episode 43 : design of rotary vacuum drum filter
TRANSCRIPT
SAJJAD KHUDHUR ABBASCeo , Founder & Head of SHacademyChemical Engineering , Al-Muthanna University, IraqOil & Gas Safety and Health Professional – OSHACADEMYTrainer of Trainers (TOT) - Canadian Center of Human Development
Episode 43 : DESIGN of Rotary Vacuum Drum Filter
Separation Theory of RVDF
Application
Advantages
𝐃𝐢𝐬𝐚𝐝𝐯𝐚𝐧𝐭𝐚𝐠𝐞𝐬Problem Statement
Equipment Design
Remarks
OUTLINE
References
Rotary Vacuum Drum Filter
Rotary Vacuum Drum Filter
Theory of Separation
Rotary vacuum drum filter (RVDF) is one of the oldest filters used in the industrial liquid-solids separation .A rotary vacuum filter consists of a large rotating drum covered by a cloth. The drum is partially immersed in liquid/solids slurry with approximately up to (25-75) % of the screen area.As the drum rotates into and out of the trough, the slurry is sucked on the surface of the cloth and rotated out of the liquid/solids suspension as a cake. When the cake is rotating out, it is dewatered in the drying zone. The cake is dry because the vacuum drum is continuously sucking the cake and taking the water out of it. At the final step of the separation, the cake is discharged as solids products and the drum rotates continuously to another separation cycle.
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During the washing stage, the wash liquid can either be
poured onto the drum or sprayed on the cake. Cake pressing
is optional but its advantages are preventing cake cracking
and removing more moisture. Cake discharge is when all the
solids are removed from the surface of the cake by a scraper
blade, leaving a clean surface as drum re-enters the slurry.
(Avinash Gupta, Hand book of Chemical Engineer calculation ) and (J. M. COULSON & J. F. RICHARDSON ,Chemical Engineering Series , V2 )
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Range of Application
As a basic separation operation, rotary vacuum drum filter is
used in a wide range of applications:
1- Dewatering slurries of food
2- Pulp
3- Pharmaceutical and chemical,
4- Applications of metallurgical
5- The treatment of waste water.
(Wikipedia website)
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Advantages
1- The rotary vacuum drum filter is a continuous and
automatic operation, so the operating cost is low.
2- The variation of the drum speed rotating can be used to
control the cake thickness.
3- The process can be easily modified (pre-coating filter
process).
4- Can produce relatively clean product by adding a
showering device.
(Wikipedia website)
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Disadvantages
1- Due to the structure, the pressure difference is limited up to
1 bar.
2- Besides the drum, other accessories, for example, agitators
and vacuum pump, are required.
3- The discharge cake contains residual moisture.
4- High energy consumption by vacuum pump.
(Wikipedia website)
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Problem statement
A drum filter as illustrated in (Fig.:1) is to be used for filtering, washing, and drying a cake having the properties given by (Figs. 2 through 5). Air rate through the cake is determined from measurements of flow rate as a function of time with a rotameter as follows:
Air rate as a function of time
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cake mass Wc (in lbm/ft2) is (7.2 L) for each inch of cake thickness, where L is cake thickness in inches; maximum submergence is 35 percent or 126◦; effective submergence is 30 percent or 108◦; maximum washing arc is 29 percent or 104◦; suction (initial drying) arc is 7.5 percent or 27◦; discharge and resubmergence arc is 25 percent or 90◦; and minimum cake thickness is 1/8 in (0.0032 m). 1-Determine the relevant design parameters for the cake thickness of 0.25 in 2- Determine the dimension of the drum for cake rate 5000 Ibm/ h
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Fig. 1: Rotary Vacuum Drum Filter
Design Procedures: 1. Calculate the cake mass, find the filtration time for a thickness of 0.25 in (0.0064 m), and determine the minimum cycle needed for cake formation.
The cake mass is given byWc = 7.2(L) = 7.2(0.25) =1.8 lbm/ft2 ( 8.8 kg/m2) From Fig. 2, filtration time is found to be 0.22 min
Fig. 2: Dry cake mass versus filtration time
Filtration rate = Wc / filtration time
Filtration rate = (1.8/0.22)(60) = 491 lbm/h. ft2 (2402.3 kg/h.m2) of drum surface.
With an effective submergence of 30 % of drum circumference, the minimum cycle based on cake formation is filtration time divided by percent of drum circumference
Minimum cycle = 0.22/0.3 = 0.73 min/r,
Which corresponds to 1.37 r/min.
