Homework #5
Do problems 11.1, 11.2, 11.3, 11.4, & 11.5 on page 265.
CBEN 408 Spring 2016
-1-
March 31, 2016
Solutions
Problem #11.1 (10 points)
A gas contract specifies that the water content must be 4 lb/MMscf or lower. What is this
specification in mg/Nm3, ppmv, and ppmw? Assume the molar mass of the gas is 18.
Solution
The table to the right shows the results for this
problem.1
Note the following for the calculations:
 The conversion for scf to Nm³ involves not only the unit conversion for ft³ to m³ but must
also take into account the change in the reference temperature (standard temperature is
60°F & the normal temperature is 0°C/32°F). The conversion is the same whether the
MMscf is on a wet or dry basis. So:
lb  
g 
mg   6 scf  
lb 

X  4
   453.59237    1000
   0.02679

   10
lb  
g  
MMscf  
MMscf 
 MMscf  
mg
 68
Nm3

The ideal gas volume fraction (or ppmv) is essentially the molar fraction. The conversion is
slightly different depending on whether the MMscf are on a dry or wet basis (though for all
practical purposes the numbers will be the same because the molecular weights are about
the same & because the water content is so low). For the MMscf on a wet basis:
lb  
lb 

 4 MMscf    18.015 lb.mol 
 
  106  84 ppmv
X

6 MMscf 
1   379.5  10
lb.mol 

For the MMscf on a dry basis we have to add the contribution of the water into the
denominator:
lb  
lb 

 4 MMscf    18.015 lb.mol 

 

X
 106  84 ppmv
lb  
lb 

6 MMscf  
1   379.5  10
 4
 18.015
lb.mol   MMscf  
lb.mol 

Note that the numbers shown may be slightly different from hand calculations because of the number of
digits retained in intermediate calculations.
1
CBEN 408 Spring 2016
-2-
March 31, 2016

Just like the ppmv, the mass fraction (or ppmw) is essentially the same whether the MMscf
is on a wet or dry basis, mostly because the molecular weigt of the hydrocarbon portion is
essentially the same as the molecular of water. Since these molecular weights are
essentially equal:
lb 

 4 MMscf 


X
 106  84 ppmw
lb
MMscf

 

 18 lb.mol    379.5 lb.mol 

 

CBEN 408 Spring 2016
-3-
March 31, 2016
Problem #11.2 (10 points)
Estimate the water content of a natural gas (molar mass of 18) at 1000 psia (69 bara) and 100°F
(38°C) by using Figure 11.1a and by using Equation 11.2. (See Appendix B for vapor pressure of
water.)
Solution
The table to the right shows the specified
values as well as intermediate & final
results.1,2
The determination of the water
content from Figure 11.1a is
pretty straightforward.
The vapor pressure of water can be determined using the equation & coefficients in Appendix B.
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 & chart values are from the Kidnay, et. al. text book.
1
CBEN 408 Spring 2016
-4-
March 31, 2016
B


vap
PH2O
 exp  A   C ln T   DT E 
T


7258.2


 exp  73.649 
 7.3037 ln T   4.1653  106 T 2 
T


7258.2
2

 exp  73.649 
 7.3037 ln 310.9  4.1653  106 310.9  
310.9


 6550 Pa  0.95 psia
So:
yH2O 
vap
PH2O
0.95

 0.00095
P
1000
and the water content (based on wet gas scf) will be:
X H2O 
yH2O MH2O  0.00095118.015

 45.1 lb/MMscf
Vstd
379.49  106


Note that the water content based on dry gas basis will be:
X H2O 
0.00095118.015
yH2O MH2O

 45 lb.2/MMscf
1  yH2O Vstd 1  0.000951 379.49 106


Note: this value can also be estimated from appropriate software, such as HYSYS. Using a
methane/ethane mixture as shown & the Peng-Robinson EOS the water content will be:
X H2O 
0.00121518.015
yH2O MH2O

