1590701599PTE_526-Natural_Gas_Engineering (1).pptx

PTE 526
NATURAL GAS ENGINEERING
BY
ENGR. J. O. OWOLABI
CHEMICALAND PETROLEUM ENGINEERING
DEPARTMENT
AFE BABALOLA UNIVERSITYADO-EKITI,
EKITI STATE, NIGERIA

LEARNING MATERIALS
1. Natural Gas Production Engineering, C.U.Ikoku.
2. Gas Production Operations, H. Dale Beggs, Gulf Pub.
Company,
Houston, Texas, USA. 1984.
3. Gas Processing and Conditioning, Vols 1 & 2, J.M.
Cambell.

PHYSICAL CONSTANTS OF HYDROCARBON
GASES

PSEUDO-CRITICAL PROPERTIES OF GASES
AND CONDENSATES

WORKED EXAMPLE 1
Calculate the specific gravity and the gas deviation factor of
Salford Field gas from its composition. Reservoir pressure and
temperature are 3250 psia and 213oF respectively.
Component Mol. Fraction Mol. Wt Pc Tc
Methane 0.8612 16.04 673 343
Ethane 0.0591 30.07 708 550
Propane 0.0358 44.09 617 666
Butane 0.0172 58.12 550 766
Pentanes 0.0050 72.15 400 846
CO2 0.0010 44.01 1070 548
N2 0.0207 28.02 492 227
TOTAL 1.0000
NOTE: The Tc is in Rankin = oF + 460

SOLUTION
Component Mol
Fract (xi)
Mol. Wt
(Mi)
Mixi Pc xiPc Tc xiTc
Methane 0.8612 16.04 13.81 673 579.59 343 295.39
Ethane 0.0591 30.07 1.78 708 41.84 550 32.51
Propane 0.0358 44.09 1.58 617 22.08 666 23.84
Butane 0.0172 58.12 1.00 550 9.46 766 13.18
Pentanes 0.0050 72.15 0.36 400 2.45 846 4.23
CO2 0.0010 44.01 0.04 1070 1.07 548 0.55
N2 0.0207 28.02 0.58 492 10.18 227 4.70
Total 1.0000 19.15 666.67 374.40

SOLUTION
 The specific gravity of the gas:
γg = 19.15 /28.997 = 0.661
 The pseudocritical pressure and temperature are
666.67 psia and 374.40 R
Pr = 3250 / 666.67 = 4.87
Tr = 673 / 374.40 = 1.80
 The gas deviation factor (z) from chart = 0.91

DETERMINATION OF VISCOSITY
 By Correlations
   
r
rT
P
f
and
T
M
f 
 


1
2 ,
(a) Compute AMW of the gas yiMi
(b)Determine the µgi of the gas mixture at 1 atmosphere using
chart. If the pressure of the mixture is above 1 atm then
1. Calculate pseudo critical temperature/pressure
2. Calculate reduced temperature/pressure
3. Determine the viscosity ratio (g/ gi) from chart
4. Calculate g
5. If the gas contains non-HC components, then they should
be corrected using chart.

EXERCISE 1
Calculate the value of the viscosity for the dry gas
with a specific gravity of 0.818 at reservoir
temperature of 220oF and reservoir pressure of 2100
psig.

ESTIMATION OF GAS RESERVES
Three methods are available
1.Volumetric Method
2. Material Balance Method
3. Pressure Decline Method
VOLUMETRIC METHOD
This is applied in a new field for rough estimates. No
production history is required. We need only geologic
data like porosity, water saturation etc.

Volumetric Method Contd
Where: A = Area of the reservoir (Acres)
• h = Formation thickness (ft)
•  = Porosity
• Swi = Initial water saturation
• Bgi = Initial gas formation volume factor
• G = Initial Gas in place

Volumetric Method Contd
Gas produced at any given condition (Gp) is
 











g
gi
wi
p
B
B
S
Ah
G
1
1
1
43560 
Recovery Factor
g
gi
p
g
B
B
G
G
E 

 1

This is used for reservoir that has produced long
enough
There are two cases:
Without water production/influx
 With water production/influx
MATERIAL BALANCE METHOD
 
g
gi
g
p
B
B
B
G
G


 
g
w
p
e
g
gi
g
p
B
B
W
W
B
B
B
G
G





Material Balance in Gas Reservoirs
The conservation of mass may be applied to gas reservoirs to
give the following MB:
The MB may be made in terms of moles of the gas as:
If: Vi = initial gas pore volume in cu.ft,
pi = final pressure after producing Gp SCF of gas
We = water encroached into the reservoir (cu.ft)

We = water encroached into the reservoir (cu.ft)
Wp = water produced (cu.ft)
Vf = final volume after producing Gp SCF of gas
Then,
Replacing eqns (1) and (2) by their equivalents using the gas
law, we have
Or

But,
Hence,
Replacing Vi by its equivalent G/Big the equation becomes:
Dividing through by Bgi and expanding

If Bgf and Bgi are expressed in cu.ft/SCF instead of SCF/cu.ft,
they will be in the numerator and the equation reduces to:
GpBgf is the volume of the produced gas at the pressure pf,
G(Bgf – Bgi) is the change in volume of the initial gas when
expanded from pi to pf and We and BwWp are the volumes of
water influx and production respectively.

