Mini Design Slide

Chemical Properties

Chemical names

Carbon dioxide Carbonic anhydride Dry ice Carbonic acid gas Carbonic acid anhydride

Molecular weight(lb/mol) 44.01

Specific gravity 1.555

Molecular formula C02

Molecular structure

Critical temperature(psia) 1071.0

Melting point -69.9 °F

Boiling Point -109.2 °F

Liquid Density @ 70°F (lb/ft3) 47.64

Gas Density @ 70°F 1 atm

(lb/ft3)

0.1144

Physical properties of 1-Tetradecene

State Colourless

Odor Odorless

Solubility in water Soluble

APPLICATIONSAPPLICATIONS

• Multi – industry uses for Carbon Dioxide (CO2)• Metal Industry

• Manufacturing and Construction• Chemical, Pharmaceuticals and Petroleum

• Food and Beverages• Environmental uses

PROCESS DESCRIPTION

Production of Carbon Dioxide using from Biomass

Advantages :Renewable sourcesDisadvantage:Research is needed to reduce the costs of Biomass based on fuelLand used for energy crops maybe in demand for other purposes such as farming, conservation, housing, resort and agriculture use.Produce harmful air pollution

Production of Carbon Dioxide from Water Gas Shift (WGS)

Advantages :Easy to operateMore efficient, resulting in a reaction yield greater than that achieved with existing catalysts.Can process at medium temperature (250-350ºC)Can achieve a complete CO conversion

PROCESS SELECTION

PROCESS DESCRIPTIONPROCESS DESCRIPTION

• The raw material that involved in this process is carbon monoxide (CO) and the water (H2O) that to produce carbon

dioxide (CO2) and hydrogen (H2). • This process involve water- gas shift(WGS) reaction that has

been traditionally used to produce hydrogen from the syngas, which is comprises CO and H2.

• The reaction is mildly in the exothermic and the equilibrium limited.

• When the temperature increases along the length of the reactor the extent of reaction will becomes limited. To achieve the desired extent reaction the two-stage process interstage

cooling needed.

STREAM TABLE

MARKET ANALYSIS

1. The current global CO2 demand is estimated to be 80M tonnes per annum, of which 50M tonnes per annum is used for EOR in North America.

3. The future potential demand for CO2 that could eventuate 2020 is estimated to be 140M tonnes per annum

2. The remaining 30M tonnes per annum represents the global demand of all uses, predominantly the mature industries of beverage carbonation and food industry uses

4. Based on Global CCS Institute, in 2009 Dakota Gasification Company’s Great Plains Synfuels Plant sold US$53.2m worth of CO2, whilst it produced 2.8M tonnes per annum, suggesting a price of US$19 per metric tonne produced, incorporating the cost of transportation

Figure 1: Approximate proportion of current CO2 demand by end use

MASS BALANCE:

REACTOR R-1301

SEPARATOR

hkmolnL OH /70.5992.6098.02

hkmolV

V

LVF

/3.13070.59190

70.59190

98% of the water leaving at the bottom product.

2% of the water leaving at the upper product

hkmol

n OH

/22.1

92.6002.02

Overall mass balance:

Inlet Outlet

Subtances ṅ(mol/h) Ĥ(kJ/mol) ṅ(mol/h) Ĥ(kJ/mol)

CH4 1050.6 Ĥ1 - -

O2 3151.9 Ĥ2 1050.7 Ĥ4

N2 7716.8 Ĥ3 7716.9 Ĥ5

CO2 - 1050.7 Ĥ6

H2O - 2101.3 Ĥ7

ENERGY BALANCE:

FIRED HEATER H-1301

CH4 (25ºC) : Ĥ1 =(∆Ĥf)= -74.85kJ/mol

O2 (25ºC) , N2 (25ºC) : Ĥ2=Ĥ3 = 0 O2 (417ºC) : Ĥ4 = 12.28 kJ/molN2 (417ºC) : Ĥ5 = 11.68 kJ/molCO2 (417ºC) : Ĥ6 = ∆Ĥf + Ĥ(417ºC)

