Exercise 1.1
Subject: Separation operations in an ethanol process.
Given: Flow sheet for the process.
Find: Operation in each block of the process
Analysis: The flow sheet is as follows:
Starting from the left side of the top row, number the equipment blocks 1-5. On the next
row, number them from right to left as 6-10.
1. The feed to the reactor contains ethylene, propylene, and water. The very
incomplete reaction effluent of gas contains, in addition:ethyl alcohol, diethyl
ether, acetaldehyde, and isopropyl alcohol.
2. Most of the chemicals, except most of the ethylene, are condensed and sent to
item 4. The remaining gas is sent to 3.
3. The ethylene is recovered from the small quantity of other chemicals in the
absorber. Most of the ethylene is recycled to 1. The water and other chemicals
are sent to 4.
4. Remaining ethylene is flashed off and sent to 5. The liquid contains most of the
hydrocarbons and is sent to 6.
, Exercise 1.1 (continued)
5. The small quantity of ethanol plus other chemicals is absorbed from the ethylene
by water. The exit gas is vented. The liquid is wastewater that will have to be
treated elsewhere.
6. The liquid from 4 is distilled to separate water from ethanol and other chemicals
that are mostly diethyl ether and acetaldehyde. The bottoms is wastewater. The
distillate is the organic chemicals that are sent to 7.
7. This unit is a fixed-bed catalytic reactor to convert acetaldehyde to ethanol. The
reactor effluent is sent to 8.
8. Diethyl ether and small amounts of the ethanol and other chemicals are removed
as vapor by distillation and sent to 9. The bottoms containing most of the ethanol
is sent to 10.
9. The diethyl ether in the vapor from 8 is recovered by absorbing the other
chemicals with water, which is recycled to 4.
10. The crude ethanol from 8 is distilled to obtain the near-azeotrope of ethanol and
water. The bottoms is wastewater.
, Exercise 1.2
Subject: Mixing vs. separation
Given: Thermodynamic principles.
Find: Explanation for why mixing and separation are different.
Analysis: Mixing is a natural, spontaneous process. It may take time, but concentrations
of components in a single fluid phase will tend to become uniform, with an increase in
entropy. By the second law of thermodynamics, a natural process tends to randomness.
The separation of a mixture does not occur naturally or spontaneously. Energy is
required to separate the different molecular species.
Exercise 1.3
Subject: Separation of a mixture requires a transfer of energy to it or the degradation of
its energy.
Given: The first and second laws of thermodynamics.
Find: Explain why the separation of a mixture requires energy.
Analysis: As an example, consider the isothermal minimum (reversible) work of
separation of an ideal binary gas mixture. Therefore, the change in enthalpy is zero.
However, there is a change in entropy, determined as follows. From a chemical
engineering thermodynamics textbook or Table 2.11, Eq. (4):
Wmin n(h T0 s) n(h T0 s)
out in
RT0
RS n y c h
ln yi ,k n j y d i UVW
T
out
k i ,k
in
i, j ln yi , j
It can be shown that regardless of values of y between 0 and 1, that Wmin is always
positive. This minimum work is independent of the process.
, Exercise 1.4
Subject : Use of an ESA or an MSA to make a separations.
Given: Differences between an ESA and an MSA.
Find: State the advantages and disadvantages of ESA and MSA.
Analysis: With an MSA, an additional separator is needed to recover the MSA. Also,
some MSA will be lost, necessitating the need for MSA makeup. If the MSA is
incompletely recovered, a small amount of contamination may result. The use of an
MSA can make possible a separation that cannot be carried out with an ESA. An ESA
separation is easier to design.
Exercise 1.5
Subject : Distillation versus liquid-liquid extraction..
Find: State differences between the two separation operations.
Analysis:
1. Distillation involves both liquid and vapor phases. Extraction involves two liquid
phases
2. Distillation uses an ESA. Extraction uses an MSA.
3. Distillation may occur over a wide range of temperature. Extraction usually
occurs over a narrow range of temperature.
4. Mass transfer in distillation is relatively rapid. Mass transfer in extraction is
relatively slow.
5. Distillation often uses more many stages. Extraction usually involves just several
stages.
6. Distillation seldom requires mechanical agitation. Extraction is best carried out
with mechanical agitation.
, 7.
Exercise 1.6
Subject: Osmotic pressure for the separation of water from sea water by reverse osmosis
with a membrane.
