PM3125: Lectures 13 to 15

1
PM3125 Lectures 13 to 15

• Content of Lectures 13 to 17
• Evaporation
• Factors affecting evaporation
• Evaporators
• Film evaporators
• Single effect and multiple effect evaporators
• Mathematical problems on evaporation

Principal reference Chapter 8 in C.J.
Geankoplis, Transport Processes and Unit
Operations, 3rd Edition, Prentice-Hall of India
2
Evaporation

• Suppose that we have a dilute solution of a
  solute (say, sugar) dissolved in a solvent (say,
  water)
• We need to remove part of the solvent (water) to
  produce a concentrated solution
• This can be achieved by heating the solution so
  as to evaporate the solvent (water)
• This process is known as evaporation.
• The industrial equipment used for this purpose
  in known as an evaporator.

3
Types of evaporators
Open kettle or pan evaporator
4
Types of evaporators
Open kettle or pan evaporator
5
Types of evaporators

• Open kettle or pan evaporator
• simplest form of evaporators
• inexpensive
• simple to operate
• very poor heat economy
• in some cases paddles and scrapers for agitation
  are used

6
Types of evaporators
Horizontal-tube evaporator
Vapour
Dilute feed
Steam inlet
Condensate
Concentrated product
7
Types of evaporators

• Horizontal-tube evaporator
• - relatively cheap
• used for non-viscous liquids having high
  heat-transfer coefficients and liquids that do
  not deposit scales
• poor liquid circulation (and therefore
  unsuitable for viscous liquids)

8
Types of evaporators
Vertical-type short-tube evaporator - Liquid is
inside the tubes - Steam condenses outside the
tubes - used for non-viscous liquids having high
heat-transfer coefficients and liquids that do
not deposit scales
9
Types of evaporators
Vertical-type short-tube evaporator
Vapour
Dilute feed
Steam inlet
Condensate
Concentrated product
10
Types of evaporators
Falling-film-type evaporator
11
Types of evaporators
More types are given in additional handouts
uploaded at the course website.
12
Factors effecting evaporation

• Concentration in the liquid
• Liquid feed to an evaporator is relatively
  dilute.
• So its viscosity is low, and heat-transfer
  coefficient high.
• As evaporation proceeds, the solution becomes
  concentrated.
• So viscosity increases and heat-transfer
  coefficient drops.
• Density and the boiling point of solution also
  increase.

13
Factors effecting evaporation

• Solubility
• As solution is heated, concentration of the
  solute in the solution increases.
• In case the solubility limit of the solute in
  solution is exceeded, then crystals may form.
• Solubility of the solute therefore determines
  the maximum concentration of the solute in the
  product stream.
• In most cases, the solubility of the solute
  increases with temperature. This means when a hot
  concentrated solution from an evaporator is
  cooled to room temperature, crystallization may
  occur.

14
Factors effecting evaporation

• Temperature sensitivity of materials
• Pharmaceuticals products, fine chemicals and
  foods are damaged when heated to moderate
  temperatures for relatively short times.
• So special techniques are employed to reduce
  temperature of the liquid and time of heating
  during evaporation

15
Factors effecting evaporation

• Foaming and frothing
• Solutions like organic compounds tend to foam
  and froth during vaporization.
• The foam is carried away along with vapor
  leaving the evaporator.
• Entrainment losses occur.

16
Factors effecting evaporation

• Pressure and temperature
• The boiling point of the solution is related to
  the pressure of the system.
• The higher the operating pressure of the
  evaporator, the higher the temperature at
  boiling.
• Also, as the concentration of the dissolved
  material in solution increases by evaporation,
  the temperature of boiling may rise (a phenomenon
  known as boiling point rise/elevation).
• To keep the temperatures low in heat-sensitive
  materials, it is often necessary to operate under
  atmospheric pressure (that is, under vacuum).

17
Factors effecting evaporation

• Scale deposition
• Some solutions deposit solid materials (called
  scale) on the heating surfaces.
• The result is that the overall heat-transfer
  coefficient (U) may drastically decrease, leading
  to shut down of the evaporators for cleaning
  purposes.

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Factors effecting evaporation

• Materials of construction
• Evaporators are made of some kind of steel.
• However many solutions attack ferrous metals and
  are contaminated by them.
• Copper, nickel, stainless steels can also be
  used.

