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Ex8 2 Membrane-Processes Solution

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Process Intensification & Hybrid Processes – Ex. 08
CHE.822UF
Process Intensification and
Hybrid Processes
WS 2021/22
Ex. 08: Membrane processes
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3: Gas permeation
β–ͺ Many of the coolants currently in commercial use have a high Global Warming
Potential (GWP). Alternatives with low GWP, such as R-1234yf, are often more
expensive and are therefore mainly used as 1:1 mixture with the commercial
R-134a. To recycle the R-1234yf from the mixture, gas permeation is used,
which takes advantage of the different permeation of the gases for purification.
a) Calculate the max. feed flow rate to achieve a purity of 0.9 with a single membrane
unit.
b) For a process with two units, determine which of the following four arrangements has
the highest recovery, if the membrane area of a unit is half of the area of point a).
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3: Process information
β–ͺ Feed stream:
π‘₯𝐹 = 0.5
𝑝𝐹 = 6 π‘π‘Žπ‘Ÿ
𝐹,ሢ π‘₯𝐹
𝑅,ሢ π‘₯𝑅
β–ͺ Permeate stream:
𝑝𝑃 = 1 π‘π‘Žπ‘Ÿ
β–ͺ Product stream:
𝑃,ሢ π‘₯𝑃
π‘₯π‘ƒπ‘Ÿπ‘œπ‘‘ = 0.9
β–ͺ Membrane:
𝐴 = 3000 π‘š²
π‘šπ‘œπ‘™
𝒫𝑅1234𝑦𝑓 = 0.07
π‘π‘Žπ‘Ÿ βˆ™ π‘š2 βˆ™ β„Ž
π‘šπ‘œπ‘™
𝒫𝑅134π‘Ž = 0.28
π‘π‘Žπ‘Ÿ βˆ™ π‘š2 βˆ™ β„Ž
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3: Assumptions
β–ͺ The difference in partial pressure can be
described by mean logarithmic difference
βˆ†π‘π‘–,𝑖𝑛 + βˆ†π‘π‘–,π‘œπ‘’π‘‘
𝑝ΰ΄₯𝑖 =
2
β–ͺ The permeability is independent of the system
pressure
𝒫𝑖 ≠ 𝒫𝑖 (𝑝)
β–ͺ Pressure loss in system is
neglected
βˆ†π‘π‘™π‘œπ‘ π‘  = 0
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𝐹,ሢ π‘₯𝐹
𝑅,ሢ π‘₯𝑅
𝑃,ሢ π‘₯𝑃
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3: Necessary equations
β–ͺ Total balance
0 = 𝐹ሢ − π‘ƒαˆΆ − π‘…αˆΆ
β–ͺ Recovery
π‘›αˆΆ π‘ƒπ‘Ÿπ‘œπ‘‘ ∗ π‘₯π‘ƒπ‘Ÿπ‘œπ‘‘
π‘Œ=
𝐹ሢ ∗ π‘₯𝐹
β–ͺ Component balance
ሢ 𝐹 − 𝑅π‘₯
ሢ 𝑅 − 𝑃π‘₯
ሢ 𝑃
0 = 𝐹π‘₯
β–ͺ Flowrate through membrane
π‘ƒαˆΆ = ෍ π½π‘–αˆΆ ∗ 𝐴
β–ͺ Flux through membrane
π½π‘–αˆΆ = 𝒫𝑖 ∗ βˆ†π‘π‘–
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β–ͺ Partial pressure (Dalton)
𝑝𝑖 = 𝑝 ∗ π‘₯𝑖
TU Graz I Institute of Chemical Engineering and Environmental Technology
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Calculation procedure
Find the max. allowed feed stream
β–ͺ Establish transport through membrane
β–ͺ Calculate permeate concentration and flow
β–ͺ Establish balances
β–ͺ Calculate max. molar flow of feed
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Necessary membrane area
Establish transport through membrane
π½π‘–αˆΆ = 𝒫𝑖 ∗ βˆ†π‘π‘–
𝐹,ሢ π‘₯𝐹
𝑅,ሢ π‘₯𝑅
β–ͺ Difference in partial pressure is linear
βˆ†π‘π‘–,𝑖𝑛 + βˆ†π‘π‘–,π‘œπ‘’π‘‘
𝑝ΰ΄₯𝑖 =
2
𝑝𝑖,𝐹 + 𝑝𝑖,𝑅
𝑝ΰ΄₯𝑖 =
