i Content
CONTENT...I
1 SYMBOLS APPEARED IN REPORT ... 1
2 PROCESS OPTIONS AND SELECTION ... 3
2.1 BLOCK SCHEMES... 3
2.1.1 Separation and Thermal Decomposition ... 3
2.1.2 Separation and Catalytic Decomposition ... 3
3 BASIS OF DESIGN ... 4
3.1 INPUT/OUTPUT DIAGRAMS... 4
4 THERMODYNAMIC PROPERTIES AND REACTION KINETICS ... 5
4.1 VAPOR PRESSURES OF EACH COMPONENT... 5
4.2 CONDENSATION TEMPERATURE DETERMINATION... 6
4.3 JOBACK GROUP CONTRIBUTION METHOD: ISOFLURANE HEAT CAPACITY COEFFICIENTS CALCULATIONS... 6
4.4 EQUILIBRIUM CONSTANTS... 7
4.5 METHOD OF HALF LIFE – THERMODYNAMICS FOR THERMAL DECOMPOSITION REACTION... 7
4.6 PROPERTIES OF COMPOUNDS... 9
4.7 SENSITIVITY ANALYSIS... 10
5 PROCESS STRUCTURE AND PROCESS DESCRIPTION ... 17
5.1 PRICE FOR CATALYSTS... 17
5.2 THE STRATEGIES LEVEL OF KRIHNA AND SIE... 17
5.3 PROCESS FLOW SCHEMES... 19
5.4 RESULTS OF MODELING... 23
5.4.1 Thermal Decomposition Simulation ... 23
5.4.2 Catalytic Decomposition Simulation in Aspen ... 24
5.5 PROCESS STREAM SUMMERY... 25
5.5.1 Separation of isoflurane and thermal decomposition... 25
5.5.2 separation of isoflurane and catalytic decomposition ... 27
5.6 UTILITY SUMMARY... 28
5.6.1 separation of isoflurane and thermal decomposition ... 28
5.6.2 separation of isoflurane and catalytic decomposition ... 29
6 PROCESS CONTROL... 30
7 MASS AND HEAT BALANCE... 31
7.1 HEAT LOAD CALCULATIONS ……….31
7.2 JOBACK GROUP CONTRIBUTION METHOD: ISOFLURANE HEAT CAPACITY COEFFICIENTS ... 32
7.4 MASS BALANCE FOR CATALYTIC DECOMPOSITION... 37
7.4.1 Mass Balance Calculation-Gas phase mass balance ... 37
7.5 MASS AND HEAT BALANCE FOR SEPARATION OF ISOFLURANE & THERMAL DECOMPOSITION... 39
7.6 MASS AND HEAT BALANCE FOR SEPARATION OF ISOFLURANE & CATALYTIC DECOMPOSITION. ... 40
8.PROCESS AND EQUIPMENT DESIGN ... 41
8.1 PROCESS DESIGN CALCULATIONS FOR SEPARATION OF ISOFLURANE... 41
8.2 BLOWER... 45
8.3 MONOLITH REACTOR... 47
8.4 COMPRESSOR DESIGN [36] ... 49
8.5 HEAT EXCHANGER... 53
8.6 PROCESS EQUIPMENT SUMMARY FOR SEPARATION OF ISOFLURANE AND THERMAL DECOMPOSITION... 55
8.6.1 Equipment data sheet – Heat Exchanger Summary... 55
8.6.2 Equipment data sheet –Pumps, Blowers & Compressors Summary... 56
8.7 PROCESS EQUIPMENT SUMMARY FOR SEPARATION OF ISOFLURANE AND CATALYTIC DECOMPOSITION... 57
8.7.1 Equipment data sheet – Heat Exchanger Summary... 57
8. 8 EQUIPMENT DATA SHEETS FOR SEPARATION OF ISOFLURANE AND THERMAL DECOMPOSITION... 58
8.8.1 specification sheet for Heat Exchanger – Specification Sheet... 58
8.8.2 specification sheet for Microwave Oven – Specification Sheet ... 59
8.8.3 specification sheet for Adsorption bed – Specification Sheet ... 60
8. 9 EQUIPMENT DATA SHEETS FOR SEPARATION OF ISOFLURANE AND CATALYTIC DECOMPOSITION... 61
8.9.1 specification sheet for Heat Exchanger (condenser ) – Specification Sheets………61
8.9.2 specification sheet for Microwave Oven – Specification Sheet ... 62
8.9.3 specification sheet for Heat Exchanger – Specification Sheet... 63
8.9.4 specification sheet for Adsorption bed – Specification Sheet ... 64
8.9.5 specification sheet for monolith equipment – Specification Sheet... 65
9 WASTES... 66
9.1 CALCULATION OF HEATING TIME... 66
10 SAFETY... 67
10.1 PROPERTIES OF ALL THE COMPONENTS IN THE SYSTEM... 67
10.2 HAZARD IDENTIFICATION AND ANALYSIS METHODS... 67
10.3 DOW FIRE & EXPLOSION INDEX... 69
10.4 NATIONAL FIRE PROTECTION ASSOCIATION (NFPA) SYSTEM... 71
10.5 EXPLOSION LIMITS... 71
10.6 SAFETY ASPECTS OF A HEATING SYSTEM... 73
11 ECONOMY ... 74
iii
11.2 SEPARATION AND THERMAL DECOMPOSITION... 75
11.3 SEPARATION AND CATALYTIC DECOMPOSITION... 76
12 GROUP CREATIVITY AND PROCESS TOOLS... 77
12.1 THERMAL PROCESS... 77
12.2 CATALYTIC PROCESS... 77
13 CONCLUSIONS ... 78
14 HOSPITAL VISIT... 79
14.1 QUESTIONS FOR HOSPITALS... 79
14.2 VISITS TO HOSPITALS... 80
14.2.1 Visit to Ziekenhuis Leyenburg, Den Haag ... 80
14.2.2 Visit to Rotterdam Erasmus Hospital ... 81
Appendix
1 Symbols appeared in report Notation
ch
A = cross sectional area of monolith channel (open) m 2 2
( )
C N O = concentration of N O 2 mol m−3
cpsi = cell density (cell per square inch) inch−2
d = diameter m
D = monolith cell pitch (square channel) m
2 2 ( ) D N O N− = binary diffusivity of N O in 2 N 2 m s2 −1 2 ( ) k
D N O = Knudsen diffusivity of N O in a cylindrical pore 2 m s2 −1
D = average diffusivity m s2 −1
eff
D = effective diffusivity m s2 −1
h = height m
v
k = rate constant per unit catalyst volume 3 3 1 g cat
m m s− −
L = characteristic catalyst dimension m
( )
w
M i = molar mass of species i kg kmol−1
ch
O = periphery of the monolith’s channel cross section m 2
( )
p N O = partial N O 2 pressure Pa
P = total pressure Pa
o
r = average pore radius m
R = universal gas constant J mol K−1 −1
Re = Reynolds number −
T = temperature K
u = average superficial velocity m s−1
V = volume m 3 cat W = catalyst mass kg 2 ( ) X N O = fractional conversion of N O 2 − 2 ( ) y N O = molar fraction of N O 2 −
Cp = Specific heat kJ/kmolK γ = Isentropic work factor -
m = Polytropic temperature exponent
Greek
δ = Monolith wall thickness m
2
φ = Generalized Thiele modulus (1storder irreversible reaction)− g
ϕ = Volumetric gas flow Nm h−3 −1
η = Catalyst effectiveness factor −
µ = Dynamic viscosity kg ms−1
ρ = Density kg m−3
2 2
(N O N )
σ − = Collision diameter for pair N O N2 − 2 A°
p
τ = Particle tortuosity factor −
P
∆ = Pressure drop in the reactor Pa
2 2
( )
D N O N
Ω − = Collision integral for the pair N O N2 − 2 − Subscripts and superscripts
cat = Catalyst
ch = Monolith channel (open area) eff = Effective
g = Gas
h = Hydrodynamic (monolith channel) m = Monolithic reactor mw = Monolithic wall o = At reactor feed obs = Observed p = Particle r = Reactor
2 Process Options and Selection 2.1 Block schemes
2.1.1 Separation and Thermal Decomposition
Fig 2.1 Flow Scheme for Separation and Thermal Decomposition 2.1.2 Separation and Catalytic Decomposition
Natural Gas 2176.98 kg/day
Anesthetic Gas stream: 0.2 bara 25°C Isoflurane: 0.6 kg/day N2O : 20.67 kg/day O2 : 10.3 kg/day Isoflurane Adsorption 0.2 atm 25 °C Isoflurane Desorption 0.01 atm 130 °C Air 56196.57 kg/day Recycled Liquid Isoflurane: 0.6 kg/day N2O : 0.00 kg/day O2 : 0.00 kg/day Condensation 1 atm 4°C Outlet Gases: N2 : 43102.01 kg/day O2 : 4416.59 kg/day N2O: 1.03 kg/day CO2 : 5986.69 kg/day H2O: 4898.20 kg/day N2O Thermal Decomposition 2 atm 1000 °C
Anesthetic Gas stream: 0.2 bara 25°C Isoflurane: 0.6 kg/day N2O : 20.67 kg/day O2 : 10.3 kg/day Isoflurane Adsorption 0.2 atm 25 °C Isoflurane Desorption 0.01 atm 130 °C Recycled Liquid Isoflurane: 0.6 kg/day N2O : 0.00 kg/day O2 : 0.00 kg/day Condensation 1 atm 4°C Outlet Gases: N2 : 13.14 kg/day O2 : 17.83 kg/day N2O Catalytic Decomposition 2 atm 1000 °C
4 3 Basis of Design
3.1Input/Output Diagrams
Fig 3.1 Input Output streams for Separation and Thermal Decomposition
Fig 3.2 Input Output streams for Separation and Catalytic Decomposition Separation and Thermal Decomposition Natural Gas, Air Electricity
Isoflurane Exhaust Gas (N2O, O2, Isoflurane) N2, CO2 O2, H2O Coolant Coolant
Separation and Catalytic Decomposition Electricity Isoflurane Exhaust Gas (N2O, O2, Isoflurane) N2, O2 Coolant
4 Thermodynamic Properties and Reaction Kinetics 4.1 Vapor pressures of each component
Antoine Equation: log ( /10 P mmHg)= −A B t c C/( /° + ) The Antoine constants for each component are listed:
Compound A B C
Isoflurane 8.056 1664.58 273.15
Nitrous Oxide 7.003 654.26 247.16
Oxygen 6.691 319.013 266.697
Calculated by Antoine Equation, we get the vapor pressures (P) at different temperatures for these three compounds.