Given: Sea water containing 0.035 g of salt/cm3 of sea water on one side of a membrane
Molecular weight of the salt = 31.5
Temperature = 298 K
Pure water on the other side of a membrane
Find: Minimum required pressure difference in kPa across the membrane
Analysis: The minimum pressure difference across the membrane is equal to the osmotic
pressure of the sea water, since the osmotic pressure of pure water on the other side is
zero. The equation given for osmotic pressure is =RTc/M.
R = 8.314 kPa-m3/kmol-K
T = 298 K
c = 0.035 g/cm3 = 35 kg/m3
M = 31.5 kg/kmol
Minimum pressure difference across a membrane =
8.314 298 35 2, 750 kPa
31.5
Exercise 1.7
Subject: Basic separation techniques.
Given: A list of 10 separation operations.
Find: The basic separation technique for each of the 10.
Analysis:
1. Phase creation: distillation, flash vaporization
2. Phase addition: absorption, adsorption, extraction, stripping
3. Barrier: dialysis, gas permeation, pervaporation, reverse osmosis
, Exercise 1.8
Subject: Methods for removing organic pollutants from wastewater.
Given: Available industrial processes:
(1) adsorption
(2) distillation
(3) liquid-liquid extraction
(4) membrane separation
(5) stripping with air
(6) stripping with steam
Find: Advantages and disadvantages of each process.
Analysis:
Some advantages and disadvantages are given in the following table:
Method Advantages Disadvantages
Adsorption Adsorbents are available. Difficult to recover pollutant.
Best to incinerate it.
Distillation May be practical if pollutant is Impractical is water is more
more volatile. volatile.
L-L extraction Solvent are available. Water will be contaminated with
solvent.
Membrane May be practical if a membrane May need a large membrane area
can be found that is highly if water is the permeate.
selective for pollutant.
Air stripping May be practical if pollutant is Danger of producing a flammable
more volatile. gas mixture.
Steam stripping May be practical if pollutant is Must be able to selectively
more volatile. condense pollutant from
overhead.
With adsorption, can incinerate pollutant, but with a loss of adsorbent.
With distillation, may be able to obtain a pollutant product.
With L-L extraction, will have to separate pollutant from solvent.
With a membrane, may be able to obtain a pollutant product.
With air stripping, may be able to incinerate pollutant.
With steam stripping may be able to obtain a pollutant product.
Subject: Separation operations in an ethanol process.
Given: Flow sheet for the process.
Find: Operation in each block of the process
Analysis: The flow sheet is as follows:
Starting from the left side of the top row, number the equipment blocks 1-5. On the next
row, number them from right to left as 6-10.
1. The feed to the reactor contains ethylene, propylene, and water. The very
incomplete reaction effluent of gas contains, in addition:ethyl alcohol, diethyl
ether, acetaldehyde, and isopropyl alcohol.
2. Most of the chemicals, except most of the ethylene, are condensed and sent to
item 4. The remaining gas is sent to 3.
3. The ethylene is recovered from the small quantity of other chemicals in the
absorber. Most of the ethylene is recycled to 1. The water and other chemicals
are sent to 4.
4. Remaining ethylene is flashed off and sent to 5. The liquid contains most of the
hydrocarbons and is sent to 6.
, Exercise 1.1 (continued)
5. The small quantity of ethanol plus other chemicals is absorbed from the ethylene
by water. The exit gas is vented. The liquid is wastewater that will have to be
treated elsewhere.
6. The liquid from 4 is distilled to separate water from ethanol and other chemicals
that are mostly diethyl ether and acetaldehyde. The bottoms is wastewater. The
distillate is the organic chemicals that are sent to 7.
7. This unit is a fixed-bed catalytic reactor to convert acetaldehyde to ethanol. The
reactor effluent is sent to 8.
8. Diethyl ether and small amounts of the ethanol and other chemicals are removed
as vapor by distillation and sent to 9. The bottoms containing most of the ethanol
is sent to 10.
9. The diethyl ether in the vapor from 8 is recovered by absorbing the other
chemicals with water, which is recycled to 4.
10. The crude ethanol from 8 is distilled to obtain the near-azeotrope of ethanol and
water. The bottoms is wastewater.
, Exercise 1.2
Subject: Mixing vs. separation
Given: Thermodynamic principles.
Find: Explanation for why mixing and separation are different.
Analysis: Mixing is a natural, spontaneous process. It may take time, but concentrations
of components in a single fluid phase will tend to become uniform, with an increase in
entropy. By the second law of thermodynamics, a natural process tends to randomness.