= (-393.5 + 17.20) kJ/mol=-376.3 kJ/mol

H2O (417ºC) : Ĥ7 = ∆Ĥf + Ĥ(417ºC)= (-241.83 + 13.87) kJ/mol= -227.96 kJ/mol

Inlet Outlet


CH4 1050.6 -74.85 - -

O2 3151.9 0 1050.7 12.28

N2 7716.8 0 7716.9 11.68

CO2 - - 1050.7 -376.3

H2O - - 2101.3 -227.96

hkJ

Q

HQout

/36.692717

)41.78637()77.771354(

nH-nHin

Inlet Outlet


CO 33200 Ĥ1 13280 Ĥ5

H2 40800 Ĥ2 60720 Ĥ6

CO2 25200 Ĥ3 45120 Ĥ7

H2O 90800 Ĥ4 70880 Ĥ8

hkmolnn

CO

inCOoutCO /1992013320013280)()(

molkJ

HHHHH

HH

OfHfCOfHfCOr

fr

/81.524

)83.241)(1()5.393)(1()0)(1()52.110)(1(

))(1())(1())(1())(1(222

REACTOR R-1301References: CO(g), H2(g), CO2(g), H2O(g) at 25ºC and 1 atm.

Extent of reaction

Standard Heat of Reaction

hkJnH

molkJTTTH

molkJTTTH

molkJTTTH

molkJTTTH

in

C

C

C

C

/22.1927294)3017.1090800()5511.825200()7884.840800()5857.1233200(

/3017.1010593.3107604.0106880.003346.0

/5511.810464.710887.210233.403611.0

/7884.8108698.0103288.01000765.002884.0

/5857.1210220.2103548.0104110.002095.0

2.320

25

3122854

2.320

25

3122853

2.320

25

3122852

2.320

25

3122851

hkJ

nH

molkJTTTH

molkJTTTH

molkJTTTH

molkJTTTH

out

C

C

C

C

C

C

C

C

/13.2528066

)3635.1470880()7728.1145120()1847.1260720()9922.1713280(

/3635.1410593.3107604.0106880.003346.0

/7728.1110464.710887.210233.403611.0

/1847.12108698.0103288.01000765.002884.0

/9922.1710220.2103548.0104110.0020905.0

2.430

25

3122858

2.430

25

3122857

2.430

25

3122856

2.430

25

3122855

Inlet enthaply

Outlet Enthalpies

hkJ

molkJhkJmolkJhmolQ

nHnHHHQout in

r

/78.9773510

/22.1893214/64.2206551)/81.524)(/19220(

Energy balance for others equipment are shown in table below:

Equipment Q (kJ/h)

Heat exchanger E-1302 510693.2

Heat exchanger E-1303 8740796

Reactor R-1302 257068.61

HEAT INTEGRATION

Stream Conditions ṁ(kg/s)

Cp

(kJ/kg·ºC)Tin (ºC) Tout (ºC)

1 Hot 1.043 Cp1 430 203

2 Hot 1.043 Cp2 260.7 185

3 Hot 1.043 Cp3 185 50

CmolkJC

CmolkJC

CmolkJC

CmolkJC

pCO

pH

pCO

pCO

/0375.0)2.430(10593.3)2.430(107604.0)2.430(106880.01046.33

/0294.0)2.430(108698.0)2.430(103288.0)2.430(1000765.01084.28

/0312.0)2.430(10220.2)2.430(103548.0)2.430(104110.01095.28

/0507.0)2.430(10464.7)2.430(10887.2)2.430(10233.41011.36

3122853

3122853

3122853

3122853

2

2

2

32 dTcTbTaCP Calculation Cp every components :