The effect of connate water expansion and formation
compaction can quite easily be incorporated into the gas
reservoir MB. It has already been shown that :
Where HCPV = GBgi
The MB for gas reservoir becomes:

For volumetric reservoir the production volume equals the
expansion volume and neglecting the effects of water
expansion and pore volume reduction, eq (7) becomes:

PRESSURE DECLINE METHOD
• A plot of P/Z versus Gp will produce a straight line of slope
(TPb/TbVi) and intercept at Gp = 0.
• Note that the reservoir gas volume can be put in units of scf
by the use of Bg.
• Thus Vi = GBgi
• Thus both G and Pi can be obtained graphically.
i
i
i
b
b
p
b
Z
P
V
T
Z
TG
P
Z
P




WORKED EXAMPLE 2
Calculate the water influx and residual gas saturation in water-
drive reservoirs. Given that:
Initial Bulk reservoir volume = 415.3 cu.ft
Average porosity = 0.172,
Average connate water saturation= 0.25
Initial pressure = 3200 psia,
Bgi = 0.005262 cu.ft/SCF
Final pressure = 2925 psia
Bgf = 0.005700 cu.ft/SCF
Cumulative water production = 15200 STB,
Bw = 1.03 bbl/STB
Gp = 935.4 MM SCF,
Bulk volume invaded by water at 2925 psia = 13.04 MM cu.ft

SOLUTION
From Eq (5),
This much water has invaded 13.04 MM cu.ft of bulk rock,
which initially contained 25% connate water. Then the final
water saturation of the flooded portion of the reservoir is:

SOLUTION CONTD
And the residual gas saturation Sgr = 32%.

COMPRESSION OF NATURAL GAS
Add energy to the gas. The type of compression station will
depend on the purpose for which the gas will be used. For
field/gathering stations (0 – 750 psig). Between field and
transmission lines (1300 psig). EOR (6000 psig).Storage
field station > 4000 psig. Distribution units (20–100psig)
Types of compressors
(1)Positive Displacement Compressors
- Reciprocating Compressors
- Rotary Blowers
(2)Continuous flow compressors
- Centrifugal Compressors
- Ejectors
Reciprocating compressors are the most commonly used in
gas industry

COMPONENTS OF COMPRESSOR
 Piston
 Cylinder
 Suction/Discharge Valves
 Connecting Rod
 Impeller
 Shaft
 Diffuser
 Volute
Compression Cycle
 Ideal Cycle
 Actual Cycle

MULTISTAGE COMPRESSOR ARRANGEMENT
s
d
P
P
Compressor Ratio r =
1 2 3
Ps Pd
For design purposes r < 6
For practical purposes r  4
- Optimum Number of stages is given by
n
s
d
P
P
r
1







Where:
Pd = discharge pressure (psia)
Ps = suction pressure (psia)
n = No of stages required

COMPRESSOR DESIGN
4
2
v
LSE
d
q


This involves
1. The determination of compressor capacity
2. The determination of power requirements
Determination of compressor capacity
Where
q = Flow capacity (ft3/min, scfd)
d = Piston diameter (ft)
L = Stroke length (ft)
S = Compressor speed (rpm)
Ev = Volumetric efficiency

The volumetric efficiency is calculated from
Z1 = gas derivation factor at suction condition
Z2 = gas derivation factor at discharge condition
K = Cp/Cv = Isentropic exponent
r = compression ratio = p1/p2
A = factor to allow for leakage, friction usually between 0.03
and 0.06.
C = clearance, which varies from 0.04 to 0.16

















 1
1
1
2
1 k
v r
Z
Z
C
A
E

DETERMINATION OF COMPRESSOR
HORSEPOWER
Theoretical HP is determined by 3 ways
 Analytical method
 Mollier diagram
 Quickie estimate

WORKED EXAMPLE 3
A single acting reciprocating compressor having a piston
diameter of 4 inches and a stroke length of 6 inches is to be
used to compress gas from 100 psig and 100oF to 400 psig.
Calculate the flow capacity in ft3/min and scfd if the
compressor runs at a speed of 500 rpm. Given that :
k= 1.3, C = 6% , A = 5%, T2 = 753oR, Z1 = 0.98, Z2 = 0.95

