separation processes -...
TRANSCRIPT
SEPARATION PROCESSES
2018
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AIM OF THE COURSE:
i. The aim of the course is to teach students principles of chosen separation
processes most frequently used in chemical and similar technologies. The
arrangements of separation processes will be generally described and then
their mathematical description will be presented.
ii. The students would acquire knowledge necessary for both verbal and
mathematical description of selected separation processes. Students will
know principles of mass and energy transport phenomema, phase
equilibria and material and energy balancing of separation processes.
iii. The students will be able to describe principles of separation processes,
will know their applications and will be able to set-up their mathematical
models.
iv. The students will know how to solve numerically the problems concerning
separation processes and to design basic dimensions of a separation
equipment.
COURSE OUTLINE:1) Introduction to separation processes. Material and energy
balances.
2) Basic numerical methods in chemical engineering
3) Basics of fluid mechanics
4) Heat and mass transfer fundamentals
5) Separation processes – rate and equilibria
6) Adsorption processes
7) Membrane separation processes
8) Ionic exchange and electrophoretic separation
9) Membrane process design
10) Gas permeation, reverse osmosis
11) Ultrafiltration, dialysis
12) Pervaporation
13) Crystallization
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ASSESSMENT CRITERIA• Homework: solution of individual problems (max 100
points).
• Final test in week 14 (the last one): a simple problem solution (max. 100 points). Repetitions possible, only the best result is considered.
• Oral examination max. 100 pts.
• Weights:
Individual problems 1/3
Final test 1/3
Oral exam 1/3
Points(total)
100 -90 89 -80 79– 70 69 – 60 59 – 50 49 - 0
Grade A B C D E F
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LITERATURE: (SHR)
• J. D. Seader, E. J. Henley, D. K. Roper: SEPARATION PROCESS PRINCIPLES. Chemical and Biochemical Operations. (Third Ed.), John Wiley & Sons, Inc., 2011. (ISBN 978-0-470-48183-7)
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On May 3rd there will be Tuesday schedule! No Separation Processes!
Material and energy balances in chemical
engineering applications
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Outline
• Balances – principles and nomenclature• Mass and molar balances without chemical reactions• How to solve balance problems – recommended
procedure• Energy balances• Examples
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System• Part of universe with real or fictitious boundaries that can
communicate with surroundings systems (pot, membrane module, chemical reactor, chemical plant, bank account …)
• Systemso Open – mass and energy exchange
o Closed – energy exchange
o Isolated – no exchange
o Adiabatic – no heat exchange
• Systemso Homogeneous – single phase
o Heterogeneous – two or more phases
• Systemso Steady state – no change in time
o Dynamical state – properties of the system change in time
• System communicates with surroundings by streamso Streams = flows of mass, energy, volume, cash or other extensive quantities
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Chemical engineering systems -examples
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a) Open complex system with fictitious boundaries
b) Fictitious differential cube – a part of chemical reactor
c) Chemical reactor. The system boundaries coincide with the reactor
walls – real boundaries.
Nodes – parts of complex systems
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Process node Mixer
Bypass Reflux
Complex system consisting of 5 nodes
Processes taking place in systems
• Mechanicalo Grinding, crushing, milling, sieving, sorting ...
• Hydromechanicalo Gas or liquid flow, filtration, stirring, sedimentation, fluidization ...
• Heat transfero Heat exchangers, evaporators, cooling towers ...
• Mass transfero Distillation, extraction, rectification, membrane processes, drying, adsorption,
absorption ...
• Chemical processes (reactions)o Batch, PFR, CSTR, and other reactors with chemical or biochemical processes
• Biological processeso Systems containing living organisms
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Physical quantities• Extensive quantities (EQ)
o EQ can be summed over parts of a system
o EQ express amount of some quantity
o EQ may be balanced
• examples: mass, molar amount, volume, energy, enthalpy, momentum, money
• Intensive quantities (IQ)o IQ cannot be summed over parts of a system
o IQ describe physical properties of systems
o IQ cannot be balanced
• examples: temperature, pressure, velocity, radioactivity, color...
