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Single Product Batch Plants Biegler/Grossmann/Westerberg: Systematic Methods of Chemical Process Design
EX: A Processing Recipe
Design and Scheduling of Batch Plants
1. Mix raw materials A and B . Heat to to 80oC C and react during 4 hours to form product C . 2. Mix with solvent D for for 1 1 hour at ambient conditions. 3. Centrifuge to separate solid product C for 2 hour. 4. Dry in a tray for 1 hour at 60oC .
Cheng-Liang Chen
PSE
LABORATORY
Department of Chemical Engineering National TAIWAN University
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Single Product Batch Plants
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Single Product Batch Plants Cycle Times
Gantt Charts (with/without Transfer Times)
M
CT =
τ j
for non-overlappin non-overlapping g operation
j =1
for overlapping overlapping operation
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Multiple Product Batch Plants
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Single/Mixed-Product Campaigns
Flowshop/Jobshop Plants ➢
Flowshop (Multi-product) Plants: all products require all stages following the same sequence of operations
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Jobshop (Multi-purpose) Plants: not all products require all stages and/or follow the same sequence
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Effect of Cleanup Time On Cycle Time
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Transfer Policies ➢
Zero-wait (ZW) transfer
➢
No-intermediate storage ( NIS)
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Unlimited storage (UIS) N
CT UIS = max
j=1,M
niτ ij
i=1
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Cycle Times for Various Transfer Policies
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Parallel Units and Intermediate Storage
ZW
NIS
We assume zero-wait transfer and that the size of the batch in each stage is the same (1000 kg).
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Parallel Units and Intermediate Storage CT = max
j=1,M
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Parallel Units and Intermediate Storage
τ j N P j
If a large number of batches are to be produced, then to produce the same amount we can reduce the batch size
The other alternative is to introduce storage between stages. This has the effect of decoubling the two stages si that each stage can operate with different cycle times and batch sizes. Stage 1 has a cycle time of 12 hours and handles batches of 1000 kg; Stage 2 has a cycle time of 3 hours and handles batches of 250 kg. The intermediate storage must hold up to three batches ( 750 kg).
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Sizing of Vessels in Batch Plants
Sizing of Vessels in Batch Plants
An Example
An Example
Problem: Design a two-stage plant to produce 500, 000 b/yr of product C (operating 6, 000 hours per year) with the following recipe
☞
➢
Size Factor S j for stage j : S j = volume vessel j required to produce 1b of final product S 1 =
Mix 1 b A, 1 b B, and react for 4 hours to form C . The yield is 40% in weight and the density of the mixture is ρ m = 60b/ft3. Add 1b solvent and separate by centrifuge during 1 hour to recover 95% of product C . Density of the mixture is ρm = 65b/ft3.
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S 2 =
14
1 ft3 60 b mix 1 ft3 65 b mix
3
(1+1) b mix ft = 0.0438b prod × (2 ×0.4 ×0.95) b prod 3
b mix ft × (2 ×(1+1+1) 0.4×0.95) b prod = 0.0604 b prod
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Sizing of Vessels in Batch Plants
Sizing of Vessels in Batch Plants
An Example
An Example
Cycle Time, # of Batches, Batch Size, Vessel Volumes CT = max{4, 1}
# Batches = Batch Size = Vessel V1 = Vessel V2 =
6000 hours 4 hrs/batch 500,000 b 1500 batches ft3 b S 1 b prod B batch 3 ft b S 2 b prod B batch
Total Vol. = V 1 + V 2
➢
Bottleneck: stage 1 ⇒ two units operating in parallel CT = max{4/2, 1}
= 4 hours = 1500 batches
# Batches =
= 333 b/batch ft3
Batch Size = 3
b = 0.0438 b prod × 333 batch = 14.6 ft
Vessel V1 =
b = 0.0604 bftprod × 333 batch = 20.1 ft3
Vessel V2 =
3
= 14.6 + 20.1 = 3 4.7 ft3
6000 hours 2 hrs/batch 500,000 b 3000 batches ft3 b S 1 b prod B batch 3 ft b S 2 b prod B batch
Total Vol. = 2V 1 + V 2
= 2 hours = 3000 batches = 166 b/batch 3
ft b × 166 batch = 0.0438b prod = 7.3 ft 3 3
ft b = 0.0604b prod = 10. ft 3 × 166 batch
= 2 × 7.3 + 10 = 24.6 ft 3
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Sizing of Vessels in Two-Product Batch Plants ➢
