Case Study No 3Case Study No 3 - Multistage Evaporator

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Evaporator

Mohd Sobri Takriff Dept. of Chemical & Process

Engineering UniversiE Kebangsaan Malaysia

EvaporaEon

•  A special case of heat transfer, where a phase change takes place.

•  Concentrate a soluEon consisEng of a nonvolaEle solute and a volaEle solvent – Why??

•  The volaEle solute evaporates leaving a more concentrated soluEon.

•  Example: removal of water from fruit juice, removal of water from dairy product,

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Evaporators

•  a unit operaEon that is used extensively in processing foods, chemicals, pharmaceuEcals, fruit juices, dairy products, paper and pulp, and both malt and grain beverages.

•  a unit operaEon which, with the possible excepEon of disEllaEon, is the most energy intensive

Evaporators

•  Evaporators are used to separate materials based on differences in their boiling temperatures.

•  Its purpose is to concentrate nonvolaEle solutes such as organic compounds, inorganic salts, acids or bases.

•  Examples of solutes are phosphoric acid, causEc soda, sodium chloride, sodium sulfate, gelaEn, syrups and urea.

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Why EvaporaEon •  Reduces transportaEon cost •  Storage costs •  Prepare for the next Unit operaEon –

drying, crystallisaEon etc. •  Reduces deterioraEve chemical reacEons •  BeVer microbiological stability •  Recovery of solvent

vaporization

energy in 2260 kJ

fusion

energy in 334 kJ

condensation

energy out 2260 kJ

solidification

energy out 334 kJ

steam (1 kg)

water (1 kg)

ice (1 kg)

Summary: Change of State

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ice

Liquid water

steam

Water temperature with heating

-100

-50

0

50

100

150

200

0.0E+00 1.0E+06 2.0E+06 3.0E+06 4.0E+06

heat input (J/kg)

tem

pera

ture

(C)

Heating Curve for Water

Water boils

Ice melts

Heat of Fusion •  When the phase change is from solid to

liquid, the sample must absorb heat; when the phase change is from a liquid to solid, the sample must release heat.

•  The heat of transformation for these phase changes is called the heat of fusion, λf.

•  Water: λf = 334 J/g = 79.5 cal/g

q =m ⋅λf

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Heat of Vaporization •  To vaporize a liquid means to change it from the liquid

state to the vapor or gas state. This process requires energy because the molecules must be freed from the liquid state.

•  Condensing a gas to a liquid is the reverse of vaporizing; it requires that energy be removed from the gas so that the molecules can cluster together instead of flying away from each other.

•  The heat of transformation for these phase changes is called the heat of vaporization, λv.

•  Water: λv = 2256 J/g = 539 cal/g

q =m ⋅λv

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Heat Transfer Terminology •  Sensible heat

– Sensible heat is heat exchanged by a body or thermodynamic system that has as its sole effect a change of temperature

•  Latent heat – The quanEty of heat absorbed or released by a substance undergoing a change of state, such as ice changing to water or water to steam, at constant temperature and pressure

qsensible =m•

cPΔT

qlatent =m•

λ

Evaporators Terminology •  SINGLE-‐EFFECT EVAPORATION Single-‐effect evaporaEon occurs when a dilute soluEon is contacted only once with a heat source to produce a concentrated soluEon and an essenEally pure water / solvent vapor discharge.

•  MULTIPLE-‐EFFECT E APORATION MulEple-‐effect evaporaEons use the vapor generated in one effect as the energy source to an adjacent effect. Double-‐ and triple-‐effect evaporators are the most common; however, six-‐effect evaporaEon can be found in the paper industry,

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Evaporators Terminology •  BOILING-‐POINT ELEVATION (BPE) -‐ expresses the difference between the boiling point of a constant composiEon soluEon and the boiling point of pure water at the same pressure.

•  Use Dühring plot to determine the boiling point of the soluEon at various concentraEon

•  For example, pure water boils at 212°F (100°C) at 1 atmosphere, and a 35% sodium hydroxide soluEon boils at about 250°F (121°C) at 1 atmosphere.

Dühring plot

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Evaporators Terminology •  CAPACITY The capacity for an evaporator is measured in

terms of its evaporaEng capability, i.e. Kg of vapor produced per unit Eme. The steam requirements for an evaporaEng train may be determined by dividing the capacity by the economy.

