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MULTIPLE EFFECT EVAPORATOR 1
The invention of the multiple effect evaporator is generally credited to Norbert Rillieux. Rillieux developed a multiple pan evaporation system for use in sugar refining. Rillieux was born in Louisiana and trained in France. Most of his working career was spent in the U.S., although he later returned to Europe where he is buried in the famous Pere Lachaise cemetery in Paris. Rillieux's achievements were little acknowledged during his lifetime, because according to the laws of the time he was "a free person of color."
http://www.cbu.edu/~rprice/lectures/evap1.html,
MULTIPLE EFFECT EVAPORATORS 2
Vapor from the first effect
Consider two evaporators connected so that the vapour line from one is connected to the steam chest of the other as shown in, making up a two effect evaporator.
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SIMPLIFIED DIAGRAM – MULTIPLE EFFECT EVAPORATORS
Vapor T1
feed, TF
Vapor T2
Vapor T3
(1)
(2)
(3)
T1
T2
T3
steam, Ts
Concentrate from first effect
Concentrate from second effect
Concentrated product
MULTIPLE EFFECT EVAPORATORS 4
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 = U1A1(Ts - T1) = U1A1 T1 where q1 is the rate of heat transfer, U1 is the overall heat transfer coefficient in evaporator 1, A1 is the heat-transfer area in evaporator 1, Ts is the temperature of condensing steam from the boiler, T1 is the boiling temperature of the liquid in evaporator 1 T1 is the temperature difference in evaporator 1, = (Ts - T1).
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MULTIPLE EFFECT EVAPORATORS Temperature drops in multiple effect evaporator • Assumption: 1. No boiling point rise (BPR) 2. No heat of solution 3. Approximate all the latent heat of the condensing steam appear as latent heat in the vapor
Second effect evaporator
MULTIPLE EFFECT EVAPORATORS Third effect evaporator
Since areas are EQUAL
Hence, temperature drops ΔT inversely proportionally to U
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MULTIPLE EFFECT EVAPORATORS If NO boiling point rise
Since ΔT1 is proportional to 1/U1, hence:
Exercise : Problem 8.5-1 A solution with a negligible BPR is being evaporated in a triple effect evaporator using saturated steam at 121.1o C (341.3 K). The pressure in the vapor of the last effect is 25.6 kPa abs. The heat transfer coefficients are U1 = 2840, U2 = 1988 and U3 =1420 W/m2.K The areas are equal. Estimate the boiling point in each evaporators.
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MULTIPLE EFFECT EVAPORATORS Capacity of multiple-effect evaporators • Capacity of 3 effect evaporator compared to single effect evaporator is
MULTIPLE EFFECT EVAPORATORS • Assumptions: 1. U same in each effect 2. A equal in each effect
If single effect evaporator used same A,U and ΔT drop gives SAME CAPACITY as for multiple effect evaporator. Hence, the increase in steam economy will reduced the evaporator’s capacity
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Calculation for multiple effect evaporator Values to be obtained: 1. Area of heating surface in each effect (normally assumed equal) 2. Amount (kg) of steam supplied per hour 3. Amount of vapor leaving each effect (last effect) 4. Total vaporized (V1+V2+V3)
Calculation for multiple effect evaporator Given values 1. Steam pressure to the first effect 2. Final pressure in the vapor space of the last effect 3. Feed condition and flow to the first effect 4. The final concentration in the liquid leaving the last effect 5. Physical properties such as enthalpies/heat capacities of the liquid and vapor 6. Overall heat transfer coefficient in each effect
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Calculation for multiple effect evaporator Calculation Steps: 1. Determine TBP of last effect (from given outlet concentration and pressure). 2. Perform overall material balance to determine V (amount of vapor evaporated).
First trial: assumed V1=V2=V3 From material balance you can get the L1, L2 and L3 Perform solid balance to get solid concentration at each effect
Calculation for multiple effect evaporator Step by step calculation: 3. Estimate the T drop using below equation
Calculate TBP at each effect In the present of BPR-determined BPR in each effects- using above equation calculate ΔT1, ΔT2 and ΔT3 –then calculate TBP at each effect.
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Calculation for multiple effect evaporator Step by step calculation: 4. Calculate V and L in each effect by performing heat and material balance. 5. Calculate q transferred in each effect – then calculate the area, A1, A2 and A3 -then calculate the average value , Am
Note that if areas are close to each other, the calculation is complete thus not required second trial
Calculation for multiple effect evaporator Step by step calculation: 6. To start 2nd trial, use new value of L1, L2, L3, V1, V2, and V3 from step 4. Calculate new solid balance of each effects
7. Obtain new balance of :
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Calculation for multiple effect evaporator Step by step calculation: 8. Using new ΔT’ from step 7 then repeat the calculation starting from step 4. Note: 2 trials sufficient to get area reasonably close to being equal
Example 8.5.1 - Evaporation of sugar solution in a triple effect evaporator
F = 22,680 kg/hr XF = 0.1 TF = 26.7 oC
V1 = 22,680 - L1
V2 = L1 – L2
T1
T2
(1)
V3 = L2 – 4536 T3
(2)
(3) 13.4 kPa
S, 205.5 kPa TS1 = 121.1 oC
TS1
T1
TS2
T2
TS3
T3
L1,
L2,
L3 = 4536 kg/hr,
x1,
x2,
X3= 0.5
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REFERENCES 19
Books : 1. Hall,
Geankoplis C. J., Transport Processes and Separation Process Principles, 4th Edition, Prentice 2003.
2.
McCabe W. M., Smith J. C. and Harriott P., Unit Operations of Chemical Engineering, 7th Ed., McGraw Hill, 2005.
Websites: http://www.nzifst.org.nz/unitoperations/index.htm
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THANK YOU.
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