Pan Section Capacity Calculation in Sugar Industry | Crystallization
Contents
Sugar crystallization process takes place in pan section of sugar plant. The equipment supply tanks, Batch/continuous pans, condensers, molasses conditioners, spray pond, crystallizers are under pan section.
Capacity of Batch/Continuous pans
The boiling time consider for A, B & C massecuites are 4 hours, 6hours and 8 hours respectively to find the pans capacity calculation. For refinery massecuite it consider 2 to 3 hours.
Example:
Crushing Capacity of the plant = 230 TCH
“A” massecuite%cane = 25 to 30%
“B” massecuite%cane = 12 to 13%
“C” massecuite%cane = 6 to 8%
” A” Massecuite Quantity = 230 x 30% = 69 T/hr = 1656 T/day
” B” Massecuite Quantity = 230 x 13% = 30 T/hr = 720 T/day
” C” Massecuite Quantity = 230 x 8% = 18.5 T/hr = 444 T/day
While consider batch pans with 60 Ton capacity each
Massecuite | Boiling Hours | No. of strikes per day per pan | Quantity of massecuite per strike in Ton | No. of pans required |
A | 4 | 24 hours/4 = 6 | 1656/6 = 276 | 276/60 ≈ 5 nos. |
B | 6 | 24 hours/6 = 4 | 720/4 = 180 | 180/60 = 3 nos. |
C | 8 | 24 hours/8 = 3 | 444/3 = 148 | 148/60 ≈ 3 nos. |
While consider Continuous pans for all massecuite boilings
For continuous pans capacity consider 10% to 20% extra on required capacity. So
” A” Continuous Pan = 69 T/hr x 110% ≈ 76 T/hr
” B” Continuous Pan = 30 T/hr x 120% ≈ 35 T/hr
” C” Continuous Pan = 18.5 T/hr x 120% ≈ 22 T/hr
Grain and Molasses ratio for A, B & C massecuites generally taken as follow as ( It is depend upon grain size and purity of material)
“A” Grain to Liquor (syrup/melt/AL) ratio – 1 : 1 to 2
“B” Grain to Liquor (A heavy) ratio – 1 : 2 to 3
“C” Grain to Liquor (B heavy/ C light) ratio – 1 : 3 to 4
B massecuite purity online calculation sheet | Sugar Technology
C massecuite final purity calculation |Grain Quantity requirement for C CVP
Massecuite | Boiling Hours | No. of strikes per day per pan | Quantity of grain required in Ton | Quantity of massecuite per strike in Ton | No. of pans required |
A Grain | 4 | 24/4 = 6 | 1656/2 = 828 | 828/6 = 276 | 138/60 = 2.3 (70T x 2 nos.) |
B Grain | 6 | 24/6 = 4 | 720/3 = 240 | 240/4 = 180 | 60/60 = 1 no. |
C Grain | 8 | 24/8 = 3 | 444/4 = 111 | 111/3 = 148 | 37/60 ≈ 1 no. |
Thumb rules for finding the capacities of batch/continuous pans
Note: It is not accurate capacity but it gives approximate value instantly
Batch pans
“A” Batch pan capacity in Ton – TCD x 0.06 ( Ex: 5000 x 0.06 = 300 T )
“B” Batch pan capacity in Ton – TCD x 0.04 ( Ex: 5000 x 0.04 = 160 T )
“C” Batch pan capacity in Ton – TCD x 0.03 ( Ex: 5000 x 0.03 = 150 T )
Continuous pans
“A” Continuous pan capacity in Ton – TCD x 0.014 ( Ex: 5000 x 0.015 = 75 T/hr )
“B” Continuous pan capacity in Ton – TCD x 0.006 ( Ex: 5000 x 0.006 = 30 T /hr)
“C” Continuous pan capacity in Ton – TCD x 0.004 ( Ex: 5000 x 0.004 = 20 T /hr)
“A” Grain pan capacity in Ton – TCD x 0.025 ( Ex: 5000 x 0.025 = 125 T )
“B” Grain pan capacity in Ton – TCD x 0.01 ( Ex: 5000 x 0.01 = 50 T )
“C” Grain pan capacity in Ton – TCD x 0.01 ( Ex: 5000 x 0.01 = 50 T )
Low Grade Massecuite Treatment in Sugar Crystallization Process
Sugar Seed Slurry Requirement Calculation for B and C massecuite
Types of Graining Techniques in sugar crystallization process | Pan Boiling
Pan Supply Tanks
“A” massecuite feeding liquor (syryp/melt/A light) consider minimum 2 hours retention time
