Design of Ammonia Section in Ammonia Synthesis Plant

1
Design of Ammonia Section in Ammonia Synthesis
Plant
United Arab Emirates University Collage of
Engineering Department of Chemical Petroleum
Engineering Industrial Training Graduation
Projects Unit Graduation Project II Course Fall
Semester 2008
Advisor Dr. Nayef Ghasem

• Rashed Khalfan Al Kindi 200235986
• Shabbeer Ali Yusuf 200337936
• El-Hassan Mohammed 200337931
• Ali Saleh Mohammed 200202794
• Hamad Al Zaidi 200200655

2
Overview

• Introduction
• Design Equipments
• Heat Exchangers
• Reactors
• Calculations of design
• Hand Calculations (Excel)
• HYSYS design
• Cost Estimation
• Operation Cost
• Capital Cost
• Cost of Manufacturing
• Results by CAPCOST

3
Introduction

• Objectives
• Design of ammonia section in ammonia synthesis
  plant for a 30 ton/day production of ammonia
• Estimate the Cost
• Material and energy balance was done in GPI

4
Introduction

• Design of process units theoretically and using
  HYSYS software
• The units designed were reactor and heat
  exchanger

5

• Special concerns related to the process was done
• corrective measures
• Safety and environmental impact of the project
  were analyzed
• Cost estimation was conducted using CAPCOST
  software

6
Process Flow Diagram
7
Reactor Design

• Assumptions
• The rate equation from HYSYS
• No catalyst, no void fraction
• No pressure drop
• Change in temperature

8
Derivation

• The design equation of PFR is
• X is the conversion
• V is the volume
• -rN2 is the rate of reaction of the limiting
  reagent, nitrogen
• FN2o is the input flow rate of the nitrogen

9
Derivation

• The rate of reaction
• kf and kb are the forward and backward rate
  constants
• PN2, PH2, PH3 are the partial pressures of
  nitrogen, hydrogen and ammonia

10
Derivation

• The partial pressure of component B
• where PBo is the intial partial pressure of
  component B
• b is the stoichiometric coefficient of the
  component B
• ?B is the ratio of flow rate of component B to
  that of flow rate of basis component.

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Derivation

• So, for our reaction,
• Po is the total input pressure which in this case
  is

12
Derivation
13
Rate vs. Reactor Volume
14
Derivation

• Change in temperature through the reactor

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Behavior
16
Derivation

• Change in temperature through the reactor

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Derivation
? d
N2 2.90E-02 2.20E-06 5.72E-09 -2.87E-12
H2 2.88E-02 7.65E-08 3.29E-09 -8.70E-13
NH3 3.52E-02 2.95E-05 4.42E-09 -6.69E-12
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Derivation

• Enter these equations to the polymath software to
  obtain the volume for 40 conversion

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Results

• The volume of the reactor obtained from Polymath
  is 1.78 m3
• Further HYSYS was used to obtain the volume of
  the reactor by setting the diameter to be 0.75 m.
  Thus the volume obtained was 0.998m3 for the
  highest conversion

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Heat Exchanger Design

• Importance ? Heat integration
• Type of heat exchanger to be designed is
  countercurrent shell and tube
• The most important factor in a heat exchanger
  design is the heat transfer area (A)

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Heat Exchanger Design

• Heat Exchanger design procedures
• Inlet and outlet temperatures are
• Flow rates mc mh 0.577 kg/s

Tc,in 49.8 oC
Tc,out 114 oC
Th,in 333.2 oC
Th,out 263.2 oC
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Heat Exchanger Design

• 1- Find the physical properties
• 2- Calculate the heat transfer rate (q)
• q 148680.4 J/s

T (C) ? (kg/m3) µ (kg/m.s) k (W/m.K) Pr Cp (J/kg.C)
Shell 81.9 18.09 0.000115 0.1381 335296.7 4016
Tube 298.2 39.69 0.000168 0.1507 417740.2 3745
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Heat Exchanger Design

• 3- also
• 4- Find (?T)LMTD
• (?T)LMTD 216 oC
• 5- Find F

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Heat Exchanger Design

• R 0.92 and P 0.25
• ? F 0.98

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Heat Exchanger Design

• 6- Assume a value for U 10 50 W/m2.C
• 7- The heat transfer area (A) is
• A 14.6 m2
• 8- choose initial values for L, Do and Di
• L 2.3 m, Do 0.04 m and Di 0.036 m

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Heat Exchanger Design

• Tube side
• 9- Calculate the area of one tube
• Atube 0.37 m2
• 10- Calculate the number of tubes (Nt),
• Nt 40 tubes
• 11- Find the fluid velocity
• uin 9.14 m/s

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Heat Exchanger Design

• 12- Find Reynolds number
• Re 97063.2 ? turbulent flow
• 13- Find Nusselt number (Nu)
• Nu 314.87
• 14- Calculate the pressure drop
• Np number of tube passes (2)
• ?Pt 12.02 kPa

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To find the tube side friction jf
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Heat Exchanger Design

• Shell side
• 15- Choose the pitch type ? Triangular
• 16- Find the bundle diameter
• Db 0.867 m
• 17- Find the shell diameter (Ds), Ds Db
  bundle diametrical clearance
• Ds 0.92 m
• 18- Calculate the baffle spacing , lB
• lB 0.184 m

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Heat Exchanger Design

• 19- Choose the tube pitch(pt), 1.25Do, and the
  baffle cuts, 25
• 20- Calculate the cross flow area As,
• As 0.0121 m2
• 21-Calculate the mass velocity ,
• Gs 47.5 kg / s.m2
• 22- Calculate the equivalent diameter ,
• de 0.043 m

