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Detection of glycol loss and utilizing new molecular seive in the glycol dehydration plant

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Detection of glycol loss and utilizing new molecular seive in the glycol dehydration plant Powered By Docstoc
					Detection of glycol loss and utilizing new molecular seive in the glycol dehydration plant
Presented by:
Student Name Sultan Rashed Saoud Ahmed Omar Al-Akbari Saleh Al Jabri Rashed Al Hakmani Student ID number 200213414 200213431 200101711 200235602 200204495

Supervised by : Dr. Nayef Gasem

Outlines
• • • • • • • Introduction Adsorption process Adsorption dehydration Glycol dehydration unit design Molecular sieve unit design Estimation cost Conclusion

Introduction
• (GP1) In order to solve Abu Hasa Plant problem (losses of glycol) we:
– studied the natural gas composition and properties. – studied the problem of water in natural gas and its affect. – studied two methods of dehydration which is liquid desiccant and solid desiccant – studied the system’s equipments – did material balances for the glycol regeneration cycle – did energy balances – used chemical engineering simulation programs “HYSYS“ – did HAZOP study.

• (GP2) In order to compare between glycol system and solid desiccant we:
– – – – do design for equipments of glycol and solid desiccant systems calculate the total capital and operating costs determine possible locations of glycol losses compare between the two systems depending on advantages and disadvantages of each system.

Adsorption of water by a solid desiccant
• Adsorption is purely a surface phenomenon • molecules from the gas are held on the surface of a solid by surface forces • function of operating temperature (↓as T↑) and pressure (↑ as P↑) • In NG industry a solid desiccant is used to remove water vapor from a gas stream

Properties of physical adsorbents in NG dehydration
1. Large surface area “ to 500-800 m2/g”. 2. Good "activity and retention" for the components to be removed. 3. High mass transfer rate “rate of removal”. 4. Easy, economic regeneration. 5. Small resistance to gas flow “small ∆P” 6. High mechanical strength “resist crushing and dust formation”. 7. Cheap, non-corrosive, non-toxic, and chemically inert

Typical types of adsorbents

Solid-desiccant dehydration process
• two or more adsorption towers are filled with a solid desiccant (on, stand-by) • wet NG is passed through from top to bottom • high-temperature dry gas stream is used to regenerate solids • There are two mechanisms:
– Chemisorption “uncommon” – Physical adsorption “common in gas dehydration”

Design of separator
• Separation units are engineered to meet the needs and requirements for:
– – – – – – – – Volume Gravity Pressure Foaming Paraffin Hydrates Impurities Corrosion

• •
•

The design standard of separation units must meet quality and reliability. sufficient instrumentation and control devices are necessary to ensure safe and continuous operation Horizontal separators are used:
– in handling high to medium gas-oil ratios – for large volumes of gas and liquids – as 3-phase separators.

Info. needed for design
• The separator used before absorption tower in Abu-Hassa Natural gas plant is used to separate condensed liquid from gas feed to absorber. • After doing our GP1 mass balance we found all information needed to design this separator.

Info. needed for design
• The information needed are:
– – – – – – – – – P, Pressure in separator = 387.3psia T, Temperature in separator = 64.4F Mwav, Average molecular weight = 29.68lb/lbmole ρL, density of liquid @STP = 26.92lb/ft3 z, compressibility factor = 0.88 q, Gas flow rate @STP = 200.6MMSCFD Area for liquid = 0.25 Area for gas F, volumetric flow rate to separator = 6612m3/d Residence time =10min

Design An Absorption Dehydration
Information Required to Design an Absorption Dehydration Plant:
   The number of stages needed in the absorber tower. Determine the column diameter. Overall height of the column.

Design An Absorption Dehydration
 Natural gas information : From Figure 4-6: Dew point of natural gas Inlet gas contains Outlet gas contains At top absorber 2 47 Ib H2O/MMscf Ib H2O/MMscf

y water content of exit gas
x lean glycol concentration

2
99.6

Ib H2O/MMscf
wt % TEG

At bottom absorber y water content of entering gas x rich glycol concentration 47 86 Ib H2O/MMscf wt % TEG

Design An Absorption Dehydration

Design An Absorption Dehydration
 The Equilibrium data for TEG water system: Gas water content Weight % TEG
50.0 45.0 40.0
(86%,47)