− 𝑝𝑖,𝑃
2
𝑃,ሢ π‘₯𝑃
β–ͺ Partial pressure according to Dalton
𝑝𝑖 = π‘₯𝑖 ∗ 𝑝
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Necessary membrane area
Establish transport through membrane
π½π‘–αˆΆ = 𝒫𝑖 ∗ βˆ†π‘π‘–
𝐹,ሢ π‘₯𝐹
𝑅,ሢ π‘₯𝑅
β–ͺ Difference in partial pressure is linear
βˆ†π‘π‘–,𝑖𝑛 + βˆ†π‘π‘–,π‘œπ‘’π‘‘
𝑝ΰ΄₯𝑖 =
2
𝑝𝑖,𝐹 + 𝑝𝑖,𝑅
𝑝ΰ΄₯𝑖 =
− 𝑝𝑖,𝑃
2
β–ͺ Partial pressure according to Dalton
𝑝𝑖 = π‘₯𝑖 ∗ 𝑝
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𝑃,ሢ π‘₯𝑃
Transport through membrane
π‘₯ + π‘₯𝑅
ሢ𝐽1 = 𝒫1 ∗ (𝑝𝐹 ∗ 𝐹
− 𝑝𝑃 ∗ π‘₯𝑃 )
2
1 − π‘₯𝐹 + 1 − π‘₯𝑅
ሢ𝐽2 = 𝒫2 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ (1 − π‘₯𝑃 ))
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Necessary membrane area
β–ͺ Calculate permeate concentration and flow
π‘₯𝐹 + π‘₯𝑅
𝐽1ሢ = 𝒫1 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ π‘₯𝑃 )
2
1 − π‘₯𝐹 + 1 − π‘₯𝑅
ሢ𝐽2 = 𝒫2 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ (1 − π‘₯𝑃 ))
2
β–ͺ Adding both fluxes
𝐽 ሢ = 𝐽1ሢ + 𝐽2ሢ
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TU Graz I Institute of Chemical Engineering and Environmental Technology
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Necessary membrane area
β–ͺ Calculate permeate concentration and flow
π‘₯𝐹 + π‘₯𝑅
𝐽1ሢ = 𝒫1 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ π‘₯𝑃 )
2
1 − π‘₯𝐹 + 1 − π‘₯𝑅
ሢ𝐽2 = 𝒫2 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ (1 − π‘₯𝑃 ))
2
β–ͺ Adding both fluxes
𝐽 ሢ = 𝐽1ሢ + 𝐽2ሢ
𝐽 ሢ = 𝑝𝐹 ∗ 𝒫1 ∗
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π‘₯𝐹 + π‘₯𝑅
π‘₯𝐹 + π‘₯𝑅
+ 𝒫2 1 −
2
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− 𝒫2 ∗ 𝑝𝑃 − 𝒫1 − 𝒫2 ∗ 𝑝𝑃 ∗ π‘₯𝑃
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Necessary membrane area
β–ͺ Calculate permeate concentration and flow
π‘₯𝐹 + π‘₯𝑅
𝐽1ሢ = 𝒫1 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ π‘₯𝑃 )
2
1 − π‘₯𝐹 + 1 − π‘₯𝑅
ሢ𝐽2 = 𝒫2 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ (1 − π‘₯𝑃 ))
2
β–ͺ Adding both fluxes
𝐽 ሢ = 𝐽1ሢ + 𝐽2ሢ
𝐡 = 0.000518
𝐽 ሢ = 𝐡 + 𝑐 ∗ π‘₯𝑃
𝑐 = 0.00021
β–ͺ Introducing it into one of the flux equations
π‘₯𝐹 + π‘₯𝑅
2
𝐡 ∗ π‘₯𝑃 + 𝑐 ∗ π‘₯𝑃 = 𝒫1 ∗ 𝑝𝐹 ∗
− 𝒫1 ∗ 𝑝𝑃 ∗ π‘₯𝑃
2
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Necessary membrane area
β–ͺ Introducing it into one of the flux equations
π‘₯𝐹 + π‘₯𝑅
2
𝐡 ∗ π‘₯𝑃 + 𝑐 ∗ π‘₯𝑃 = 𝒫1 ∗ 𝑝𝐹 ∗
− 𝒫1 ∗ 𝑝𝑃 ∗ π‘₯𝑃
2
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 a) Necessary membrane area
π‘₯𝑃 = 0.475
β–ͺ Calculate the permeate flow
𝐽1ሢ
𝐴
π‘₯𝐹 + π‘₯𝑅
ሢ
𝑃=𝐴∗
=
∗ 𝒫1 ∗ (𝑝𝐹 ∗
− 𝑝𝑃 ∗ π‘₯𝑃 )
π‘˜π‘šπ‘œπ‘™
ሢ
𝑃 = 1.65
π‘₯𝑃 π‘₯𝑃
2
β„Ž
Establish the balance
0 = 𝐹ሢ − π‘ƒαˆΆ − π‘…αˆΆ
ሢ 𝐹 − 𝑅π‘₯
ሢ 𝑅 − 𝑃π‘₯
ሢ 𝑃
0 = 𝐹π‘₯
β–ͺ Rearrange the balances
π‘ƒαˆΆ ∗ (π‘₯𝑃 − π‘₯𝑅 )
𝐹ሢ =
Purity
0.9
π‘₯𝐹 −π‘₯𝑅
π‘˜π‘šπ‘œπ‘™
𝐹ሢ = 1.75
β„Ž
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Yield
0.103
