CE 408 - SENS · 2021. 3. 5. · 7 Multi-Effect Evaporation Energy Evaluation • Assume all of the heat from Steam that enters 1st effect goes to latent heat of the “Steam” coming

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Chemical Engineering Plant Design

Lecture 10 Multi-Effect Evaporation

Instructor: David Courtemanche

CE 408

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Multi-Effect Evaporation

Single Effect Evaporation

• In single effect evaporation the energy from the steam vaporizes solvent from the liquor

• The steam is then discharged as condensate as the steam condenses.

• This is quite inefficient and leads to low Economy

• Economy is defined as pound of solvent vaporized per pound of steam fed to

evaporator

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• In multi-effect the vapor from the tube-side of of Effect 1 is fed to the shell side of Effect 2

• The energy is not discarded, but is used to vaporize more solvent from the liquor in Effect 2!

• The tube-side vapor from Effect 2 goes on as the shell-side energy source for Effect 3…

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• In order to maintain a thermal driving force (DT) we must maintain PS > P1 > P2 > P3

• This allows the liquor boiling temperature for each successive effect to be lower than the

temperature of the vapor coming off of the tube-side of the previous effect

• Note the control valves F1, F2, F3 which (along with frictional pressure drop throught the

tubes) maintain the pressure variation from effect to effect

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• A Multiple Effect Evaporator is really just a series of Single Effect Evaporators connected

together in a train

• Typically they are all identical

• It is more cost effective to specify multiple identical units

• Saves on design

• Saves on spare parts

• Identical units will have similar maintenance needs

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Energy Evaluation

• First Effect 𝒒𝟏 = 𝑨 𝑼𝟏∆𝑻

• Ignore heat going to change temperature of the feed liquor

• It is small compared to the energy that goes to phase change

• Assume that the temperature of the Condensate leaving the 2nd effect is

approximately 𝑻𝟏, the temperature of the boiling vapor from 1st effect

• Meaning that the vapor coming off of the 1st effect tube-side condenses but

does not subcool

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Energy Evaluation

• Assume all of the heat from Steam that enters 1st effect goes to latent heat of the “Steam”

coming off the tube side of the 1st effect and enters the shell side of the 2nd effect

• Because the “steam” of the 2nd effect shell side all condenses, it is transferring the same

amount of energy to the liquor in the 2nd effect as it picked up from the steam that went

though the 1st effect shell side

• The same assumptions apply to the 2nd and 3rd effects

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Energy Evaluation

• Because the heat is transferred from one effect to the next:

𝒒𝟏 ≅ 𝒒𝟐 ≅ 𝒒𝟑

𝑨𝟏 𝑼𝟏 ∆𝑻𝟏 ≅ 𝑨𝟐 𝑼𝟐 ∆𝑻𝟐 ≅ 𝑨𝟑 𝑼𝟑 ∆𝑻𝟑

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Energy Evaluation

• Because you generally design all of the effects to be identical

𝑨𝟏 = 𝑨𝟐 = 𝑨𝟑

𝑼𝟏 ∆𝑻𝟏 = 𝑼𝟐 ∆𝑻𝟐 = 𝑼𝟑 ∆𝑻𝟑 =𝒒

𝑨

• Remember that 𝒒𝟏 ≅ 𝒒𝟐 ≅ 𝒒𝟑 = 𝐪

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Energy Evaluation

• Are 𝑼𝟏 = 𝑼𝟐 = 𝑼𝟑? No, net necessarily

• Differing viscosities and compositions of the varying concentrations can affect the 𝑼𝑶

of each effect

• With all of this extra surface area we must have a large increase in capacity!