SOLUTION
r = p1/p2 = 414.7/114.7 = 3.62
Z1/Z2 = 0.98/0.95 = 1.03
Ev = 1 – 0 .05 – 0.06 [1.03 (3.62)0.77 – 1] = 0.85
q = [π(4/12)2(6/12)(500)(0.85)]/4
= 18.54 ft3/min
To convert to ft3/day
= 18.54 ft3/min x 1440 min/day = 26704 ft3/day
qsc = q/Bg
Bg = 0.0283 ZT/P = 0.0283(0.98)(560)/114.7 = 0.135 ft3/scf
qsc = 26704/0.135
= 1.97 x 105 scfd

Where:
W = Power required HP/MMscf
Pb = Pressure at standard condition (psia)
Tb = Temperature standard condition (oR)
T1 = Suction Temperature (oR)
Zs = Gas deviation factor at suction condition
   









 1
1
1
027
.
3
1
K
K
Z
r
K
K
T
T
P
W s
b
b
Analytical Method

Mollier Diagram Method
W = 0.0432 H
where
W = Power required (HP/MMSCFD)
H = Enthalpy change (BTU/Ib-mol)
Note: Enthalpy-Entropy diagrams apply only to the gaseous
state. If a gas is cooled below its dew point, condensation
occurs and the heat removed can not be determined directly
from the diagram.

WORKED EXAMPLE 4
Calculate the horsepower required to compress 1 MMscf of a
0.6 gravity gas from 100 psia and 60oF to 1600 psia using the
analytical method. Assume that a two stage compressor is
used and that the gas is cooled back to 60oF between stages.
Also find the final discharge temperature. Other data are:
K = 1.28, Pb = 14.65, Tb = 60oF
Use the Mollier diagram method to solve the above example.

SOLUTION (ANALYTICAL METHOD)
The total compression ratio is 1600/100 = 16. Hence, r = 4
for each compressor stage.
First Stage:
P1 = 100, P2 = 400, Z1 = 0.985, Z1(k – 1)/k =0.22
w1 = 75.8 Hp/MMscfd

Solution (Analytical Method) Contd
The gas temperature after compression from 100 to 400 psia
is:
The gas is cooled to 80oF before entering the second stage.
Second Stage
P1 = 400, P2 = 1600, Z1 = 0.94, Z1(k – 1)/k =0.21

Solution (Analytical Method) Contd
w2 = 71.6 hp/MMscfd
T2 = Tfinal = 273oF
The total power requirement is:
Hp = q(w1 + w2)
= 1 MMscfd (75.8 + 71.6)
= 147.4

SOLUTION (MOLLIER DIAGRAM METHOD)
First Stage:
The first stage compresses the gas from 100 to 400 psia at
constant entropy. Enter p1=100 and T1= 80oF and read H1=
380 Btu/Ib-mole. Proceed vertically (ΔS = 0) to p2 = 400
psia and read H2= 1990, T2 = 260oF. Follow the 400 psia
line to T3 = 80oF and read H3 = 220.
The power required for first stage is:
w = 0.0432 ΔH1-2 = 0.0432(H2 – H1)
w = 0.0432 (1990 – 380)
= 70 Hp/MMscfd

Solution (Mollier Diagram Method) Contd
Second Stage:
From point 3 where p = 400, T = 80 and H = 220 proceed
vertically to the final pressure
of 1600 psia and read T4 = 274oF, H4 = 1920 Btu/Ib-mole.
The power required for the second stage is:
w2 = 0.0432 (H4 – H3)
= 0.0432 (1920 – 220)
= 73.4 Hp

The total horsepower required is then:
Hp = q(w1 + w2)
= 1 MMscfd (70 + 73.4)
= 143.4

WORKED EXAMPLE 5
• What is the amount of heat that must be removed in
cooling 1 Ib-mole of 0.6 gravity gas at 200 psia from
600oF to 100oF ?
• If this gas is then heated from 100oF to 300oF, how much
heat is required per Ib-mole of gas?

SOLUTION
• At 200 psia/600oF, H = 6100 Btu/Ib-mole
Follow constant pressure to 100oF, H = 500 Btu/Ib-mole.
Heat removed = 6100 – 500 = 5600 Btu/Ib-mole.
• At 200 psia/300oF, H = 2600 Btu/Ib-mole
Hence Heat Added = 2600 – 500 = 2100 Btu/Ib-mole.

QUICKIE CHARTS
Used for making quick estimates. They give higher values
than
other methods. Should not replace accurate methods.
 
