3.4 Working point in a pipe system 3.5 Particular characteristic curves
3.4 & 3.5 Material Balances & Flow Sheet
description
Transcript of 3.4 & 3.5 Material Balances & Flow Sheet
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3.3 CHEMISTRY & SEPARATION
PART I
Separation Processes
• Separation processes are needed for feed pretreatment, product recovery and waste processing
• Most separations are based on moving a component from one phase to another and then segregating the two phases– Driven by activity gradient as phases try to reach equilibrium– Affected by rates of mass transfer and heat transfer
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Separation Specifications
• Recovery: How much of the desired component made it to the stream it was supposed to be in:
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F, zA, zB
R, xA, xB
P, yA, yB Product enriched in A
AA
A
A
A
xRyPyP
zFyP
A ofRecovery
Separation Specifications
• Purity: The concentration of desired component in the stream it was supposed to be in:
©2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
F, zA, zB
R, xA, xB
P, yA, yB Product enriched in A
Purity of A in product = yA
Impact of Separation Specifications• Tighter specifications lead to higher cost:
• Final product must meet purity specifications– Set by ASTM, USP, etc.
• Recycles sometimes have purity specifications– e.g. to protect catalyst from contaminants or poisons
• Product that is not recovered is lost profit and also increased waste cost: separation recovery factors into plant yield
Purity or Recovery)%(
Cost
90 99 99.9 99.99
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Vapor-Vapor SeparationsMembrane
Based on differences in relative permeability of gases
Used for H2/CH4, CO2 removal,air separation
AbsorptionUsing a liquid solvent
in an absorber-stripper loop
Used for acid gases, drying, water wash
AdsorptionAdsorb components selectively on a solidRegenerate sorbent by temperature swing
(TSA) or pressure swing (PSA)Used for air separation, H2/CH4, most
separations involving low concentrations
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Adsorption• One component from vapor phase preferentially adsorbs onto the
surface of a solid adsorbent• Two types of adsorption:
– Reversible: • Usually physisorption• Adsorbed component can be released by decreasing pressure or increasing
temperature• Sorbent can be regenerated and used in multiple cycles, hence temperature-swing
adsorption (TSA) and pressure-swing adsorption (PSA)
– Irreversible:• Usually chemisorption• Adsorbed component usually reacts irreversibly with solid• Low concentrations can be achieved, but solid is difficult to regenerate• Used for contaminant removal guard beds
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Membrane Separation• Thin membranes of polymer or metal can be used to separate
gases:– Different species diffuse through a thin membrane at different rates:– Different gases have different solubility in metal or polymer
• Permeate passes through the membrane and becomes enriched in faster or more soluble species
• Retentate does not pass through membrane and becomes enriched in slower or less soluble species
• Membranes have relatively low cost, but cannot obtain high purity or recovery
• Membranes are therefore widely used for bulk separation of gases
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Dense layer 0.1 to 1.0 m
(not to scale)
Porous support
0.1 to 1.0 mm
Asymmetric Membrane
• Polymer membranes are usually cast as asymmetric membranes• Thin, dense, active layer is supported on a thicker stronger porous layer• Backing cloth is used in some cases as support for active layer
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Feed
Permeate
Retentate
One hollow fiber (of thousands)
Potting
Hollow Fiber Membranes
• Membranes are cast as long thin fibers• A bundle of fibers is set into a resin (potting) that effectively forms a
tubesheet• Feed is fed shell-side and permeate withdrawn from inside the fibers
Membrane cross section
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Flat Sheet (Spiral Wound) Membranes
• Membranes are cast as sheets• Sheets are glued back-to-back along edges to form an envelope
and attached to a perforated tube• The assembly is then rolled up into a spiral-wound module
Retentate
Feed
Permeate
Enlarged cross section
DensePorousPorousDense
Membrane envelopes
Permeate
Perforated tube
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• SEM, TEM, STEM can be used for
microscopic analysis
• Note asymmetric structure– Thin selective skin
– Porous support layer
UOP 5565M-12
Gas Separation Membranes
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Membrane Modules
UOP Separex modules for rejecting CO2 from natural gas
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Vapor-Liquid SeparationsFlash
Single stage thermal &phase eqbm
DistillationMultiple stage separation
between identified light key &heavy key components
EvaporationSingle stage removal of
volatile solute or solvent
StrippingMulti-stage removal
of volatile solute from solvent
AbsorptionRemoval of vaporcomponent using
non-volatile solvent
FractionationSeparation of
multicomponentmixture into fractions by boiling ranges (e.g. in oil
refining) ©2012 G.P. Towler / UOP. For educational use in conjunction with
Towler & Sinnott Chemical Engineering Design only. Do not copy
Mul
ti-st
age:
see
next
lect
ure
See heat exchange lectures
Vapor-Liquid Flash Drums
• Flash or knockout drums are widely used in chemical plants:– Downstream of condensers and coolers– Upstream of compressors and between compressor stages– As reflux drums on columns– In relief systems
• Design function is to separate liquid drops from vapor and prevent vapor blowing out into liquid-filled lines by maintaining liquid level control
• There will usually be ~1 to 2% liquid entrainment in the vapor from a knockout drum unless a demister is used.