Table 4.1 Vapor pressures at different temperatures
K P isoflu P N2O P O2 K P isoflu P N2O P O2 303.15 0.483327 57.75645 543.2148 283.15 0.197877 37.84527 454.207 302.15 0.46352 56.63078 538.6857 282.15 0.188607 36.9896 449.8543 301.15 0.444401 55.51912 534.1639 281.15 0.179711 36.14679 445.5122 300.15 0.425951 54.4214 529.6497 280.15 0.171175 35.31679 441.1806 299.15 0.408151 53.33759 525.1432 279.15 0.162988 34.49951 436.8599 298.15 0.390983 52.26764 520.6445 278.15 0.155138 33.69488 432.5502 297.15 0.374429 51.21149 516.1538 277.15 0.147613 32.90284 428.2516 296.15 0.358471 50.1691 511.6713 276.15 0.140403 32.12331 423.9644 295.15 0.343092 49.14042 507.197 275.15 0.133496 31.35621 419.6886 294.15 0.328274 48.1254 502.7311 274.15 0.126882 30.60146 415.4245 293.15 0.314002 47.12398 498.2739 273.15 0.120551 29.859 411.1722 292.15 0.30026 46.13611 493.8253 272.15 0.114493 29.12873 406.9319 291.15 0.28703 45.16172 489.3857 271.15 0.108698 28.41059 402.7038 290.15 0.274298 44.20078 484.955 270.15 0.103157 27.70449 398.4881 289.15 0.262049 43.25321 480.5336 269.15 0.09786 27.01034 394.2849 288.15 0.250267 42.31895 476.1215 268.15 0.092799 26.32807 390.0944 287.15 0.238938 41.39795 471.719 267.15 0.087964 25.65758 385.9169 286.15 0.228049 40.49015 467.3261 266.15 0.083348 24.99881 381.7524 285.15 0.217584 39.59548 462.943 265.15 0.078942 24.35164 377.6012 284.15 0.207531 38.71387 458.5699 264.15 0.074738 23.71601 373.4634
The vapor pressures of nitrous oxide and oxygen are most like two linear lines. Vapor pressure of oxygen is very high. But we can find the vapor pressure of isoflurane increase faster at higher temperature than at lower temperature. Draw two rough slopes to indicate this change and find the turning point is around 10 °C. Above the “turning point”, the vapor pressure has sensitive response to the temperature, while below the point less
6
sensitive response. That means more isoflurane liquid can be got by cooling the gas mixture, but after the “turning point” cooling became less effective and much more costly.
4.2 Condensation Temperature Determination
The main reason of finding the proper condensation temperature is to save energy consumed by condensation and get more isoflurane liquid.
1. The gas remaining in the condenser is recycled and mixed with the process inlet gas flow, so the more remaining gas, the more energy we need to supply to them in the next cycles. We want to increase the isoflurane liquid
2. Low condensation temperature asks for expensive coolants and more critical operation conditions than higher temperatures.
To make a clear view of the vapor pressure curve, we increase isoflurane vapor pressures 2000 times and nitrous oxide 10 times. Actually, oxygen is out of question, for the temperatures are higher than its critical temperature.
Fig.4.1.1.
Simulate the condensation process by “superpro designer”, and find the amount of isoflurane liquid we may get above 5°C is less than 70% of the isoflurane existing in the inlet gas flow to our process.
4.3 Joback group contribution method: isoflurane heat capacity coefficients calculations
Joback group contribution offers a simple method to learn the thermodynamic properties of a compound.
A molecule is considered to be composed of several groups and each group has contribution on the thermodynamic properties of the molecule. The thermodynamic properties of the molecule may be written as the sum of each group’s contribution multiplied by the number of the group.
Vapor pressures at different temperatures
50 150 250 350 450 550 650 260 265 270 275 280 285 290 295 300 305 T(k)
Vapor Pressure (atm)
Iso * 2000 N2O* 10 O2
Take isoflurane as an example. The molecule is CF3-CHCl-O-CHF2, consequently, 5 –F
groups, 2 -CH- groups, 1 -O- group, 1 -C≡ group, and 1 –Cl group. 4.4 Equilibrium constants K of reaction 1 at 1273K, 9 bara ln * G K R T = − +
K = EXP (4.49 / -8.314E-3/1273) = 1.244 SEC-1.(GMOL/LIT)
4.5 Method of Half life – Thermodynamics for Thermal Decomposition Reaction The half-life of a reaction, t1/2 is defined as the time it takes for the concentration of the
reactant to fall to half of its initial value. By determining the half-life of a reaction as a function of the initial concentration, the reaction order and specific reaction rate can be determined. [30]
For a constant volume reaction system
2 2 2 2N O→2N +O ---(4.5.1) 2 A N O A dC r kC dt α − = − = ---(4.5.2)
Integrating with the limits; Initial condition CA = CA0, when t =0, we can find; 1 0 1 0 1 1 ( 1) A A A C t C kC α α α − − = − − ---(4.5.3) 1 0 1 0 1 1 ( 1) A A A C t C kC α α α − − = − − ---(4.5.4) The half-life is defined as the time required for the concentration to drop to half of its initial value; i.e.
1/ 2 t t= when 1 0 2 A A C = C substitute CA in equation (4.5.4) 1 1 1 2 0 2 1 1 ( 1) A t k C α α α − − − = − ---(4.5.5) Taking natural log of both the sides of equation (4.5.5) we get,
1 1 0 2 2 1 ln ln (1 ) ln ( 1) A t C k α α α − − = + − − --- (4.5.6) From the plot of ln t1/2 as a function of ln CA0 is equal to 1 minus the reaction order
i.e.