•  ECONOMY This term is a measure of steam use and is expressed in pounds of vapor produced per pound of steam supplied to the evaporator train. –  For a well-‐ designed evaporator system the economy will be about

10% less than the number of effects; thus, for a triple-‐effect evaporator the economy will be roughly 2.7.

Falling Film Evaporator

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Forced CirculaEon Evaporator

Single Effect Evaporator

•  Single-‐effect evaporaEon occurs when a dilute soluEon is contacted only once with a heat source to produce a concentrated soluEon and an essenEally pure water / solvent vapor discharge

•  Heat source is supplied by suitable heaEng media

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Evaporator


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Falling Film Evaporator

Forced CirculaEon Evaporator

Agitated Film Evaporator


mS

•

mC

•

mV

•

mF

•

, xF

m•

, x

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mS

•

mC

•

mV

•

mF

•

, xF

m•

, x

•

mF =mass flow rate of feedxF = solid/solute fraction in feed streamHF = enthalpy of feed stream•

mV =mass flow rate of vaporHV = enthalpy of vapor stream•

m =mass flow rate of productx = solid/solute fraction in product streamH = enthalpy of product stream•

mS =mass flow rate of steamHS = enthalpy of steam•

mC =mass flow rate of condensateHC = enthalpy of condenstae

Mass balance for heaEng media

Mass balance for process fluids

•

mS =•

mC

•

mF =•

m+•

mV

Mass Balance for single effect evaporator

mS

•

mC

•

mV

•

mF

•

, xF

m•

, x

•

mF =mass flow rate of feedxF = solid/solute fraction in feed streamHF = enthalpy of feed stream•

mV =mass flow rate of vaporHV = enthalpy of vapor stream•

m =mass flow rate of productx = solid/solute fraction in product streamH = enthalpy of product stream•

mS =mass flow rate of steamHS = enthalpy of steam•

mC =mass flow rate of condensateHC = enthalpy of condensateλS = latent heat of vaporization of steamqS = rate of heat transfer through heating mediaq = rate of heat transfer from heating surface to liquid

Energy balance for the heaEng media qS =m

•

S HS −HC( ) =m•

S λ S

Energy balance for fluid

q =m•

H +m•

V HV −m•

F H F

In the absence of heat loss:

qS = q

m•

S λ S=m•

H + mF

•

−m•"

#$

%&'HV −m

•

F H F

Energy Balance for single effect evaporator

m•

S λ S=m•

H −HV( )+m•

F H −H F( )

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Example 1

A single effect evaporator is to concentrate 20,000 lbh-‐1 (9700 kgh-‐1) of a 20% soluEon of sodium hydroxide to 50% solids. The gauge pressure of the steam is to be 20 psi and the absolute pressure in the vapor space is to be 100 mm Hg. The overall coefficient is esEmated to be 250 btun-‐2h-‐1oF-‐1 (1400 Wm-‐2oC-‐1. The feed temperature is 100oF (37.38oC). Calculate the amount of steam consumed, the economy and the heaEng surface required.

Example 1 SoluEon Basis 1 lb NaOH Feed: (80/20)= 4 lb of water per lb of NaOH Product: (50/50)= 1.0 lb of water per lb of NaOH Water evaporated: (4-‐1.0)*20000lb/hr*0.20

=8000 lb/hr à capacity

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Example 1 (Ref: Unit OperaEon in Chemical Engineering, McCabe, Smith & Harriot)

Steam consumpEon: •  Boiling Point ElevaEon

–  Boiling point of water at 100mmHg = 124oF (App. 7) –  Boiling point of 50% soluEon = 197oF (figure 16.3) –  BPE = BP50%NAOH – BPwater= 73oF

•  Enthalpy of feed and thick liquor –  Feed 20% solids, 100oF: Hf= 55 btu/lb (fig 16.6) –  Thick liquid 50% solids at 197oF: H=221 btu/lb

Example 1

•  Enthalpy of vapor leaving evaporator, Hv –  Enthalpy of saturated steam at 197oF=1149 Btu/lb (App. 7) –  Heat capacity of vapor at 197oF = 0.45 btu/lboF (App. 14) –  Hv= 1149 btu/lb + (0.45 btu/lboF) (73oF)= 1,149 btu/lb