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Vertical Flash Drum• Vessel diameter is chosen to give
vapor velocity that is less than terminal velocity of drops
• Use 0.15 ut if there is no demister• Allow 1 diameter above feed and at
least 0.6 diameters below feed for settling, also allow 0.4 diameters for demister
• Height of liquid depends on level control
ut 0.07[)L v(/v]1/2
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Liquid Level Control Bands• Level control needs to allow for some natural variation in
liquid level due to splashing, etc.• Alarms must not be set too close to normal operating level
or they will be a nuisance (& will be ignored)• Operators need time to respond to alarms before
shutdown
LAHH shutdown trip
LAH alarm
LAL alarm
LALL shutdown trip
Normal operating band
•A typical assumption is about 2 A typical assumption is about 2 minutes between alarm and trip, so minutes between alarm and trip, so allow 10 minutes of liquid residence allow 10 minutes of liquid residence
time below feedtime below feed•But note: midpoint of normal But note: midpoint of normal
operating band should be > operating band should be > DDvv/2 /2 below feed point, so if 5 mins of below feed point, so if 5 mins of
liquid holdup gives height < liquid holdup gives height < DDvv/2, /2, use half a diameter to the midpoint use half a diameter to the midpoint
and 5 min holdup belowand 5 min holdup below . .
5 min of liquid
5 min of liquid or Dv/2
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Horizontal Flash Drum
• Bigger area for settling + more space for liquid holdup• Trade-off is higher plot space and stronger foundations needed to support vessel• Often used when process control demands some liquid inventory, e.g. reflux
drums• Design is more complex than vertical drum – see Ch16
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Liquid-Liquid Separations
ExtractionMulti-stage
contacting of two liquid phases
DecantingSingle stage thermal
&phase eqbm
Mixer-SettlerSingle theoretical stage
extraction processOften 2 or 3 stages are still
cheaper than a column
MembraneBased on differences in relative
permeability of componentsMembrane can be used to keep
two solvents from mixing
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Feed
Light liquid
Dispersion band
Drain
Vent
Heavy liquid
Horizontal Decanter
• Design is similar to knockout drum: allow droplets to settle and provide adequate holdup for level control – see Chapter 16
• Siphon take-off can control level without instruments if densities are constant
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Multistage Extraction: Sulfolane Process
• Used for L/L extraction of benzene and toluene from gasoline using sulfolane solvent
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3.4 UNIT RATIO MATERIAL BALANCE
3.5 DETAILED FLOW SHEET
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Process Classification• Chemical processes can be classified as batch,
continuous or semi-batch and as either transient or steady state
• Batch process is one in which the feed is charged into the system at the beginning of the process, and the products are removed all at once some time later
• Continuous process is when the inputs and outputs flow continuously across the boundaries throughout the duration of the process.