1 Slope
8 The corresponding rate law is;
2 2 N O A r kC − = ---(4.5.8) For the thermal decomposition of N2O at approximately 800°C i.e. 1073K, the data for
partial pressure of N2O is;
p0 (mm Hg) 52.5 139 290 360 t1/2 (sec) 860 470 255 212 0 0 0 A A P P C RT RT
= = For an ideal gas
Method of Half-life y = -0.7288x + 9.678 R2 = 0.994 0 1 2 3 4 5 6 7 8 3 3.5 4 4.5 5 5.5 6 6.5 ln p0 ln t1/2
Fig 4.2 Method of half-life
Order of the reaction: From equation (4.5.7) 1 0.728 1.728
α = + =
For the value of 52.5 mm Hg p0 we can calculate CA0
For 6840 mm Hg p0 from the extrapolation of the graph Fig 4.2 we can get the value of
t1/2 = 25.60 sec 0 6840*1.013 5 1.173 5 / 760*8314*1073 A E C = − = E− gmol lit
Rate constant of the reaction at 9 bara and 1073 K
1.728 1 (1 1.728) 1 0.728 (2 1)*(1.173 5) 0.780sec *( / ) 6840*(1.728 1) E k gmol lit − − − − − − = = − Rate of reaction: 2 1.728 0.708 N O A r C − =
4.6 Properties of compounds
Component name Technological data Healthy & Safety data
Design Syste-matic Fomula Mole Weigh t Phase Boil- ing point [1] Melt-ing point [1] Flash Point Liq- uid Den-sity[2] Va- por Den-sity[2] Auto- Ignatio n Temp [1] Flam-mable Limits LEL [3] UE L [3] LC50 In air /water MAC Value (risk) LD50 Oral [3] Chemical Reactivity Notes g/mol V/L oC oC oC kg/m3 kg/m3 oC % vol In air % % mg/m3 mg/m3 mg/kg nitrous
oxide Oxidizer N2O 44.01 V -88.5 -90.9 n,a 743 1.799 n,a n,a n,a n,a 160 46 n,a Not Flammable [4]
Iso- flurane Ether aliphatic Halogen s C3H2Cl
F5O 184.5 V 49.00 n,a n,a 1500 1.5 n,a n,a n,a n,a n,a 15.1 4771
Stable at
normal [5] [6]
oxygen Oxygen O2 32 V -183 -219 n,a 435.33 1.308 n,a n,a n,a n,a n,a n,a n,a No
methane Hydro- Carbon CH4 16 V -162 -182.5 -221 161.74 1.819 537 5-15 5 15 10.8 9.0- n,a n,a
Stable. Extremely flammable
nitrogen Nitrogen N2 28 V -196 -210 n,a 314.22 1.145 n,a n,a n,a n,a n,a n,a n,a No
water water H2O 18 L 100 0 - 1000 0.7363 - - - - - - - No carbon dioxide Acid Anhydri de
CO2 44 V -78.5 n,a n,a 713.54 1.950 n,a n,a n,a n,a n,a n,a n,a Unstable
Notes: [1] At 101.325kpa
[2] Density at 25oC, unless other specified.
[3] Oral in mg/kg of rabbit [4] Exposure for 6 hours for rat
o
Project ID Number: CPD3301 Completion Date: Octo 29th 2003
Project ID Number: CPD3301 Completion Date: Jan 9th 2004
10 4.7 Sensitivity analysis
Sensitivity analysis is carried out to observe the effect of moles of N2O and O2 in the
heating system. Higher amount of exhaust flow rate i.e. 8 lit/min is a worst-case scenario for this process, as this could be the maximum flow rate the exhaust gas can have. The amount of natural gas can be controlled and saved, when we pass N2O in the heating
system. The analysis has been done to calculate the amount of natural gas, which can be saved during the whole operation. Minimum and maximum flow rates of exhaust gas have been considered i.e. 3 lit/min to 8 lit/min, which is a data for this assignment.
The values for the N2O and O2 will vary, as there will be a change in inlet flow rate
conditions. This change can be seen in table 4.2, calculated from Aspen.
Table 4.2 Mass flow rate of N2O and O2 when exhaust flow rate changes from 30lit/min to 80
lit/min
N2O
gm/min 31.32 36.54 41.76 46.98 52.2 57.42 62.64 67.86 73.08 78.3 83.52
O2
gm/min 15.6 18.2 20.8 23.4 26.00 28.6 31.2 33.8 36.4 39 41.63
The molar flow rate will also change. From the calculation of the heat balance of heating system, we know that how much amount of natural gas can be saved. e.g. for 1 mole of N2O, 0.089 moles of natural gas can be saved.
Table 4.3 Moles of N2O and O2 when exhaust flow rate changes from 30lit/min to 80 lit/min
N2O mol/min 0.69 0.81 0.93 1.05 1.17 1.28 1.40 1.52 1.64 1.76 1.88 O2 mol/min 0.53 0.58 0.66 0.74 0.81 0.89 0.97 1.05 1.11 1.20 1.28 CH4 mol/min 0.06 0.07 0.08 0.09 0.10 0.11 0.13 0.14 0.15 0.16 0.17
N2O is not a flammable gas and it doesn’t have any flammability limit, but it can form
explosive limits. The LEL and UEL percentage range is not mentioned. [37] The use of N2O as a fuel is not recommended, as this is only a fuel enhancer and will provide
oxygen for burning. For instance, the amount of N2O is increased heavily in the future,
then according to our concept the impression could be we don’t have to use natural gas anymore. This is not true though, as we are assuming the worst case and it is possible to use this concept for that case. If in future the use of N2O increases a lot, which is not true,
then we have to reconsider the concept again. If the amount of N2O increases we can use
a storage tank and regulate the flow rate, but the explosion limits of natural gas should be kept in mind.
The analysis of molar change in the system can be shown on the graph and the linear behavior can be observed.
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 0.06 0.07 0.08 0.09 0.10 0.11 0.13 0.14 0.15 0.16 0.17 Moles CH4 saved
Moles N2O and O2
in exhaust
O2 moles N2O moles
Fig. 4.3 Sensitivity analysis of moles N2O and O2 with respect to moles of CH4 in feed to
heating system
The change in molar concentration of N2O will also change the heat exerted by the
decomposition reaction. There are two ways by which the heat can be controlled in the heating system;
o Closing of fuel inlet valve o Closing of air supply
There are separate pipes for the fuel and air inlets and the mixing takes place at the combustion zone. According to the process scheme it is assumed that the mixture of natural gas and air is mixed with N2O and compressed. The mixture of natural gas and air
will be transported to the combustion chamber so does the N2O gas. The controls of these
gases flow rates and operation of system at safe conditions is discussed in the process control section [ref chapter 6]
Additions of N2O in the heating system will not only support the burning by providing
oxygen, but also the decomposition of N2O is an exothermic reaction and this will
produce some amount of heat energy in the heating system. The heat energy produced by the N2O moles is studied and an estimation of heat that is produced by the moles of
12
Heat senitivity of N2O decomposd
0.00 20.00 40.00 60.00 80.00 100.00 120.00 140.00 160.00 0.69 0.81 0.93 1.05 1.17 1.28 1.40 1.52 1.64 1.76 1.88 Mole of N2O Heat en erg y p ro d u ced kJ/mo le
Fig 4.4 Heat produced by the moles of N2O in the heating system (Results from Excel) Energy that natural gas produced after N2O decomposed
337680 337700 337720 337740 337760 337780 337800 337820 337840 378 .93 378. 92 378. 91 378. 90 378. 89 378. 88 378. 87 378. 86 378 .85 378. 84 378. 83
Moles Natural Gas input
Enegry produced by methane(kJ/mole)
Fig 4.5 Heat produced by natural gas after moles N2O decomposed in the heating system
(Results from Excel)
From Aspen simulation also some results are obtained. The mass flow rate and the heat energy content of vent gas is studied from Aspen and the results can be seen in fig 4.6
Sensitivity for mass flow 160000 170000 180000 190000 200000 210000 220000 230000 30 35 40 45 50 55 60 65 70 75 80
Exhaust gas (lit/min)
V
ent gas flow (gm/min)
Fig4.6 Mass flow discharge change with respect to change in exhaust gas flow rate
Heat sensitivity analysis
0 200 400 600 800 1000 1200 1400 1600 1800 30 35 40 45 50 55 60 65 70 75 80
Exhaust flow rate (lit/min)
Vent gas heat (kj/kg)
Fig4.7 Heat of vent gases with respect to change in exhaust gas flow rate
It is seen that the amount of heat produced by the N2O decomposition is compensated by
the amount of heat generated by the natural gas burning. This shows that the control of natural gas inlet with respect to the N2O moles inlet tries to maintain the heat energy of
the system at constant level. Catalytic Decomposition
Thermodynamic properties and reaction kinetics are tested by Aspen for the temperature range from 500K to 1500K.
Table 4.4 Sensitivity of monolithic variables for Temperatures difference
Variable Tmin (500K) Tmax(1500K)
D_eff (m2/s) -1.62E-08 -3.01E-08
K_obs (1/s) 1.69E+03 5.27E+03
R_obs (mol/m3/s) 3.10E+00 9.68E+01
Residence time (s) 2.73E-03 8.74E-04
14
Function of Pressure Drop with Velocity
0.00E+00 2.00E-07 4.00E-07 6.00E-07 8.00E-07 1.00E-06 1.20E-06 1.40E-06 1.60E-06 25 30 35 40 45 50 55 60 65 70 75 80 85 90
Mass Flow of N2O (g/min)
Pressure drop (Pa)
Figure 4.8. Pressure drop in the monolith reactor as a function of mass flow
Function of Pressure Drop with Average Superficial Velocity
0.00E+00 2.00E-07 4.00E-07 6.00E-07 8.00E-07 1.00E-06 1.20E-06 1.40E-06 1.60E-06 0.01 0.011 0.012 0.013 0.014 0.015 0.016 0.017 0.018 0.019 0.02 Average Superficial Velocity (m/s)
Pressu
re Dro
p
(Pa)
Figure 4.9 Pressure drop as a function of the flow velocity
As can be seen here is that the reaction happens very fast. In almost 0.5 seconds the reaction is done.