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
Parallel
Serial Permeate
β–ͺ Membrane:
𝐴 = 1500 π‘š²
Serial Retentate
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Recycle Permeate
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Calculation procedure
Determine the recovery
β–ͺ Establish balances
β–ͺ Establish transport through membrane
β–ͺ Calculate streams
β–ͺ Determine recovery
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: parallel
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: parallel
β–ͺ Establish blalances
0 = 𝐹ሢ − π‘ƒαˆΆ − π‘…αˆΆ
ሢ 𝐹 − 𝑅π‘₯
ሢ 𝑅 − 𝑃π‘₯
ሢ 𝑃
0 = 𝐹π‘₯
β–ͺ Establish transport through membrane
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π‘ƒαˆΆ =
𝐴
π‘₯𝑃
π‘ƒαˆΆ =
𝐴
1−π‘₯𝑃
∗ 𝒫1 ∗ 𝑝𝐹 ∗
π‘₯𝐹 +π‘₯𝑅
2
…(1)
…(2)
− 𝑝𝑃 ∗ π‘₯𝑃
…(3)
π‘₯𝐹 +π‘₯𝑅
)
2
…(4)
∗ 𝒫2 ∗ 𝑝𝐹 ∗ (1 −
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− 𝑝𝑃 ∗ 1 − π‘₯𝑃
4 unknowns
4 equations
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: parallel
β–ͺ Same process as with point a)
β–ͺ Rearrange and insert the equations
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Purity
0.982
Yield
0.050
TU Graz I Institute of Chemical Engineering and Environmental Technology
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: parallel
β–ͺ Same process as with point a) (1) into (2)
ሢ 𝐹 − 𝑃π‘₯
ሢ 𝑃
𝐹π‘₯
π‘₯𝑅 =
𝐹ሢ − π‘ƒαˆΆ
β–ͺ Eleminating π‘ƒαˆΆ in equation (3)&(4)
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: serial permeate
p=3 bar
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: serial permeate
β–ͺ Same process with all shemas
Purity
Yield
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0.634
0.891
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: serial retentate
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: serial retentate
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Purity
0.760
Yield
0.491
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: recycle permeate
p=3 bar
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
β–ͺ Four different shemas: recycle permeate
β–ͺ Solved in an iterative way
Purity
Yield
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0.637
0.8837
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-3 b) Comparing process shemas
Parallel
Serial Permeate
Purity
Yield
0.982
Purity
0.050
Yield
Serial Retentate
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0.634
0.891
Recycle Permeate
Purity
0.76
Purity
Yield
0.491
Yield
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0.637
0.8837
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4: Pervaporation
β–ͺ Pervaporation is used to overcome the azeotropic point of ethanol and water
and produce pure ethanol. The distillate coming from a rectification column with
a mass flow of 30 kmol/h and a purity of 82% is fed into a pervaporation cell.
