Homework #4
Do problems 10.1, 10.3, & 10.5 on pages 234-235.
CBEN 408 Spring 2016
-1-
March 19, 2016
Solutions
Problem #10.1 (10 points)
Feed gas enters an amine treating unit at a rate of 100 MMscfd (2.68 × 106 m3/d). The gas contains
5% CO2 and no H2S. The treater uses 18 wt% MEA solution. Assume the lean amine contains 0.12
mol CO2/mol of amine and the rich amine contains 0.45 mol CO2/mol amine. To meet the pipeline
specification of 2% CO2, what minimum lean amine flow rate in gpm (m3/h) is required? Amine
properties are given in Appendix B.
Solution
The table to the right shows the
specified values, intermediate results, &
final result (686 gpm) for this
problem. 1,2
Inlet
Outlet
Absorbed
Outlet Specifications
CO2 (mol%)
2.0%
Composition (mol%)
Sweet Gas
CO2
Total
95.0%
5.0%
100.0%
98.0%
2.0%
100.0%
Sweet Gas
CO2
Total
95.00
5.00
100.00
95.00
1.94
96.94
CO2
MEA
9.150
3.548
Gas Rate (MMscfd)
3.06
3.06
Rate (lb.mol/min)
Loadings (mol acid gas/mol amine)
Rich
Lean
Pickup = Rich - Lean
Lean Amine (wt%)
Lean Amine density (lb/gal)
Mol Wt
0.45
0.12
0.33
18.0%
8.40
MEA
Lean Amine Rate (lb/min)
Lean Amine Rate (gal/min)
5.602
16.975
61.08
5,760.3
685.7
There are a couple calculations that should be noted:
• The allowable amount of CO2 in the outlet gas can be directly calculated knowing that the
amount of hydrocarbons remains the same (i.e., they are not absorbed). So:
=
yCO2
NC CO2
yCO2 C
C
⇒ N
=
N(C
CO2
C
C
1 M yCO2
N(C + NCO2
=
0.02
=
( 95) 1.94 MMscfd
1 M 0.02
Note that the numbers shown may be slightly different from hand calculations because of the number of
digits retained in intermediate calculations.
2 The pure component parameters used are from Tables B.1 & B.15 as well as Figure B.1 in the Kidnay, et. al.
text book.
1
CBEN 408 Spring 2016
-2-
March 19, 2016
•
•
•
The amount of CO2 absorbed in the amine is the difference between that in the Inlet gas &
that in the Outlet gas.
The pickup by the amine solution is the difference between the rich loading & the lean
loading (i.e., the residual CO2 in the lean amine solution after stripping as much CO2 as
possible).
The required molar flowrate of the amine (in the MEA/water solution) is found from the
amount of CO2 absorbed & the pickup:
NC CO2
NC MEA =
LCO2/MEA
•
M1
M1

scf  
min 
6 scf CO2  
 379.46
  24 ⋅ 60
 3.06 ×10

day  
lb.mol  
day 
lb.mol MEA
=
16.975
mol CO2
min
0.33
mol MEA
The required mass flowrate of the total lean amine solution (MEA & water) is found from
the this molar amount of MEA:
 Solvent =
m
N MEA MMEA
wMEA
lb.mol MEA 
lb 

 16.975
 61.08

lb
min
lb.mol 
=

5760
0.20
min
and the required standard liquid volumetric flowrate is found using the liquid density:
lb
5760