“B” massecuite feeding liquor (A Heavy) consider minimum 3 hours retention time
“C” massecuite feeding liquor ( B heavy/ C light) consider minimum 4 hours retention time
Example:
Crushing Capacity of the plant = 230 TCH
Syrup % cane – 25 to 30%
Melt % cane – 12 to 14%
A light %cane- 2 to 3%
A heavy%cane- 12 to 15%
B heavy%cane – 6 to 7%
C light%cane – 2 to 3%
Syrup + melt + A light = 43% (average) = 230 x 43% ≈ 100 T/hr
High grade massecuite supply tanks capacity = 100 x 2 hours = 200 / 1.25(density) = 160 M^{3} = 1600 HL
A heavy molasses quantity = 230 x 15% = 34.5 T/hr
A heavy supply tanks capacity = 34.5 x 3 hours = 103.5 / 1.3(density) ≈ 80 M^{3} = 800 HL
B heavy + C light molasses quantity = 230 x 10% = 23 T/hr
A heavy supply tanks capacity =23 x 4 hours = 92 / 1.3(density) ≈ 70 M^{3} = 700 HL
Thumb rules for finding the capacities of supply tanks in pan section
High grade massecuite feed materials ( Syrup + melt + A light ) supply tanks capacity in HL = TCH x ( 7 to 8)
Low grade massecuite feed materials ( A Heavy + B heavy + C light ) supply tanks capacity in HL = TCH x ( 7 to 8)
Molasses Conditioners capacity
Its capacity consider extra 10 to 20% on molasses production
Example:
Crushing Capacity of the plant = 230 TCH
A heavy%cane- 12 to 15%
B heavy%cane – 6 to 7%
C light%cane – 3 to 4%
A heavy molasses quantity = 230 x 15% = 34.5 T/hr
A heavy molasses conditioner capacity = 34.5 x 110% = 38 T/hr
B heavy molasses quantity = 230 x 7% = 16.1 T/hr
B heavy molasses conditioner capacity = 16.1 x 110% = 18 T/hr
C light molasses quantity = 230 x 4% = 9.2 T/hr
C Light molasses conditioner capacity = 9.2 x 110% = 11 T/hr
Capacity calculation of crystallizers
Crystallizers are used for cooling and holding of the massecuite. Air cooled type crystallizers are used for high grade massecuites, receiving crystallizers of continuous pans and for seed crystallizer. Water cooled crystallizers are used for low grade massecuites for proper cooling and better exhaustion.
A – Massecuite – ( 2 hours cooling purpose + 2 hours curing purpose) – Air cooled
B – Massecuite – ( 6 to 8 hours cooling purpose + 3 hours curing purpose) – Air cooled + water cooled
C – Massecuite – ( 20 to 24 hours cooling purpose + 4 hours curing purpose) – Air cooled + water cooled
The each crystallizer capacity should be 10 to 15% more than the existing pan capacity.
For example if consider 60 T (42 m³) pan, then the capacity of crystallizer can be taken as 70 T (48 m³).
If consider air cooled type crystallizers for “C’ massecuite then cooling time can be go upto 72 hours. So proper design of cooling element used in crystallizers then cooling time come down upto 18 to 24 hours.
Crystallizers Application in Sugar Industry | Crystallizer Capacity Calculation
Example:
Crushing Capacity of the plant = 230 TCH
“A” massecuite%cane = 25 to 30%
“B” massecuite%cane = 12 to 13%
“C” massecuite%cane = 6 to 8%
” A” Massecuite Quantity = 230 x 30% = 69 T/hr
” B” Massecuite Quantity = 230 x 13% = 30 T/hr
” C” Massecuite Quantity = 230 x 8% = 18.5 T/hr
Crystallizer capacity for “A” massecuite
Quantity of “A” massecuite for (2+2) hrs = 69 x 4 = 276 Tons ≈ 300 Ton
Total volume of “A” crystallizers = 300 /1.45 = 206 M^{3} = 2060 HL ( sp.gr =1.45 )
So total capacity split into no of crystallizers and each capacity having 10 to 15% more than the capacity of pan. Generally A massecuite crystallizers total capacity equals to total capacity of “A” pans.