Problem: Design a two-stage plant to produce 500, 000 b/yr of product A and 300, 000 b/yr of product B (operating 6, 000 hours per year). Processing Time (hr.) Stage 1 Stage 2 A B
8 6
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Sizing of Vessels in Two-Product Batch Plants ➢
Single product campaign; Production of A and B in each campaign: for A: for B:
Size Factor (ft3/b prod) Stage 1 Stage 2
3 3
0.08 0.09
0.05 0.04
➢
500,000 b 6 300,000 b 6
(equating the batch sizes and constraining production times to 992 hours)
Bi = production i # batches i 83,333 tA/8
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V j = max {V ij } i=1,N
Volumes V ij = S ij Bi
➢
Inventory: A trade-off
Produc Rate: Inv Profile:
Stage 1 Stage 2 A
77.9
48.7
B
87.7
39.0
V j
87.7
48.7
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Sizing of Vessels in Two-Product Batch Plants Demand Rate:
V ij = S ij Bi
tB = 308 hours
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Sizing of Vessels in Two-Product Batch Plants Vessel volumes (select largest volume in each stage)
50,000 tB /6
= 992
tA = 684,
⇒
= production i ti/CT i =
tA + tB
➢
= 50, 000 b
Batch size B i of product i; # of batches, total production time t i
Cleanup Time = 4 hours A to B , B to A Length of production cycle = 1000 hours (repeat 6 times) 1000 − 4 − 4 = 992 hr.s for production operation (effective time)
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= 83, 333 b
Max Inventory Ave Inventory
dA dB pA pB pA − dA −dA −dB pB − dB I Amax I Bmax I A I B
Invent Cost:
production cycle ⇔ inventories
b = 83,333 1000 hr = 83.3 b/hr 50,000 b = 1000 hr = 50.0 b/hr b = 83,333 = 121.8 b/hr 684 hr 50,000 b = = 162.3 b/hr 308 hr = 121.8 − 83.3 = 3 8.5 b/hr (accum rate, 0 ∼ 684 hrs) = −83.3 b/hr (depletion rate, 684 ∼ 1000 hrs) = −50.0 b/hr (depletion rate, 0 ∼ 688 hrs) = 162.3 − 50 = 112.3 b/hr (accum rate, 688 ∼ 996 hrs) = 38.5 (b/hr) × 684 (hr) = 26334 (b) = 112.3 (b/hr) × 308 (hr) = 34598 (b) = 1000(26334) = 13167 (b) 2(1000)
=
1000(34600) 2(1000)
= 17300 (b)
C inv = (13167 + 17300) (b) × $1.25/ (b-yr) = $38048/yr
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Synthesis of Flowshop Plants
Sizing of Vessels in Two-Product Batch Plants
Three Major Decision Levels ➢
Structural Level
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Assignments of tasks to equipment Merge adjacent tasks whose sum of processing times does not exceed cycle time (construction materials must meet all required functions)
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Synthesis of Flowshop Plants
Synthesis of Flowshop Plants
Three Major Decision Levels
Three Major Decision Levels ➢
Structural Level (cont.)
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Number of parallel units or intermediate storage Use parallel units whenever there is a stage with a much larger processing times (a smaller # of bigger equip. ↔ a larger # of smaller pieces)
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Multiproduct Batch Plant Design
Three Major Decision Levels
Problem Given a multiproduct batch plant that consists of three processing stages: mixing, reaction, and centrifuge separation. Two products, A and B, are to be manufactured in such a plant using production campaigns of single products. The data for processing times, size factors for the units, demands, and cost data are given in the next page. Assuming continuous sizes, and that the plant is operated with single-product campaigns, determine the sizes of the units required at each processing stage, as well as the number of units that ought to be operating in parallel to minimize the investment cost. Furthermore, resolve the problem for the following cases:
Equipment sizing To select the same batch size for all products
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Scheduling Level
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Synthesis of Flowshop Plants Sizing Level
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Nature of production campaigns , transfer policies single products if cleanup or transition times are large; ZW, NIS, or UIS ?
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Length of production cycles
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Sequencing of products
(requires detailed evaluation and optimization)
1. The demands of A and B are increased by 20%. 2. For the above demands the time available for production is increased from 6000 hours to 7500 hours.