19
Method of operation of evaporators
Single-effect evaporation

• When a single evaporator is used ,the vapor from
  the boiling liquid is condensed and discarded.
  This is called single effect evaporation.
• It is simple but utilizes steam ineffectively.
• To evaporate 1 kg of water from the solution we
  require 1-1.3 kg of steam.
• Increasing the evaporation per kg of steam by
  using a series of evaporators between the steam
  supply and condenser is called multiple effect
  evaporation

Multiple-effect evaporation
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Calculation methods for single-effect evaporators
Feed F mass flow rate xF mass fraction of
solute in feed TF temperature of feed hF
enthalpy of feed
Vapour leaving the evaporator V mass flow
rate yV mass fraction of solute in vapour T1
temperature of vapour HV enthalpy of vapour
Concentrate leaving the evaporator L mass
flow rate xL mass fraction of solute in
concentrate T1 temperature of concentrate hL
enthalpy of concentrate
Steam S mass flow rate PS steam
pressure TS steam temperature HS enthalpy
of steam hS enthalpy of condensate
P pressure in the evaporator T1 temperature
in the evaporator
21
Calculation methods for single-effect evaporators
Overall material balance F L V
Solute balance F xF L xL V yV
If the vapour is free of solute F xF L xL
Heat balance F hF S HS L hL V HV S hS
Rewriting F hF S (HS - hS) L hL V HV F hF
S ? L hL V HV
where ? HS - hS
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Calculation methods for single-effect evaporators
Energy lost by the steam q S ? S (HS
hS) In case of no energy loss to the
environment, q amount of energy gets transferred
from steam to the solution through the tube wall
of area A and overall heat transfer coefficient
U. Therefore, q U A ?T U A (TS T1)
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Example 1 A continuous single-effect
evaporator concentrates 9072 kg/h of a 1.0 wt
salt solution entering at 38ºC to a final
concentration of 1.5 wt . The vapor space of
the evaporator is at 101.325 kPa (1.0 atm abs)
and the steam supplied is saturated at 150 kPa.
The overall coefficient U 1704 W/m2.K.
Calculate the amounts of vapor and liquid
products and the heat-transfer area required.
Assumed that, since it its dilute, the solution
has the same boiling point as water.
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Calculation methods for single-effect evaporators
Data provided F 9072 kg/h xF 1 wt 0.01
kg solute / kg feed TF 38ºC xL 1.5 wt
0.015 kg solute / kg liquid product P 101.325
kPa (1.0 atm abs) PS 150 kPa U 1704 W/m2.K
T1 saturated temperature at P ( 101.325 kPa)
100ºC
TS saturated temperature at 150 kPa 111.4ºC
25
Calculation methods for single-effect evaporators
Available equations
Data provided F 9072 kg/h xF 0.01 kg solute
/ kg feed TF 38ºC xL 0.015 kg solute / kg
liquid product P 101.325 kPa T1 100ºC PS
150 kPa TS 111.4ºC U 1704 W/m2.K
Overall material balance F L V
Solute balance F xF L xL (no solute in the
vapour)
Heat balance F hF S ? L hL V HV
where ? HS hS
q S ? U A ?T U A (TS T1)
Amounts of vapor and liquid products ? F, xF
and xL are known, and therefore L 6048 kg/h
and V 3024 kg/h
26
Calculation methods for single-effect evaporators
Available equations
Data known F 9072 kg/h L 6048 kg/h, V
3024 kg/h TF 38ºC P 101.325 kPa T1 100ºC
PS 150 kPa TS 111.4ºC U 1704 W/m2.K
Heat balance F hF S ? L hL V HV
where ? HS hS
q S ? U A ?T U A (TS T1)
Heat transfer area A S ? / U (TS T1) ?
S ? L hL V HV F hF (F V) hL V
HV F hF F (hL hF) V (HV hL )
F Cp (T1 - TF) V (Latent heat of
vapourization at 101.325 kPa )
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Calculation methods for single-effect evaporators
Data known F 9072 kg/h L 6048 kg/h, V
3024 kg/h TF 38ºC P 101.325 kPa T1 100ºC
PS 150 kPa TS 111.4ºC U 1704 W/m2.K
Heat transfer area A S ? / U (TS T1) ?
S ? F Cp (T1 - TF) V (Latent heat of
vapourization at 101.325 kPa) F, T1 , TF and V
are already known. Cp 4.14 kJ/kg.K
(assumed) Latent heat of vapourization at 101.325
kPa 2256.7 kJ/kg Therefore, S ? (9072)
(4.14) (100 38) (3024) (2256.7) kJ/h
9152862 kJ/h
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Calculation methods for single-effect evaporators
Data known F 9072 kg/h L 6048 kg/h, V
3024 kg/h TF 38ºC P 101.325 kPa T1 100ºC
PS 150 kPa TS 111.4ºC U 1704 W/m2.K
Heat transfer area A S ? / U (TS T1) ?
S ? 9152862 kJ/h 9152862 1000 / 3600 W T1
and TS are known U 1704 W/m2.K Therefore, A
S ? / U (TS T1) 9152862 1000 / 3600 /
1704 (111.4 100) 130. 9 m2
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Effects of processing variables on evaporator
operation