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Composition of a mass within a system
• To be balanced, all components have to be identified in the system
• The contents of components are expressed as concentrations or fractions
• Mass (𝜚𝑖), molar (𝑐𝑖)or volume (𝜑𝑖) concentrations
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𝑚𝑖 - mass of a component in the system𝑛𝑖 - molar amount of a component in the system𝑉𝑖 - volume of a component in the system𝑉 - total volume of the system
Composition of mixtures• Mass (𝑤𝑖) and mole (𝑥𝑖) fractions
• Relative mass (𝑊𝑖) and relative (𝑋𝑖) mole fractionso are used in processes with inert components (amount of inert component does not
change within the system)
• Each identified component in the system can be balanced we can formulate n linearly independent balances for ncomponents
This is not valid in simple dividers and mixers (compositions of all inlets and outlets are identical)
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𝑚 – total mass of the system𝑛 - total molar amount in the system
𝑚𝑖′- mass of an inert component𝑛𝑖′- molar amount of an inert component
Principles of balancing
• Inputs – BQ that comes into the system through boundaries
• Outputs - BQ that leaves the system through boundaries
• Sources – internal formation or extinction of BQo Positive – interests, formation of a component by chemical reaction
o Negative – transaction fees, decay of a chemical component
• Accumulation – the difference between final and initial amounts of BQ within the system
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Inputs + Sources = Outputs + Accumulation
Bank account
salary
rente
flat etc.
shopping
Balancing
• Sources and accumulation can be treated as fictitious inputs or outputso Positive sources = fictitious inputs
o Negative sources = fictitious outputs
o Accumulation
• Initial amount of a quantity in the system – fictitious input
• Final amount of a quantity in the system – fictitious output 19
Inputs + Fictitious Inputs = Outputs + Fictitious Outputs
Bank account
salary
rente
rent
shopping
initial value
final value
interests
transaction fees
Material balances
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Stream #1
Stream #4
Stream #2
Stream #3
Three chemical components: A, B, C
𝑚𝐴1 +𝑚𝐴2 = 𝑚𝐴3 +𝑚𝐴4
𝑚1𝑤𝐶1 +𝑚2𝑤𝐶2 = 𝑚3𝑤𝐶3 +𝑚4𝑤𝐶4
𝑚1𝑤𝐵1 +𝑚2𝑤𝐵2 = 𝑚3𝑤𝐵3 +𝑚4𝑤𝐵4
𝑚1𝑤𝐴1 +𝑚2𝑤𝐴2 = 𝑚3𝑤𝐴3 +𝑚4𝑤𝐴4
𝑚1 +𝑚2 = 𝑚3 +𝑚4
𝑤𝐴1 +𝑤𝐵1 + 𝑤𝐶1 = 1
𝑤𝐴2 +𝑤𝐵2 + 𝑤𝐶2 = 1
𝑤𝐴3 +𝑤𝐵3 + 𝑤𝐶3 = 1
𝑤𝐴4 +𝑤𝐵4 + 𝑤𝐶4 = 1
Mass balance of a component
Can be rewritten by means of mass fractions
Four balances obtained – three of them are linearly independent
Sum of all mass fractions in a stream is always 1
One nodeFour streams
Process node
Material balances – recommended procedure
1) Draw flowsheet – nodes, streams.2) Identify all (chemical) components in the system to be balanced.3) Choose an extensive quantity to be balanced (BQ) – mass, molar
amount, volume, …4) Evaluate/identify compositions of streams (whenever possible).5) Compose an input matrix.6) When no extensive quantity is given (e.g., mass of a stream or
mass of a component in a stream), then an extensive quantity is chosen/defined for one chosen stream.
7) Determine number of unknowns in the input matrix = Degrees ofFreedom Analysis (DOF).
8) Set-up balance equations.9) Set-up auxiliary equations, number of independent equations is
equal to the number of unknowns (DOF).10) Solve the set of equations = algebraic problém.
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DoF Analysis(Degrees of Freedom)
• useful tool to assess solvability of a balance problem• assessment of number of unknowns to be evaluated vs. number of equations
and data
DoF vars eqnsN N N
number of allvariablesdescribing theproblem
number of allindependent equations
number ofDegrees ofFreedom = number ofdata that havebe specified
Ex.: A single-phase stream with C components.NDoF = C + 1 + 1 + 1 = C + 3
concentrations T P amount
SHR:
Xylene, styrene, toluene and benzene are to be separated with the array of distillationcolumns shown in figure (right), where F, D, B, D1, B1, D2 and B2 are the molar flow rates(mol/min). Concentrations are given in molar%.
