lecture 2.pdf

Gas Dehydration
Dr Sourav Poddar
Department of Chemical Engineering
National Institute of Technology, Warangal

Gas absorption – a process when separate gas (vapor)
mixes components are captured by the entire volume
of sorbent liquid (or less frequently – a solid) creating
a solution.
Gas adsorption – Mechanism of gas adsorption,
where gas molecules are adsorbed through van der
Waals forces on the surface of a material without
forming a chemical bond.

Problem
For the given feed stream, estimate the product stream compositions for 98%
propane recovered in the overhead product with a maximum iso-butane content of
the overhead stream of 1%.

Computation Method
In order to determine the design parameters for a fractionation problem, the following method is
recommended:
1. Establish feed composition, flow rate, temperature, and pressure.
2. Make product splits for the column and establish condenser temperature and column pressure. From
column pressure, calculate the reboiler temperature.
3. Calculate minimum number of theoretical stages from the Fenske equation.
4. Calculate minimum reflux rate from the Underwood equations.
Once  is determined, the minimum reflux ratio is:

5. Obtain theoretical
stages/operating reflux relation from

6. Adjust actual reflux for feed vaporization if necessary

• Actual trays at 1.3 times the minimum reflux ratio
Problem:

Determine the minimum number of trays

Correct for change in relative volatility by using
(A)

Find the minimum reflux, Rm
and

Problem
Evaluate the tray efficiency for the system in the previous problem

Solution Steps
The system in previous problem required 21 theoretical stages including the reboiler. The total actual trays is:

Static pressure of reboiler leg

Problem: Calculate the GPM of the Alberta gas given in Table below. Computation of the GPM requires summation of
the product of the number of moles of each component in 1,000 scf of gas by the gallons of liquid per mole for that
component.

Solutions
The Gal/mole for butanes
was taken as the average of
isobutane and n-butane;
the value for C5+ was taken
to be that of pure n-
pentane. The resulting GPM
for this gas is 4.34.

Problem: Estimate the hydrate-formation temperature at 325 psia (22.4 bar) for the gas with the composition in
Table below.
Compare the results from Figure below and equation below. The specific gravity = molar massgas /molar massair =
20.08/28.96 = 0.693.

FIGURE Pressure−temperature curves
for estimation of hydrate-formation
conditions as a function of gas gravity.
These curves should be used for
approximation purposes only. (Adapted
from Engineering Data Book, 2004d.)

Solutions
Figure gives 50°F, and the experimentally reported value for a 0.7-gravity gas is 50F

Problem: A sweet gas with a specific gravity of 0.73 leaves a gas-liquid separator at
100F and 600 psia saturated with water. The gas drops to 35F before reaching the
next booster station at 500 psia. Assume no hydrocarbon condensate has formed in the
line. Calculate how much methanol must be added to prevent hydrate formation
between the separator and the booster station per MMscf. Repeat the calculation for
ethylene glycol, which is added in an 80 wt% mixture with water.

Solutions
Use of above Equation, shows that the hydrate-formation temperature at 600 psia is
59F. This value means a 59F – 35F = 24F subcooling into the hydrate region, and
hydrate formation is probable without inhibition.
Determination of the inhibitor rate requires:
1. Determination of the amount of liquid water formed
2. Calculation of the required amount of inhibitor in the water phase
3. Calculation of the required amount of inhibitor in the gas phase
1. The water content of the gas leaving the separator is 95 lb/MMscf and 13
lb/MMscf at 35F and 600 psia. (From figure). Therefore, 95 lb – 13 lb = 82 lb of
water per MMscf drops out in the line, assuming worse-case conditions.

Methanol Requirement
2. To estimate the concentration of methanol, rearrange to give ln xW = -t(F)/129.6 =
-24/129.6 = -0.185
xW = 0.831 or xMeOH = 0.169 mole fraction. This value is (0.169*32)/(0.831*18 + 0.169*32) = 26.6 wt% or 0.362 lb methanol/lb
water. Thus, the methanol needed in the water phase is 0.362 * 82 = 29.7 lb/M Mscf.
3. To estimate the methanol in the vapor phase use Figure -a below which, at 35F and 600 psia, gives 1.22 lb methanol
vaporized per wt% methanol in the aqueous phase, or 1.22*26.6 = 32 lb methanol per MMscf. Thus, the vapor phase consumes
more methanol than does the aqueous phase. The total amount of methanol required is (26 + 32) = 58 lb/MMscf. If a
condensate phase was present as well, the losses estimated by use of Figure – b would have to be added into that phase. This
amount of methanol will be much less than what goes into the aqueous
and vapor phase in gas lines.