There it will be refined to a purity of 99.9 % ethanol.
a) Calculates the stream of pure ethanol gained by the secondary treatment.
b) How high is the energy demand of the membrane unit to keep all streams at
the same constant temperature?
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 Further information
Process:
β–ͺ Temperature
πœ— = 78 °πΆ
β–ͺ Distillate pressure
𝑝 = 1.0 π‘π‘Žπ‘Ÿ
β–ͺ Permeate pressure
𝑝 = 0.1 π‘π‘Žπ‘Ÿ
Membrane:
β–ͺ Membrane permeability
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𝑅,ሢ π‘₯𝑅
𝐷,ሢ π‘₯𝐷
𝑃,ሢ π‘₯𝑃
kmol
β–ͺ
Ethanol
𝑃1 = 0.00003 barΞ‡m2Ξ‡h
β–ͺ
Water
𝑃2 = 0.0016 barΞ‡m2Ξ‡h
kmol
cp
Δhvap78°C
Ethanol
2.65 kj / kgK
851 kj / kg
Antoineparameters
Water
4.19 kj / kgK
2308 kj / kg
(bar, °K, ln)
Substance data
π‘„αˆΆ
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A
B
C
12.2917
3803.98
- 41.68
11.6839
3816.44
- 46.13
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 Assumptions
β–ͺ Raoult's and Dalton's laws may be used for
the calculation of partial pressures.
𝑝𝑖 = 𝑝𝑖𝑆 ∗ π‘₯𝑖
𝑝𝑖 = 𝑝 ∗ 𝑦𝑖
β–ͺ The difference in partial pressure can be
assumed to be linear
βˆ†π‘π‘–,𝑖𝑛 + βˆ†π‘π‘–,π‘œπ‘’π‘‘
𝑝ΰ΄₯𝑖 =
2
β–ͺ The enthalpy of the mixture can be calculated
by adding up the pure substance enthalpies.
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π‘„αˆΆ
𝐷,ሢ π‘₯𝐷
𝑃,ሢ π‘₯𝑃
β„Žπ‘šπ‘–π‘₯ = ෍ π‘₯𝑖 ∗ β„Žπ‘–
𝑖
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𝑅,ሢ π‘₯𝑅
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 Necessary equations
β–ͺ Total balance
0 = 𝐹ሢ − π‘ƒαˆΆ − π‘…αˆΆ
β–ͺ Component balance
ሢ 𝐹 − 𝑅π‘₯
ሢ 𝑅 − 𝑃π‘₯
ሢ 𝑃
0 = 𝐹π‘₯
β–ͺ Antoine equation
𝑝𝑖𝑆
=
𝐡
𝐴−𝐢+𝑇
𝑒
β–ͺ Energy balance
ሢ 𝐹 − π‘…β„Ž
ሢ 𝑅 − π‘ƒβ„Ž
ሢ 𝑃 + π‘„αˆΆ
0 = πΉβ„Ž
β–ͺ Flowrate through membrane
π‘ƒαˆΆ = ෍ 𝒫𝑖 ∗ βˆ†π‘π‘– ∗ 𝐴
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 a) Calculation procedure
Stream of pure Ethanol (99.9%)
β–ͺ Establish transport through membrane