MMcfd
bHp
T
T
VP
b
b
/
4
.
14
1
bHp = OR bHp =
 
 
MMcfd
bHp
T
T
qP
b
b
/
10
4
1
Where
V = Inlet capacity of compressor (mmcfd)
q = Inlet capacity of compressor (cfm)
Pb = Standard pressure (Psia)
Tb = Standard temperature (oR)
T1 = Inlet temperature of compressor (oR)
(bHp/MMcfd) = Factor determined from chart
bHp = Power requirement (HP)

NATURAL GAS METERING
• To determine volumetric flow rate
• To determine pressure loss for a particular flow rate
Equipment used
• Orifice meter
• Turbine meter
• Pilot tube
• Critical flow prover

CHOICE OF MEASURING EQUIPMENT
Factors to consider in the choice of measuring equipment are:
1. Volume
2. Accuracy
3. Useful life
4. Range of flow
5. Temperature
6. Maintenance requirement
7. Availability of power
8. Nature of fluid
9. Cost of operation
10. Safety of the device

ORIFICE METERING
f
hwp
C
qsc 
Means of measuring the PD caused by a change in velocity of
the gas as it passes through a restriction placed in the pipe.
 Gas flow rate in scf/hr is given as:
Where:qsc = gas flowrate scf/hr
C = Orifice constants
hw = differential press across the orifice (inches H2O)
pf = flowing pressure (Psia) and
C = Fb *Fpb *Ftb *Fg* Ftf*Fr*Fpv*Fm*Y
The orifice constant depends primarily on the basic orifice factor Fb.
Many of the other terms are negligible or essentially equal to one.

Basic Orifice Factor (Fb):
The value of Fb is found from tables. Two standard are
provided in gas measurement – flange taps and pipe taps.
Pressure-Base Factor (Fpb):
The Fpb factor corrects the value of Fb for cases where the
pressure base used is not 14.73 psia. It may be determined by
the equation Fpb = 14.73/Pb.
Temperature-Base Factor (Ftb):
The Ftb factor corrects for any contract wherein the basic
temperature is not 520oR (60oF). This factor may be computed
by the formula Ftb = Tb/520.

Specific-Gravity Factor (Fg):
• The Fg is to correct the basic orifice equation for those
cases where the specific gravity of the gas is other than
1.00. The equation is Fg = γg
-0.5 .
Flowing Temperature Factor (Ftf):
• For flowing temperature correction if the flowing
temperature of gas is other than 600F
Reynolds Number Factor (Fr):
• It accounts for variation of the discharge coefficient with
Reynolds number. Value obtained from chart
5
.
0
520 






f
T
Ftf

Supercompressibility factor (Fpv):
It corrects the variation from the ideal gas law:
  5
.
0
1


pv
F
Expansion Factor (Y):
It accounts for change in gas density as the pressure changes across the
orifice. The change is usually small and therefore ignored
Manometer Factor (Fm):
For correction when mercury-type manometer is used. It accounts for
different heads of gas above the two legs of the manometer. Usually
small value and therefore insignificant.

WORKED EXAMPLE 6
A meter run that is equipped with flange taps and a 3000 inch
orifice has an inside diameter of 6.065 inches. The static
pressure obtained from the downstream tap reads 80 psia and
the average differential pressure is 49.5 inches of water. If the
pressure and the temperature bases are 14.9 psia and 60oF
respectively, calculate the flow rate in standard cubic feet per
hour. The gas gravity is 0.60 and the flowing temperature is
65oF.

SOLUTION
Calculate C’ = Fb *Fpb *Ftb *Fg* Ftf*Fr*Fpv*Fm*Y
• From tables, for d= 3.000 and D= 6.065,
• Fb= 1891.9
• Fpb= 14.73/Pb = 14.73/14.9 = 0.989
• Ftb= 1.000
• Fg= (γg
)-0.5= (0.6)-0.5 = 1.291
• Ftf= (520/Tf)0.5 = (520/525)0.5 = 0.995

• From the tables, b= 0.0332
• Fr= 1+ b/(hw*pf)1/2= 1+ 0.0332/(49.5*80)1/2= 1.001
• For γg = 0.6, T = 65oF, P= 80 psia, the value calculated for Z
is 0.990.
• Fpv= (Z)-0.5 = (0.990)-0.5 = 1.005
• Fm= 1.000
• To determine Y,
β= d/D = 3.000/6.065 = 0.495
hw/Pf = 49.5/80 = 0.619

SOLUTION
• From the tables, Y = 1.0037 (requires interpolation)
C’=1891.9(0.989)(1)(1.291)(0.995)(1.001)(1.005)(1.000)(1.0037)
= 2426.9
• qsc= C’√(hwPf) = 2426.9√(49.5)(80)
= 152,721 scf/hr
= 3.665 MMscfd

WORKED EXAMPLE 7
A metering system is required to measure approximately 8.5
MMscfd of 0.62 gravity gas at a line pressure of 250 psig. The
meter run is to be made of 8 inches pipe (7.981 in ID).
Determine the size of the orifice plate to give a differential of
about 50 inches. Flowing temperature averages about 80oF.