 58 lb/MMscf
1  yH2O Vstd 1  0.001215 379.49 106
CBEN 408 Spring 2016


-5-
March 31, 2016
Problem #11.3 (5 points)
Determine the correction needed for molar mass of the gas in Exercise 11.2 using Figure 11.1b.
What is the additional correction needed if the gas is in contact with produced water containing 2%
by weight brine?
Solution
The table to the right shows the specified
values as well as intermediate & final
results.1,2
The dry gas’s specific gravity is nearly
0.6 so there is very little correction to
the figure’s value. The correction for
the brine’s salinity is greater, &
reduces the water content to
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 & chart values are from the Kidnay, et. al. text book.
1
CBEN 408 Spring 2016
-6-
March 31, 2016
Problem #11.4 (15 points)
A field TEG unit is required to produce a gas (molar mass of 22) that will not condense water at
40°F and 500 psia (4°C and 34 bara). Gas to the glycol unit is water saturated at 110°F and 300 psia
and the flow rate is 80 MMscfd (25 N-m³/s).
a. Using a TEG rate of 3 gal/lb (0.025 m³/kg) of water removed, what will be the minimum
glycol circulation rate in gpm (m3/h)?
Utility costs are $0.04 per kwh for the compressor motor and $5.00 per MMBtu ($4.75/GJ) for fuel
gas.
b. The TEG unit has a regeneration duty equal to the heat required to vaporize the removed
water with an overall efficiency of 33%. Assume TEG regeneration temperature of 400°F
(204°C).
Compute the daily heater fuel energy cost.
Solution
The table to the right shows the specified
values as well as intermediate & final
results.1,2 Note the following about the
given data & the results:
 Though not clearly called out the
rate given, 80 MMscfd, will be used
for the dry gas portion of the gas.
The contribution for the water in
the gas needs to be added on top of
this.
 The water heat of vaporization is
calculated specifically at the
regeneration temperature, 400°F.
 There is no compression in this
process so these cost contributions
are absent.
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 & chart values are from the Kidnay, et. al. text book.
1
CBEN 408 Spring 2016
-7-
March 31, 2016
The first step is to determine the
water contents at the inlet & dried
conditions. The copy of Fig. 11.1a on
the right shows that these water
contents are 200 & 14.5 lb/MMscf,
respectively.
The figure below 11.1a shows the
specific gravity corrections. The
correction at the inlet temperature,
110°F, can be read from the chart.
However, the expected curve at the
dried condition is not on the chart; it
is expected to be 1.
Note: these values can also be estimated form appropriate software, such as HYSYS. Using a
methane/ethane mixture as shown & the Peng-Robinson EOS the water content at the initial
condition (dry gas basis) will be:
X H2O 
0.004520 18.015
yH2O MH2O

 215.6 lb/MMscf
1  yH2O Vstd 1  0.004520  379.49 106


and the dried target will be:
X H2O 
0.000278 18.015
yH2O MH2O

 13.2 lb/MMscf
1  yH2O Vstd 1  0.000278  379.49 106
CBEN 408 Spring 2016


-8-
March 31, 2016
Now that we know the water contents we need to determine how much water is to be removed.
Noting that the gas flowrate is on a dry basis:
mabs  196.8  14.580  14580 lb/day
The TEG flowrate is based on the given rate of 3 gal TEG per lb H2O:
gal  
lb H2O   1 day 
gal EG

*
.
VTEG
 3
  14580
  30.4

day   24  60 min 
min
 lb H2O  
The heat required to boil off the absorbed water & regenerate the TEG (as well as ultimately
figuring out how much fuel gas is needed to do this) is based on the heat of vaporization of water at
the regeneration temperature, 400°F. From the steam tables in the back of the textbook the
enthalpy of saturated vapor & liquid are 1202.2 Btu/lb & 375.3 Btu/lb, respectively; the heat of
vaporization will be the difference between these two values, 826.9 Btu/lb. So the heat needed to
regenerate the TEG & the associated cost is:
Qregen