•  Latent heat of vaporizaEon –  At 20 psi: λs=939 btu/lb (App7)

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Example 1 •  Heat transfer

q = mf −m( )Hv +mH −mfH f

= 20000lbh−1 −8000lbh−1( )(1149btulb−1)+ 8000lbh−1( ) 221btulb−1( )− 20000lbh−1( ) 55btulb−1( )=14, 446, 000btuh−1

m•

s =14, 456, 000btuh−1

939btulb−1=15, 400lbh−1

•  Steam consumpEon:

q =msλs = mf −m( )Hv +mH −mfH f

Example 1 •  Economy

–  pounds of vapor produced per pound of steam supplied

economy = 12,000lbh−1

15, 400lb−1= 0.78

q =UAΔT

A = qUΔT

A = 14, 456, 000btuh−1

250btuft−1 oF−1h−1( ) 259−197( ) oF= 930 ft2

•  HeaEng Surface:

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Example 2 A single effect evaporator is required to concentrate a soluEon from 10% solids to 30% solids at the rate of 250 kg of feed per hour. If the pressure in the evaporator is 77kPa absolute and if the steam is available at 200 kPa gauge, calculate the steam required per hour and the area of heat transfer surface if the overall heat transfer coefficient is 1700 Jm-‐2s-‐1oC-‐1. Assume that the temperature of the feed is 10oC and the boiling point of the soluEon under pressure of 77kPa absolute is 91oC. Assume also that the specific heat of the soluEon is the same as water that is 4.186x103Jkg-‐1oC-‐1, and the latent heat of vaporizaEon of the soluEon is the same as that for water under the same condiEons.

MulEple-‐effect Evaporator •  Water / solvent is boiled in a sequence of vessels, each held at a lower pressure than the last.

•  Because the boiling point of water / solvent decreases as pressure decreases, the vapor boiled off in one vessel can be used to heat the next

•  Generally the first vessel (at the highest pressure) requires an external source of heat

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MulE Effect Evaporator

MulEple Effect Evaporator

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Forward feed Backward feed

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Mixed feed Parallel feed


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MulE-‐Effect Evaporator First Effect

q1 = A1U1ΔT1

Second Effect

q2 = A2U2ΔT2

q1 and q2 are nearly equal, all the heat that is required to create the vapor in the first effect must be given up when the same vapor condenses in the second effect

A1U1ΔT1 = A2U2ΔT2

Same argument may be extended for addiEonal effects

A1U1ΔT1 = A2U2ΔT2 = ......... = AnUnΔTn

MulEple-‐Effect Evaporator

q1 = A1U1ΔT1q2 = A2U2ΔT2q3 = A3U3ΔT3

Consider triple effects evaporator

Total capacity is proporEonal to total rate of heat transfer

q1 + q2 + q3 = A1U1ΔT1 + A2U2ΔT2 + A3U3ΔT3

qT = AU ΔT1 +ΔT2 +ΔT3( ) =UAΔT

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Example 4

EsEmate the requirement of steam, heat transfer surface ad the evaporaEng temperature in each effect for a triple evaporator, evaporaEng 500 kgh-‐1 of 10% soluEon up to a 30% soluEon. Steam is available at 200kPa gauge and the pressure is the evaporaEon space in the final effect is 60kPa. Assume that the overall heat transfer coefficient are 2270, 2000 and 1420Jm-‐2s-‐1oC-‐1 in the first, second and third effects, respecEvely. Neglect sensible heat. Assume no boiling point elevaEon and also equal heat transfer in each effect.

Example 4

Mass balance

Item Solid (kgh-‐1) Liquid (kgh-‐1) Total (kgh-‐1)

Feed 50 450 500

Product 50 117 167

EvaporaEon 333

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Example 4 Energy balance: From steam table the temperature of steam at 200 kPa gauge (300 kPa absolute) is 134oC and latent heat of 2164 kJkg-‐1. EvaporaEng temperature at the final effect under 60kPa absolute is 86oC and since there is no BPE the latent heat is 2294 kJkg-‐1.