Conversion of N2O versus temperature 0 20 40 60 80 100 120 500 502 504 506 508 510 512 514 516 518 520 522 Temperature (K) Conv er si on of N2 O [-]
Figure 4.10 Conversion as a function of the feed temperature at 0.2 bar and the average flow rate of N2O 14.35 g/min.
The temperature of the gas mixture influences the conversion of decomposition. There is a sharply increase around 503K.
Table 4.5, 4.6 and 4.7 gives the relation of conversion of N2O with feed temperature for
different conditions.
Table 4.5: The feed temperature of the reactor that gives significant conversion at the maximum flowrate: N2O 82.6g/min And pressure of 1 bar
K C Conversion 500 226.85 3.14 501 227.85 4.14 502 228.85 7.52 503 229.85 100 504 230.85 99.99 505 231.85 100 506 232.85 100
The conversion is poorly reached when the feed temperature is 502K but one Kelvin higher in the temperature makes the conversion complete to 100%. The activation energy is set to 170 J/Kmol
Table 4.6: The feed temperature of the reactor that gives significant conversion at the maximum flowrate possible; N2O:82.6g/min and pressure of 0.2 bar
K C Conversion 554.85 281.85 3.57 555.85 282.85 4.26 556.85 283.85 5.35 557.85 284.85 7.84 558.85 285.85 99.99 559.85 286.85 100
16
Table 4.7: The feed temperature of the reactor that gives significant conversion at the minimum flow rate of N2O:31.31g/min and pressure of 0.2 bar
K C Conversion 522.85 249.85 3.48 523.85 250.85 4.40 524.85 251.85 6.55 525.85 252.82 99.99 526.85 253.85 100
5 Process Structure and Process Description 5.1 Price for catalysts
Table 5.1 Price of catalyst
Catalysts Configuration Mass (g) Price (Euro)
Rhodium, 1% on alumina pellet Pellet 10 207.60 Rhodium, 1% on alumina pellet Pellet 50 1000.00 Rhodium, 5% on alumina pellet Pellet 2 99.50 Rhodium, 5% on alumina pellet Pellet 10 399.00 Rhodium, 5% on alumina powder Powder 25 735.80 Rhodium, 5% on alumina pellet Pellet 50 1709.00 Rhodium, 0.5% on alumina pellet Pellet 25 96.50 All prices is checked in Alfa Aesar
The price of the coating layer on the wall is not known.
The design if the reactor is based on a monolith reactor with the very thin layer of
catalyst coated on the wall. The prices of the wash-coated catalyst, rhodium on alumina is not known. The prices of the monolithic wash coated catalyst and or extruded monolith catalyst depend on the price of the metal catalyst used. The price of Cu/Fe/Pb is almost negligible while Rh is expensive. The price of the ceramic monolith material is about 3 Euro per liter and can be assumed negligible.
5.2 The strategies level of Krihna and Sie
First we have to consider the process benefits, which are mentioned in table 2, the well-known ‘musts’ and ‘wants’.
Table 5.2: The considerations to select the proper reactor configuration.
‘ Musts’ ‘Wants’
• Operability within
temperature, pressure and residence time range • Safe operation and without
runaway
• Easy operability
• Maximum possible conversion of the feedstock (N2O)
• Maximum possible selectivity of reactants.
• Easy operability
• As low as possible capital and operating costs.
At level 1 of the strategies level of Krishna and Sie, the catalyst size and shape is not known yet because that depends on the reactor volume, the conversion, and the adiabatic temperature rise. But from literature (1) we know that some catalyst gives high conversion, and high stability (500 – 900K).
18
Level 2: injection and dispersion strategies. As in this work we have an exhaust gas stream from the operation room of a hospital, and after separation we know the gas flow rate range, there are some reactor type that can be used for this stream. There is thus one incoming stream. We have decided not to mix with another stream, e.g. air because the cost will increase due to the increase of reactor volume and the addition of extra stream. As we do not earn money from this catalytically treatment we can directly treat this gas stream that only contains nitrous oxygen (52% N2O) and oxygen (38% O2).
Level 3: If we use a fixed bed or a monolithic reactor we can assume in both cases a plug flow model. Because the flow rate is not high, even in a fixed bed we can assume a plug flow.
For gas phase applications, typically using high space velocities, structured catalysts are very convenient. Structured catalysts can achieve decoupling of the hydrodynamics, kinetics, and transport phenomena.
Structured and fixed-bed reactors are the best reactor types for gas phase reactions. Table 5.3: Comparison between Monolithic, Slurry and Packed-Bed Reactors
Reactor type Characteristic
Monolith Reactor Slurry Reactor Trickle-Bed Reactor Energy Input Low Medium (stirring) High (pressure drop) Catalyst efficiency High, thin active layer High, small particles Low, large particles
required for pressure drop
Safety High, self-draining
reactor
Medium, easy cooling, difficult to separate catalyst from liquid
Low, difficult cooling, catalyst bed retains liquid Catalyst separation Easy Costly filtering
necessary Easy
Preparation Medium, new technology, methods have now been developed.
Easy Easy
Catalyst loading Medium, very open structure, for wash coated systems: low
Low–medium High, dense bed
Catalyst
replacement Difficult, shut down required, monoliths have to be carefully Stacked
Easy, continuous
during operation Medium–difficult, Shut down required Experience Gas-phase: extensive;
liquid and multiphase: very limited
Separation and Thermal Decomposition Recycled Isoflurane Exhaust Gas from operation theatre A01 A02 P01 TC1 F02 C01 V01 M01 K01 F01 TC 3 PC 25 0.2 2 25 0.2 1 25 1 8 130 1 4 Natural Gas N2 and O2 to atmosphere 25 1 9 TC wc TC 2 PC 4 1 6 4 5 1 2 10 1 25 7 3 25 0.2 113 V1 V2 V3 V5 V7 V6 1 11 110 Boiling water V4
Process Equipment Summary Designers Process Flow Schem- Improved Solution
A 01 : Adsorber A02 : Adsrober C01 : Condenser F01 : Heating System F02 : Microwave Oven K01: Compressor M01: Mixer P01 : Blower Chilukuri Vanisudha Shindgikar Nikhil Kisoen Dinesh Zhao Ying Liu Yiling Ma Chunyu
Project : Removal of N2O and iso-flurane from the exhaust
stream of the operation rooms in hospitals Project ID number: CPD3301
Completion Date : January 14, 2004
22 Separation and Catalytic Decomposition
Recycled Isoflurane Exhaust Gas from operation theatre A01 A02 TC 1 F02 C01 V01 25 0.2 2 25 0.2 1 130 1 4 N2 and O2 to atmosphere wc TC 2 PC 4 1 6 4 5 1 3 25 0.2 V1 V2 V3 R01 TC 3 F03 TC 3 V5 E01 1129 9 0.2 1130 10 0.2 250 7 0. 2 135 4 8 0.2 V4
Process Equipment Summary Designers Process Flow Schem- Improved Solution
A 01 : Adsorber A02 : Adsrober C01 : Condenser E01 : Heat exchanger
F02 : Microwave Oven F03 : Electric Heater R01 : Monolith reactor Chilukuri Vanisudha Shindgikar Nikhil Kisoen Dinesh Zhao Ying Liu Yiling Ma Chunyu
Project : Removal of N2O and iso-flurane from the exhaust
stream of the operation rooms in hospitals Project ID number: CPD3301
Completion Date : January 14, 2004
5.4 Results of modeling
5.4.1 Thermal Decomposition Simulation Table 5.4.1 Results from Aspen
10EXST HS-FEED MIXER NG-IN VENT
Temperature C 1000 25 116.7955 25 25 110
Pressure bar 9 0.2 2 1 1 1.03
Vapor Frac 1 1 1 1 1 1
Mole Flow mol/min 5793.203 2.128126 5792.595 5792.5948 5790.467 5793.203 Mass Flow gm/min 162304.7 83.488 162304.7 162304.66 162221.2 162304.7 Volume Flow l/min 68284.51 263.6197 169368.7 717500.53 143530 183834.4 Enthalpy MMkcal/hr -2.02497 0.001506 0.642515 -0.4027049 -0.40421 -4.51104
Mass Flow gm/min
N2O 2.819117 56.38234 56.38234 56.382336 0 2.819117 O2 12265.95 27.10566 36434.24 36434.24 36407.13 12265.95 CH4 0 0 6063.327 6063.327 6063.327 0 CO2 16633.41 0 0 0 0 16633.41 H2O 13617.67 0 0 0 0 13617.67 N2 119784.8 0 119750.7 119750.71 119750.7 119784.8 Mass Frac N2O 1.74E-05 0.675335 0.000347 0.0003474 0 1.74E-05 O2 0.075574 0.324665 0.224481 0.2244806 0.224429 0.075574 CH4 0 0 0.037358 0.0373577 0.037377 0 CO2 0.102483 0 0 0 0 0.102483 H2O 0.083902 0 0 0 0 0.083902 N2 0.738024 0 0.737814 0.7378144 0.738194 0.738024
Mole Flow mol/min
N2O 0.064052 1.281042 1.281042 1.2810417 0 0.064052 O2 383.3254 0.847084 1138.613 1138.6127 1137.766 383.3254 CH4 0 0 377.9479 377.94787 377.9479 0 CO2 377.9479 0 0 0 0 377.9479 H2O 755.8957 0 0 0 0 755.8957 N2 4275.97 0 4274.753 4274.7532 4274.753 4275.97 Mole Frac N2O 1.11E-05 0.601958 0.000221 0.0002212 0 1.11E-05 O2 0.066168 0.398042 0.196564 0.1965635 0.196489 0.066168 CH4 0 0 0.065247 0.0652467 0.065271 0 CO2 0.06524 0 0 0 0 0.06524 H2O 0.13048 0 0 0 0 0.13048 N2 0.738101 0 0.737969 0.7379686 0.73824 0.738101
Table 5.4.2Results from Excel
Component mol/min mol/min gm/min gm/min
IN OUT IN OUT N2O 378.958 0.000 6063.327 0.000 O2 1137.723 380.439 36407.134 12174.046 CH4 4276.811 4278.075 119750.714 119786.102 CO2 0.000 378.958 0.000 16674.150 H2O 1.284 0.021 56.512 0.903 N2 0.000 757.916 0.000 13642.486 Total 5794.776 5795.408 162277.686 162277.687