NO! No necessarily…

• Let’s assume 𝑼𝟏 = 𝑼𝟐 = 𝑼𝟑

𝒒𝒕𝒐𝒕𝒂𝒍 = 𝑼 𝑨 ∆𝑻𝟏 + ∆𝑻𝟐 + ∆𝑻𝟑 = 𝑼 𝑨 ∆𝑻𝒕𝒐𝒕𝒂𝒍

• Capacity is determined ~ 𝒒𝒕𝒐𝒕𝒂𝒍

• For the same incoming steam pressure and same final vapor pressure, ∆𝑻𝒕𝒐𝒕𝒂𝒍 is the

same no matter how many effects one has!

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Why have Multiple Effects if you don’t get extra capacity?

• Steam Economy

• Lower Energy Cost

• You are not discarding the enthalpy of the vapor you are boiling off

Balance of Energy Savings versus Increased Capital Cost

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• There are many different possible configurations…

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• Multi-Effect may increase capacity if the heat transfer coefficient U is increased for lower

concentration

• More surface area for the low concentration and therefore more heat transfer

• Boiling Point Elevation is more of a concern for Multi-Effect units

• 1 Effect ∆𝑇𝑡𝑜𝑡𝑎𝑙 = 105 ℉• 2 Effects ∆𝑇𝑡𝑜𝑡𝑎𝑙 = 85 ℉• 3 Effects ∆𝑇𝑡𝑜𝑡𝑎𝑙 = 79 ℉

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Design Steps

• You know the steam temperature and the Final (3rd Effect) Temperature

• Based on the operating pressures

• Assume values for the boiling temperatures in the 1st and 2nd effects

• Do the enthalpy balances for rates of steam/vapor/liquor from effect to effect

• Calculate the required heating surface area for each effect

• If the areas are not approximately equal then you must estimate new boiling

temperatures and iterate

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Evaporation

Example Calculation

• A solution of organic colloids is to be concentrated from 10 to 50 % solids

• Feed to evaporator is 55,000 lb/hr thin liquor

• Steam is available at 15 psig (temperature is 249 F)

• A pressure of 4 inch Hg absolute is maintained in the vapor space the third effect

• Water boils at 125 F at this pressure 𝑻 = 𝟏𝟐𝟓℉• Solution has negligible elevation of boiling point and negligible heat of dilution effects

• Temperature of feed is 𝑻𝑭 = 𝟕𝟎℉

• Specific heat of liquor is 𝑪𝒑𝒇 = 𝟏. 𝟎𝑩𝑻𝑼

𝒍𝒃𝒎 °𝑭at all concentrations

• Latent heat of condensation at 15 psig: 𝝀𝑺 = 𝟗𝟒𝟔𝑩𝑻𝑼

𝒍𝒃𝒎

• Calculate the required

• Steam consumption

• Economy

• Heat transfer surface

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Evaporation

Example Calculation, Solution

• Feed contains Τ9010 ൗ𝑙𝑏𝑚𝑤𝑎𝑡𝑒𝑟

𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠 = 9 ൗ𝑙𝑏𝑚𝑤𝑎𝑡𝑒𝑟𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠

• Thick liquor contains Τ5050 ൗ𝑙𝑏𝑚𝑤𝑎𝑡𝑒𝑟

𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠 = 1 ൗ𝑙𝑏𝑚𝑤𝑎𝑡𝑒𝑟𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠

• There are 0.10 ൗ𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠𝑙𝑏𝑚 𝑡ℎ𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟 ∗ 55,000 ൗ𝑙𝑏𝑚 𝑡ℎ𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟

ℎ𝑟 = 5,500 ൗ𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠ℎ𝑟

• The amount of water that is to be removed is

9 − 1 ൗ𝑙𝑏𝑚𝑤𝑎𝑡𝑒𝑟

𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠 ∗ 5,500 ൗ𝑙𝑏𝑚 𝑠𝑜𝑙𝑖𝑑𝑠ℎ𝑟 = 44,000 ൗ𝑙𝑏𝑚𝑤𝑎𝑡𝑒𝑟