Stream 1:T=430.2ºC

Components Cp (kJ/mol·ºC) MW Cp (kJ/kg·ºC) Mole Fraction, X

Cpi X

CO2 0.0507 44.01 1.1515 0.2375 0.2735

CO 0.0312 28.01 1.1138 0.0699 0.0779

H2 0.0294 2.016 14.5894 0.3196 4.6628

H2O 0.0375 18.016 2.0838 0.3731 0.7775

Total 5.7916

Components Cp (kJ/mol·ºC) MW Cp (kJ/kg·ºC) Mole Fraction, X

Cpi X

CO2 0.0436 44.01 0.9903 0.2375 0.2352

CO 0.0299 28.01 1.0691 0.0699 0.0747

H2 0.0290 2.016 14.3943 0.3196 4.6004

H2O 0.0352 18.016 1.9547 0.3731 0.7293

Total 5.6396

T=230ºC

CkgkJC p /152.06396.57916.51

CskJCm p

/16.0152.0043.1

Stream Conditions ṁCp

(kJ/s·ºC)Tin (ºC) Tout (ºC)

1 Hot 0.16 430 203

2 Hot 0.052 260.7 185

3 Hot 0.077 185 50

For stream 1:

Repeat the same method for stream 2 and stream 3 .

Step 1 : Minimum approach temperature ∆Tmin

The value 10 ºC is chosen in this system.

Step 2: Temperature Interval Diagram

Streams 1 2 3

ṁCp

(kW/ºC)0.16 0.052 0.077

430.2 420.2

260.7 250.7

203 193

185 175

50 40

kWQ

kWQ

kWQ

kWQ

TCmQ

D

C

B

A

p

40.10)50185(077.0

32.2)185203(077.0)185203(052.0

23.12)2037.260(052.0)2037.260(16.0

12.27)7.2602.430(16.0

Step 3: Cascade Diagram

Cascade diagram

It can be shown in this system it does not have pinch because heat is only cascade downward and rejected to cold utility.

In this system, there is no need to supply energy from the hot utility to the process.

This case is refer as threshold problem with respectively no cold utility or hot utility and without pinch point.

Threshold problems only need a single thermal utility either hot or cold but no both and over a range of minimum temperature difference ranging from zero to threshold temperature.

HEAT EXCHANGER(E-1303)

From table 11.11, the following heuristics is used:

Rule 1 - F=0.9 for shell and tube heat exchanger, there is no phase change.

Rule 6 - ∆T = 10 ⁰C

Rule 7 - cooling water inlet is 30˚C and maximum outlet 40˚C

Rule 8: Heat transfer coefficient, for water to liquid, U= 850 W/m2 ⁰C

The cold outlet is 50⁰C

The hot inlet is 185⁰C

∆T hot = 185⁰C - 40˚C = 145 ⁰C ∆T cold = 50⁰C - 30˚C = 20 ⁰C ∆Tlm = ∆T hot - ∆T cold / ln [∆T hot / ∆T cold] ∆Tlm = (145⁰C) – (20 ⁰C ) / ln [145 ⁰C /20 ⁰C]= 63.01˚C

Q= 242800 W Q=UAF∆Tlm

242800W = (850 W/m2.˚C) A (0.9) (63.01)

A = 5.04 m2

FURNACE (H-1301)

From table 11.11, use the following heuristic :

Rule 13 – Equal heat transfer in radiant and convective sections

Radiant rate = 37. 6 kW/ m2, Convective rate = 12.5 kW/ m2

Duty =192.42 kW

Area radiant section = (0.5)(192.42)/(37.6) = 2.56 m2

Area convective section = (0.5)(192.42)/(12.5) = 7.70 m2

REACTOR


Rule 1 – The rate of reaction in very instance is established on the laboratory

Rule 2 – Dimensions of catalyst in packed bed 2 – 5 mm (powder)

Rule 13- The value of a catalyst may improve the selectivity.