SOLUTION
For hw = 50,
For an approximation, all the terms that make up the value of
C’ will be ignored in this case, except Fg and Ftf.
From the Fb table for flange taps, for D = 7.981, we have:
d=3.500 for Fb of 2537.7 and d= 3.375 for Fb of 2352.0,
.

To determine Y,
β= d/D = 3.000/6.065 = 0.495
hw/Pf = 49.5/80 = 0.619
hence: a 3.500 inch orifice plate would be selected to obtain
an hw reading of approximately 50 inches at the design flow
rate

TRANSPORTATION OF NATURAL GAS
Mainly by (1) Pipeline
(2) Liquefaction
• Pipeline:
Trans-Sahara pipeline (Algeria to Europe)
West African Gas Pipeline (WAGP)
Soviet Union to Europe
Alaska to Southern USA
• Liquefaction
LNG Plants

PIPELINE DESIGN
Factors to consider:
1. Terrain – land, swamp, sea
2. Location
3. Distance of transportation
4. Nature and volume of gas
5. Use of gas
6. Elevation of route
7. Scope of expansion
8. Reserves available
9. Design software packages are available

GAS FLOW IN PIPELINES
The Value of C depends on the units used
 T d L q C
Psia 0R in Mi scfd 77.54
Psia 0R in ft scfd 56.38
Psia 0R in ft MMscfd 5.638 x 10-3
Kpa 0K M m M3/d 1.149 x 106
Where
f = friction factor
d = Diameter of pipeline (inches)
L = Length of Pipeline (miles)

EVALUATION OF FRICTION FACTOR
3
1
032
.
0
D
For Weymouth Equation =
For Panhandle A Eequation =
For Panhandle B Equation =
147
.
0
Re
/
085
.
0 N
f
183
.
0
Re
015
.
0
N
Pipeline flow equation without ‘f’ becomes:
 
5
1
4
3
2
21
2
1
2
1 a
a
g
a
a
b
b
d
y
L
Z
T
T
E
a
q















 












Where the values of a1- a5 are

a1 a2 a3 a4 a5
Weymouth: 433.50 1.000 0.5000 0.5000 2.667
Panhandle A: 435.87 1.0788 0.5394 0.4604 2.618
Panhandle B: 737.00 1.0200 0.5100 0.4900 2.530
q = ft3/day
T = 0R
 = psia
L = miles
d = inches
Weymouth equation is used for pipeline diameter of d 16
inches

WORKED EXAMPLE 8
Using the following data, calculate the flow capacity of the
pipeline using the Weymouth equation and Panhandle B
equation.
P1= 847 psia, P2 = 600 psia,
d = 25.375 in, L = 100 miles,
Gas gravity = 0.67, T = 505oR,
Z = 0.846, Tb= 520oR,
Pb = 14.7 psia, E = 1.0

SOLUTION
Tb/Pb = 520/14.7 = 35.374
Weymouth
q = 433.5(35.374)1.0(8.366)0.5(1/0.67)0.5(25.375)2.667
= 301,610,000 ft3/day
Panhandle B
q = 737.0935.374)1.02(8.366)0.51(1/0.67)0.49(25.375)2.53
= 359,732,857 ft3/day

PIPELINES IN SERIES



N
i i
i
e
d
L
d
L
1
333
.
5
333
.
5
If a pipeline or gathering line consists of sections of different diameter pipe, the
flow capacity of the entire pipeline can be calculated by first determining an
equivalent length of a line of some arbitrary diameter that would have the same
flow capacity as the system. For a series pipeline:
T = 1 + 2 + 3
If the friction factor is not required, the Pressure Drop becomes:
Where a3 and a5 are obtained from the previous table. For example, if
Weymouth equation is used, since a3 = 0.5 and a5 = 2.667, the equation
becomes:
ΔP1 ΔP2
ΔP3

PIPELINES IN PARALLEL
D1L1q1
qt
qt
D3L3q3
D2L2q2
For pipes in parallel, the total flow rate is the sum of the rates in the
individual pipes or
qT = q1 + q2 + q3
If the length of the individual pipes are the same, the total flow
capacity is calculated from:

PIPELINES IN PARALLEL
For pipes in parallel, the total flow rate is the sum of the rates
in the individual pipes or
qT = q1 + q2 + q3
If the length of the individual pipes are the same, the total flow
capacity is calculated from:
When applied without friction factor, the above equation
becomes:



N
i i
i
T
f
d
C
q
1
5
.
0
5
.
2

FIELD HANDLING OF NATURAL GAS
A typical well stream is a high velocity, turbulent and constantly
expanding mixture of gases, hydrocarbon liquids, water
vapour, free water, solids and other contaminants. Field
processing of NG consists of four basic processes.
1. Separation of the gas from free liquids and entrained solids.
2. Processing the gas to remove condensable and recoverable
hydrocabon vapours.
3. Processing the gas to remove condensable water vapour to
avoid hydrate formation.
4. Processing the gas to remove other undesirable components
such as hydrogen sulphide and carbon dioxide.

DEHYDRATION OF NATURAL GAS
• Removal of water vapour from the natural gas.
• Maximum allowable is 7 Ib of H20/MMscf of gas
Two processes are used:
- Absorption (gas-liquid) Process
- Adsorption (gas-solid) Process

(ABSORPTION PROCESS)

TEG is used as the absorbing agent. Process same as acid gas
removal

(ADSORPTION PROCESS)
Drying of gas using molecular sieve beds to adsorb water to
prevent ice & hydrate formation at low temperatures. This is a
gas-solid process.
Adsorption Process
Adsorption describes any process in which molecules from the
gas are held on the surface of a solid by surface forces.
Adsorbents
Bauxite
Alumina
Molecular Sieves
Silica gel
Carbon (Charcoal)

DESIGN CAPACITY FOR ECONOMIC LIFE
• Bauxite: 4 – 6 kg water per 100 kg desiccant
• Alumina: 4 – 7 kg water per 100 kg desiccant
• Molecular Sieves : 9 – 12 kg water per 100 kg desiccant
• Silica gel: 7 – 9 kg water per 100 kg desiccant
In normal operation, outlet dewpoint will be below -40oC
until water breakthrough.

79 of
52
Regen.
Gas
supply
Regen Gas
to U-1100
NG from
U-1100
DEHYDRATION
UNIT
U 1200
NG to
U-1300
Two adsorbers on
line anytime
and one on
regeneration/stand-by
in rotating sequence.
C-1201 B C-1201 C
C-1201 A

DESIGN INPUT CRITERIA FOR ABSORBER
1 Gas flow rate (MMscfd)
2 Specific gravity of gas
3 Gas inlet pressure (psig)
4 Maximum working pressure of absorber (psig)
5 Gas inlet temperature
6 Outlet gas water content required (Ibm/MMscfd)
7 Vap/Liquid equilibrium correlation for TEG-Water
system

WHAT YOU NEED
• Gas capacity/Absorber size chart
• Water content chart
• TEG mol % vs mol fraction of water (x)
• Activity coefficient (γ)chart
• Absorber efficiency (Ea)
• Absorption factor (A)

ABSORBER GAS CAPACITY
(TRAYED COLUMN)

ABSORBER GAS CAPACITY
(PACKED COLUMN)

USING THE RELATIONSHIP:
Ea = Absorption efficiency
K = 1.33 x 10-6 (W) (γ)

WATER CONTENT OF NATURAL GAS
(OILFIELD UNIT)

WATER CONTENT OF NATURAL GAS (SI UNIT)

Rewriting the above equation in terms of water content we
have:
Where:
Wo = (W) (γ) (xo)
γ = activity coefficient for water in TEG system (from chart)
Xo = mole fraction water in TEG-Water system (from chart)
Wo = water content at saturation
WN+1 = Inlet gas water content
W1= exit gas water content
Ea = absorption efficiency (ie fraction of component absorbed)
K = 1.33 x 10-6 (W) (γ)

WATER DEW POINT VS TEMPERATURE

Concentration of Rich TEG leaving the bottom of
the Contactor is:
Lw = glycol to water circulation rate (gal TEG/Ib water)
ρi = specific gravity x (8.34)
ρi = density of lean TEG solution (Ibm/gal)
Rich TEG = conc of TEG in rich solution from contactor, %/100
Lean TEG = conc of TEG in lean solution to contactor, %/100
Number of trays
Actual No of trays = No of theoretical trays / tray efficiency
Where tray efficiency = 25% for bubble cap & 33.3% for valve
tray

Stripper sizing
L = Lw x Wi x q / 24
Where L= glycol circulation rate (gal/hr)
Lw = glycol to water circulation rate (gal TEG/Ib water)
Wi = water content of inlet gas (Ib water/MMscfd)
q = gas flow rate (MMscfd)
Reboiler duty
Ht = 2000 x L
Ht = total heat load on reboiler (Btu/hr)
Overall size of reboiler
A = Ht / 7000 (based on heat flux of 7000 Btuh/ft2)
(Max ht flux across fire box is 8000 Btu/h/ft2)
A = total area of firebox