lb H2O  
Btu 
 14580 day   826.9 lb H2O 
Btu



 36.5  106
0.33
day

Cost    36.5

MMBtu  
$

  $183 per day
5
day   MMBtu 
There is no recompression is the TEG dehydration process so there are no associated costs, either.
CBEN 408 Spring 2016
-9-
March 31, 2016
Problem #11.5 (20 points)
A field 4A molecular sieve adsorbent system is required to produce a gas (molar mass of 22) that
will not condense water at 40°F and 500 psia (4°C and 34 bara). Gas to the unit is water-saturated
at 110°F (43°C) and 300 psia (21 bara) and the flow rate is 80 MMscfd (89 x 103Nm³/h).
The loading averages 10% (by weight) of water on the adsorbent bed, the superficial velocity in the
adsorbent vessel is 40 feet per minute (0.20 m/s). The dehydrator uses an adsorption time of 8
hours, and heats and cools in 4 hours each.
a. What mass of adsorbent is needed to remove the water?
b. What are the dimensions of the adsorbent bed?
For the adsorbent system, assume the vessel wall thickness is 2” (50 mm) and uses steel with a
density of 500 lb/ft³ (6.4 kg/m³). The heat capacity of the steel is 0.12 Btu/lb-°F (0.5 kJ/kg-K).
Ignore the vessel heads but include 5 feet (1.5m) of extra shell for bed supports and screens. Ignore
heat losses to the surroundings and assume the heat of adsorption is equal to the heat of
vaporization of water.
c. What is the required regeneration duty to heat the bed to 450°F (230°C)?
The adsorbent system has a pressure drop of 15 psi (1 bar). Utility costs are $0.04 per kWh for the
compressor motor and $5.00 per MMBtu ($4.75/GJ) for fuel gas. Assume the regeneration gas
heater has an efficiency of 75%.
d. Compute the daily energy cost for both heater fuel and compression (see Chapter 9 for
compression).
Solution
The table to the right shows
the initial calculations to
determine the amount of
water to be adsorbed.1,2 Note
the following about the given
data & the results:
 Though not clearly
called out the rate
given, 80 MMscfd, is
for the dry gas portion
of the gas. The
contribution for the
water in the gas needs
to be added on top of
this.
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 & chart values are from the Kidnay, et. al. text book.
1
CBEN 408 Spring 2016
- 10 -
March 31, 2016
The first step is to determine the
water contents at the inlet & dried
conditions. The copy of Fig. 11.1a on
the right shows that these water
contents are 200 & 14.5 lb/MMscf,
respectively.
The figure below 11.1a shows the
specific gravity corrections. The
correction at the inlet temperature,
110°F, can be read from the chart.
However, the expected curve at the
dried condition is not on the chart; it
is expected to be 1.
Note: these values can also be estimated form appropriate software, such as HYSYS. Using a
methane/ethane mixture as shown & the Peng-Robinson EOS the water content at the initial
condition (dry gas basis) will be:
X H2O 
0.004520 18.015
yH2O MH2O

 215.6 lb/MMscf
1  yH2O Vstd 1  0.004520  379.49 106


and the dried target will be:
X H2O 
0.000278 18.015
yH2O MH2O

 13.2 lb/MMscf
1  yH2O Vstd 1  0.000278  379.49 106
CBEN 408 Spring 2016


- 11 -
March 31, 2016
The next step is to determine the bed
characteristics as shown in the table on
the right. Note the following about the
given data & the results:
 The actual flowrate is used to
determine the superficial
velocity in the bed, not the
standard velocity. The actual
flowrate can be calculated from
the standard flowrate by using
pressure & temperature ratios
(actual to standard) but must
also include the water content.
The calculation for the actual flowrate is:
 P*
Vact  V * 
 Pact
  Tact
 *
 T
 14.69596 psia   110  459.67 °R 

   80.33 MMscfd   300 psia  60  459.67 °R   4.13 MMcfd




and the minimum bed diameter to meet the superficial velocity spec is:
Amin
 P*
V
 act  V * 
vmax
 Pact
Amin 
  Tact
 *
 T
ft³
4.13  106

day
 71.7 ft²

  40 ft   24  60 min 
 min 
day 


4 71.7 ft²
4 Amin
 2
Dmin  Dmin 

 9.56 ft
4


In the U.S. vessel dimensions are usually in ½ ft increments, so the actual diameter will be 10 ft.
CBEN 408 Spring 2016
- 12 -
March 31, 2016
Next, the mass amount of mole sieve material is used to determine the bed height:
msieve
 14580 lb 