• Effect of feed temperature
• - The inlet temperature of the feed has a large
  effect on the evaporator operation.
• When feed is not at its boiling point, steam is
  needed first to heat the feed to its boiling post
  and then to evaporate it.
• Preheating the feed can reduce the size of
  evaporator heat-transfer area.

30
Effects of processing variables on evaporator
operation

• Effect of pressure
• Pressure in the evaporator sets the boiling
  point of the solution (T1).
• Steam pressure determines the steam temperature
  (Ts)
• Since q U A (TS T1), larger values of (TS
  T1) will help reduce the heat-transfer area
  needed and hence the cost of evaporator.
• Vacuum can be maintained in the solution side
  using a vacuum pump.
• For example, if the pressure in Example 1 is
  reduced to 41.4 kPa, boiling point of water
  reduces to 349.9 K and that would increase the
  (TS T1) from 10 K to 33.3 K. A large decrease
  in heat-transfer area would be obtained.

31
Effects of processing variables on evaporator
operation

• Effect of steam pressure
• High pressure provides high Ts values, and hence
  TS T1 will increase.
• High pressure steam is however more costly.
• Therefore, overall economic balances must be
  considered to determine the optimum steam
  pressure.

32
Boiling point rise of solutions

• In example 1, the solution is assumed to be
  dilute enough to be considered to have the same
  thermal properties as water. It is not true
  always.
• For concentrated solutions, heat capacity and
  boiling point are quire different from that of
  water.
• Duhrings rule is an empirical law that relates
  the boiling point of a solution to the boiling
  point of the solvent at different pressures for a
  solution of given concentration.

33
Boiling point rise of solutions (an example)

Duhring plot for boiling point of sodium chloride
solutions
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Enthalpy-concentration charts of solutions