-r carbon dioxide = ko exp ( −EaRT )P methanol

-r carbon dioxide = 296.81 exp ( −40.739ሺ8.314ሻ(698.1)) 1

-r carbon dioxide= 296.51

V= 𝐹𝑎𝑜−𝑟 𝑐𝑎𝑟𝑏𝑜𝑛 𝑑𝑖𝑜𝑥𝑖𝑑𝑒 1𝑑𝑥0.60

V = 190000ሺ44ሻ1.98296.51 (0.6− 0)

V=8486L

VESSEL (V-1301)


Rule 3 – Gas-liquid phase separators are usually vertical vessel

Rule 4 – L/D between 2.5 and 5 with optimum at 3.0

Rule 5 – Liquid holdup time is 5 min based on ½ volume of vessel

Rule 9 – Gas velocity in gas/liquid separators, u=√𝑝𝑙𝑝𝑣 − 1 , k=0.0305

Rule 12 – Good performance can be expected at velocities of 30% - 100% of those calculated

with given k, with 75% is popular.

From the stream table

Vapor flow = Stream 9 =2680.98 kg/h P=12.7 bar T = 50°C

Liquid flow = Stream 10 = 1075.50 kg/h P= 11.7 bar T = 50°C

ρv = 0.630 kg/m3 ρl = 958 kg/m3

From the rule 9

u = 1.18 m/s

uact = (0.75)(1.18m/s) = 0.885 m/s

mass flow rate of vapor = uρvπD2/4 = 2680.98/h x 1h/3600s = 0.744kg/s

solving for D , D= 1.33 m

From the rule 5

Volume of liquid = 0.5LπD2/4 = 0.895L

5min of liquid flow = [(5)(60)(1075.50)]/[958/3600] = 0.094m3

Equating the two results above, L= 0.11 m3

From the rule 4

L/D should be in the range 2.5 to 5 but in this case L/D = 0.08 (not in ranged)

So change L=2.5D = 3.33 m.

Therefore, V-1301 should be vertical vessel with D = 1.33m, L= 3.33 m

Bare Module Cost of EquipmentBare Module Cost of EquipmentEquipment

Cp˚ (2001)

($)

CBM (2001)

($)

CBM ˚ (2001)

($)

Heat Exchanger

E-1301

E-1302

E-1303

57 650.9

28 133.8

57 777.6

196 160.0

95 725.3

196 588.3

189 671.5

92 560.2

190 088.3

Fired Heater

H-1301

169 572

440 289.3

440 289.3

Vessel

V-1301

4 618.8

24 307.8

18 798.5

Reactor

R-1301

R-1302

91 891.0

69517.1

367 564.0

278 068.4

367 564.0

278 068.4

Total 479 161.2 1 598703.1 1 577 040.2

CEPCI 2014 = 576.1

CBM (2014) = CBM (2001) (576.1/394)

CBM (2014) = (1 598703.1) (576.1/394) = $ 2 337 596.08

•VESSELV-1301Vertical

Carbon steelL = 1.83 m D = 0.61 m

Maximum pressure rating of 19 bar From Table A.4;Volume = 0.5348 m3 B1 = 2.25

B2 = 1.82CBM = Cp˚ [ B1 + B2 FP FM ]

From Appendix A Cp˚/ V = $ 8 636.4 / m3

Cp˚(2001) = x 0.5348 m3 = $ 4 618.8

Equation A.2: From Table A.3;FP = for P>-0.5 barg FM = 1

FP = = 1.6554

CBM (2001) = Cp˚ (2001) [ B1 + B2 FP FM ] = $ 4 618.8 [ 2.25 + 1.82 (1.6554)(1)]

= $ 24 307.7CBM (2014) = CBM ˚ (2001) (576.1/394) = $ 24 307.7 (576.1/394)

= $ 35 542.3

• HEAT EXCHANGERE-1301

Floating head, carbon steel, shell and tube designProcess stream in tubes

A = 67.2 m2

Q = 1580 MJ/hMaximum pressure rating of 19 bar

CBM = Cp˚ [ B1 + B2 FP FM ] From Table A.4;From Appendix A Cp˚/ A = $ 857.9 / m2 B1 = 1.63Cp˚(2001) = x 67.2 m2 = $ 57 650.9 B2 = 1.66

Equation A.3: From Table A.3;

log FP = C1 + C2 log P + C3 (log P)2 FM = 1From Table A.2;

C1 = 0.03881, C2 = -0.11272, C3 = 0.08183log FP = 0.03881 - 0.11272 log 19 + 0.08183 (log 19)2