HEAT LOSS FROM REBOILER/STRIPPER
Hl = 0.24 x As x (Tv – Ta)
where
Hl = overall ht loss from reboiler/stripper (Btu)
As = total exposed surface area of reboiler/stripper (ft2)
Tv = temp. of fluid in the vessel (oF)
Ta = min. ambient air temperature (oF)
0.24 = ht loss from large insulated surfaces (Btuh/ft2)

WORKED EXAMPLE 9
Calculate the circulation rate of 98.7 wt% lean TEG
needed to dry 106 std m3/d (1762 kg mol/hr) of natural
gas at 7.0 MPa and 40oC in a six tray absorber (1.5
theoretical tray) to achieve an exit gas water content
of 117 kg /106 std m3. The inlet water content is 1100
kg/ 106 std m3.

SOLUTION:
Given: Wt % of TEG-Water system:
1) Determine mol fraction of water in TEG system using chart. (xo
= 0.099).
2) Determine the activity coefficient (γ) using chart. (γ = 0.66)
3) Determine water content of the gas if not given.
4) Calculate water content at saturation Wo = (1100)(0.66)(0.099)
= 71.9kg/106 std m3.
5) Calculate absorption efficiency Ea.
6) Determine the absorption factor A, knowing the number of
theoretical trays using chart. For N = 1.5, and Ea = 0.956, A =
7.

7) Calculate the TEG circulation rate (L)
L = (A)(K)(VN+1)
but K = 1.33 x 10-6 (W) (γ)
= 1.33 x 10-6 (1100) (0.66)
= 0.000967
and VN+1 = 1739 kmol/h
Hence L = (7.3)(0.000967)(1739) = 12.3 kmol/h.
8) Mol wt of the lean TEG solution
(Mol fract. of water x Mol wt of water + Mol fract. of TEG
x Mol wt of TEG)
(0.099 x 18) + (0.901 x 150) = 137 kg/kmol

9) L = 12.3 x 137 = 1685 kg TEG/h
10)Density of TEG = 1.12 g/cm3 = 1.12 kg/ litre
TEG circulation rate = 1685 / 1.12 = 1500 litre/h
Amount of water absorbed in 1 hr = (1100 – 117) / 24 = 41 .0 kg
Hence TEG to water circulation rate = 1500 / 41.0 = 36.6
litre / kg H2O

WORKED EXAMPLE 10
Size a trayed-type glycol dehydrator for a field installation to meet
the following requirements:
Gas flow rate = 10.0 MMscfd
Gas specific gravity = 0.7
Operating line pressure = 1000 psig
Maximum working pressure of contactor = 1440 psig
Gas inlet temperature = 100oF
Outlet gas water content = 7 Ib H2O/MMscf
Glycol to water circulation rate = 3.0 gal TEG/Ib H2O
Lean glycol concentration = 99.5% TEG
No of theoretical trays = 1.5

SOLUTION
1) Determine the size of the contactor using chart:
For P= 1000 psig and Qg = 10.0 MMscfd;
the contactor diameter = 24-in OD.
2) Determine inlet water content from chart:
For P = 1000 psig = 1014.7 psia and T = 100oF
Inlet water content = 61 Ib H2O/MMscf
3) Determine the density of lean glycol:
ρ = (S.G)( density of water)
= (1.111)(8.34)
= 9.266 Ib/gal

4) Calculate the concentration of TEG out of contactor:
= 96.0%
5) No of trays = No. of theo. Trays / tray efficiency
= 1.5 / 0.333
= 4.50
= 5.0 (next whole number of tray should be used).

6)Calculate glycol circulation rate:
L = Lw x Wi x q / 24
= (3)(61)(10) / 24
= 76.25 gal/hr
7) Calculate Reboiler heat load: Ht = (L)(ρ)(CP)(ΔT)
Ht = 2000 x L
= (2000) ( 76.25)
= 152,500 Btu/hr

ADSORBER DESIGN CONSIDERATIONS
1 Cycle time (< 1hr to 8hr)
2 Gas flow rate
3 Desiccant capacity
4 Required dewpoint
5 Total amount of water to be removed
6 Regeneration requirements
7 Pressure drop limitations
8 Adsorption performance of the adsorber