Evaluate all unknown component concen-trations and fluxes.
There is:
7 streams 7 (4+1) = 35 variables
6 unknown flow rates (B, D, B1, D1, B2, D2)8 unknown concentrations (streams B and D)=====================================14 unknowns
35 - 14 = 21 d.o.f. – we have to provide 21 values
21 values specified in the schema
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Example: Natural gas composed of CH4 and C2H6 is combusted with excess amount of the air. The amounts of H2O and CO2 in combustion gasses are in molar ratio 9:1. After removing of these two components the composition of resulting gas is: 8.3 mol.% of CO, 6.5 mol.% of 02 and 85,2 mol.% of N2. Calculate: a) composition of the natural gas (in mol. %), b) amount of the combustion gasses evolved by combustion of 1 mole of
the natural gas,c) what was the per cent excess of the oxygen?Hint: Consider these three reactions taking place within the burner:
2CH4 + 3O2 2CO + 4H2O2C2H6 + 5O2 4CO + 6H2O
2CO + O2 2CO2
Results: a) Composition of natural gas: 85.5 mol.% of CH4 and 14.5 mol.% of C2H6, b) 13,3 mole of combustion gasses per 1 mole of natural gas, c) Excess of the oxygen: 11.7 %
CH4 A
C2H6 B
O2 C
N2 D
CO E
CO2 F
H2O G
Reaction # Imaginary streams
2 CH4 + 3 O2 2 CO + 4 H2O (1) 5,6
2 C2H6 + 5 O2 4 CO + 6 H2O (2) 7,8
2CO + O2 2CO2(3) 9,10
BURNER
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2
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4
natural gas
air
CO2 + H2O
CO + O2+ N2
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#1 #2 #3 #4 #5 #6 #7 #8 #9 #10
n [mol] 1 ? ? ? ? ? ? ? ? ?
xA CH4 ? --- --- --- --- 2/5 --- --- --- ---
xB C2H6 ? --- --- --- --- --- --- 2/7 --- ---
xC O2 --- 0.21 --- 0.065 --- 3/5 --- 5/7 --- 1/3
xD N2 --- 0.79 --- 0.852 --- --- --- --- --- ---
xE CO --- --- --- 0.083 2/6 --- 4/10 --- --- 2/3
xF CO2 --- --- ? --- --- --- --- --- 1 ---
xG H2O --- --- ? --- 4/6 --- 6/10 --- --- ---
There are 13 unknowns we need 13 equations
We can formulate:2x sum of molar fractions for streams #1 and #33x ratios of molar amounts of imaginary streams (given by reaction stoichiometries)7x balance equations1x ratio of molar fractions in stream #3: xG3/xF3 = 9
The problem is solvable as there is 13 equations for 13 unknowns.
DoF vars eqns 13 13 0N N N
#1 #2 #3 #4 #5 #6 #7 #8 #9 #10
n [mol] 1 11.78 2.384 10.927 2.564 2.137 0.726 0.5085 0.2384 0.3575
xA CH4 0.8547 --- --- --- --- 2/5 --- --- --- ---
xB C2H6 0.1453 --- --- --- --- --- --- 2/7 --- ---
xC O2 --- 0.21 --- 0.065 --- 3/5 --- 5/7 --- 1/3
xD N2 --- 0.79 --- 0.852 --- --- --- --- --- ---
xE CO --- --- --- 0.083 2/6 --- 4/10 --- --- 2/3
xF CO2 --- --- 0.1 --- --- --- --- --- 1 ---
xG H2O --- --- 0.9 --- 4/6 --- 6/10 --- --- ---
RESULTS
Ratio of amount of combustion gases to amount of natural gas: (n3+n4)/n1 = 13.31
Excess of the oxygen:excess=(xC2 n2 -(xC6 n6 + xC8 n8 + xC10 n10))/(xC6 n6 + xC8 n8 + xC10 n10) = 0.40
oxygen entering with
stream #2
oxygen necessary to burn al methane and
all ethane
Problem #1:
Mass balance of a crystallizer
1000 kg/h of 10 w% sodium carbonate solution in water is mixed with a recyclestream. The obtained mixture is continuously introduced into an evaporator,where 30 w% sodium carbonate solution is produced. This solution is then ledinto a crystallizer where crystals of sodium carbonate dodecahydrate areproduced together with 20 w% sodium carbonate solution. Certain part of thissolution is recycled. The other part is withdrawn from the system. Yield of theprocess is 90 %, i.e. 90 % of sodium carbonate introduced into the system isobtained in the form of crystals. Determine mass flow rate of the reflux stream.