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• Semi-batch process is a process in which its inputs are nearly instantaneous but the outputs are continuous or vice versa
• If the values of all process variables in a process do not change with time, the process is said to be operating at steady state. If any changes with time, transient or unsteady state operation exists
• One of the main responsibilities of chemical engineers is to create/construct/ analyse chemical processes (or, at least, to understand the existing processes)
• The layout of a chemical process is called “process flow sheet (PFS)” or “process flow diagram (PFD)” PFS or PFD can be for just a single process unit or for the whole process, either simple or complicated process.
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Examples of PFS or PFD
PFD for a water-softening by ion-exchange process
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PFD for Ammonia Synthesis Plant27
Normally, a PFS or a PFD comprises:• All major process equipments/units• Lines entering or leaving the process/unit and/or lines
connecting two or more process equipments/units (these lines are called “streams”)
• Flow rate of each stream• Composition of each stream• operating conditions of each stream and/or
unit/equipment (e.g., T, P)• Energy/heat needed to be added to and/or removed
from any particular part of the process or the entire process
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Some important symbols of process equipments
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Material Balance Equation
• The law states that mass can neither be created nor destroyed
• Material balance equations are the manifestation of the law TOTAL MASS INPUT = TOTAL MASS OUTPUT
• The design of a new process or analysis of existing one is not complete until it is established that the inputs and outputs of the process satisfy the material balance equation.
Material balance are based on :Law of Conservation of Mass
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Material Balance Equation• Suppose methane, is a component of both input and
output of a process
• If the flow rates of input and output are found to be different. Possible explanations are .…1. methane is leaking2. methane is consumed or generated in a reaction 3. methane is accumulating in the process vessel4. wrong measurement
Process unit
qin(kg CH4/h) qout(kg CH4/h)
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General Material Balance Equation
• A balance on a material in a process system may be written as:
Input + generation - output - consumption = accumulation• The equation may be written for any material that
enters or leaves any process system• It can be applied to the total mass or total moles of this
material or to any atomic species involved in the process
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EXAMPLE: The General Balance Equation• Each year 50,000 people move into a city, 75,000 people
move out, 22,000 are born, and 19,000 die. • Write a balance on the population of the city.• SOLUTION Let P denotes people:• Input + generation - output - consumption = accumulation
• Each year the city's population decreases by 22,000 people.
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• Two types of balances may be written for any system; – differential balances and – integral balances
• Differential balances indicate what is happening in a system at an instant of time. Each term is a rate and has a unit of quantity unit per time
• Integral balances describe what happens between two instant of time. Each term of the equation is an amount of the quantity with a corresponding unit
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Balances on Steady- State Processes
• The process is said to be operating at steady-state when all process variables do not change with time.
• The accumulation term in a balance must equal to zero to ensure that the amount/mass of material in the process do not change with time
STEADY STATE means ACCUMULATION = 0
Input + generation - output - consumption = 0
Input + generation = output + consumption
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• The generation and consumption terms are applied only when chemical reaction is involved
• if there is no reaction, Input =output
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• One thousand kilograms per hour of a mixture of benzene (B) and toluene (T) containing 50% benzene by mass is separated by distillation into two fractions. The mass flow rate of benzene in the top stream is 450 kg B/h and that of toluene in the bottom stream is 475 kg T/h. The operation is at steady state. Write balances on benzene and toluene to calculate the unknown component flow rates in the output streams.
Balances on Steady- State Continuous Processes (Continuous Distillation Process)
450 kg B/hrq1 (kg T/hr)
500 kg B/hr500 kg T/hr
475 kg T/hrq2 (kg B/hr)
Distillation
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450 kg B/hrq1 (kg T/hr)
500 kg B/hr500 kg T/hr
475 kg T/hrq2 (kg B/hr)
Distillation
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no reaction, rate of input = rate of output
EXAMPLE: Balances on a Batch Mixing Process• Two methanol-water mixture are contained in separate
flasks. The first mixture contains 40 wt % methanol, and the second contains 70% methanol. If 200 g of the first mixture are combined with 150 g of the second, what are the mass and composition of the product.