Crystallizer capacity for “B” massecuite
Quantity of “B” massecuite for (7+3) hrs = 30 x 10 = 300 Tons
Total volume of “B” crystallizers = 300 /1.5 = 200 M^{3} = 2000 HL ( sp.gr =1.5 )
The total capacity split into water cooled and air cooled crystallizers in the ratio of 7 : 3 or 8 : 2
Crystallizer capacity for “C” massecuite
Quantity of “C” massecuite for (24 +4) hrs = 18.5 x 28 = 518 Tons ≈ 550 Ton
Total volume of “C” crystallizers = 550 /1.5 = 370 M^{3} = 3700 HL ( sp.gr =1.5 )
The total capacity split into water cooled and air cooled crystallizers in the ratio of 8 : 1
Vertical Crystalliser Design Calculation for Sugar Massecuite Cooling
Vertical Crystallizer Concepts in Sugar Plant | Mono Vertical Crystallizer
Vacuum crystallizers :
Each Vacuum crystallizer consider equal capacity of existing batch pan capacity and no of crystallizers consider as per no of massecuites ( i.e 3 nos. for A,B & C massecuite boiling)
Capacity of condenser
Evaporation coefficients rate for batch pans depend on the purity of material and hydro-static head of the massecuite. So the massecuite level increased in pan then evaporation rate will be decreased.
As per the Mr. E.Hugot the evaporation rates in kg/m² /hr as follow as
Initial | Final | |
Footing Pan | 85 | 17 |
A-Masseccutie | 71 | 32 |
B-Masseccutie | 46 | 11 |
C-Masseccuite | 36 | 17 |
Condenser capacity calculation purpose consider batch pan evaporation rates 50 to 60 in kg/m² /hr and for continuous pans evaporation rate 20 to 30 kg/m² /hr
Average evaporation rate in Batch Pans A m/c – 60 , B m/c – 55 and C m/c – 50
Average evaporation rate in Continuous pans A m/c – 30 , B m/c – 25 and C m/c – 20
Example:
60T Batch pan having 282 m² heating surface then condenser capacity required
282 m² x 50 kg/m² /hr = 14100 kg/hr ≈ 14 T/hr
35 T/hr capacity “B” continous pan having the heating surface 650 m² then condenser capacity required
650 m^{2} x 25 kg/m² /hr = 16200 kg/hr ≈ 16 T/hr
Injection water System and Condensers
The vapour condensation quantity consider of pan section and evaporator last effect vapour. So calculate total vapour produced from both systems and also water required to condense that vapour.
Vapour produced from pan section = 18 to 25% on cane ( For back-end refinery plants its go upto 28% cane)
Vapour produced from last effect evaporator body = 5 to 8 % on cane
Water required for condensing the vapour calculated on the basis of cooling water ratio.
= Total heat of the vapour = 621 Kcal/kg @ 55 ^{0}C
Definitions in Steam Properties and Online Steam Table For Saturated steam
To = Condenser outlet warm water temperature in ^{0}C
Ti = Condenser inlet cold water temperature in ^{0}C
Example:
Crushing Capacity of the plant = 230 TCH
To = Condenser outlet warm water temperature = 47 ^{0}C
Ti = Condenser inlet cold water temperature = 36 ^{0}C
So Total vapour quantity for condensing = 230 x 33% = 80 T/hr
Cooling water ratio = (621 – 47) / (47 – 36) = 52.2 T/hr
i.e One ton of vapour required for 52.2 tons of water
Total water required for condenses = 80 x 52.2 = 4176 T/hr
Condenser System for vacuum creations and its types with design criteria
Injection water pump capacity
Operating Injection water pump capacity = 4000 M^{3}/hr
Installed Injection water pump capacity = 50 % more then requirement = 4000 x 150% = 6000 M^{3}/hr
( Split the total capacity as per the stand by operation)
Spray pond capacity
Theoretically 750 kg/hr of warm water requires 1 m² of area of spray pond.
But practically good efficiency design 900 to 1000kg/hr of warm water required 1 m^{2} area of spray pond.
Spray Pond area required = 4000000 / 900 = 4444 m^{2} ≈ 4500 m^{2}
Sugar Plant Capacity Calculation
Pan Section
Centrifugal section
Sugar House
4 thoughts on “Capacity Calculation of Pan Section in Sugar Industry | Boiling House Capacity”
Vinod Kumar Mishra
(July 29, 2018 - 4:15 pm)Very very good sir
siva alluri
(August 13, 2018 - 2:37 pm)Thank you Mr.Vinod Kumar Mishra
Aravind
(August 11, 2018 - 2:16 pm)Dear sir your website is very useful for sugar workers . I calculate evaporator vapour bleeding caculation but I get X valve -3.14 then how to further caculation plz help me sir
siva alluri
(August 13, 2018 - 2:46 pm)Dear Mr.Aravind
X value comes “-ve” means availability of vapour is less when compare to bleeding vapour so it is not possible.
i.e As per our bleed vapour not matching to vapour generation
So we can do for getting positive value
1. Increase the juice % cane ( increase imbibition % or use DCH .. like)
2. Increase the evaporator syrup brix ( i.e evaporation % to be increased across the set)
3. Decrease the bleed vapour ( Like reduced the steam demand for pans section )