FP = 1.0678

CBM (2001) = Cp˚ (2001) [ B1 + B2 FP FM ] = $ 57 650.9 [ 1.63 + 1.66 (1.0678)(1)]

= $ 196 160CBM (2014) = CBM ˚ (2001) (576.1/394) = $ 196 160 (576.1/394)

= $ 286 821.77

•Fixed Capital Investment (FCI)

CTM = 1.18 = 1.18 ($ 1 598 703.1) (576.1/394)

= $ 2 758 363.4

CGR = CTM + 0.50 = $ 2 758 363.4 + 0.50 ($1 577 040.2) (576.1/394)

= $ 3 911 323.9

•Cost of Operating Labor (COL)

NOL = ( 6.29 + 31.7P2 + 0.23Nnp )0.5

P = 0, because no handling particulate solid NOL = ( 6.29 + 31.7(0)2 + 0.23(6) )0.5

= 2.77The number of operations required per shift = 2.77

Operating labor = 4.5(2.77) = 12.5 (round to the nearest integer yields 13 operators)

Labor Cost = 13 x $ 59 580 = $ 774 540/year

Equipment No of equipment Nnp

Reactors

Heaters

Exchangers

Vessels

2

1

3

1

2

1

3

-

Total 6

No Types Description Price Price/year

1. High pressure steamP – 19 bar

Q – 1580 MJ/h $ 17.70 /GJ $ 232 846.5


Q – 3764 MJ/h$ 17.70 /GJ $ 67 021.2


Q – 455 MJ/h$ 17.70 /GJ $ 554 434.9

4. Natural gasP – 19 bar

Thermal efficiency – 80%$ 13.88 /GJ $ 92 407.5

Total = $ 946 728.1 / year

•Fired Heater H-1301Duty = 800 MJ/h Cost = $ 13.88 / GJ

Thermal efficiency = 75% t = 0.95

Yearly Cost = Q Csteam t = x x x x x 0.95 = $ 92 407.5 / year

•Heat Exchanger E-1301

Duty = 1 580 MJ/h Cost = $ 17.70 / GJt = 0.95

Yearly Cost = Q Csteam t = x x x x x 0.95

= $ 232 864.5 / year

•Cost of Waste Treatment (CWT)Assume the waste treatment primary (filtration)

Component of water = 1059.1 kg/h Density = 988.04 kg/m3

Yearly Cost = (Yearly flowrate)(Cost per unit mass)(Density) = x x x x x 0.95

= $ 365.7 / year

•Cost of Raw MaterialsComponent

Mass Flowrate

(kg/h)

Cost of Material

($/kg)

Yearly Cost

($/year)

CO 929.93 0.028 294 077.3

CO2 1109.05 0.16 1 476 722.3

H2 82.42 0.0025 1 714.7

H2O 1621.35 5.7x10-3 76 909.4

Total = $ 1 849 423.7 / year

•Cost of Manufacturing (COMd)COMd = 0.18FCI + 2.73COL + 1.23 (CUT + CWT + CRW)

= 0.18($ 3 911 323.9/year) + 2.73($ 774 450/year) + 1.23 ($ 946 728.1/year + $ 365.7/year + $ 1 849

423.7 /year) = $ 6 258 249.03 / year

•Cost of LandLand Cost = 10 acres x = $ 10 000 000

•Working CapitalWorking Capital (WC) = x FCI = 0.2 x $ 3 911 323.9

= $782 264.8

•Assumptions: Land cost, L = $ 10 000 000

Total Fixed Capital Investment, FCIL = $ 3 911 323.9Fixed Capital Investment during year 1 = $ 2 058 823.9Fixed Capital Investment during year 2 = $ 1 852 500

Plant start-up at end year 2Working Capital = $782 264.8

Yearly Sales Revenue, R = $ 10 000 000Cost of Manufacture without Depreciation, COMd = $ 6 258