ℎ𝑟

• The flow rate of thick liquor is 55,000 ൗ𝑙𝑏𝑚ℎ𝑟 − 44,000 ൗ𝑙𝑏𝑚

ℎ𝑟 = 11,000 ൗ𝑙𝑏𝑚ℎ𝑟

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• The process described on the previous two slides was solved in Lecture 09 as a single effect evaporator

• Now we will look at this a triple effect evaporator

• Overall heat transfer coefficients are 𝑼𝟏 = 𝟓𝟓𝟎𝑩𝑻𝑼

𝒇𝒕𝟐𝒉𝒓 °𝑭, 𝑼𝟐 = 𝟑𝟓𝟎

𝑩𝑻𝑼

𝒇𝒕𝟐𝒉𝒓 °𝑭, 𝑼𝟑 = 𝟐𝟎𝟎

𝑩𝑻𝑼

𝒇𝒕𝟐𝒉𝒓 °𝑭

• Flow rate of feed, thin liquor: ሶ𝑚𝑓

• Flow rate of steam: ሶ𝑚𝑆

• Flow rate of liquor out of stage I: ሶ𝑚1

• Flow rate of liquor out of stage II: ሶ𝑚2

• Flow rate of liquor out of stage III: ሶ𝑚3

• Due to low elevation of boiling points

assume that boiling temperature in the

tube side of one effect equals the

condensing temperature in the shell side

of the following effect

• The latent heat of condensation in shell

side of each effect equals latent heat of

vaporization in the tube side of previous

effect


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• Heat Balance Equations𝒒𝟏 = ሶ𝒎𝑺 𝝀𝑺 = ሶ𝒎𝒇 − ሶ𝒎𝟏 𝝀𝟏 + ሶ𝒎𝒇 𝒄𝒑𝒇 𝑻𝟏 − 𝑻𝒇

𝒒𝟐 = ሶ𝒎𝒇 − ሶ𝒎𝟏 𝝀𝟏 = ሶ𝒎𝟏 − ሶ𝒎𝟐 𝝀𝟐 + ሶ𝒎𝟏 𝒄𝒑𝟏 𝑻𝟐 − 𝑻𝟏𝒒𝟐 = ሶ𝒎𝟏 − ሶ𝒎𝟐 𝝀𝟐 = ሶ𝒎𝟐 − ሶ𝒎𝟑 𝝀𝟑 + ሶ𝒎𝟐 𝒄𝒑𝟐 𝑻𝟑 − 𝑻𝟐

• Notice that the “vapor” from one effect is the “steam” in the next one

• The total temperature drop will be 249 F – 125 F = 124 F

• Looking at the relative heat transfer coefficients, we will

estimate

• ∆𝑇1 = 38 𝐹• ∆𝑇2 = 33 𝐹• ∆𝑇3 = 53 𝐹

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• Heat Balances (all cp = 1.0)

• 𝒒𝟏 = 𝟗𝟒𝟔 ∗ ሶ𝒎𝑺 = 𝟗𝟕𝟏 ∗ 𝟓𝟓, 𝟎𝟎𝟎 − ሶ𝒎𝟏 + 𝟓𝟓, 𝟎𝟎𝟎 ∗ 𝟐𝟏𝟏 − 𝟕𝟎• 𝒒𝟐 = 𝟗𝟕𝟏 ∗ 𝟓𝟓, 𝟎𝟎𝟎 − ሶ𝒎𝟏 = 𝟗𝟗𝟏 ∗ ሶ𝒎𝟏 − ሶ𝒎𝟐 + ሶ𝒎𝟏 ∗ 𝟏𝟕𝟖 − 𝟐𝟏𝟏• 𝒒𝟐 = 𝟗𝟗𝟏 ∗ ሶ𝒎𝟏 − ሶ𝒎𝟐 = 𝟏𝟎𝟐𝟐 ∗ ሶ𝒎𝟐 − 𝟏𝟏, 𝟎𝟎𝟎 + ሶ𝒎𝟐 ∗ 𝟏𝟐𝟓 − 𝟏𝟕𝟖