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Problem #2:Pyrite calcination
1000 kg of pyrite composed of 85 mass % of FeS2 and 15 mass % of rubbish is calcined with 100 % excess of the air. The calcination is described by chemical reaction
4 FeS2 + 11 O2 = 2 Fe2O3 + 8 SO2
The rubbish does not react with the oxygen and does not release any gasses.The solid material (waste) after calcination contains 2 mass % of FeS2.
Calculate:
a) concentration of Fe in solid waste, b) volume of air (at 20 oC and 0.0981 MPa) required for the calcination, c) volume of the gas evolved by calcination (at 300 oCand 0.0981 MPa) and its composition in mole fractions.
Results:
a) the waste contains 54.9 mass % of Fe, b) 4610 m3 of air, c) volume of evolved gasses is 8760 m3 and its composition : 7.7 mol.% SO2, 11.0 mol.% O2 and 81.3 mol.% N2.
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Components:A ... pyriteB ... rubbishC ... O2
D ... N2
E ... Fe2O3
F ... SO2
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pyrite
air
gaseousproducts
solid products
2 Fe2O3 + 8 SO2
4 FeS2 + 11 O2
HINTS:
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Problem #3:Sulphur is combusted with stoichiometric amount of air in a furnace and sulphurdioxide is produced. SO2 is then partly oxidized with the excess amount of dry air to SO3 in the first catalytic chamber. The sulphur trioxide is completely removed from the gas phase in the first absorber. Resulting gas containing 3.3 mol.% of SO2
and 8.3 mol.% of O2 is conveyed to the second catalytic chamber and the second absorber. In the second chamber all SO2 is oxidized to SO3. All SO3 is removed from the gas in the second absorber.Calculate:a) what fraction of the sulphur entering the process was converted to SO3 in the
first chamber, b) b) what is the excess of oxygen used for sulphur combustion, c) c) composition of the gas phase leaving the second absorber (volumetric per
cent).
Results:a) 70.5 % of total sulphur, b) b) excess of oxygen is 39.5 %, c) c) gas phase consists of 7 vol.% of O2 and 93 vol.% of N2
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HINTS:
furnace catalytic chamber I
catalytic chamber II
absorber I absorber II1
2
3 5
6
7 9
10
11
16 17 12 13 14 15
S +
O2
SO2
2 SO
3
2 SO
3
2 SO
2 +
O2
2 SO
2 +
O2
Components: A ... sulphur, B ... O2, C ...N2, D ... SO2, E ... SO3
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Computer (algorithmic) solution of materialbalances – example in Mathematica
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ENTHALPY BALANCING
INTERNAL ENERGY• The first law of thermodynamics
o U – internal energy
o Q – heat
o 𝑊 – work
• The internal energy represents energy of the system that can be transformed into heat or work.
• Absolute/total value of internal energy of systems isunknown. We can, however, determine its changes.