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• Draw a flowchart of the process, using boxes or other symbols to represent process units (reactors, mixers, separation units, etc.) and lines with arrows to represent inputs and outputs.
Flowchart
100 mols/hr C2H6
2000 mols/hr Air
0.21 mol O2/ mol
0.79 mol N2/ mol
2100 mols/hr
0.0476 mol C2H6/ mol
0.200 mol O2/ mol
0.752 mol N2/ mol
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The chart must be fully labeled with values of known variables at the locations of the streams
For example a stream containg 21 mole % O2 and 79%N2 at 320ºC and 1.4 atm flowing at a rate 400 mol/h might be labeled.
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• the total amount or flow rate of the stream and the fractions of each component,
• Or directly as the amount or flow rate of each component.
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• Assign algebraic symbols to unknown streams [such as ṁ (kg solution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write their associated units on the chart
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• If a volumetric flow rate of a stream is given, convert to mass or molar flow rate since balances are not normally written in volumetric quantities
• When labeling component mass or mole fractions of a stream the last one must be 1 minus the sum of the others.
• If you are given that the mass of stream 1 is half that of stream 2, label the masses of these streams m and 2m rather than ml and m2;
• if you know that there is three times as much nitrogen (by mass) in a stream as oxygen, label the mass fractions of O2 and N2 y(g O2/g) and 3y(g N2/g) rather than yl and y2.
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Degree of Freedom Analysis• Draw and label flow chart• Count the unknown variables on the flow chart, nunknowns
• Count the independent equations relating them, nindep eqns
• ndf = nunknowns - nindep eqns
– If ndf = 0, the problem is solvable– If ndf>0, the problem is underspecified, need to provide
more information/equations.– If ndf˂0, the problem is overspecified, more equations
than unknowns, redundant and possibly inconsistent information.
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• An experiment on the growth rate of certain organisms requires an environment of humid air enriched in oxygen. Three input streams are fed into an evaporation chamber to produce an output stream with the desired composition.
• A: Liquid water, fed at a rate of 20.0 cm3/min• B: Air (21 mole% O2, the balance N2)• C: Pure oxygen, with a molar flow rate one-fifth of the molar
flow rate of stream B. The output gas is analyzed and is found to contain 1.5 mole % water. Draw and label a flowchart of the process, and calculate all unknown stream variables.
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Flowchart Scaling and Basis of Calculation
• The procedure of changing the values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions unchanged is referred to as scaling the flow chart.
• Scaling up: if the final stream quantities are larger than the original quantities,
• Scaling down: if they are smaller.
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• Suppose you have balanced a process and the amount or flow rate of one of the process streams is n1.
• You can scale the flowchart to make the amount or flow rate of this stream n2 by multiplying all stream amounts or flow rates by the ratio n2/n1.
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• It is desired to achieve the same separation with a continuous feed of 1250 lb-moles/h. Scale the flowchart accordingly.
• The scale factor is:
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Before scaling
After scaling
Material balance on single unit process
General Procedure for Material Balance Calculations1. Choose as a basis of calculations an amount or flow rate of one of
the process streams2. Draw a flowchart of the process. Include all the given variables on
the chart and label the unknown stream variables on the chart 3. Write the expressions for the quantities requested in problem
statement4. Convert all mass and molar unit quantities to one basis5. Do the degree of freedom analysis. For any given information that
has not been used in labeling the flowchart, translate it into equations in terms of the unknown variables
6. If nDF = 0, write material balance equations in an order such that those involve the fewest unknowns are written first
7. Solve the equations and calculate the additional quantities requested in the problem statement
8. Scale the quantities accordingly 54
• Example :An aqueous solution of NaOH contains 20% NaOH by mass. It is desired to produce an 8% NaOH solution by diluting a stream of 20% solution with a stream of pure water.
• Calculate the ratios (liters H2O/kg feed solution) and (kg product solution/kg feed solution).
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NaOH balance (input = output).
Total mass balance (input = output).
Ratios requested in problem statement.
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