• The last two equations can be solved to give:

• ሶ𝒎𝟏 = 𝟒𝟏, 𝟑𝟖𝟎 𝒍𝒃/𝒉𝒓• ሶ𝒎𝟐 = 𝟐𝟔, 𝟔𝟔𝟎 𝒍𝒃/𝒉𝒓

• Then the first equation gives

• ሶ𝒎𝑺 = 𝟐𝟐, 𝟏𝟖𝟎 𝒍𝒃/𝒉𝒓

Temperature, F 𝝀 (BTU/pound)

Steam 249 = TS 946 = 𝜆𝑆

Feed to I 70 = Tf

Liquor in I and steam to II 249 – 38 = 211 = T1 971 = 𝜆1

Liquor in II and steam to III 211 – 33 = 178 = T2 991 = 𝜆2

Liquor in III 178 – 53 = 125 = T3 1022 = 𝜆3

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• Solving the heat balances with the calculated mass flows:• 𝒒𝟏 = 𝟗𝟒𝟔 ∗ 𝟐𝟐, 𝟏𝟖𝟎 = 𝟐𝟎, 𝟗𝟖𝟎, 𝟎𝟎𝟎 𝑩𝑻𝑼/𝒉𝒓• 𝒒𝟐 = 𝟗𝟕𝟏 ∗ 𝟓𝟓, 𝟎𝟎𝟎 − 𝟒𝟏, 𝟑𝟖𝟎 = 𝟏𝟑, 𝟐𝟐𝟓, 𝟎𝟎𝟎 𝑩𝑻𝑼/𝒉𝒓• 𝒒𝟐 = 𝟗𝟗𝟏 ∗ 𝟒𝟏, 𝟑𝟖𝟎 − 𝟐𝟔, 𝟔𝟔𝟎 = 𝟏𝟒, 𝟓𝟖𝟖, 𝟎𝟎𝟎 𝑩𝑻𝑼/𝒉𝒓

• The areas required for each stage can be calculated

• 𝑨𝒊 =𝑸

𝑼𝒊 ∆𝑻𝒊

• 𝑨𝑰 =𝟐𝟎,𝟗𝟖𝟎,𝟎𝟎𝟎

𝟓𝟓𝟎 ∗𝟑𝟖= 𝟏, 𝟎𝟎𝟒 𝒇𝒕𝟐

• 𝑨𝑰𝑰 =𝟏𝟑,𝟐𝟐𝟓,𝟎𝟎𝟎

𝟑𝟓𝟎 ∗𝟑𝟑= 𝟏, 𝟏𝟒𝟓 𝒇𝒕𝟐

• 𝑨𝑰𝑰𝑰 =𝟏𝟒,𝟓𝟖𝟖,𝟎𝟎𝟎

𝟐𝟎𝟎 ∗𝟓𝟑= 𝟏, 𝟑𝟕𝟓 𝒇𝒕𝟐

• These do not match very well

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• Looking at how the computed areas compare to the average we come up with a new set

of temperature drops:• ∆𝑇1 = 32 𝐹• ∆𝑇2 = 31 𝐹• ∆𝑇3 = 61 𝐹

Temperature, F 𝝀 (BTU/pound)

Steam 249 = TS 946 = 𝜆𝑆

Feed to I 70 = Tf

Liquor in I and steam to II 249 – 32 = 217 = T1 967 = 𝜆1

Liquor in II and steam to III 217 – 31 = 186 = T2 987 = 𝜆2

Liquor in III 186 – 61 = 125 = T3 1022 = 𝜆3

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• Heat Balances (all cp = 1.0)

