 1030
0.021. .02 x 103
log (
)
0.01670
log (
N=
N = 9.17 10
stages
4. Propane is to be stripped from non-volatile oilby steam in a counter current
tower. Four moles of steam will be supplied at the bottom of the tower for
every 100 mol of oil-propane feed at the top The oil originally 2.5 mole
percent and this concentration must be reduced t0 0.25 mole percent. The
tower is maintained at 280F and 35 psia. The molecular weight of the heavy
oil is 300 and the molecular weight of propane is 44. The equilibrium
relationship is y* = 33.4xi. How many equilibrium stages are required? If the
pressure is increased to 70 psia, how many equilibrium stages are required?
Solution:
L2 = 100
X2 = 0.025
Y* = 33.4x
X1 = 0.025
v1= 4
Y1=0
N=
y1 = 0
y 2 y 2
y 1 y 2
y 1 y 2
log
1
y 1 y
log
V = v1 (1- y1) = 4
L = L2 (1- x2) = 100 (1- 0.025) = 97.5
V (Y1 Y2) = l (x1 x2)
4(
y2
0.0025
0.025
)
=
97.5
(
1 y 2
10.0025 10.025 )
y2 = 0.3606
y*= 33.4x
y1*= 33.4 x = 33.4 (0.0025) = 0.0835
y2*= 33.4x2 = 33.4 (0.025) = 0.835
00.0835
)
0.36060.835
00.3606
log(
)
0.08350.835
log (
N=
2.37 3
stages
35
00.0418
)
0.36060.4175
00.3606
log (
)
0.04180.4175
log (
N=
7.52
stages
TOP
0
0.001
BOTTOM
0.08
0.009
Yb = 0.009
Xb = 0.08
Gm
Hoy = Hy + (M ( Lm ))Hx
Z = Hoy Noy
0.009
MB: G ( 10.009
G Gm
=
L Lm
0.001
0.08
0 )
10.001 ) = L ( 10.08
= 10.76
YbYa 0.0090.001
=
YLm
0.004
=2
L'
L'
Kya = 10
Calculate the height of the tower.
Solution:
V & L : solvent
Z = HTU (NTU)
HTU
NTU
V=
V1=
v
S Kya(1 y) lm )
y 2Y 2
Y 1Y 1
Y 2Y 2
ln
Y 1Y 1
Y 1Y 2
V 1+V 2
2
(1-y)lm=
1 y 1
1 y 2
ln( )
( 1 y 1 ) (1 y 2)
Y1 = 0.05 =
Y1
1 y 1
= 273.69lbmol/hr
; y1 = 0.0476
V'
1 y 2
; y2 =0.0005 =
=
260.66
=
10.0005
273.69+ 260.79
2
m
A
S=
Y2
1 y 2
; y2= 0.0005
260.79
= 267.24
= 13.03 ft2
10.0476
10.0005
ln ()
( 10.0476 )(10.0005)
(1-y) lm =
HTU =
267.24
13.03 ( 10 ) (0.976)
Y1= 0.8xi x1 =
y1
0.8
2.10
0.0476
0.8
= 0.0595
L'
( V ' )min. =
L'
( V ' )min=
Y 1Y 2
X 1X 2
0.050.0005
0.0595
0
10.0595
= 0.7824
L'
( V ' )min = 1.2 (0.7824) = 0.9389
L'
Y1 Y2 = ( V ' ) (x1 - 0)
x1
NTU =
0.04760.04
(0.00050)
0.0005
(0.04760.04)
ln
0.04760.0005
=18.05
ft =
38
ft
= 0.976
X1 =0
xa =8
L =1
Let x =cream
Y = steam
10x = y
10x1 = y1 ; X1 = ?
y1 = ya = 10x1
y2 = 0 (no taint in the entering steam.)
xa = 8 ppm.
Basis is 1 kg of cream
1/0.75 = 1.33 is the ratio of cream flow rate to steam flow rate = L/V.
If n = 1,
y2 = x1L/V + ya - xaL/V
y2 = 0 = x11.33 + 10x1 - 8 x 1.33
x1 = 10.64/11.33
=
0.94 ppm
y1
P1= 1 atm ; T= 80F
y7=0.30 ft3
X=6
We find yNH3 = 1.414 x NH3 at 80 F.
x 0xN
xN
xN
1(Mv / L)
Mv
1
N +1
L
( )
V
=
L
30 scf Air
1 lbwater
x 0xN
xN
xN
1lbmol Air
18 lbwater
1(Mv / L)
Mv
1
N +1
L
( )
Where:
xN =X6 = ?
x0= 0.001 ; m= 1.414 ; b= 0
xN = y7/m = 0 ; V/L = 1.43 ; N = 6
xN =
1(Mv / L)
Mv
1
N +1
L
xN =
1.414
1.43
( )
1
1(1.414 (1.43))
( )
(0.001)
HG =
Nsc =
HL =
D
0.226
1.34
0.958
4.069
( 0.660 )0.5 ( 6.782 )
D ab
0.357
fp
1.866 x 105
1.166 (0.1670 x 104)
=
0.2426
m(0.796ft)
-0.5
0.8007 x 103
995.68(2.270 x 1009)
Nsc
372
0. 5
0.958
Gx /
6.782
0.8937 x 103
354.3
0.3
0.5424
0.678 )0.35 =
HL =
0.3
0.357
1.34
354.3
372
)0. 5 (
0.2306 m
(0.759ft)
4.069
3
0.8007 x 10
6.782
0.8937 x 103
0.2322
m
10. A tray tower is to be designed to absorb SO2 from an air stream by using
pure water at 293 K. the entering gas contains 20 mol % SO2 and that
leaving 2 mol 5 at a total pressure of 101.3 kPa. The inert flow rate is 150 kg
water/ h-m2 . Assuming an overall tray effieciency of 25%., how many
theoretical trays and actual trays needed?
Solution:
V = 150 / 29 = 5.18 kg mol inert air/ h-m2
L = 6000/18.0 = 333 kg mole inert water/ h-m2
YN+1 = 0.20, y1 = 0.02 and x0 = 0
0
333 ( 10 )+ 5.18
0.20
xN
0.002
10.20 ) = 333 ( 1xN )+ 5.18 ( 10.02 )
XN = 0.00355
0
333 ( 10 )+ 5.18
YN+1 = 0.07
yn+1
xN
0.002
1 yn+1 ) = 333 ( 1xN )+ 5.18 ( 10.02 )
0 + 5.18
0.07
xN
0.002
10.07 )= 333 ( 1xN ) + 5.18 ( 10.02 )
XN = 0.000855
YN+1 = 0.13 ; XN = 0.00201
No. of theoretical trays
No. of actual trays = 2.4/
2.4
trays
=
9.6
trays
0.25 =