24
5.4.2 Catalytic Decomposition Simulation in Aspen ‘Plug flow model reactor with very high a/v ratio’.
Temperature in the monolith reactor versus length
Length (m) Temperature oC 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 500 750 1000 1250
1500 N2O 14.35 g/min, O2 7.16 g/min at 0.2 bar
Figure 5.1: Temperature inside the monolith reactor by length.
Block B2: Temperature
Residence time sec
Temperature C
0 0.5 1 1.5 2 2.5 3 3.5
500
1000
1500
N2O 14.35 g/min, at 0.2 bar
5.5 Process Stream Summery
5.5.1 Separation of isoflurane and thermal decomposition
STREAM Nr: 1 IN 2 3 4 5
Name: Process-original-feed Str.1 mixed with Str.5 After A01(A02) Outlet of F02 Recycle from V01
COMP MW kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day
Isoflurane 184.50 0.60 0.0033 0.75 0.0041 0.00 0.0000 0.75 0.0041 0.15 0.0008 N2O 44.00 20.67 0.4698 20.72 0.4710 20.67 0.4698 0.05 0.0012 0.05 0.0012 O2 32.00 10.30 0.3218 10.41 0.3253 10.31 0.3221 0.11 0.0035 0.11 0.0035 CO2 44.00 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 N2 28.00 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 H2O 18.00 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 CH4 16.00 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 Total 31.57 0.7949 31.88 0.8003 30.98 0.7919 0.92 0.0088 0.31 0.0054 Enthalpy kJ/day 34487 33583 38545 -4982 -905 Phase V V V V V Press. Bara 0.2 0.2 0.2 1.0 1.0 Temp °C 25.0 24.8 24.8 130.0 4.0 STREAM Nr: 6 OUT 7 8 IN 9 7+8 10
Name: Recovered Isoflurane After P01 Nature Gas Mixture of Str.7 & 8 High pressure gas
COMP MW kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day
Isoflurane 184.50 0.60 0.0033 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 N2O 44.00 0.00 0.0000 20.67 0.4698 0.00 0.0000 20.67 0.4698 20.67 0.4698 O2 32.00 0.00 0.0000 10.31 0.3221 13107.06 409.5956 13117.37 409.9177 13117.37 409.9177 CO2 44.00 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 N2 28.00 0.00 0.0000 0.00 0.0000 43089.51 1538.9111 43089.51 1538.9111 43089.51 1538.9111 H2O 18.00 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 CH4 16.00 0.00 0.0000 0.00 0.0000 2176.98 136.0613 2176.98 136.0613 2176.98 136.0613 Total 0.60 0.0033 30.98 0.7919 58373.55 2084.5679 58404.53 2085.3599 58404.53 2085.3599 Enthalpy kJ/day -4058.44 38545.50 -10139284.35 -10100738.85 -10100737.63 Phase L V V V V Press. Bara 1.0 1.0 1.0 1.0 2.0 Temp °C 4.0 25.0 25.0 25.0 113..34
26
STREAM Nr: 11 OUT
Name: Discharge F01
COMP MW kg/day kmol/day
Isoflurane 184.50 0.00 0.0000 N2O 44.00 1.03 0.0235 O2 32.00 4416.59 138.0183 CO2 44.00 5986.69 136.0612 N2 28.00 43102.01 1539.3575 H2O 18.00 4898.20 272.1224 CH4 16.00 0.00 0.0000 Total 58404.53 2085.5830 Enthalpy kJ/day -119341455 Phase V Press. Bara 1.0 Temp °C 110.0
Overall Component Mass Balance & Stream Heat balance
STREAM Nr. : 1+8 IN 6+11 OUT OUT-IN
Name : Total Plant Total Plant Total Plant COMP MW kg/day kmol/day kg/day kmol/day kg/day kmol/day isoflurane 184.50 0.60 0.0033 0.60 0.0033 0.00 0.0000 N2O 44.00 20.67 0.4698 1.03 0.0235 -19.64 -0.4463 O2 32.00 13117.36 409.9175 4416.59 138.0183 -8700.77 -271.8991 CO2 44.00 0.00 0.0000 5986.69 136.0612 5986.69 136.0612 N2 28.00 43089.51 1538.9111 43102.01 1539.3575 12.50 0.4464 H2O 18.00 0.00 0.0000 4898.20 272.1224 4898.20 272.1224 CH4 16.00 2176.98 136.0613 0.00 0.0000 -2176.98 -136.0613 Total 58405.12 2085.3628 58405.13 2085.5863 0.01 0.2234 Press. Bara Temp oC Enthalpy KJ/day -10104797 -119345514 109240716
5.5.2 separation of isoflurane and catalytic decomposition
STREAM Nr. : 1 IN 2 3 4 5
Name :
COMP MW kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day isoflurane 184.50 0.60 0.0033 0.75 0.0041 0.00 0.0000 0.75 0.0041 0.15 0.0008 N2O 44.00 20.67 0.4698 20.72 0.4710 20.67 0.4698 0.05 0.0012 0.05 0.0012 O2 32.00 10.30 0.3218 10.41 0.3253 10.31 0.3221 0.11 0.0035 0.11 0.0035 N2 28.00 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 Total 31.57 0.7949 31.88 0.8003 30.98 0.7919 0.92 0.0088 0.31 0.0054 Enthalpy KJ/day 34487 33583 38545 -4982 -905 Phase V V V V V Press. Bara 0.2 0.2 0.2 1.0 1.0 Temp oC 25.0 24.8 24.8 130.0 4.0 STREAM Nr. : 6 OUT 7 8 9 10 Name :
COMP MW kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day kg/day kmol/day isoflurane 184.50 0.60 0.0033 0.00 0.0000 0.00 0.0000 0.00 0.0000 0.00 0.0000 N2O 44.00 0.00 0.0000 20.65 0.4694 0.00 0.0000 0.00 0.0000 0.00 0.0000 O2 32.00 0.00 0.0000 10.32 0.3226 17.83 0.5573 17.83 0.5573 17.83 0.5573 N2 28.00 0.00 0.0000 0.00 0.0000 13.14 0.4694 13.14 0.4694 13.14 0.4694 Total 0.60 0.0033 30.97 0.7920 30.97 1.0267 30.97 1.0267 30.97 1.0267 Enthalpy KJ/day -4058.44 38514.27 38514.27 0.00 0.00 Phase L V V V V Press. Bara 1.0 0.2 0.2 0.2 0.2 Temp oC 4.0 253.0 1354.3 1128.0 1129.0
Overall Component Mass Balance & Stream Heat balance
STREAM Nr. : 1 IN 6+11 OUT OUT-IN Name : Total Plant Total Plant Total Plant COMP MW kg/day kmol/day kg/day kmol/day kg/day kmol/day isoflurane 184.50 0.60 0.0033 0.60 0.0033 0.00 0.0000 N2O 44.00 20.67 0.4698 0.00 0.0000 -20.67 -0.4698 O2 32.00 10.30 0.3218 17.83 0.5572 7.53 0.2354 N2 28.00 0.00 0.0000 13.14 0.4693 13.14 0.4693 Total 31.57 0.7949 31.57 1.0297 0.00 0.2348 Press. Bara Temp oC Enthalpy KJ/day 34487 -4058 38546
28 5.6 Utility summary
5.6.1 separation of isoflurane and thermal decomposition
SUMMARY OF UTILITIES
EQUIPMENT UTILITIES
Heating Cooling Power REMARKS
Nr. Name Load Consumption (kg/day) Load Consumption (kg/day) Actual Consumption (kg/h, kWh/day)
Steam Hot
cooling water Air
Nacl
brine Load Steam (kg/day) Electr. kJ/day LP MP HP Oil kJ/day kJ/day HP MP kWh/day
C01 Condenser 250 8.425 A01 Adsorber. F01 Heating system F02 Micro wave 474 0.1316 K01 Compressor 4146768 1144.50 M01 Mixer P01 Blower 3283 0.9062 TOTAL 250 8.425 4150525 1145.54 Project ID Number : CPD3301 Completion Date: 9th January 2004
5.6.2 separation of isoflurane and catalytic decomposition
SUMMARY OF UTILITIES
EQUIPMENT UTILITIES
Heating Cooling Power REMARKS
Nr. Name Load
Consumption
(kg/day) Load Consumption (kg/day) Actual Consumption (kg/h, kWh/day)
Steam Hot
Coolin g water Air
Nacl
brine Load Steam (kg/day) Electr. kJ/day LP MP HP Oil kJ/day kJ/day HP MP kWh/day
C01 Condenser 250 8.425
A01 Adsorber.