β–ͺ Calculate permeate concentration and flow
β–ͺ Establish balances
β–ͺ Calculate the mass flow of pure Ethanol
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Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 a) Flowrate pure ethanol
Establish transport through membrane
π‘ƒπ‘–αˆΆ = ෍ 𝒫𝑖 ∗ βˆ†π‘π‘– ∗ 𝐴
π‘„αˆΆ
𝐷,ሢ π‘₯𝐷
β–ͺ Introducing the linear difference in partial
pressure
𝑝𝑖,𝐹 − 𝑝𝑖,𝑃 + 𝑝𝑖,𝑅 − 𝑝𝑖,𝑃
π‘ƒπ‘–αˆΆ = 𝒫𝑖 ∗
∗𝐴
2
β–ͺ Introducing Dalton`s and Raoult`s law
𝑃,ሢ π‘₯𝑃
𝑝𝑖𝑆 ∗ π‘₯𝑖,𝐹 + 𝑝𝑖𝑆 ∗ π‘₯𝑖,𝑅 − 2 ∗ 𝑝𝑃 ∗ π‘₯𝑖,𝑃
π‘ƒπ‘–αˆΆ = 𝒫𝑖 ∗
∗𝐴
2
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𝑅,ሢ π‘₯𝑅
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 a) Flowrate pure ethanol
P_1=........ = P/xp
1)
P_2=........= P/1-xp
2)
Calculate permeate concentration and flow
β–ͺ Dividing the two permeate flows
π‘₯𝑃
𝒫1 ∗ (𝑝1𝑆 ∗ π‘₯𝐹 + 𝑝1𝑆 ∗ π‘₯𝑅 − 2 ∗ 𝑝 ∗ π‘₯𝑃 )
=
1 − π‘₯𝑃 𝒫2 ∗ (𝑝2𝑆 ∗ (1 − π‘₯𝐹 ) + 𝑝2𝑆 ∗ (1 − π‘₯𝑅 ) − 2 ∗ 𝑝 ∗ (1 − π‘₯𝑃 ))
Dividing P_1/P_2 =
(P/xp)/(P/1-xp) = (1-xp)/(xp) = ......
β–ͺ Rearranging the Equation
π‘₯𝑃 ∗ 𝒫2 ∗ 𝑝2𝑆 ∗ 1 − π‘₯𝐹 + 𝑝2𝑆 ∗ 1 − π‘₯𝑅 − 2 ∗ 𝑝 ∗ 1 − π‘₯𝑃
= 1 − π‘₯𝑃 ∗ 𝒫1 ∗ 𝑝1𝑆 ∗ π‘₯𝐹 + 𝑝1𝑆 ∗ π‘₯𝑅 − 2 ∗ 𝑝 ∗ π‘₯𝑃
2 ∗ 𝒫2 ∗ 𝑝 ∗ π‘₯𝑃2 + 𝒫2 ∗ 𝑝2𝑆 ∗ 1 − π‘₯𝐹 + 𝑝2𝑆 ∗ 1 − π‘₯𝑅 − 2 ∗ 𝑝 ∗ π‘₯𝑃
= 2 ∗ 𝒫1 ∗ 𝑝 ∗ π‘₯𝑃2 − 𝒫1 ∗ 𝑝1𝑆 ∗ π‘₯𝐹 + 𝑝1𝑆 ∗ π‘₯𝑅 + 2 ∗ 𝑝 ∗ π‘₯𝑃 ∗ π‘₯𝑃 + 𝒫1
∗ 𝑝1𝑆 ∗ π‘₯𝐹 + 𝑝1𝑆 ∗ π‘₯𝑅
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TU Graz I Institute of Chemical Engineering and Environmental Technology
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 a) Flowrate pure ethanol
Calculate permeate concentration and flow
Quadratic equation
𝑆
𝑆
2
2 ∗ (𝒫2 − 𝒫1 ) ∗ 𝑝 ∗ π‘₯𝑃 + (𝒫1 ∗ 𝑝1 ∗ π‘₯𝐹 + 𝑝1 ∗ π‘₯𝑅 + 2 ∗ 𝑝 + 𝒫2
∗ 𝑝2𝑆 ∗ 1 − π‘₯𝐹 + 𝑝2𝑆 ∗ 1 − π‘₯𝑅 − 2 ∗ 𝑝 ) ∗ π‘₯𝑃 − 𝒫1 ∗ 𝑝1𝑆 ∗ π‘₯𝐹 + 𝑝1𝑆 ∗ π‘₯𝑅 = 0
β–ͺ Calculate the vapor pressure with Antoine
𝑝𝑖𝑆
=𝑒
𝐡
𝐴−𝐢+𝑇
Ethanol
0.9998 bar
Water
0.4369 bar
π‘₯𝑃 = 0.679
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TU Graz I Institute of Chemical Engineering and Environmental Technology