0.0 100.0

5.0 98.0

10.0 96.0

37.5 85.0

Gas water content

35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 99.0 97.0 95.0 93.0
Weight % TEG

(99.6%,2) Equilibrium Data Curve

91.0

89.0

87.0

85.0

Design An Absorption Dehydration
 Determine the column diameter:
 Diameter may be estimated by cross-section area:
Gas flow rate velocity

Cross

 sec tion

area 

Superficia l

 Superficial velocity is estimated:
   V   C SB  L  V  
1/ 2

V max

 The density of vapor and liquid:
 P . MW avg v    Z . R .T      1 . 61  Ib   3  ft      

 L  62 . 4  

 Ib   3   ft 

Design An Absorption Dehydration

Design An Absorption Dehydration
 Gas volumetric flow rate of 162.3 MMscfd :
Flow P   Z rate  V sc   sc      P   Z sc  ft 3 rate  4327   min        T   T   sc    

Flow

 Diameter may be estimated by cross-section area:
Cross  sec tion area  Gas flow rate velocity  64 ft Superficia l

 
2



Absorber diameter:
D  4 Area .   9 . 03  ft   

Design An Absorption Dehydration
 The height of absorber tower:
 There is relationship between the height and number of trays.

Height  ( Number Heigt  9 Height Height Total for for column Tray (

of 2 ft Tray

Trays )( )  18  ft  scurbber inlet , mist

2 ft Tray

)

int ernal glycol

 5  ft  extractor at top  5  ft 

height  28  ft 

Heat Exchanger Design
– Objectives sought in H.E. design :
1. Heat load. 2. Define the area or size (dimensions) and length of tube. of the H.E. 3. Number of tube. 4. Velocity. 5. Drop in pressure.

Shell and tube heat exchangers

TC

TD

ms

Shell m t Tube TB

TA

Fig. Shell and tube heat exchangers

Where the symbols in the equations refer to the following parameters:

•
• • • • •
• •

Rate of heat transfer Re Reynolds number  Density of TEG  Viscosity of TEG u Velocity  T Different in temperature  T lm Log mean temperature difference f Friction factor

Q

• • • • • •

Overall heat transfer coefficient A Area of heat transfer L Length of tube d Diameter of tube n Number of tubes  P Pressure drop inside tube
U

Step1: Calculate heat transfer rate:

Q  m C pT Q  173626 . 03 J s



Step2: Calculate the log mean temperature difference:
Q  UA  T lm L . M .T . D  ln (T1  t 2 ) /( T 2  t1 ) 


(T1  t 2 )  (T 2  t1 )

 T lm  9 . 594 C

Step 3 Choose the tube length and diameter Capacity unit
Area (m2) Tube Dia. (m) Length (m) Temp (oC)

Max. value
1000 0.0254 6.6 400

Min. value
10 0.019 2.5 -30

• Tube length = 6 m , Tube diameter = 0.0254 m

• Step 4 Calculate the area and number of tube:
Q  UA  T lm A  Q U  T lm 78 . 67 m
2

 T lm  9 . 594 C


A  n  dL n  A

U  230

W m . C
2 

 dL

 164

• Step 5 Calculate the velocity:
u 



m

 An
m4


u 

 d n
2

 0 . 04

m s

• Step 6 Calculate the pressure drop inside the tube:
L u P  8 f ( ) d 2 Re  39 . 09 , Re  , f  0 . 02 ud 



 P  1265 . 4 Pa  0 . 0125 atm

Reboiler design

•

Step1: Required heat load :
Q R 1  M lean C p  T  69333.8 Btu/hr

•

Step2: Heat to vaporize water picked up in absorber :
QR2  M
H 2O

h fg Btu/hr

 143234.3

•

Step3: Heat to vaporize reflux water in still. 25% is returned
Q R 3  0.25 x Q R2  35808.6 Btu/ hr

•

Step4: Heat losses from still :

Q R 4  10000 Btu/hr

•

Total heat load:

Q RT  Q R 1  Q R 2  Q R 3  Q R 4 Q RT  258376 . 7 Btu hr

• •

Using rule of thumb equation: Regenerator duty :
Q  900  966 m Q  reqenerato r duty Btu/lb m  gal TEG/lb
H 2 O removed H 2 O removed

 Q  900  ( 966  0 . 5825 )  1462 . 65 Btu/lb
H 2 O removed

Mole sieve
• Calculate the vessel diameter, weight of desiccant, vessel height, pressure drop, thickness of the vessel and weight of the vessel. • Regeneration design
– Estimate heat required for regeneration – Total regeneration cooling – Estimation time of regeneration