E01 Heat Exchanger
R01 Monolith Reactor F02 Micro wave 474 0.1316 TOTAL 250 8.425 474 0.1316 Project ID Number : CPD3301
30 6 Process Control
7 Mass and Heat Balance
7.1 Heat load calculations (heat capacity of isoflurane are got from app 7.2): Table 7.1 Temperature and pressure changes in each step
Steps inlet conditions final conditions ∆T ∆ P Adsorption 25 C 0.2bar 25 C 0.2bar 0 0 Desorption 25 C 0.2bar 130 C 0.01bar 105 0.2 Condensation 130 C 0.01bar 8 C 1 bar 122 1 And the heat capacity coefficients of each component are given below in a table. The temperature range is from 100K to 1500K.
Table 7.2 heat capacity coefficients
Compound A B C D E Vaporization Heat (J/mol)
Isoflurane 41.17 4.29E-1 -3.09E-4 6.96E-7 - 31.77e03 N2O 23.219 6.1984E-2 -3.7989E-5 6.9671E-9 8.1420E-13 11.07e03
O2 29.596 -8.8999E-3 3.8083E-5 -3.2629E-8 8.8607E-12 0
Zeolites also have contribution on heat load, and because of the lack of published Cp information of zeolites, we assume the heat capacity of zeolites does not change by temperature and pressure and we use the heat capacity at 400 K.
Step 1 Adsorption:
As no temperature and pressure change during this step and no reaction take place, the heat load is 0.
Step 2 Desorption:
Heat load for gas are calculated by:
1 0 , T j p T Q m C dT Where = ⋅
∫
2 3 4 pC = + ⋅ + ⋅A B T C T + ⋅D T + ⋅ and mE T j is the mole of each component.
The heat capacity of zeolite is 0.7 J/g /k [31]
The mass of each component is got from “SuperPro Design” stimulation. And the heat distributed to each compound is listed:
Table 7.3 Heat required by each compounds
Mass (mol/day) Desorption (kJ/day)
Isoflurane 4.08 74.50 N2O 1.20 5.11 O2 3.54 11.11 Zeolite 5200 (g/day) 382.2 Total (kJ/day) 474.019 Step 3 Condensation:
The initial temperature of the gas stream is 130°C and the desired temperature is 8°C. The temperature change of zeolite is from 130°C to 25°C.
32
Table 7.4 Heat required to be removed in condenser and released to air by zeolite
Mass (mol/day) Condensation (kJ/day)
Isoflurane 4.08 -216.87 (-129.62)
N2O 1.20 -19.35 (-13.284)
O2 3.54 -13.30 (0)
Total ( kJ/day) -249.52
Zeolite 5200 (g/day) -382.2
Values in brackets are phase change heats.
When design a condenser, only consider the heat needs to be removed from gases.
7.2 Joback group contribution method: isoflurane heat capacity coefficients
Joback group contribution offers a simple method to learn the thermodynamic properties of a compound.
A molecule is considered to be composed of several groups and each group has contribution on the thermodynamic properties of the molecule. The thermodynamic properties of the molecule may be written as the sum of each group’s contribution multiplied by the number of the group.
Take isoflurane as an example. The molecule is CF3-CHCl-O-CHF2, consequently, 5 –F
groups, 2 -CH- groups, 1 -O- group, 1 -C≡ group, and 1 –Cl group. Special heat capacity of isoflurane is estimated:
Normally, we use 2 3
p
C = + ⋅ + ⋅A B T C T + ⋅ to calculate the heat capacity. By this D T method, the equation is rewrite into:
2 3 0 ( ) 37.93 ( ) 0.21 ( ) 3.91 4 ( ) 2.06 7 k k k k k k k k p N CpAk N CpBk T N CpCk e N CpDk e T T C = + ⋅ + ⋅ + ⋅ − + − − − −
∑
∑
∑
∑
The A, B, C, D values are calculated by comparing these two equations:
37.93 0.21 3.91 4 2.06 7
(
)
(
)
(
)
(
)
k k k k k k k k A B C e D eN CpAk
N CpBk
N CpCk
N CpDk
= = + = − = − − − −∑
∑
∑
∑
Table 7.6 Parameters of Isoflurane
Table 7.5 Heat Capacities by groups J/mol/k
Group Number CpAk CpBk CpCk CpDk
CH 2 -23 0.204 -2.65E-04 1.20E-07
C 1 -66.2 4.27E-01 -6.41E-04 3.01E-07
O 1 25.5 -6.32E-02 1.11E-04 -5.48E-08
F 5 26.5 -9.13E-02 1.91E-04 1.03E-07
Cl 1 33.3 -9.63E-02 1.87E-04 -9.96E-08
( # )
k k
N Cp k
∑
79.1 2.19E-01 8.20E-05 9.02E-07Compound A B C D
7.3 Heat balance for the Heating system
Normally natural gas is used for burning, which produces energy to heat the water in the heating system. Natural gas is at room temperature (25°C, 2 bar). The compressor is used to put the natural gas in the burning section. Air is mixed with the natural gas in the burning section.