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Multiproduct Batch Plant Design
Multiproduct Batch Plant Design
Data
Notation
Demand A = 200, 000 kg
N = number of products to be produced
Demand B = 150, 000 kg Horizon time = 6, 000 hr Processing Time (hrs)
Size Factors (L/kg)
Mixer Reactor Centrifuge
Mixer Reactor Centrifuge
A
8
20
4
B
10
12
3
Mixer cost = 250V
A
2
3
4
B
4
6
3
0.6
Reactor cost = 500V 0.6 Centrifuge cost = 340V 0.6
Min size =
250 L
Max size = 2500 L
M Bi S ij tij H Qi αi, β j V jL, V jU T Li N j Θi ni
= = = = = = = = = = = =
number of stages in the batch plant size of product i at the end of the M stages size factor of product i in stage j (L/kg) processing time of product i in stage j (hr) horizon time (hr) demand of product i (kg) cost coefficient and cost exponent for unit j lower and upper bounds of volumes cycle time for product i number of parallel units in stage j production time number of batches
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Multiproduct Batch Plant Design
Formulation
Formulation
Volume V j can process all products
➢
Example: N 1 = N 2 = N 3 = 1 T LA = max{8, 20, 4}
i = 1, · · · , N ; j = 1, · · · , M
= 20 hr
T LB = max{10, 12, 3} = 12 hr
Cycle time T Li = max ∀j
⇒
T Li ≥
tij N j
tij N j
i = 1, · · · , N ; j = 1, · · · , M
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Multiproduct Batch Plant Design
V j ≥ S ij Bi ➢
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Multiproduct Batch Plant Design
Multiproduct Batch Plant Design
Formulation
Formulation
Example: N 1 = N 3 = 1, N 2 = 2 T LA = max{8, 20/2, 4}
➢
Production time
= 10 hr
Θi = niT Li =
T LB = max{10, 12/2, 3} = 10 hr
N
⇒
N
Θi =
i=1
➢
i=1
Qi T Li Bi
Qi T Li ≤ H Bi
Investment cost
M
C =
j=1
βj
N j αj V j
i = 1, · · · , N
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Multiproduct Batch Plant Design
Multiproduct Batch Plant Design
Summary
Given Maximum K Parallel Units L U ≤ T Li ≤ T Li T Li
M
min C =
x∈Ω
βj
N j αj V j
BiL ≤ Bi ≤ BiU
j=1
N
=
Ω
x
i=1
V j ≥ S ij Bi t T Li ≥ N ijj Qi T Li ≤ H Bi
V jL ≤ V j ≤ V jU Bi ≥ 0, T Li ≥ 0 i = 1, · · · , N ; j = 1, · · · , M
tij K U T Li = max {tij }
= {Bi, V j , N j , T Li; i = 1, · · · , N ; j = 1, · · · , M }
X
L T Li = max
j = 1, · · · , M j = 1, · · · , M
L QiT Li H V jU = min j S ij
BiL =
BiU
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Multiproduct Batch Plant Design
Multiproduct Batch Plant Design
Variable Transformation ⇒ Convex Constraints
MINLP with Convex Constraints
M
C =
let vj ≡ ln V j
nj ≡ ln N j
bi ≡ ln Bi
tLi ≡ ln T Li
βj
N j αj V j
C =
j=1
αj exp (nj + β j vj )
x
K
(ln k)yjk
k=1
V j ≥ S ij Bi
⇒ ⇒
o, t Li ≥ (ln tij ) − nj
yjk = 1 j = 1, M
k=1
i
vj
e
vj (tLi−bi )
Qie
bi
≥ S ij e
≥ ln S ij + bi ∀i, ∀ j ≤ H
αj exp (nj + β j vj )
= {bi, vj , nj , tLi, yjk ; i = 1, N ; j = 1, M }
N j ∈ {1, 2, · · · , K }
K
j=1
j=1
nj = ln N j nj =
x∈Ω
M
⇒
M
min C =
Ω
=
x
vj ≥ ln S ij + bi, tLi ≥ ln tij − nj Qiexp (tLi − bi) ≤ H
i
K
nj =
k=1
K
(ln k)yjk ,
yjk = 1
k=1
ln V jL ≤ vj ≤ ln V jU ln BiL ≤ bi ≤ ln BiU L U ln T Li ≤ tLi ≤ ln T Li yjk ∈ {0, 1}, nj ≥ 0
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* INITIAL POINT * Binary variables set for 3 units per stage Y.L(K,J) = 0.; Y.L(’3’,’MIXER’) = 1; Y.L(’3’,’REACTOR’) = 1; Y.L(’3’,’CENTRIFUGE’) = 1; N.L(J) = SUM( K , COEFF(K)* Y.L(K,J) ) * Batch sizes are set at the mid-point between bounds B.L(I) = (B.UP(I) + B.LO(I))/2 V.L(J) = SMAX(I , LOG(S(I,J)) + B.L(I)) TL.L(I) = SMAX(J , LOG(T(I,J))-N.L(J)) MODEL
BATCH
OPTIONS LIMCOL OPTIONS ITERLIM OPTION OPTCR=0;
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Multiproduct Batch Plant Design Optimal Design ; ; ; ;
/ ALL / ; = 0 = 10000
SOLVE BATCH USING MINLP MINIMIZING COST
; ;
optimal inve. cost Product A: batch numb batch size Product B : batch numb batch size
;
* Converting variables into original form ACTV(J) = EXP(V.L(J)) ; ACTB(I) = EXP(B.L(I)) ; ACTTL(I) = EXP(TL.L(I)); ACTN(J) = EXP(N.L(J)) ; * Optimal Design DISPLAY ACTV ; DISPLAY ACTB ; DISPLAY ACTTL ; DISPLAY ACTN ;
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Thank You for Your Attention Questions Are Welcome
= 167, 428 = 320 = 625 kg cycle time = 10 hr = 467 = 321.5 kg cycle time = 6 hr