Ethylene Glycol Requirement
2. Use Equation with the constant 3,225 and a molar mass of ethylene glycol of 62 to obtain the
mass fraction of pure glycol required:
However, the glycol is diluted to 80 wt%. To obtain the mass of inhibitor solution added per unit mass of free water present
to obtain the desired concentration, we use X0 /(Xi – X0), where X0 is the weight fraction of inhibitor in the solution to be
added. The amount of inhibitor solution added per pound of free water initially present is then 0.8/(0.8 – 0.40) = 2.00, and
the total amount of ethylene glycol solution added is 2.00*82 = 164 lb/MMscf. At these conditions, glycol loss into the
vapor phase is negligible, so the total amount of solution required is 164 lb/MMscf.

Problem: Assume an ideal gas, and compute the ratio of heat capacities for the southwest Kansas gas in Table * at
100F. Assume butanes to be all n-butane and C5+ fraction to be hexane. (Table -*)

The CVM for the mixture is CPM - R = 9.43 – 1.99 = 7.44
Btu/lb-mol-R and g = 9.43/7.44 = 1.33.
Solutions

Problem: Assume ideal gas behavior, and compute the
reversible work required to compress a natural gas mixture
from 10 to 60 psig, both isothermally and adiabatically, with
an initial temperature of 80F. Also, calculate the exit
temperature for adiabatic compression. The molar mass of the
gas is 18, and the ratio of heat capacities is 1.15. Assume the
compressor is 100% efficient for this calculation.

Isothermal Compression
Solutions
Adiabatic Compression
To compute the work, use

Problem: Determine the isentropic efficiency of a compressor if the
polytropic efficiency is 77% and the gas has g = 1.15. Assume the molar
mass of the gas is 18, the inlet temperature is 80F, and the compression
ratio is 3.0.

Solutions
Conversion from polytropic to adiabatic efficiency requires the following steps:
1. Calculate ideal polytropic work.
2. Calculate the actual work by dividing polytropic work by polytropic efficiency.
3. Calculate isentropic work.
4. Calculate isentropic efficiency from ratio of isentropic work to polytropic work
Step 1. Both g and hP are given so Equation can be used to determine k for use in
Equation

Step 2. The actual work required is wS = -72.1/0.77 = -93.6 Btu/lb
Step 3. The isentropic work required is
Step 4. The isentropic efficiency is then -70.3/-93.6 = 0.75 or 75%, two percentage points below the polytropic
efficiency.

Problem: Compute the overall thermal efficiency of a gas
turbine that has a heat rate of 8,500 Btu/hp-h (12,000
kJ/kWh).

Problem: Calculate the water content of the sweet natural gas shown in Table below at 300 psia (20.7 bar) and 80F
(26.7C) by use of Equation –a and Figure -**
Equation -a
Figure-**a,b

From Table, the MW of the gas mixture is 18.41 and the specific gravity is
From Figure - **
WSat = 85 lb/MMscf (1,400 mg/Sm3). Correct for specific gravity by obtaining C from Figure -**b, (C = 0.99), and
multiplication gives
The values differ by 4%. Increasing the pressure to 1,000 psia (69 bar). Equation –a is 27% below the value obtained
from Figure - **a.

Problem: An existing 4A molecular sieve bed has been processing 80 MMscfd
on a 12-hour cycle with two beds. Exit gas goes to a cryogenic turboexpander
section. Gas flow is increased to 100 Mscfd. Estimate the increased pressure
drop and determine whether the bed capacity allows continued operation on a
12-hour cycle or the cycle time should be changed. The gas enters the bed at
120°F and 950 psig. Water content is 60% of saturation at 120°F. The molar
mass of the gas is 18.5, with a viscosity of 0.014 cP and a compressibility factor
of 0.84.
The adsorption bed contains 41,000 lbs of 1/8-inch diameter beads with a bulk density of 44
lb/ft3. The inside wall diameter of the bed is 7.5 ft. The absorbent was installed 2 years ago.