Assumptions:
• Natural gas 378.987 mol/min
• 50% excess air in natural gas and air mixture • Heating system temperature 1273 K
• Exhaust gas feed temperature 298 K • Vent gas temperature 393 K
• No heat loss in the system Conditions
1. The temperature
Table7.7 Overall temperature of the heating system In and Out
T IN (K) T Reaction (K) T OUT (K)
298 1273 393
2. Feed in the Furnace
Tabel 7.8 Molar flow of the natural gas
Component mol /min IN mol/min OUT
CH4 378.9579544 0 O2 1136.873863 378.9579544 N2 4276.811199 4276.811 CO2 0 378.958 H2O 0 757.916
3. Heat constants for system components
Table 7.9 Specific heat constants for the components [32]
Reactions CH4+2O2→2H O CO2 + 2 Equations: ( 2 2) P C = A BT CT+ + +DT− × ---(7.2.1) R * in T f T p H n C dT ∆ =
∫
*( ( ) 1 ( 2 2) 1 ( 3 3) ( 1 ) ( 1 )) 2 3in end in end in end
in end n A T T B T T C T T D T T = − + − + − + − − − --(7.2.2) Component A B C D CH4 1.702 0.009 0.000 0.000 CO2 5.457 0.001 0.000 -115700.000 H2O 3.470 0.001 0.000 12100.000 O2 3.639 0.001 0.000 -22700.000 N2 3.280 0.001 0.000 4000.000
34 1237 0 i i in T f f T P H H C dT ∆ = ∆ +
∫
---(7.2.3) 1237 1237 1237 i i i f i f rxn = (m h )products- (m h )reactants H ∆ Σ ∆ Σ ∆ ---(7.2.4) Calculations:1. Heat of stream IN: Temperature range from 298K to 1273K Table 7.10 Heat IN by the feed stream Ref [ Eq. 7.2.1,7.2.2]
2. Heat of the stream OUT: The temperature range is 1273 K- 393K Table 7.11 Heat OUT by the stream Ref [Eq. 7.2.1,7.2.2]
3. Heat of reaction (The amount of heat produced by the natural gas oxidation only) Table 7.12 Total heat by the Natural gas oxidation reaction Ref [ Eq. 7.2.3,7.2.4]
Total energy produced by natural gas oxidation in a heating system:
Q = In – Out + Formed ---(7.2.5) = 194014.564 - 174240.92 - 357656.6193 = -337882.98kJ/min
The heat of N2O decomposition
Conditions
1. Temperatures for N2O decomposition
Component ∆Hf kJ/mol n IN mol/min HIN kJ/min
CH4 71.61885257 378.958 27140.5339 CO2 32.23511414 1136.874 36647.2587 H2O 48.41686123 0.000 0 O2 37.61959912 0.000 0 N2O 30.44950203 4276.811 130226.771 N2 71.61885257 378.958 0.000 Total IN 194014.564kJ / min
Component ∆Hf kJ/mol mol/min n OUT HOUT kJ/min
CH4 -68.2309238 0.000 0 CO2 -29.6822644 378.9579544 -11248.33 H2O -45.025384 378.958 -17062.727 O2 -34.743554 757.916 -26332.692 N2O -27.9640994 4276.811 -119597.17 N2 -68.2309238 0.000 0
Total OUT -174240.92kJ/min
Component Moles IN mol/min Moles OUT mol/min
1237 f H ∆ kJ/mol 1237 rxn H ∆ kJ/min CH4 378.9579544 0.000 6.585466461 2495.6149 CO2 757.9159088 378.9579544 35.49354872 26901.1252 H2O 0 378.958 -337.6731025 -255927.82 O2 0 757.916 -190.870945 -72332.063 Total -357656.6193kJ /min
Table 7.13 Temperatures for N2O decomposition
T IN (K) T Reaction (K) T OUT (K)
298 1273 393
2. The feed of N2O decomposition
Table 7.14 Feed of N2O decomposition
Component mol /min IN mol/min OUT
N2O 1.284 0.021
O2 0.8490625 1.480990818
N2 0 1.264
3. Heat constants for components
Table 7.15 Specific heat constants for the components [32]
Reaction
2N O2 →2N2+O2 Calculations
1. Heat IN by exhaust gas (Temperature from 298K to 1273K) Table 7.16 Heat IN by exhaust gas Ref [Eq. 7.2.1,7.2.2]
2. Heat OUT by N2O decomposition reaction (Temperature from 1273K to 393K)
Table 7.17 Heat OUT by exhaust gas Ref [Eq. 7.2.1,7.2.2]
Heat of N2O decomposition reaction (1273K)
Table 7.18 Heat by N2O decomposition reaction Ref [Eq. 7.2.3,7.2.4]
Component A B C D
O2 3.639 0.001 0.000 -22700.000
N2O 5.328 0.001 0.000 -92800.000
N2 3.280 0.001 0.000 4000.000
Component n IN mol/min ∆Hf IN kJ/mol HIN kJ/min
N2O 1.284 48.9365893 62.85314542
O2 0.8490625 32.23511414 27.3696266
N2 0 0 0
Total 90.22277203kJ/min
Component n OUT mol/min ∆Hf h OUT
kJ/mol HOUT kJ/min
N2O 0.021 -45.4538177 -0.932836304
O2 1.480990818 -29.6822644 -43.95916101
N2 1.264 -27.9640994 -35.34261264
36
Total energy produced N2O decomposition in a heating system:
Q = In – Out + Formed ---(7.2.6) =90.22277203-80.23460995--110.547=100.5438912kJ/min
The water that can be heated by a natural gas Assumptions:
• Incoming water temperature = 333K • Outgoing water temperature = 393K Conditions: (393 ) 84.888 / . 0.084.88 / . * 0.084.88*(393 333) 5.09328KJ/mol in p K T p T C KJ Kmol K KJ mol K H n C dT = = ∆ =
∫
= − = Calculations:In – Out + Reaction = Heat removed by water
= -337882.98KJ/min / 5.09328KJ/mol=66338.97606mol/min
The amount of circulating water is 66338.97606mol/min through the heating system, which can be heated by the natural gas oxidation.
The water that can be heated by N2O thermo decomposition:
100.5438912KJ/min /5.09328KJ=19.7495mol/min
The N2O decomposition concern
When 1.284 moles of N2O are decomposed, the natural gas consumption is reduced.
The relationship between the amount of N2O used and the amount of natural gas used is
shown in Table 7.19.
Table 7.19 Evaluation of natural gas saving
Component N2O used Natural Gas saved
mol /min 1.00 0.0893
mol/min 11.195 1
Effectively the usage of N2O is 1.28 mol/min in the system, so this will lead to the saving
of 0.114 mol/min of natural gas i.e. 1.88264gm/min.
The amount of natural gas required before N2O gas put in the heating system is
378.987 mol/min*16g/mol =6063.3273 gm/min.
The total amount of natural gas could be required, till the N2O is passed to the heating
system will be:
6063.3273 – 1.88264= 6061.505gm/min. Component
Moles IN mol/min
Moles OUT
mol/min ∆Hf kJ/mol H react kJ/min
N2O 1.264 0 138.7695272175.404682
O2 0 0.631928 35.4935487222.4293785
N2 0 1.264 33.5665888442.4281683
7.4 Mass balance for catalytic decomposition
Fig 7.1 Examples of cordierite monolith structures of different cell density (cells per square inch, cpsi). Indicated are the channel size and geometric surface area.
7.4.1 Mass Balance Calculation-Gas phase mass balance
Consider a differential element of volume dV of the reactor (see figure) . The mass/mole balance for component i for the gas phase is:
rate of rate of i rate of i rate of i
accumulation entering dV leaving dV leaving dV of i by bulk flow by bulk flow by mass transfer
= − − ---(1)
Rate of accumulation of i in the gas phase = A z GCi G, t
ε
∂ ∆
∂
Moles of i entering the gas phase by bulk flow = volumetric flow * concentration of i = (A U C⋅ G⋅ i G z, )
Moles of i leaving the gas phase by bulk flow = (A U C⋅ G⋅ i G z, )+∆z Moles of i leaving the gas phase by mass transfer = k aGS GS(Cb−C A zs) ∆ (Equal to rate of reaction at the surface).
a : the catalytic surface area per unit volume of reactor, A : the cross-sectional area normal to the direction of gas flow ε : holdup
UG : Volocity of gas
k : mass transfer coefficient per unit area.
38
{
}
{
}
, , , ( ) ( ) ( ) G i G G i G z G i G z z GS GS b s C A z AU C AU C k a C C A z t ε +∆ ∂ ∆ = − + − ∆ ∂ --- (2)Dividing equation (2) by A z∆ , and at steady state there is no accumulation term. , ( G i G) GS GS( b s) 0 d U C k a C C dz − − = --- (3) in terms of volume , ( G i G) c A d AU C k aC dV − = --- (4) Where / 0 0 c k aV b A C =C e− ν
The reaction takes place at the surface of the catalysts. The correlation(1) to calculate the
mass transfer coefficient is
if p f D d k sh= ⋅ if f f D Sc ⋅ = ρ µ
For monoliths reactor:
45 . 0 1 Re ) 1 ( L d Sc B Nu Sh= + ⋅ ⋅ H ∞
In appendix … the meaning of the symbols is reported. Mass balance on the catalyst surface
For a monolithic reactor and fixed bed reactor we assume a plug flow behavior of the flow. The dispersion in the axial direction is assumed negligible; the flow is fully laminar.