PROCEDURE FOR ADSORBER SIZING
CALCULATIONS
There are a series of dependent variables. Some must be fixed for
design purposes. The general procedure used are:
1. Estimate an allowable superficial velocity
2. From the inlet water content, calculate the load water to be
adsorbed per cycle, assuming all inlet water is removed.
3. Determine vessel diameter. Choose OD of vessel to be used.
Subtract vessel wall thickness x 2 to give ID. Then subtract for
internal insulation (if used) to find desiccant bed diameter
4. Find actual “Vg” from equation. Check to see if it falls within
proper velocity limits. If not, change diameter so that it does and
find new “Vg”

5 Determine “x” from equation and charts for xs
6 Determine minimum bed length hT
7 Check tB from equation to see if cycle length chosen is
satisfactory. If not, adjust and repeat steps 1 – 6. Note water
loading “q” is independent of cycle length.
8 Use the results of the above for regen. Calculations
Given:
Gas flow rate
Inlet water content
Operating Temp and Pressure
Gas composition or other gas properties

We can determine the followings:
(1) Minimum Bed Diameter
Where Vg = 540 Dp / ρg (Dp = avg particle diameter)
(2) Determination of Bed Length (ht)
ht = [(127.4) (wt of water adsorbed/cycle) /(ρB) (D2)(x)]
Where D = bed diameter ( ft or m)
x = useful capacity wt.of water (Ib or kg)
ρB = bulk density of desiccant (Ib/ft3 or kg/m3)
If wt of water adsorbed is not known, use the relationship:

(3) Desiccant capacity
X = desiccant useful capacity kg water per 100 kg desiccant
xs = capacity at saturation kg water per 100 kg desiccant
ht = bed length
hz = zone length (depends on the characteristics of the adsorbent)
For Silica gel :
Molecular Sieve: hz = 0.6 hz (Si gel)
Alumina hz = 0.8 hz (Si gel)
A = constant (141 or 375)
hz = zone length ( cm or in)
q = water loading (kg/h/m2 or Ib/h/ft2)
Vg = superficial velocity (m/min or ft/min)
R.S.= % relative saturation of inlet gas(assume 100% if saturated)

RELATIVE WATER SATURATION CHARTS

(4) Total Water Load (q):
q = 0.054[(Flowrate)(W) / D2]
Where q = water loading (Ib/h/ft2 or other units)
Flowrate = gas rate (MMscfd)
D = bed diameter ( ft)
W = water content (Ib/MMcsfd)
(5) Breakthrough Time (tB)
tB = [(0.01) (x) (ρB ) ( ht )] / q
where tB = breakthrough time ( hr)
ρB = bulk density of desiccant (Ib/ft3)
ht = minimum bed length (ft or m)
q = water loading (Ib/h/ft2 or other units)

REGENERATION SYSTEM
The total heat load is the sum of
1. heat to desiccant
2. heat to hydrocarbon and water
3. heat to vessel
4. heat to inert ball
NB: Take Sp Ht of steel = 0.5 kJ/kg/oC
Sp Ht of balls = 1.0 kJ/kg/oC
If vessel is insulated, omit (3) and add 10% to the calculated
1,2 and 4.

(1) Heat load
If “m” is the regeneration gas flow rate (kg/h) the available
energy to heat the vessel and its contents to the required temp
is:
Q = m x Sp Ht x TH – T1
T1 = temp of inlet wet gas
TH = (19 – 38oC) > Tm (Values above 315oC should not
be used)
Tm = maximum regen temp ( 175 – 260oC)
(2)Fuel gas required
V = Q / (0.7 x Gas Heating Value)

WORKED EXAMPLE 11
0.27 x 106 std m3 (10 MMscf/d) of a 0.6 relative density
natural gas is to be dehydrated. The wet gas enters saturated
at 6.9 Mpa (1000 psia) and 38oC (100oF). The vendor
proposes a unit composed of 2.76cm (30 in) O.D. towers
containing silica gel beds 4.57 m (15 ft) in length. After
allowing for shell and internal insulation thickness, the bed
diameter is 64.8 cm (25.5 in). Does this meet company
criteria including a gas superficial velocity not exceeding 9.15
m/min (30 ft/min)? The water content of the inlet gas is 1021
kg/ 106 std m3 (61 Ib/MMscf). Z = 0.88 and bulk density of
gel is 721 kg/m3.

SOLUTION
1) Calculate the total water load per cycle.
Water absorbed = (0.27)(1021) / 3 = 91.9 kg/cycle
2) Calculate the useful capacity (x)
3) Calculate hz
4) Calculate the gas velocity (vg)

5) Calculate the water loading (q)
Hence hz

xs = (16)(0.9) = 14.4
6) Calculate the Capacity at saturation (from chart)
7) Calculate the Desiccant Capacity
8) Determine the minimum bed length (hB)

9) Determine the breakthrough time (θB)

1590701599PTE_526-Natural_Gas_Engineering (1).pptx

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