Solutions
Pressure-Drop Calculation
Determine the height of the adsorption bed. The volume of adsorbent V = mass/adsorbent density = 41,000/44 =
932 ft3. The cross-sectional area of the bed A = πD2/4 = π(7.5)2/4 = 44.2 ft2. The bed height then is V/A = 932/44.2
= 21.1 ft.
Gas density,
ρ = MW P/(zRT) = 18.5 × (950 + 14.7)/[ 0.84 × 10.73 (460 + 120)] = 3.41 lb/ft3.
To obtain superficial velocity, we first need the increased actual volumetric flow rate, Q

Use of ΔP/L (psi/ft) = BμVS + CrVS
2, with B = 0.0560 and C = 0.0000889 gives
∆P/L = (0.0560)(0.014)(22.4) + (0.000089)(3.41)(22.4)2= 0.170 psi/ft.
Total bed pressure-drop is 0.170 psi/ft  21.1 ft = 3.58 psi, which is in a good operating range
Bed Capacity Calculation
To determine the capacity, first calculate the capacity per pound of absorbent. The bed has aged, so at 75F the
absorbent should hold 13 lb water/100 lb of sieve if the entering gas is saturated with water vapor. However, the gas
enters at 60% of saturation. With either , the absorbent holds only 97% of capacity or 12.6 lb
water/100 lb sieve. The temperature effect from Figure below is to reduce capacity to 88%, so the bed should hold
12.6  0.88 = 11.1 lb water/100 lb sieve in the equilibrium zone.
The length of the MTZ is calculated by either Equation (above)
LMTZ (ft) = 2.5 + (0.025  22.4) = 3.1 ft.

This value represents 15% (3/21) of the bed height. Assume the MTZ holds 50% of the equilibrium
loading; the bed should hold
11.1/100  41,000(0.85 + 0.15  0.5) = 4,210 lb of water.
From Figure -**a, the water content at saturation for a gas at 950 psig and 120F is 100 lb/MMscf. Essentially,
all water into the bed must be removed. The gas enters at 60% of saturation, which is 0.6  100 = 60 lb/MMscf.
For a 12-hour cycle, the water adsorbed is 60 *100  0.5 = 3,000 lb water. The bed is slightly oversized but can
remain on a 12-hour cycle

Determine the total bed pressure-drop of the adsorption
bed. The volume of adsorbent V = mass/adsorbent density =
41,000/44 = 932 ft3. The cross-sectional area of the bed A =
πD2/4 = π(7.5)2/4 = 44.2 ft2. The bed height then is V/A =
932/44.2 = 21.1 ft.
Given: B = 0.0560 and C = 0.0000889
Problem

Solutions
Gas density,
r = MW P/(zRT) = 18.5 *(950 + 14.7)/[ 0.84 * 10.73 (460 + 120)] = 3.41 lb/ft3.
To obtain superficial velocity, we first need the increased actual volumetric flow rate, Q

Problem: Estimate the heat required to regenerate an adsorption bed that
holds 40,000 lb of 4A molecular sieve and 4,400 lb of water. The vessel contains
55,000 lb of steel, and the regeneration temperature is 600°F. The bed
operates at 100°F. Also estimate the gas flow rate under the assumption that
the CP for the gas is 0.68 Btu/lb-°F, that 60% of the regeneration time involves
heating the bed, and that the bed is on an 8-hour cycle. The gas leaves the
regeneration gas heater at 650°F.

Solutions
Use Equations
through
Where, qw = heat required to desorb the water , qsi = heat the adsorbent, and qst = heat the vessel walls
to compute the heat load. The temperature Trg = 600 - 50 = 550F.