Mole sieve

specific gravity 

 gas  air
)*(



Mw Mw

ave air

 n  (163 . 1 * (

1 1440

10

6

)  298 . 455 lb mole / min  135 . 37 mol / min

379 . 5

  V  n ZRT / P Tr  Pr  T Tc P Pc Z  0.89   525 480 . 5 387.3 701.4475  0 . 552  1.19

from figure 4.1

Calculate the viscosity, density and pressure drop


  0 . 011 cp

from GPSA figure 23-26

 Density

 

P * Mw Z * R *T



272 . 7 * 23 . 44 0 . 89 * 10 . 732 * ( 65  460 )

 1 . 288 lb / ft

3

 Pressure drop

 P / L  ( B *  * v )  (C *  * v )
2

 0 . 322 psi / ft

Estimate vessel internal diameter


 Gases volumetric flow:  V max =34 ft/min  Vessel area: vessel  Vessel ID= 2.12m

v



n* Z * R * T p

(fig.9-11)
area  V / v max


 10 . 5 m

2

• Estimate water loading for 8 hr cycle
Inlet gas contains 42.5 lbH2O /MMscf from figure 4.4

Water loading(m

w

)  (163.1 * 42.5 * 8)/24  2310.58  1047 kg H 2 O/cycle

lb H2O /cycle

Estimate weight of desiccant required in vessel md Dynamic capacity at 65 °F= dynamic capacity at 75 °F *CT*Css 65 °F = 13*0.98*1 = 12.74 lbH2O / 100 lb sieve
lb sieve required saturation  2310.58 0 . 1274 Depth of mass transfer zone : L MT Z  ( v 35 )
0 .3

zone  water loading /dynamic

capacity

 18136.42

lb sieve

*Z  (

102 35

)

0 .3

* 1 . 8  2 . 48 ft

where v calculated  v  v /(

from :


4

* D )  34 . 05 ft / min
2

Vessel height
Mole sieve required for MTZ 


4

* D * L MTZ *  b 
2


4

* 6 . 94 * 2 . 48 * 44  41129 . 5 lb Sieve
2

m d  ( 41129 . 5  18136 . 4 )  22265 . 8 lb sieve  10100kg

(

4

) * ( ID ) * ( L ) * (  )  md
2

L  bed

md


4

* D * b
2



20932 . 6


4

* 6 . 94

2

 12 . 57 ft

* 44

lenght  12 . 57 ft height  L  ID  19 . 5 ft height  19 . 5  12  31 . 5 ft  9 . 63 m

vessel vessel

 P  0.333 * 31.5  10 . 48 psia

 72kPa

Regeneration design
• Estimate heat required for regeneration – heating of water to 250 °F
Q hw  m w Cp w  T  2310 . 58 * 1 * ( 250  65 )  4 . 27 * 10 BTU
5

– Vaporizing water
Q vw   m w  2310 . 58 * 1800  4 . 16 * 10 BTU
6

– Heating of water from 250 °F to 550 °F
Q hw  m w Cp w  T  2310 . 58 * 1 * ( 550  250 )  6 . 93 * 10 BTU
5

– heating vessel
Q hv  m v Cp d  T  70177 * 0 . 12 * ( 500  65 )  3 . 66 * 10 BTU
6

Regeneration design
• Estimate heat required for regeneration – heat desiccant bed to 500 °F
Q hd  m d Cp d  T  22265 * 0.25 * (500 - 65)  2 . 4 * 10 BTU
6

– Total regeneration heat:
Q rh 

 Q  13 .32 * 10

6

BTU  11 . 8 * 10 kJ
6

Cost Estimation
• Cost plays an active role for the engineering life • Design engineer needs to be able to make quick cost estimates to decide between alternative designs and for the project evaluation.