• 𝒒𝟏 = 𝟗𝟒𝟔 ∗ ሶ𝒎𝑺 = 𝟗𝟔𝟕 ∗ 𝟓𝟓, 𝟎𝟎𝟎 − ሶ𝒎𝟏 + 𝟓𝟓, 𝟎𝟎𝟎 ∗ 𝟐𝟏𝟕 − 𝟕𝟎• 𝒒𝟐 = 𝟗𝟔𝟕 ∗ 𝟓𝟓, 𝟎𝟎𝟎 − ሶ𝒎𝟏 = 𝟗𝟖𝟕 ∗ ሶ𝒎𝟏 − ሶ𝒎𝟐 + ሶ𝒎𝟏 ∗ 𝟏𝟖𝟔 − 𝟐𝟏𝟕• 𝒒𝟐 = 𝟗𝟖𝟕 ∗ ሶ𝒎𝟏 − ሶ𝒎𝟐 = 𝟏𝟎𝟐𝟐 ∗ ሶ𝒎𝟐 − 𝟏𝟏, 𝟎𝟎𝟎 + ሶ𝒎𝟐 ∗ 𝟏𝟐𝟓 − 𝟏𝟖𝟔

• The last two equations can be solved to give:

• ሶ𝒎𝟏 = 𝟒𝟏, 𝟑𝟖𝟎 𝒍𝒃/𝒉𝒓• ሶ𝒎𝟐 = 𝟐𝟔, 𝟕𝟐𝟎 𝒍𝒃/𝒉𝒓

• Then the first equation gives

• ሶ𝒎𝑺 = 𝟐𝟐, 𝟒𝟕𝟎 𝒍𝒃/𝒉𝒓

• Then

• 𝒒𝟏 = 𝟐𝟏, 𝟐𝟓𝟔, 𝟎𝟎𝟎 𝑩𝑻𝑼/𝒉𝒓• 𝒒𝟐 = 𝟏𝟑, 𝟏𝟕𝟏, 𝟎𝟎𝟎 𝑩𝑻𝑼/𝒉𝒓• 𝒒𝟐 = 𝟏𝟒, 𝟒𝟔𝟗, 𝟎𝟎𝟎 𝑩𝑻𝑼/𝒉𝒓

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The areas required for each stage can be calculated

• 𝑨𝒊 =𝑸

𝑼𝒊 ∆𝑻𝒊

• 𝑨𝑰 = 𝟏, 𝟐𝟎𝟖 𝒇𝒕𝟐

• 𝑨𝑰𝑰 = 𝟏, 𝟐𝟏𝟒 𝒇𝒕𝟐

• 𝑨𝑰𝑰𝑰 = 𝟏, 𝟏𝟖𝟓 𝒇𝒕𝟐

• These match fairly well

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• Economies

• First Effect𝑚𝑓 −𝑚1

𝑚𝑆=55,000 − 41,380

22,470= 0.606 𝑙𝑏 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑒𝑑 𝑝𝑒𝑟 𝑙𝑏 𝑠𝑡𝑒𝑎𝑚

• Second Effect𝑚1 −𝑚2

𝑚𝑓 −𝑚1=41,380 − 26,720

55,000 − 41,380= 1.076 𝑙𝑏 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑒𝑑 𝑝𝑒𝑟 𝑙𝑏 𝑠𝑡𝑒𝑎𝑚

• Third Effect𝑚2 −𝑚3

𝑚1 −𝑚2=26,720 − 11,000

41,380 − 26720= 1.072 𝑙𝑏 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑒𝑑 𝑝𝑒𝑟 𝑙𝑏 𝑠𝑡𝑒𝑎𝑚

• Over-all Economy

𝑚𝑓 −𝑚3

𝑚𝑆=55,000 − 11,000

22,470= 1.96 𝑙𝑏 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑒𝑑 𝑝𝑒𝑟 𝑙𝑏 𝑠𝑡𝑒𝑎𝑚

Documents

CE 408 - SENS · 2021. 3. 5. · 7 Multi-Effect Evaporation Energy Evaluation • Assume all of the heat from Steam that enters 1st effect goes to latent heat of the “Steam” coming