Water: qW = (1,800 Btu/lb)(lbs of water on the bed) = 1,800 *4,400 = 7,920 Btu
Sieve: qsi = (lb of sieve)(0.24 Btu/lb-F)(Trg - T1) = 40,000*0.24 (550 - 100) = 4,320 Btu
Steel: qst = (lb of steel)(0.12 Btu/lb-F)(Trg - T1) = 55,000 *0.12 (550 -100) = = 2,970 Btu
Use Equation to obtain the total heat requirement
q = 2.5 (qW + qsi + qst) (1.10) = 2.5 (7,920 + 4,320 + 2,970)(1.10) = 41,800 Btu.
Regeneration Gas Flow Rate Calculation
Time allocated for heating, , is 0.6  8 = 4.8 hours. Then the gas flow rate is
m = q/[Cp (tHot – tB )] = 41,800/[0.68  4 .8 (650 – 100)] = 23,300 lb/h.

Problem
Determine the saturated water content for a sweet lean hydrocarbon
gas at 150°F and 1,000 psia.
For a gas in equilibrium with a 3% brine,

Solutions:
W = 220 lb/MMscf
For a gas,

First the composition must be converted for use with

MEMBRANE: FUNDAMENTALS
From thermodynamics, the driving force for movement through the membrane is the difference in chemical potential,
μ, for a given component on the two sides of the membrane. If subscript i is the diffusing component, then
μi,feed > μi,permeate
where the material is moving from feed side to the permeate side. To make the equation physically comprehensible in
gas systems, fugacity, f, which is proportional to the chemical potential is used and gives
fi,feed > fi,permeate
If ideal behavior is assumed for the diffusing gas, then the fugacities can be replaced by the partial pressures and
yi,feed Pi,feed > yi,permeate Pi,permeate,
where y is the mole fraction and P is the total pressure. This equation shows what terms affect the driving force across
the membrane. The equation can be rearranged to obtain
This relation tells us that the separation achieved (y i,permeate/y i,feed) can never exceed the pressure ratio (P feed/P permeate)

OPERATING CONSIDERATIONS
Flow Patterns
A non-optimized single-stage CO2/CH4 membrane separation process

A non-optimized
two stage CO2/CH4
membrane
separation process

Flow Rate
A maximum acceptable feed gas rate per unit area applies to the membrane, and required membrane area is directly
proportional to the flow rate. Membrane units perform well at reduced feed rates, but their performance drops when
design flow rates are exceeded. Additional modules are added in parallel to accept higher flow rates.
Operating Temperature
Increased operating temperature increases permeability but decreases selectivity. Because membranes are organic
polymers, they have a maximum operating temperature that depends upon the polymer used. Exceeding this
temperature will degrade membrane material and shorten the useful life of the unit
Operating Pressures
Increased feed pressure decreases both the permeability and selectivity, but at the same time, the pressure difference
across the membrane (the driving force in the membrane flux equation) is increased, which results in a net increase in
flow through the membrane.

ADVANTAGES AND DISADVANTAGES OF MEMBRANE SYSTEMS
Advantages
• Low capital investment when compared with solvent systems
• Ease of operation: process can run unattended
• Ease of installation: Units are normally skid mounted
• Simplicity: No moving parts for single-stage units
• High turndown: The modular nature of the system means very high turndown ratios can be achieved
• High reliability and on-stream time
• No chemicals needed
• Good weight and space efficiency
Disadvantages
• Economy of scale: Because of their modular nature, they offer little economy of scale
• Clean feed: Pretreatment of the feed to the membrane to remove particulates and liquids is generally
required
• Gas compression: Because pressure difference is the driving force for membrane separation, considerable
recompression may be required for either or both the residue and permeate streams.
For natural gas systems, the following disadvantages also apply:
• Generally higher hydrocarbon losses than solvent systems
• H2S removal: H2S and CO2 permeation rates are roughly the same, so H2S specifications may be difficult to meet
• Bulk removal: Best for bulk removal of acid gases; membranes alone cannot be used to meet ppmv specifications

Schematic of a two-bed adsorption unit. Valving is set to have absorber #1 in drying cycle and absorber #2 in regeneration
cycle.

Process flow diagram for amine treating by use Monoethanolamine (MEA) Contactor commonly operates at pressures up to
1,000 psi (70 bar). Flow rates to reclaimer are 1 to 3% of amine circulation rate.

Process flow diagram for hot potassium carbonate process.