Estimating the total capital cost of a plant
• Direct project expenses include:
– equipment cost – materials required for installation – labor to install equipment and material
– Indirect cost:
• • • • freight, insurance, and taxes construction overhead contractor engineering expenses contingency

Factors affecting the costs associated with Capital Cost evaluation of chemical plants
• • • • • • • • • • • • • • • (1) Direct Project Expenses Equipment free on board cost Materials required for installation Labor to install equipment and material (2) Indirect Project Expenses: Transportation costs, insurance, and taxes Construction overhead (vacation, sick leave and salaries) Contractor engineering expenses (salaries and project management) (3) Contingency and Fee: Contingency (loss of time due to storms, strikes, and small changes in the design). Contractor fee (depend on type of plant) (4) Auxiliary Facilities: Site development (civil engineering work) Auxiliary Buildings Off-sites and Utilities

Lang Factor Technique
C TM  F Lang

C
i 1

n

P ,i

Type of Chemical Plant Fluid processing plant Solid-fluid processing plant Solid processing plant

Lang Factor = Flang 4.74 3.63 3,10

Estimating the manufacturing (operating) cost of a plant • Direct Manufacturing Costs
1. Variable costs:
• • • • Operating and labor Row materials Pollution control (air, water, and solid waste) Utilities:
» » » Electricity Fuels Water

2.

Semi-variable costs
• • • Laboratory charges Maintenance Overhead (plant and salaries)

• •
– – – –

Fixed Manufacturing Costs General Expenses
Management Sales Financing Research functions

Estimated Operating labor Cost
N OL  ( 6 . 29  31 . 7 P  0 . 23 N np )
2 0 .5

•

Where NOL is the number of operators per shift, P is the number of processing steps involving the handling of particulate solids
2 0 .5

Equipment Type Compressor Exchangers

Number of Equipment 0 5

Nnp 0 5

Heaters/Furnaces
N OL  ( 6 . 29  31 . 7 ( 2 )  0 . 23 ( 7 ))  11 . 6

1 2
0

1 0

Pumps*
Reactors

Towers
Operating Labor

1 3
TOTAL

1 7

 ( 4 . 5 ).( 11 . 6 )  52 . 12  53

Vessels*

Labor Costs

 ( 53 ).($ 50 , 000 )  $ 2 , 650 , 000 / yr

For new plant
N OL  ( 6 . 29  31 . 7 ( 2 )  0 . 23 ( 5 ))
2 0 .5

 11 . 58

Operating Labor  ( 4 . 5 ).( 11 . 58 )  52 . 12 Labor Costs  ( 53 ).($ 50 , 000 )  $ 2 , 650 , 000 / yr

Equipment Type Compressor Exchangers Heaters/Furnaces Pumps* Reactors Towers Vessels*

Number of Equipment 1 0 1 1 0 3 -

Nnp 1 0 1 0 3 -

TOTAL

5

Estimated utility cost for mole sieve plant
 The utility cost for fan:
– Duty
 kJ     589 . 7  hr 

– Cost of electricity $0.06/kW.hr. And the efficiency of electricity 0.9. – calculate the electricity power
 output power





589.7 0.9

 655.2kW

– Yearly Cost
 Q .( C steam ).( t )  ( 655 . 2 ).( 0 . 06 ).( 24 )( 365 )( 0 . 95 )  $ 32 , 7155 / yr

The utility cost for heater:
• Duty • Cost of noncreative process $6.0/GJ with the efficiency 0.9. • Yearly Cost
 Q .( C steam ).( t )  ( 4 . 37 ).( 6 . 0 ).( 24 )( 365 )( 0 . 95 )  $ 242 , 448 / yr
 GJ     4 . 37  hr 

The utility cost for Pump (Original plant):
• The shaft Power is 5.92 kW. The efficiency of an electric drive is about 85%. 5 . 92  • Electric power      6 . 96 kW
 0 . 85 

• The Cost of Electric is $0.06/kWh • Yearly Cost
 6 . 96 kW ( 24 )( 365 )( 0 . 95 )( $ 0 . 06 kWh )  $ 3 , 475 / yr

The utility cost for fan (original plant) :
• Duty • Cost of electricity $0.06/kW.hr. And the efficiency of electricity 0.9. • calculate the electricity power
 output power
 kJ     182  hr 





182 0.9

 2 02 .2kW

• Yearly Cost
 Q .( C steam ).( t )  ( 202 . 2 ).( 0 . 06 ).( 24 )( 365 )( 0 . 95 )  $ 100 ,963 / yr

The utility cost for heater (Original plant):
• Duty  hr   0 .27 • Cost of noncreative process $6.0/GJ with the efficiency 0.9. • Yearly Cost
GJ

 Q .( C steam ).( t )  ( 0 . 27 ).( 6 . 0 ).( 24 )( 365 )( 0 . 95 )  $ 15 ,124 / yr