m
Solvent
min 686 gal
*
V
=
=
=
Solvent
*
lb
min
ρSolvent
8.40
gal
CBEN 408 Spring 2016
-3-
March 19, 2016
Problem #10.3 (10 points)
The 18 wt% MEA used in the treating unit in Exercise 10.1 is replaced with 31 wt% DEA solution. If
the lean amine contains 0.05 mol CO2/mol of amine and the rich amine contains 0.38 mol of
CO2/mol of amine, how much will the required lean amine solution flow rate change (based on
density at 60°F (15°C))?
Solution
The table to the right shows the
specified values, intermediate results, &
final result lean solvent flowrate for this
problem. 1,2 Since the flowrate found in
Problem #1 was 686 gpm of 18% MEA,
the flowrate will change by 22 gpm less
31% DEA (a decrease of 3%).
Inlet
Outlet
Absorbed
Outlet Specifications
CO2 (mol%)
2.0%
Composition (mol%)
Sweet Gas
CO2
Total
95.0%
5.0%
100.0%
98.0%
2.0%
100.0%
Sweet Gas
CO2
Total
95.00
5.00
100.00
95.00
1.94
96.94
CO2
DEA
9.150
3.548
Gas Rate (MMscfd)
3.06
3.06
Rate (lb.mol/min)
Loadings (mol acid gas/mol amine)
Rich
Lean
Pickup = Rich - Lean
Lean Amine (wt%)
Lean Amine density (lb/gal)
Mol Wt
0.38
0.05
0.33
31.0%
8.67
DEA
Lean Amine Rate (lb/min)
Lean Amine Rate (gal/min)
5.602
16.975
105.14
5,757.4
664.1
There are a couple things to note about the calculations:
• The amount of CO2 absorbed in the amine is the difference between that in the Inlet gas &
that in the Outlet gas.
• Even though the rich & lean loadings are different between the two solvents the pickup will
be the same, 0.33 mol CO2/mol amine. This means that the molar flowrate of each amine
within its total solvent will be the same.
• The required mass flowrate of the total lean amine solution is essentially the same between
the two cases. The 3% reduction in volumetric flow comes from the 3% difference in the
mass densities.
Note that the numbers shown may be slightly different from hand calculations because of the number of
digits retained in intermediate calculations.
2 The pure component parameters used are from Tables B.1 & B.15 as well as Figure B.1 in the Kidnay, et. al.
text book.
1
CBEN 408 Spring 2016
-4-
March 19, 2016
Problem #10.5 (20 points)
The gas entering an MDEA treating unit contains 0.5 mol% H2S and 8 mol% CO2. Inlet flow rate is
120 MMscfd (3.22 × 106 m³/d). The treater inlet pressure is 1000 psig (70 barg). The plant
dehydrates the gas and then sells it as pipeline-quality gas. The pipeline specification requires that
the gas contain no more than 0.25 g/100 scf (6 mg/Nm3) of H2S and a maximum of 2% mol CO2.
a. Assuming that the rich MDEA holds 0.4 mol of acid gas/mol of MDEA and the lean amine
contains essentially no H2S and 0.005 mol of CO2/mol amine, what is the minimum amine
circulation required (gpm [m3/h]) to obtain pipeline-quality gas?
b. How much sulfur (LTD (tonnes/day)) would be produced from the acid gas leaving the
treater, assuming 99.95% sulfur recovery? (See Chapter 16 for a discussion of sulfur
recovery.)
Solution
The table to the right shows the
specified values, intermediate
results, & final results. 1,2 The key
results are:
• Minimum MDEA flow is
1033 gpm.
• The amount of sulfur
produced from the SRU
(Sulfur Recovery Unit) is
23 LTPD (long tons per
day, where 1 long ton is
2200 lb).
Note that the numbers shown may be slightly different from hand calculations because of the number of
digits retained in intermediate calculations.
2 The pure component parameters used are from Tables B.1 & B.15 as well as Figure B.1 in the Kidnay, et. al.
text book.
1
CBEN 408 Spring 2016
-5-
March 19, 2016
There are a couple calculations that should be noted
• The H2S spec on the outlet gas is a ratio of mass H2S to the total moles of gas. The mass
units of “grains” is not one normally used; a grain is 1/7000 of a lb. This can be converted to
a molar ratio:
−1
lb  
scf 
 0.25

lb H2S  34.082

  379.49

lb.mol  
lb.mol 
 7000

y=
= 3.98 ×10−6 ,
H2S
100 scf
•
essentially 4.0 ppmv.
The allowable amounts of CO2 & H2S in the outlet gas can be calculated after first equating
the hydrocarbon rate in both the inlet sour gas & the outlet sweet gas (i.e., we assume that
the hydrocarbons are not dissolved in the amine solution). From the inlet:
NC (C =
yCO2 yCO2 )Sweet NC Sweet =
yCO2 yCO2 )Sour NC Sour
(1 −−
(1 −−
NC (C =
(1 M yCO2 M yCO2 )Sweet NC Sweet =
(1 M 0.08 M 0.005)(120 MMscfd )
NC Sour
=
= 109.8 MMscfd
(109.8 MMscfd )
NC (C
=
(1 M yCO2 M yCO2 )Sour
(1 M 0.02 M 0.00000398 )
= 112.04 MMscfd
(Note that the amount of H2S is so small that it does not significantly affect the overall
amount of sweet gas is produced.) So the molar flow of the acid gases out of the absorber
will be:
=
NC CO2,Sweet yCO2,Sweet
=
NC Sweet Gas
=
N
y=
(2S,Sweet
(2S,Sweet NSweet
•
=
(0.02
)(112.04 )
2.24 MMscfd
=
(0.00000398
)(112.04 )
0.00045 MMscfd
The required amount of MDEA solution is from an additive effect for the pickup of both the
CO2 & the H2S:
CBEN 408 Spring 2016
-6-
March 19, 2016
NC CO2
=
NC MDEA
•
LCO2/MDEA
+
NC H2S
LH2S/MDEA


scf H2S  
6 scf CO2 
0.5996 ×106
M1
M1
  7.36 ×10



day  
day   
scf  
min 

=
+
379.46
24
⋅
60
 

mol CO2
mol H2S

 
lb.mol  
day 
0.395
0.40


mol MDEA
mol MDEA


lb.mol MDEA
= 36.836
min
The required mass flowrate of the total lean amine solution (MDEA & water) is found from
this molar amount of MDEA & the concentration of the lean amine. The concentration was
not given in the problem statement. We will use the highest normal concentration (50 wt%)
since that will lead to the lowest circulation rate:
 Solvent =
m
N MDEA MMDEA
wMDEA
lb.mol MDEA 
lb 

 36.836
 119.16

lb
min
lb.mol 
=

8779
0.50
min
The required standard liquid volumetric flowrate is found using the liquid density:
lb
8779

m
gal
Solvent
min

V
=
=
= 1,033
*
lb
min
ρSolvent
8.50
gal
The SRU (Sulfur Recovery Unit) converts H2S to sulfur on a mole-to-mole basis. So the mass
rate of sulfur recovered uses the molecular weight of sulfur & the recovery efficiency of the
SRU:
*
Solvent
•
−1

lb.mol S 
lb  
min 
lb
long ton

S =
m
0.9995 ⋅  1.097
23
 32.06
 24 ⋅ 60
 2200
 =
min
lb.mol
day
long
ton
day





The required standard liquid volumetric flowrate is found using the liquid density:
CBEN 408 Spring 2016
-7-
March 19, 2016