See the handout.
35
Example 2 An evaporator is used to concentrate
4536 kg/h of a 20 NaOH solution entering at 60ºC
to a product of 50 solids. The pressure of the
saturated steam used is 170 kPa and the vapor
space pressure of the evaporator is at 12 kPa.
The overall coefficient U is 1560 W/m2.K.
Calculate the steam used, the steam economy (in
kg vapourized / kg steam used) and the heating
surface area.
36
Calculation methods for single-effect evaporators
Data provided F 4536 kg/h xF 20 wt 0.2
kg solute / kg feed TF 60ºC xL 50 wt
0.5 kg solute / kg liquid product P 12 kPa
0.12 bar PS 170 kPa 1.7 bar U 1560 W/m2.K
T1 ? saturated temperature at P ( 0.12 bar)
49.4oC
TS saturated steam temperature at 1.7 bar
115.2oC
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Calculation methods for single-effect evaporators
Amounts of vapor and liquid products ? F, xF
and xL are known, and therefore L 1814 kg/h and
V 2722 kg/h
Steam used ? S ? L hL V HV - F hF
hF enthalpy of 20 solution at 60oC 214 kJ/kg
(using the enthalpyconcentration chart)
hL enthalpy of 50 solution at T1 ? (using
the enthalpyconcentration and boiling-point rise
charts)
Saturated temperature at P ( 0.12 bar)
49.4oC Using the boiling-point rise chart, we get
89.5oC ( read against 49.4oC and 50 wt) as the
boiling point of the solution. That is T1
89.5oC
hL enthalpy of 50 solution at 89.5oC 505
kJ/kg (using the enthalpyconcentration
chart)
HV enthalpy of superheated steam at 89.5oC and
0.12 bar 2667 kJ/kg (using the superheated
steam table)
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Calculation methods for single-effect evaporators
Amounts of vapor and liquid products ? F, xF
and xL are known, and therefore L 1814 kg/h and
V 2722 kg/h
Steam used S ? S ? 1814 x 505 2722 x
2667 4536 x 214 7204940 kJ/h
? latent heat of vapourization of water at 1.7
bar and 115.2oC 2216 kJ/kg (using the steam
table) Therefore S 7204940 / 2216 kg/h 3251
kg/h
Steam economy kg vapourized / kg steam used V
/ S ? Steam economy 2722 / 3251 0.837
A S ? / U (TS T1) 7204940 1000 / 3600
/ 1560 (115.2 89.5) 49. 9 m2
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Example 3 (Repeat Example 2 assuming that the
thermal properties of the liquid in the
evaporator can be approximated by those of
water) An evaporator is used to concentrate
4536 kg/h of a 20 NaOH solution entering at 60ºC
to a product of 50 solids. The pressure of the
saturated steam used is 170 kPa and the vapor
space pressure of the evaporator is at 12 kPa.
The overall coefficient U is 1560 W/m2.K.
Calculate the steam used, the steam economy (in
kg vapourized / kg steam used) and the heating
surface area.
40
Calculation methods for single-effect evaporators
Data provided F 4536 kg/h xF 20 wt 0.2
kg solute / kg feed TF 60ºC xL 50 wt
0.5 kg solute / kg liquid product P 12 kPa
0.12 bar PS 170 kPa 1.7 bar U 1560 W/m2.K
T1 saturated temperature at P ( 0.12 bar)
49.4oC
TS saturated steam temperature at 1.7 bar
115.2oC
41
Calculation methods for single-effect evaporators
Amounts of vapor and liquid products ? F, xF
and xL are known, and therefore L 1814 kg/h and
V 2722 kg/h
Steam used ? S ? L hL V HV - F hF (F
V) hL V HV F hF F (hL hF) V (HV hL )
F Cp (T1 - TF) V (Latent heat of
vapourization at 0.12 bar) (4536)
(4.14) (49.4 60) (2722) (2383) kJ/h (Cp
4.14 kJ/kg.K is assumed) 6301078 kJ/h
? latent heat of vapourization of water at 1.7
bar 2216 kJ/kg Therefore S 6301078 / 2216
kg/h 2843.5 kg/h
Steam economy kg vapourized / kg steam used V
/ S ? Steam economy 2722 / 2843.5 0.957
A S ? / U (TS T1) 6301078 1000 / 3600
/ 1560 (115.2 49.4) 17 m2
42
Calculation methods for single-effect evaporators
Compare the solutions of Examples 2 and 3 and
discuss the importance of considering the boiling
pint rise and enthalpy change of concentrated
solution.
43
Double-effect evaporators
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Calculation methods for double-effect evaporators
If liquid is to be evaporated in each effect,
and if the boiling point of this liquid is
unaffected by the solute concentration, then
writing a heat balance for the first evaporator
q1 U1 A1 ?T1 U1 A1 (TS T1)
Similarly, in the second evaporator, remembering
that the "steam" in the second is the vapour from
the first evaporator and that this will condense
at approximately the same temperature as it
boiled, since pressure changes are small, q2
U2 A2 ?T2 U2 A2 (T1 T2)
If the evaporators are working in balance, then
all of the vapours from the first effect are
condensing and in their turn evaporating vapours
in the second effect. Also assuming that heat
losses can be neglected, there is no appreciable
boiling-point elevation of the more concentrated
solution, and the feed is supplied at its boiling
point,                 q1 q2
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Calculation methods for double-effect evaporators
If the evaporators are working in balance, then
all of the vapours from the first effect are
condensing and in their turn evaporating vapours
in the second effect. Also assuming that heat
losses can be neglected, there is no appreciable
boiling-point elevation of the more concentrated
solution, and the feed is supplied at its boiling
point,                 q1 q2 That is, U1 A1
?T1 U2 A2 ?T2 Further, if the evaporators are
so constructed that A1 A2, the foregoing
equations can be combined.   U2 / U1 ?T1 /
?T2
That is, the temperature differences are
inversely proportional to the overall heat
transfer coefficients in the two effects. This
analysis may be extended to any number of effects
operated in series, in the same way.
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Example 4 Estimate the requirements of steam
and heat transfer surface, and the evaporating
temperatures in each effect, for a triple effect
evaporator evaporating 500 kg h-1 of a 10
solution up to a 30 solution. Steam is
available at 200 kPa gauge and the pressure in
the evaporation space in the final effect is 60
kPa absolute. Assume that the overall heat
transfer coefficients are 2270, 2000 and 1420 J
m-2 s-1 C-1 in the first, second and third
effects, respectively. Neglect sensible heat
effects and assume no boiling-point elevation,
and assume equal heat transfer in each effect.
Source http//www.nzifst.org.nz/unitoperations/
evaporation2.htm