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 a) Flowrate pure ethanol
Establish balances
β–ͺ mass balance
0 = 𝐹ሢ − π‘ƒαˆΆ − π‘…αˆΆ
β–ͺ Component balance
ሢ 𝐹 − 𝑅π‘₯
ሢ 𝑅 − 𝑃π‘₯
ሢ 𝑃
0 = 𝐹π‘₯
𝐹ሢ ∗ (π‘₯𝐹 − π‘₯𝑃 )
π‘…αˆΆ =
π‘₯𝑅 − π‘₯𝑃
35
TU Graz I Institute of Chemical Engineering and Environmental Technology
π‘˜π‘šπ‘œπ‘™
π‘…αˆΆ = 13.22
β„Ž
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 b) Calculation procedure
Calculate energy demand of unit
β–ͺ Establish heat balance
β–ͺ Calculate all enthalpies
β–ͺ Calculate the energy demand
36
TU Graz I Institute of Chemical Engineering and Environmental Technology
Graz, 16.12.2021
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 b) Energy demand heating
β–ͺ Establish heat balance
ሢ 𝐹 − π‘…β„Ž
ሢ 𝑅 − π‘ƒβ„Ž
ሢ 𝑃 + π‘„αˆΆ
0 = πΉβ„Ž
β–ͺ Determine all the enthalpies
β„ŽπΉ = 𝑐𝑝,1 ∗ 𝑀1 ∗ π‘₯𝐹 + 𝑐𝑝,2 ∗ 𝑀2 ∗ 1 − π‘₯𝐹
π‘„αˆΆ
𝐷,ሢ β„Žπ·
∗πœ—
𝑃,ሢ β„Žπ‘ƒ
β„Žπ‘… = 𝑐𝑝,1 ∗ 𝑀1 ∗ π‘₯𝑅 + 𝑐𝑝,2 ∗ 𝑀2 ∗ 1 − π‘₯𝑅
∗πœ—
78°πΆ
β„Žπ‘ƒ = πœ— ∗ 𝑐𝑝,1 + Δh78°πΆ
∗
𝑀
∗
π‘₯
+
πœ—
∗
𝑐
+
Δh
1
𝑃
𝑝,2
π‘£π‘Žπ‘,1
π‘£π‘Žπ‘,2 ∗ 𝑀2 ∗ 1 − π‘₯𝑃
37
TU Graz I Institute of Chemical Engineering and Environmental Technology
Graz, 16.12.2021
𝑅,ሢ β„Žπ‘…
Process Intensification & Hybrid Processes – Ex. 08
Ex. 08-4 b) Energy demand heating
hF
8502.66 kj/kmol
hR
9504.57 kj/kmol
hP
48260.19 kj/kmol
β–ͺ Calculate the energy demand
π‘„αˆΆ
𝐷,ሢ β„Žπ·
𝑃,ሢ β„Žπ‘ƒ
ሢ 𝑅 + π‘ƒβ„Ž
ሢ 𝑃 − πΉβ„Ž
ሢ 𝐹
π‘„αˆΆ = π‘…β„Ž
π‘„αˆΆ = 188.99 kW
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TU Graz I Institute of Chemical Engineering and Environmental Technology
Graz, 16.12.2021
𝑅,ሢ β„Žπ‘…
Process Intensification & Hybrid Processes – Ex. 08
CHE.822UF
Process Intensification and
Hybrid Processes
WS 2021/22
Ex. 08: Membrane processes
39
TU Graz I Institute of Chemical Engineering and Environmental Technology
Graz, 16.12.2021
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