The separation of a mixture does not occur naturally or spontaneously. Energy is
required to separate the different molecular species.
Exercise 1.3
Subject: Separation of a mixture requires a transfer of energy to it or the degradation of
its energy.
Given: The first and second laws of thermodynamics.
Find: Explain why the separation of a mixture requires energy.
Analysis: As an example, consider the isothermal minimum (reversible) work of
separation of an ideal binary gas mixture. Therefore, the change in enthalpy is zero.
However, there is a change in entropy, determined as follows. From a chemical
engineering thermodynamics textbook or Table 2.11, Eq. (4):
Wmin n(h T0 s) n(h T0 s)
out in
RT0
RS n y c h
ln yi ,k n j y d i UVW
T
out
k i ,k
in
i, j ln yi , j
It can be shown that regardless of values of y between 0 and 1, that Wmin is always
positive. This minimum work is independent of the process.
, Exercise 1.4
Subject : Use of an ESA or an MSA to make a separations.
Given: Differences between an ESA and an MSA.
Find: State the advantages and disadvantages of ESA and MSA.
Analysis: With an MSA, an additional separator is needed to recover the MSA. Also,
some MSA will be lost, necessitating the need for MSA makeup. If the MSA is
incompletely recovered, a small amount of contamination may result. The use of an
MSA can make possible a separation that cannot be carried out with an ESA. An ESA
separation is easier to design.
Exercise 1.5
Subject : Distillation versus liquid-liquid extraction..
Find: State differences between the two separation operations.
Analysis:
1. Distillation involves both liquid and vapor phases. Extraction involves two liquid
phases
2. Distillation uses an ESA. Extraction uses an MSA.
3. Distillation may occur over a wide range of temperature. Extraction usually
occurs over a narrow range of temperature.
4. Mass transfer in distillation is relatively rapid. Mass transfer in extraction is
relatively slow.
5. Distillation often uses more many stages. Extraction usually involves just several
stages.
6. Distillation seldom requires mechanical agitation. Extraction is best carried out
with mechanical agitation.
, 7.
Exercise 1.6
Subject: Osmotic pressure for the separation of water from sea water by reverse osmosis
with a membrane.
Given: Sea water containing 0.035 g of salt/cm3 of sea water on one side of a membrane
Molecular weight of the salt = 31.5
Temperature = 298 K
Pure water on the other side of a membrane
Find: Minimum required pressure difference in kPa across the membrane
Analysis: The minimum pressure difference across the membrane is equal to the osmotic
pressure of the sea water, since the osmotic pressure of pure water on the other side is
zero. The equation given for osmotic pressure is =RTc/M.
R = 8.314 kPa-m3/kmol-K
T = 298 K
c = 0.035 g/cm3 = 35 kg/m3
M = 31.5 kg/kmol
Minimum pressure difference across a membrane =
8.314 298 35 2, 750 kPa
31.5
Exercise 1.7
Subject: Basic separation techniques.
Given: A list of 10 separation operations.
Find: The basic separation technique for each of the 10.
Analysis:
1. Phase creation: distillation, flash vaporization
2. Phase addition: absorption, adsorption, extraction, stripping
3. Barrier: dialysis, gas permeation, pervaporation, reverse osmosis
, Exercise 1.8
Subject: Methods for removing organic pollutants from wastewater.
Given: Available industrial processes:
(1) adsorption
(2) distillation
(3) liquid-liquid extraction
(4) membrane separation
(5) stripping with air
(6) stripping with steam
Find: Advantages and disadvantages of each process.
Analysis:
Some advantages and disadvantages are given in the following table:
Method Advantages Disadvantages
Adsorption Adsorbents are available. Difficult to recover pollutant.
Best to incinerate it.
Distillation May be practical if pollutant is Impractical is water is more
more volatile. volatile.
L-L extraction Solvent are available. Water will be contaminated with
solvent.
Membrane May be practical if a membrane May need a large membrane area
can be found that is highly if water is the permeate.
selective for pollutant.
Air stripping May be practical if pollutant is Danger of producing a flammable
more volatile. gas mixture.
Steam stripping May be practical if pollutant is Must be able to selectively
more volatile. condense pollutant from
overhead.
With adsorption, can incinerate pollutant, but with a loss of adsorbent.
With distillation, may be able to obtain a pollutant product.
With L-L extraction, will have to separate pollutant from solvent.
With a membrane, may be able to obtain a pollutant product.
With air stripping, may be able to incinerate pollutant.
With steam stripping may be able to obtain a pollutant product.