The Estimated Cost of Equipment: Absorption tower cost:
Equipment Absorption Tower Carbon steel

MOC
2.75 Diameter (m) 7.9 Height (m) 27 Pressure (bar) Stainless steel sieve MOC of Tray 9 Number of tray

Calculate the purchased cost :
 Absorption tower:
• For the Tower:
A A

D
4

2

V 
2

D
4

2

L
2

  ( 2 . 75 )
4

 5 . 94 m

2

V 

  ( 2 . 75 )
4

 7 . 9  46 . 9 m

3

C BM  C F BM
o p

log 10 C p  K 1  K 2 log 10 (V )  K 3 (log
o

(V )) 10
K3

2

Equipment Type Tower

Equipment Description Absorption

K1 3.4974

K2 0.4485

0.1074

log 10 C p  3 . 4974  0 . 4485  log 10 ( 46 . 9 )  0 . 1074  (log
o

( 46 . 9 )) 10

2

C p  35227 $( purchased cost)
o

F BM  B 1  B 2 F p F M
Equipment Type Tower Equipment Description Distillation B1 2.25 B2 1.83

F BM  B 1  B 2 F p F M ( P  1) D Fp  2 850  0 . 6 ( P  1)  0 . 0063  0 . 00315

F p  7 .8 FM  1 for vertical tower Carbon Steel

F BM  16 . 4 C BM C BM
( Tower ) ( Tower )

 C p F BM
o

 577723 $

• For the Tray:
log 10 C p  K 1  K 2 log 10 ( A )  K 3 (log
o

( A )) 10

2

Equipment Type Tower
D
4
2

Equipment Description Distillation
  ( 2 . 75 ) 2  2     5 . 94 m 4  

K1 2.9949

K2 0.4465

K3 0.3961

Atray 

log 10 C p  2 . 9949  0 . 4465  log 10 ( 5 . 94 )  0 . 3961  (log
o

10

( 5 . 94 ))

2

C p  3781 $
o

log 10 Fq  K 1  K 2 log 10 ( N )  K 3 (log

10

( N ))

2

Equipment Type Tower

Equipment Description Distillation

K1

K2

K3

0.4771

0.08516

-0.3473
2

log

10

F q  0 . 4771  0 . 08516 log

( 9 )  0 . 3473 (log 10

( 9 )) 10

F q  5 . 24 F BM  1 . 83 , From table C BM C BM C BM
( tray )

 C p F BM NF q
o

( tray )

 3781  1 . 83  9  5 . 24  326311 $

( tray )

– Total cost of absorption tower is:
C BM C BM
( total )

 C BM

( tray )

 C BM

( tower )

( total )

 326311  577723  904034 $

Purchased Cost for other equipment:
Equipment Absorption tower Makeup pump Heat Exchanger Heat exchanger Cost 904034 $ 181036 $ 79821 $ 78582 $

Fired heater
Scrubber Flash drum Total

1017575 $
217654 $ 10375 $ 2582077 $

Conclusion
• Natural gas processing is an essential part of chemical engineering. • Natural gas dehydration is an important process due to:
– the pipe line specifications – water problems in natural gas transportation and processing – requirements of the users

Conclusion
• The possible places of glycol losses are located in
– accumulator-reboiler system – flash drum

• Advantages of Glycol dehydration
– – – – Lower installed cost It is a continuous process. Low pressure drop (5-10 psi) Require less regeneration heat per pound of water vapor removed (low operating cost).

• Disadvantages of Glycol dehydration
– Water dew-point temperature is limited to temperature value higher then -25 oF – For lower temperature than -25 oF, a stripping gas is required with very high concentrated lean glycol solution. – Glycols are corrosive when decomposed or contaminated.

Conclusion
• Advantages of molecular sieves:
– – – – Very low dew point and water content can be obtained Best suited for large volumes of gas under very high pressure Dehydration of very small quantities of natural gas at low cost insensitive to moderate changes in gas temperature, flow rate, and pressure. – They are relatively free from problems of corrosion, foaming, etc. – Some types can be used for simultaneous dehydration and sweetening

• Molecular Sieves disadvantages
– – – – – The most expansive adsorbents The regeneration temperature is very high (operating cost). Pressure drop is too high High space and weight required Mechanical breaking and contamination of liquid, oil and glycol are possible


				
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