• Change of the internal energy (dU) is equal to the heat (dQ) and work (dW) delivered into system
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System
dUdQ dW
d d d dU Q W dQ p V
ENTHALPY• There are volume, electric, magnetic and other works (W) that can be delivered to
or produced by the system
• Volume work is given by a volume change - compression or expansion
• We can define a new extensive quantity – enthalpy
• Total differential of enthalpy reads
• After substitution from the first law of thermodynamics, we obtain
• In isobaric systems (dp = 0), change of the enthalpy is equal to heat delivered to or removed from the system
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ENTHALPY BALANCING
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Enthalpy of Material
Inputs
Material Stream #1
Material Stream #4
Material Stream #2
Material Stream #3Processnode - H
Heat sourceQs
Heat loss Ql
𝐻1 + 𝐻2 + 𝑄𝑠 − 𝑄𝑙 = 𝐻3 + 𝐻4 + ∆𝐻𝑠
Accumulation
Heat Sources
Enthalpy of MaterialOutputs
Enthalpy Accumulationwithin System
+ +=
Specific/molar enthalpy
• Enthalpy of material stream H (heat contained in material) can be expressed as a product of mass/molar amount (m/n) and specific enthalpy/molar enthalpy (hm/hn)
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𝐻 = 𝑚ℎ𝑚
𝐻 = 𝑛ℎ𝑛
hm or hn are given in [J/kg] or [J/mol]
In the following slides, we will make use of single symbol h for both specific and molar enthalpies (hm/hn) – (the meaning of h will result from the context)
Enthalpy of mixtures• Specific enthalpy of a mixture is given by mass contributions
of all chemical components forming the mixture:
• ℎ𝑖 is specific enthalpy of the component i
• ∆ℎ𝑚𝑖𝑥 is specific enthalpy of mixing (values can be found in tables for particular mixtures)
• ∆ℎ𝑚𝑖𝑥=0 for an ideal mixture
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ℎ =
𝑖
𝑤𝑖ℎ𝑖 + ∆ℎ𝑚𝑖𝑥
Enthalpy of a chemical component
• There is no absolute value of enthalpy
• Enthalpy value is always related to the reference state
• Reference state can be chosen arbitrarily
• We evaluate/study an increase or a decrease of enthalpy with respect to the reference state
• The enthalpy increase or decrease are equal to the amount ofheat consumed by the system or released from the system
• Change of the enthalpy = „How much heat I have to deliver to the system or to remove from the system“
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Reference state• The reference state: enthalpy of a chemical component equals
zero at the reference state
• The reference state is defined by stating:o Temperature of the reference state
o State of matter (g,l,s) for each chemical component
• Proper selection of the reference state may significantly simplify the solutions of the problems:
o The reference temperature would be chosen as a temperature in the studied system, e. g., temperature of a stream,
o The state of matter of a chemical component would be the same as one state of the component in the system,
o Availability of physical and chemical data has to be taken into account.
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• Change of specific/molar enthalpy with temperature is called specific/molar heat capacity
• Specific/molar heat capacity tells us how much heat has to be supplied in order to warm one kilo/mole of the component by 1 Kelvin
• Specific/molar heat capacity depends on temperature
• Change of enthalpy can be evaluated as the integral of a polynomial
• For engineering estimation of enthalpy, specific/molar heat capacity is considered to be a constant evaluated at the mean temperature 𝑇
Enthalpy as function oftemperature
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Enthalpy as a function of the state of matter
• Change of state of matter is always accompanied with a largechange of enthalpy value
• This enthalpy change depends on temperature and can usually be found in tables etc.
• The most common changes of state of the matter are:o Vaporization – condensation
o Sublimation – desublimation
o Melting - freezing
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∆ℎ𝑖,𝑝ℎ𝑎𝑠𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = ℎ𝑖,𝑛𝑒𝑤 𝑝ℎ𝑎𝑠𝑒 − ℎ𝑖,𝑜𝑙𝑑 𝑝ℎ𝑎𝑠𝑒 = −∆ℎ𝑖,𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑝ℎ𝑎𝑠𝑒 𝑐ℎ𝑎𝑛𝑔𝑒
∆ℎ𝑖, 𝑣𝑎𝑝 = ℎ𝑖,𝑔 − ℎ𝑖,𝑙 = −∆ℎ𝑖,𝑐𝑜𝑛𝑑
∆ℎ𝑖, 𝑠𝑢𝑏 = ℎ𝑖,𝑔 − ℎ𝑖,𝑠 = −∆ℎ𝑖,𝑑𝑒𝑠𝑢𝑏
∆ℎ𝑖, 𝑚𝑒𝑙𝑡 = ℎ𝑖,𝑔 − ℎ𝑖,𝑠 = −∆ℎ𝑖,𝑓𝑟𝑒𝑒𝑧
Dash-dotted line route
Dashed line route
boiling
Solid line route
Combined dependencies of enthalpy on temperature and state of matter (vaporization as an example) – enthalpy is state variable
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Point A reference state, a liquid at temperature 𝑇𝑟𝑒𝑓
Point B – vapor at temperature 𝑇
The total enthalpy change between A and B does not depend on the routefrom the point A to the point B.
boilboil boil
vap
vap
boilboil
Enthalpy balance of a process mixer
10 kg of water at 25 °C has been mixed with 3 kg of ice at 0 °C with 5 kg of water at 100 °C. What is temperature of the obtained mixture? Is there some ice in the resultingmixture? Consider specific heat capacity of water to be approximately constant within the relevant temperature interval.