Currently only two methods are available for
dealing with large quantities of H2S:
• Disposal of the gas by injection into
underground formations
• Conversion of the H2S into a usable product,
elemental sulfur
Sulfur Recovery
PROPERTIES OF SULFUR

SULFUR RECOVERY PROCESSES
CLAUS PROCESS
Basic Chemistry
The Claus process consists of the vapor-phase oxidation of hydrogen sulfide to form water and elemental sulfur,
according to the overall reaction:
The above overall reaction does not represent the reaction mechanism or show intermediate steps. In practice, the
reaction is carried out in two steps:

For H2S concentrations in the range 25 to 40%, the split flow configuration can be utilized.
In this scheme, the feed is split, and one third or more of the feed goes to the furnace and the remainder joins the
furnace exit gas before entering the first catalytic converter. When two thirds of the feed is bypassed, the combustion
air is adjusted to oxidize all the H2S to SO2, and, consequently, the necessary flame temperature can be maintained.
The split-flow process has two constraints
1. Sufficient gas must be bypassed so that the flame temperature is greater than approximately 1,700F (927C).
2. Maximum bypass is two thirds because one third of the H2S must be reacted to form SO2.
If air preheating is used with the split-flow configuration, gases with as little as 7% H2S can be processed.
Generally, the sulfur recovery in the conventional plants discussed above varies from 90 to 96% for two catalytic
converters. It increases to 95 to 98% for three catalytic converters.

CLAUS TAIL GAS CLEANUP
The processes for this final sulfur removal are generally divided into three categories
• Direct oxidation of H2S to sulfur
• Sub-dew point Claus processes
• SO2 reduction and recovery of H2S
Basic Chemistry

Sub-Dew Point Claus Processes
Total sulfur recoveries
greater than 99% can be
obtained with
CBA processes
Cold bed adsorption (CBA)
process because it is the
most widely used sub-dew
point process.

Reduction of SO2 and Recovery of H2S
The SCOT process is an example of the process that reduces the SO2 in the Claus plant offgas back to H2S.

• The process can produce an exit gas that contains 10 to 400 ppmv of total sulfur, while increasing total sulfur
recovery to 99.7% or higher.
• The feed, offgas from the Claus unit, is heated to 575°F (302°C) in an inline burner, along with a reducing gas,
H2 or a CO and H2 mixture. The reducing gas is supplied either from an outside source or generated by partial
oxidation in an inline burner. The mixture then flows to the SCOT catalytic reactor (cobalt-molybdenum on
alumina), where the sulfur compounds, including SO2, CS2, and COS are reduced to H2S and water. The gas
that leaves the reactor goes to a waste-heat exchanger, where it is cooled to about 320F (160C) and produces
low-pressure steam.
• The gas from the waste-heat exchanger then flows through a quench tower, where it is cooled to
approximately 100°F (38°C) by externally cooled recycle water in countercurrent flow. The water from the
tower is condensed, and the excess water is sent to a sour water stripper. Gas from the quench tower then
contacts an aqueous amine solution in the absorption column. The amine is generally methyl diethanolamine
(MDEA) or diisopropylamine (DIPA) to absorb H2S while slipping CO2. The gas that exits the top of the absorber
contains very little H2S (10 to 400 ppm) and is sent to an incinerator. The rich amine that leaves the bottom of
the absorber flows to the regenerator, where heat is applied to strip the H2S from the amine solution. The
overhead from the regenerator is cooled to condense the water, and the H2S is recycled to the Claus unit. Lean
amine is cooled and returned to the absorber.

SULFUR STORAGE
A brief summary, adapted from Johnson and Hatcher (2003), is given below:
• Sulfur fires. Uncommon, but they can produce large amounts of SO2.
• H2S. Any H2S dissolved in the molten sulfur from the condensers may be a significant hazard if appropriate
degassing techniques are not used.
• Corrosion. A wet sulfidic atmosphere can lead to severe corrosion of carbon steel.
• SO2, Highly toxic and it forms highly corrosive sulfurous acid in the presence of water.
• Static discharge. Because of the excellent insulating properties of molten sulphur, static discharge may occur
under certain conditions and lead to possible fires or explosions.

Important Properties, Safety, and Environmental Concerns of Molten Sulfur

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