Equal heat transfer rate in each effect

q1 = q2 = q3U1A1ΔT1 =U2A2ΔT2 =U3A3ΔT3and

ΔT1 +ΔT2 +ΔT3 = (134oC −86oC) = 48oC

Example 4 Assuming:

Then

A1 = A2 = A3

ΔT2 =U1ΔT1U2

ΔT3 =U1ΔT1U3

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Example 4

ΔT1 1+2270Jm−2s−1 oC−1

2210Jm−2s−1 oC−1 +2270Jm−2s−1 oC−1

1420Jm−2s−1 oC−1

#

$%

&

'(= 48oC

ΔT1 =12.9oC

Thus:

ΔT1 1+U1U2

+U1

U3

"

#$

%

&'= 48oC

Example 4

ΔT2 =U1ΔT1U2

=2270Jm−2s−1 oC−1

2210Jm−2s−1 oC−1 (12.9oC) =14.6 oC

ΔT3 =U1ΔT1U3

=2270Jm−2s−1 oC−1

1420Jm−2s−1 oC−1 (12.9oC) = 20.6 oC

ΔT2 =U1ΔT1U2

ΔT3 =U1ΔT1U3

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Example 4

Effect EvaporaCng ToC Latent heat

1 (134oC-‐12.9oC=121.1oC 2200 kJkg-‐1

2 (121.1oC-‐14.6oC=106.5oC 2240 kJkg-‐1

3 (106.5oC-‐20.6oC=86oC 2294 kJkg-‐1

* Latent heat from steam table

Example 4 Steam consumpEon EquaEng heat transfer rate

m•

1λ1 =m•

2 λ2 =m•

3 λ3 =m•

s λs

m•

1+m•

2+m•

3 = 333kgh−1

m•

2 =m•

1λ1λ2

m•

3 =m•

1λ1λ3

m•

s =m•

1λ1λs

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Example 4 Total mass evaporated

m•

1+m•

2+m•

3 = 333kgh−1

m•

1 1+ λ1λ2+λ1λ3

!

"#

$

%&= 333kgh−1

m•

1 =333kgh−1

1+ 2200kJkg−1

2240kJkg−1+2200kJkg−1

2294kJkg−1"

#$

%

&'

=113kgh−1

Example 4

m•

2 =m•

1λ1λ2

=113kgh−1 2200kJkg−1

2240kJkg−1"

#$

%

&'=111kgh−1

m•

3 =m•

1λ1λ3=113kgh−1 2200kJkg

−1

2294kJkg−1"

#$

%

&'=108kgh−1

m•

s =m•

1λ1λs=113kgh−1 2200kJkg

−1

2164kJkg−1"

#$

%

&'=115kgh−1

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Example 3 Steam consumpEon

•  115kg steam is required to evaporate 333 kg of water

steam consumption = 115kg333kg

= 0.35kgsteam / kg water evaporated

Example 4 Heat exchange surface

q1 =U1A1ΔT1 =m•

1λ1

2270Jm−2s−1 oC−1( ) 12.9 oC( )A1 = 113kgh−1

3600h s

"

#

$$

%

&

''2.2x106 Jkg−1( )

A1 = 2.4m2 = A2 = A3

Total heat exchange surface

AT = A1 + A2 + A3 = 7.2m2

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Stream SpecificaCons

Steam T=300 oC P=120 kPa Flow rate: 600 kg/hr ComposiEon: water= 100%

NaOH SoluEon T=85 oC P=300 kPa Flow rate: 200 kg/hr ComposiEon: water= 90%; NaOH=10%

Heat Exchanger Del P tube = 120 kPa Del P shell = 30 kPa UA=3000

Example

Stream SpecificaCons

Steam T=300 oC P=120 kPa Flow rate: 600 kg/hr ComposiEon: water= 100%

NaOH SoluEon T=85 oC P=300 kPa Flow rate: 200 kg/hr ComposiEon: water= 90%; NaOH=10%

Heat Exchanger Del P tube = 120 kPa Del P shell = 30 kPa UA=3000

Example: MulE-‐effect evaporator

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Example: MulE-‐effect evaporator

E-‐101 V-‐101 E-‐102 V-‐102 Feed

Steam

Condensate

Product

Vapor

Example

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Case Study No 3

1.  To develop an a general algorithm for mulEstage evaporator

2.  To develop a matlab® code for solving the above algorithm

3.  To demonstrate the applicaEon of the algorithm and the matlab® code base on a selected system

4.  To validate the matlab® code by comparing it with iCON® simulaEon

Documents

Case Study No 3Case Study No 3 - Multistage Evaporator