  8 hr 
24 hr 


 48620 lb Sieve
lb H2O
0.1
lb Sieve
Vsieve 
hmin 
48620 lb Sieve
 1080 ft³ Sieve
lb Sieve
45
ft³
1080 ft³ Sieve
0.25 10 ft 
2
 13.8ft  h  14 ft
Now we can determine the heat
required for regeneration. Note
the following about the given data
& the results:
 Per guidelines there is a
50°F difference between
the regeneration
temperature & the regen
gas to the mole sieve.
We’ll assume the
temperature given is the
regen temperature & the
temperature of the regen
gas (not requested) is
50°F above this.
 Heat must not only be
provided to vaporize the
water off of the adsorbent
at the regen temperature.
But we must also heat up
the water, vessel, & sieve
from the bed adsorption
temperature to the
regeneration temperature
(110°F to 450°F).


Per the guidelines only 40% of the heat from the hot regen gas is actually absorbed by the
water, vessel, & mole sieve material.
The number of cycles per day is not controlled by the heating time, but rather the combined
heating & cooling time (which will match up to the time for adsorption). So there will be 3
cycles per day.
CBEN 408 Spring 2016
- 13 -
March 31, 2016

The actual amount of mole sieve material is slightly different from the minimum amount
calculated because of the change of minimum bed dimensions to ones at ½ ft increments.
The volume of steel (neglecting the heads) will be found from the thickness of the shell:


2
 2

2
2
Douter  Dinner
H
Dinner  2t  Dinner
H   tDinner  t 2 H   Dinner  t tH
4
4
.
2  2 

   10 
ft
ft  14  5 ft   101.1 ft³
12 

 12 
Vsteel 







and the volume of the mole sieve bed is:
 2
Dinner H
4
.

2
 10 ft  14 ft   1100 ft³
4
Vsieve 
The total heat needed to bring the vessel & sieve up to the regeneration temperature (per cycle)
will be:
Qsolid   sieveVsieveC p ,sieve  vesselVvesselC p ,vessel Tregen  Tads 
 lb 
Btu  
lb 
Btu  


  45  1100ft³  0.25
  500  101.1ft³  0.12
 450  110 °F  .

°F lb  
ft³ 
°F lb  


 ft³ 
 6.27  106 Btu
The heat needed to bring the water to the regen temperature & then vaporize (per cycle) will be:


Qwater  mwater  H Tregen   H Tads    H vap  


  4840 lb 351.88  775.41 Btu/lb   5.48  106 Btu
The total heat needed to the regen gas will be:
Qsolid  Qwater  Qloss
0.4

Btu 
Btu
Btu
 2.5 6.27  5.48  106
 29.4 106
 88.1 106

cycle 
cycle
day

Qregen 
with the daily total coming from 3 cycles per day. So the associated cost is:

Cost    88.1

MMBtu  
$

  $441 per day .
5
day   MMBtu 
CBEN 408 Spring 2016
- 14 -
March 31, 2016
The compression costs will come from
recompressing the dried gas back to the
inlet pressure. Note the following about
the given data & the results:
 The gas’s heat capacity ratio was
not given. A typical value of 1.3
was used.
 The compressor’s efficiency was
not given. A typical value of 70%
was used.
 No information was given about
the outlet temperature from the
mole sieve unit. It was assumed
that there was only a minimal
change, so it is the same as the
inlet.
The isentropic work of compression was calculated as:
 1 / 

  P2 
ˆ

Ws   RT1 
 1
 
  1  P1 



 1.9859 110  459.67 
1.31 /1.3

1.3  300 

1


1.3  1  285 

 58.4 Btu/lb.mol
At 70% efficiency for the compression:
Wˆ
58.4
Wˆ s ,act  s 
 83.4 Btu/lb.mol
IS 0.7
So, the power needed is:

scf   1
lb.mol   
Btu 
Btu
P  N Wˆ s ,act   80  106
 17.6  106
  83.4



day   379.49 scf   
lb.mol 
day

The other power values were determined by converting the units as appropriate.
CBEN 408 Spring 2016
- 15 -
March 31, 2016