249.03Taxation rate = 25% = 0.25

Interest rate = 2.96% = 0.0296Plant salvage value, S = $ 0

Depreciation by 5 year MACRSPlant life = 10 years

Year Investment dkFCI - ∑dk R COMd

(R -COMd -

dK) (1-t) + dkCash Flow

Cumulative Cash

Flow

0 -10000000.0 - 3911323.9 - - - -10000000.0 -10000000.0

1 -2058823.9 - 3911323.9 - - - -2058823.9 -12058823.9

2 -2634764.8 - 3911323.9 - - - -2634764.8 -14693588.7

3 - 782264.8 3129059.1 10000000 6258249.03 3001879.4 3001879.4 -11691709.3

4 - 1251623.6 1877435.5 10000000 6258249.03 3119219.1 3119219.1 -8572490.2

5 - 750974.2 1126461.3 10000000 6258249.03 2994056.8 2994056.8 -5578433.4

6 - 450584.5 675876.8 10000000 6258249.03 2918959.4 2918959.4 -2659474.0

7 - 450584.5 225292.3 10000000 6258249.03 2918959.4 2918959.4 259485.4

8 - 225292.3 0.0 10000000 6258249.03 2862636.3 2862636.3 3122121.7

9 - 0.0 0.0 10000000 6258249.03 2806313.2 2862636.3 5984758.0

10 - 0.0 0.0 10000000 6258249.03 2806313.2 2862636.3 8847394.3

11 - 0.0 0.0 10000000 6258249.03 2806313.2 2862636.3 11710030.6

12 2282264.8 0.0 0.0 10000000 6258249.03 2806313.2 5088578.0 16798608.6

Payback Period (PBP) = 1 years

Cumulative Cash Position (CCP) = $ 16.8M

Cumulative Cash Ratio (CCR) = 2.14

Rate of Return on Investment (ROROI) = 43%

Non-discounted After-Tax Cash Flows

Year Non-discounted Cash Flow Discounted Cash Flow Cumulative Cash Flow

0 -10000000.0 -10000000.0 -10000000.0

1 -2058823.9 -1999634.7 -11999634.7

2 -2634764.8 -2485448.6 -14485083.3

3 3001879.4 2750347.9 -11734735.4

4 3119219.1 2775695.0 -8959040.3

5 2994056.8 2587720.5 -6371319.9

6 2918959.4 2450286.4 -3921033.4

7 2918959.4 2379843.1 -1541190.4

8 2862636.3 2266824.5 725634.1

9 2862636.3 2201655.5 2927289.7

10 2862636.3 2138360.1 5065649.7

11 2862636.3 2076884.3 7142534.0

12 5088578.0 3585700.7 10728234.7

Discounted Payback Period (DPBP) = 1.3 years

Net Present Value (NPV) = $ 10.73M

Present Value Ratio (PVR) = 1.74

Discounted Cash Flow Rate of Return (DCFROR) = 12.15%

Discounted After-Tax Cash Flows

The value of discounted cash flow rate of return (DCFROR) found at NPV = $ 0. Therefore, interpolation method was used.

DCFROR = 10.00 + 5(0.43) = 12.15 %

Interest or Discount Rate (%) NPV ($ M)

0

2.96

10.00

15.00

20.00

16.80

10.73

1.62

-2.15

-4.67

62.115.262.10

00.1000.1500.10

DCFROR

Conclusion Conclusion

•In term of the time basis, the payback period increases as the discount rate increases. It shows that the payback period increases from 1 to 1.3 years.

•In term of the cash basis, when replacing the cash flow with the discounted cash flow decreases the value at the end of the project. It shows that the

value at the end of the project dropped from $ 16.80M to $ 10.73M.•In term of the cash ratios, discounting the cash flow gives a lower ratio. It

shows that the ratio dropped from 2.14 to 1.74.

Comparison Non-discounted Cash Flow Discounted Cash Flow

PBP / DPBP

CCP / NPV

CCR / PVR

ROROI / DCFROR

1 year

$16.80M

2.14

43%

1.3 years

$10.73M

1.74

12.15%

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Mini Design Slide