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Heat Exchanger Design

• 23- Calculate the Reynolds number
• Re 17607
• 24- Calculate the Nusselt number
• Nu 1317.6
• 25- Calculate the pressure drop
• ?Ps 128.7 kPa

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To find shell side heat transfer factor, jh
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To find shell side friction factor, jf
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Heat Exchanger Design

• Overall heat transfer coefficient
• 26- Find the local heat transfer coefficient
• hin 1054.5 W/m2.C and ho 4255.3
  W/m2.C
• 27- Calculate the overall heat transfer
  coefficient U
• 28- Use Goal Seek
• Set U 50 by changing L ? L 2.35 m

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Heat Exchanger Design

• Main Results

Area (m2) 14.6
Number of tubes 40
Tube in-diameter (m) 0.036
Tube out-diameter (m) 0.04
Shell diameter (m) 0.92
Length (m) 2.3
Pressure drop in the tube side, ?Pt (kPa) 12.02
Pressure drop in the shell side, ?Ps (kPa) 128.7
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Special Concerns

• Special concerns are out of normal operating
  conditions.
• Specific justification required else dont use
• Normal conditions
• Pressure between 1 10 bar
• Temperatures between 40 C 260 C

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Concerns in Pressure

• Pressures up to 10 bars without much additional
  capital investment
• Higher pressures
• Thicker walls
• More expensive equipment
• In vacuum conditions
• Large equipment
• Special construction techniques
• Higher cost

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Concerns in Temperature

• At high temperatures common construction
  materials like carbon steel lose their physical
  strength drastically
• high temperature - economic penalty
• more complicated processing equipment
• refractory-lined vessels
• exotic materials of construction

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Operating Conditions
40
Reactor

• High Pressure of 196 bars
• Justification
• Thermodynamically
• Kinetically

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Kinetic Justification

• Rate given by
• Concentration becomes by
• Substituting by partial pressure

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Thermodynamic Justification

• _at_ constant temperature

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Heat Exchanger

• Used Heat Integration
• If ?Tlm gt100C
• Else wastage of usable energy

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Cost Estimation

• Profitability of the project
• The feasibility of any project proposal should
  pass the stage of preliminary cost estimation
  even before any further study can be done on the
  technical aspects
• Type of costs
• - Capital
• - Operating

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Capital Cost

• Cost of the plant ready for start-up
• Includes
• Design, and other engineering and construction
  supervision
• All items of equipment and their installation
• All piping, instrumentation and control systems
• Buildings and structures
• Auxiliary facilities, such as utilities, land and
  civil engineering work

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Operating Cost

• Cost involved in the day to day operation of the
  plant.
• Includes
• Direct cost
• Raw Materials
• Utilities
• Operating Labor
• Fixed cost
• Insurance
• Local Taxes
• General Expenses
• Administration cost
• Distribution and selling cost

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Equipments

• Compressors
• Heat Exchangers
• Reactors

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Effect of Capacity on purchased cost

• Cb is the purchased cost of the equipment with
  base capacity Ab
• Ca is the purchased cost of the equipment with
  required capacity Aa
• n is the cost exponent

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Effect of time on purchased cost

• C is purchase cost
• I is the cost index
• 1 refers to base time when the cost is known
• 2 refers to time when cost is desired

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Bare Module cost estimation technique

• CBM is bare module equipment cost
• FBM is bare module cost factor
• CPo is the purchase cost for the base condition,
  i.e. atmospheric pressure and material of
  construction is carbon steel

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Pressure factor - Fp

• C1, C2 and C3 are constants for each equipment
  type

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Material FM
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Table A3
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Cost of manufacturing

• Total direct manufacturing costs
• (COM) CRM CWT CUT 1.33COL 0.3COM
  0.069FCI
• CRM is the cost of raw material
• CWT is the cost of waste treatment
• CUT is the cost of utilities
• COL is the cost of operating labor
• FCI is the fixed capital investment

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CAPCOST
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Results

• - Equipment cost
• 1- Compressor

Equipment Cost
C-101 138000
C-102 380000
C-103 34100
C-104 17200
Total 569300
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Results, cont..

• - Equipment cost
• 2- Heat Exchanger

Equipment Cost
E-101 14200
E-102 12600
E-103 7330
E-104 3380
E-105 17900
Total 55410
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Results, cont..

• - Utility cost

Equipment Utility used Cost
E-101 Cooling water 1560
E-102 Cooling water 2490
E-103 Cooling water 1380
E-104 Cooling water 160
Total 5590
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Results, cont..

• - Material cost

Material Classification Price (/kg) Consumption (kg/h) Material Costs (/y)
Hydrogen Raw Material 2.700 242 5,437,595
Nitrogen Raw Material 0.500 1160 4,826,760
Ammonia Product 5.000 805 33,496,050
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Results, cont..

• The land cost is estimated to be 1,250,000
• The operating labor cost is estimated to be
  700,000 per year
• Total cost 12,844,655

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Results, cont..

• The cost index for 2006 is 478.7

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Safety and Environmental Impact

• The exposure limit for ammonia is 25 ppm for 8
  hours exposure and 35 ppm for a 15 minutes
  exposure
• Noise
• low-noise let-down valves
• silencers
• Toxic hazard
• Explosion are not extremely dangerous

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Conclusion

• Volume of the reactor
• 1.78 m3 (theoretical calculation)
• HYSYS value was 0.998 m3
• Heat exchanger
• shell diameter of 1m and 60 tubes
• area of heat transfer 39.48 m2 with a length of
  2.35 m.
• Total capital cost was 15,548,000

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Thank you