The mass balanced for component i at the surface of the catalyst is given by:
rate of i rate of i entering consumed by the by mass transfer reaction at surface
= --- (5)
rate of i entering by mass transfer is : k aGS GS(Cb−Cs)
rate of i catalyst rate
catalyst catalyst consumed effectiveness of
density holdup by reaction factor reaction
= × × × =
η
ε ρ
s s iv R
--- (6) Equation (5) becomes ( ) GS GS b s k a C −C =η
ε ρ
s s iv R
--- (7)Substituting equation (7) in equation (2), the unsteady state mass balance for gas phase can be rewritten as:
, , ( ) G i G G i G s s i C U C v R t z ε η ε ρ ∂ = − ∂ − ∂ ∂ --- (8) And the steady state balance is:
, , ( ) 0 G i G G i G s s i C U C v R t z ε η ε ρ ∂ + ∂ + = ∂ ∂ --- (9)
7.5 Mass And Heat balance for Separation of Isoflurane & Thermal Decomposition
TOTAL HEAT & MASS BALANCE FOR STREAMS
IN OUT
EQUIPMENT EQUIPM. EQUIPMENT Plant
Mass Heat Mass Heat Stream IDENTIF. Stream Mass Heat Mass Heat kg/day kj/day kg/day kj/day Nr. Nr. kg/day kj/day kg/day kj/day
31.88 33583 31.88 33583 <2> A01 <3> 30.98 38545 90 90 F02 <4> 0.92 -4982 31.88 33673 Total 31.90 33563 0.92 -4982 <4> C01&V01 <5> 0.31 -905 <6> 0.60 -4058 0.60 -4058 -250 -250 0.92 -4982 Total 0.91 -5213 30.98 38545 <3> P01 <7> 30.98 38546 30.98 38545 Total 30.98 38546 30.98 38546 <7> M01 <9> 58404.53 -10100739 58373.55 -10139284 58373.55 -10139284 <8> 58404.53 -10100739 Total 58404.53 -10100739 58404.53 -10100739 <9> K01 <10> 58404.53 -10100738 58404.53 -10100739 Total 58404.53 -10100738 58404.53 -10100738 <10> F01 <11> 58404.53 -119341455 -109240717 -109240717 58404.53 -119341455 Total 58404.53 -119341455 58404.53 -119341455 58405.43 -119346328 Total 58405.13 -119345763 OUT - IN : -0.30 565
40
7.6 Mass And Heat balance for Separation of Isoflurane & Catalytic Decomposition.
TOTAL HEAT & MASS BALANCE FOR STREAMS
IN OUT
EQUIPMENT EQUIPM. EQUIPMENT Plant
Mass Heat Mass Heat Stream IDENTIF. Stream Mass Heat Mass Heat kg/day kj/day kg/day kj/day Nr. Nr. kg/day kj/day kg/day kj/day
31.88 33583 31.88 33583 <2> A01 <3> 30.98 38545 90 90 F02 <4> 0.92 -4982 31.88 33673 Total 31.90 33563 0.92 -4982 <4> C01&V01 <5> 0.31 -905 <6> 0.60 -4058 0.60 -4058 250 250 0.92 -4982 Total 0.91 -4713 30.98 38545 <3> E01 <7> 30.97 38514 30.98 38545 Total 30.97 38514 30.97 38514 <7> R01 <8> 30.97 0 38514 38514 30.97 38514 Total 30.97 38514 30.97 0 31.88 33673 Total 31.57 34706 OUT - IN : -0.31 1033 Project ID Number : CPD3301 Completion Date : 9th January 2004
8.Process and Equipment design
8.1 Process design calculations for separation of Isoflurane
Total heat load for condensation of gases (stream 4) = -249.52 kj/day
We have assumed that desorption and condensation process will operate just for two hours per day. The inlet & outlet temperature of gases for condensation is 130 oC and 4 oC respectively. Condensation of isoflurane will take place at 4 oC
So the total load per hour of condensation operation Q 249.52 2
:= kj /hr Inlet and Outlet temperature of coolant are -9 and 0 0C
(From super pro design simulation)
Specific heat of nacl brine=0.786 BTU/lb-F @ 60 F
(From http://www.armstrong-intl.com/common/allproductscatalog/cg-53.pdf) SPnacl:=3.2908 kj /kg c ∆Tbrine:=0−( )−9 c brinetol Q SPnacl⋅∆Tbrine := brinetol=4.212 kg /hr brineyr brinetol 2⋅ 365⋅ 1000 := ton /yr
brineyr=3.075 ton /yr
brinecost :=brineyr 300⋅
brinecost =922.519 euro /yr
Calculation of Total Heat Transfer Area:
Value of Overall heat transfer coefficient for condensation between organic gases and brine solution is U = 40 j/m2 /oC/s (ref: process heat transfer by kern,D.Q)
U 403600 1000 ⋅ := kj /m2/oC/hr U=144 kj /m2/oC/hr Design of Condenser Amount Of Coolant Needed:
42
Log mean temperature =∆ Tlm =(∆Τ1−∆T2)/ln(∆Τ1/∆T2)
∆T1:=130−( )−0 ∆T2:=4−( )−9 ∆Tlm ∆T1 ∆T2− ln ∆T1 ∆T2 := ∆Tlm 50.812=
The heat transfer for a surface condenser is Q=U *A*∆Tlm
Asurface Q U⋅∆Tlm := Asurface =0.017 m2 ~0.1829 ft2 Adesign :=Asurface 1.2⋅ Adesign =0.02 m2 ~0.2045 ft2
Cost of condenser equipment= 700 $ ~840 euro (Vertical tube, small ..[37]
∆pb hb 150⋅µg⋅us dp2 1−εb
(
)
2 εb3 ⋅ 1.75ρg us 2 ⋅ dp ⋅(
1−εb)
εb3 ⋅ + ⋅ := DESIGN OF ADSORBER Pressure drop across Adsorber bed calculations∆Pb= hb((150µg*us(1-εb)2/(d2p*ε3b))+(1.75ρg*us2*(1-εb)/(dp*ε3b))) From (catalyzed N2O activation by perez Ramirez,pg184)
Height of bed,
Dynamic viscosity of gas= Average superficial velocity Porosity of bed= Diameter of particle= Density of gas= hb := .2 m µg := 1.8 10⋅ − 5 kg m/s m /s us:=0.0041 εb := 0.45 m dp := 3 10⋅ − 3 kg/m3 ρ g := 1.64 ∆pb =0.836 pa
Inlet pressure=1.013 bar
Outlet pressure= 1.013-∆Pb bar
Pout:=0.2−∆pb 10⋅ −5
Pout=0.2 bar
Design of adsorber calculations
Amount of zeolite bed that can absorb 90 gms of isoflurane 100 min=620gms
Amount of zeolite bed required to adsorb 594 gm isoflurane at a flow rate of 55l/min for 360 min =4092gms=4.092kg
tbatch:=360 min
Mass of zeolite mzeolite 620 tbatch 90 1000⋅
⋅ ⋅1.65
:=
mzeolite 4.092= kg
We considered density of bed as ρzeolite:=700 kg/ cm3 Volume of zeolite bed required will be: v mzeolite
ρzeolite :=
v=5.846 10× −3 m3
We assumed for calculations the length/diameter for adsorber bed=1 Design volume =vdesign=1.1v
vdesign :=1.1v vdesign =6.43 10× −3 m3 l 4 22⋅ vdesign7⋅ 1 3 := So the length = l 0.202= m d:=l
Height of adsorber bed is assumed as1.2 times the calculated length ldegn:=1.2 l⋅
44 Length of bed= ldegn =0.242
Surface area of bed=S=Π*D(L+D/2)
Sarea π d⋅ ldegn d 2 + ⋅ := Sarea=0.217 From (http://www.epa.gov/ttn/catc/products.html#aptecfacts) Approximate cost of zeolite per lb is 40$ ~ 48 euro
Cost of zeolite =Cv= 48*mzeolite*2.204 Cads :=48 mzeolite⋅ ⋅2.204
Cads =432.901
Vessel Installation cost =(M&S/280)*101.9*D1.066*H0.802*(2.18+FC) $ Where D,H are diameter and height of bed in ft,& FC=1
(ref: conceptual design of chemical process by douglas pg574)
costintern =20.419 $
adsober costintern adcost:= + +Cads
adsober=904.242 $ ~1080 euro adcost 726 280⋅101.90.662 1.06 ⋅ ⋅0.79390.802⋅(2.18 1+ ) := adcost =450.922 $ Internals cost=(M&S/280)*4.7*D1.55*H*FC $ costintern 726 2804.7 .662 1.55 ⋅ ⋅0.7939⋅(1.8+2.2) :=
8.2 Blower
A graph is plotted for the flow rate and cost of the blower. A centrifugal multi stage Turbo blower is chosen. [34]
Normal maximum speed (rpm) = 1000 Normal maximum capacity (m3/h) = 8500 Normal maximum pressure (differential) (bar) Single stage = 0.35
Multiple stage = 1.7 Cost:
Table 8.1 Flow rate and operating cost (including motor cost) [35] Flow rate (ft 3/min) Flow rate (lit/min) Cost (Euro)
2.000 56.632 410.91 2.176 61.618 572.55 2.301 65.156 709.28 2.398 67.900 760.16 2.477 70.142 849.08 2.602 73.680 975.50 2.699 76.424 1083.71 2.778 78.666 1145.09 2.845 80.562 1200.00 2.903 82.204 1316.29 2.954 83.652 1356.40 3.000 84.948 1375.35