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Problem #4:
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Cooling a pipe by external flowA warm liquid is flowing through a horizontal pipe with a flow rate F (kg/s). The interaldiameter of the pipe is D and its length is L (cf. the figure). A cooling fluid is flowing aroundthe tube in a perpendicular direction. The cooling fluid has constant temperature Tw. Warm liquid enters the tube with the temperature Tini. The density and heat capacity ofthe warm liquid are and cp, respectively. The overall heat transfer coefficient betweenthe warm liquid and the cooling fluid is U (W/(m2K)).Find the temperature of the warm liquid as function of time and position within the tube, i.e., T(z,). Suppose plug-flow within the pipe and neglect heat conduction resistances in radial direction.
Problem #4:
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INTRODUCTION TO SEPARATIONS
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o Mixture: „a basic form“ of an existence ofmatter.
o Mixtures may form spontaneously but usuallyan energy input is needed and speeds-up themixture formation.
A SEPARATION of a mixture to individualcomponents or less complex mixtures requires an
energy input!
1
C
i
i
Mixture component
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Separation of a mixture to components (or to lesscomplex mixtures) represents basic task in chemical
and similar technologies
(analytical separations?)
SEPARATIONPROCESS
FEED MIXTURE
PRODUCT 1
PRODUCT 2
PRODUCT N-1
PRODUCT N
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MIXTURES AND THEIR PROPERTIES – factors affectingseparations
Single phase mixtures × multiphase mixtures
Homogeneous mixtures × heterogeneous mixtures
Molecular, thermodynamic and transport properties
Size and shape of particles
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Classification of separation methods
(a) phase creation
(b) phase addition
(c) solid barriers
(d) solid agents
(e) force field or gradient
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(a,b) phase creation or phase addition
When the feed to be separated is a single-phase then second separable phase must be developed before separation of thecomponents can be achieved.
Creation of the second phase?
Energy separating agent (ESA) or a mass-separating agent (MSA)
ESA – heat transfer to/from mixture or shaft work to-frommixture
MSA – partially or fully immiscible with one or more constituents of the mixture
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Full list of separations based on phase creation or additiongives Table 1.1 in Seader-Henley-Roper text-book:
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(c) solid barriersMacro-, micro- and nano-porous barriers are used forseparation according to mixture constituents properties
Components are separated either by theirselective retention by the barrier (filtrationprocesses) or due to their different rates ofpermeation through the barrier (solution-diffusion mechanism)
Liquid membranes – a modification of separations by solid barriers
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(d) solid agents
A particulate/granularsolid agent is mixed (in a proper way) with a gas, vapour or liquid
mixture.
Mixture constituentsselectively bind to the
solid agent.
Phase equilibrium maybe achieved (contact
time vs. mass-transfer rate).
Solid agent maybe covered by a
thin layer ofselective agent (inert carrier + agent) or the
solid agent itselfbinds the
components.
Adsorption –chromatography– ion exchange.
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(e) force field or gradient
External fields are used to separate mixture constituents dueto their various responses to applied field:
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Typically, a sequence (a combination) of separation methodshas to be used to achieve required separation and/or to meet legal regulations etc.
Heuristics:
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Selection of feasible separation method(s)
A) Feed conditions1. Composition, particularly of species to be recovered2. Flow rate3. Temperature4. Pressure5. Phase state (s, l or g)
B) Product conditions1. Required purities2. Temperatures3. Pressures4. Phases
Factors influencing selection of separation operation(s):
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c) Property differences that may be exploited1. Molecular2. Thermodynamic3. Transport
D) Characteristics of of separation operation1. Ease of scale-up2. Ease of staging3. Temperature, pressure and phase-state requirements4. Physical size limitations5. Energy requirements
E) Economics1. Capital costs2. Operating costs