2. Check to see if initial drying or washing can be done within the time available during the minimum cycle from step 1:
Minimum suction time elapses during passage through 27◦ So 27/360 = 0.075
Drying time = drying section percent X min. cycle of cake formation = 0.075 (0.73) = 0.06 min Now td/wc = 0.06/1.8 = 0.033
Where td is drying time Based on (Fig. 3), the dewatered but unwashed (D/u)
cake will have a moisture content of 30%.
Fig. 3: correlating factor for cake moisture content
Then with a wash ratio of 1.5 (given),
liquid in D/u cake = (30/70)(1.8) = 0.77 lbm/(ft2)(r), and quantity of wash =1.5(0.77) = 1.155 lbm/(ft2)(r) = (5.65 kg/m2 .r)at density of 8.33 lbm/gal,
quantity of wash = 1.155 /8.33 quantity of wash = 0.14 gal/(ft2)(r) = (5.7 litter /m2. r)
WcVw = 1.8(0.14) = 0.25.
Where: Vw is quantity of wash and Wc is the cake mass
From (Fig. 4), the required wash time is 0.15 min
Fig. 4: Cake-wash time correlation with mass of dry solid Wc and wash Vw per unit of area
Wash time = 1.5 min
This corresponds to an arc of
0.15/0.73 = 0.21
That means that 21 % of the circumference.
Since up to 29 % of the circumference can be used, washing
offers No problems.
3. Check the drying time and determine the cycle time.
For a final moisture content of 25%,
From (Fig. 3) td/Wc is (0.3)
Fig. 3: correlating factor for cake moisture content
With Wc = 1.8td /Wc = 0.3 td /1.8 = 0.3 td = 0.54 min
Percentage of drying arc = 0.54/0.73 = 0.739
which takes up nearly three-quarters of the circumference
since 25 percent of the arc is needed for discharge and resubmergence
the maximum arc for initial drying ,washing and final drying is given by 75 − (cake-formation arc)= 75 − 30 = 45 %
Using the originally calculated washing plus drying times of
0.54 + 0.15 = 0.69 min
then
the minimum cycle based on drying and washing = 0.69/0.375 = 1.84 min/r
and the washing arc = 0.15/1.84
which is 8.15 percent or 29◦.
4- Calculation of initial drying time:
initial drying time = the minimum cycle based on drying and washing x initial drying percentage =1.84(0.075) = 0.14 min
So td/wc = 0.14/1.8 = 0.08.
From (Fig. 3), D/u moisture is 27%
and accordingly, the liquor in the
D/u cake is (27/73)(1.8) = 0.67 lbm/(ft2)(r).
The quantity of wash becomes 1.5(0.67) = 1.0 lbm/(ft2)(r) = 4.9 Kg/m2.r
Fro density 8.33 lbm/gal The quantity of wash = 0.12 gal/(ft2)(r). = 4.9 litter/m2.r
Then WcVw = 1.8(0.12) = 0.22, and from (Fig. 4), the wash
time becomes 0.14 min.
Cycle time summary:
Wash time = 0.14 min
Initial drying time = 0.14 min
Final drying time = 0.54 min
5. Summarize the filtration cycle.
The total cycle time is
The total cycle time = (0.14 + 0.14 +0.54) /0.45 = 1.82 min/r = 0.55 r/min The required effective submergence = = (0.22/1.82)(100) = 12 %This is much less than the 30 percent available
6. Calculate the efficiency of solute recovery:
For wash ratio 1.5
and from (Fig. 5).
With wash ratio j = 1.5
the fraction remaining is 0.145 To be on the safe side use
a value of 0.2.
Fig. 5: Cake – wash curve
The following calculations are needed:
Solute in feed = (60/40)(0.02) = 0.03 lb solute per pound of feed
Solute in D/u cake = (27/73)(0.02) = 0.0074 lb solute per pound of cake
Solute in washed cake = 0.0074(0.2) = 0.0015 lb solute per pound of washed cake
Efficiency = (0.030 − 0.0015)/0.03 = 0.95
7. Calculate Dimension of the drum:
Taking into account the effective submergence of 12% we calculate the filtration rate as 491(0.12) = 59 lbm/(h)(ft2) = 0.08 kg/(s)(m2)
For cake rate 5000 Ibm/ h The required surface are of the drum = 5000/59 = 84.7 ft2 = 7.87 m2Area of drum = ∏ x D x L
Assume L/D = 2.5
84.7 = (2.5) D2
D = 3.28 ft = 1m
L = 8.2 ft = 2.5 m
L= 2.5 m
D= 1 m
8. Calculate the air rate.
The air rate can be calculated based on the data previously presented
and shown in (Fig. 5). During the 0.14 min of initial drying, the average rate is
found to be 4.9 (ft3/min)/(ft2)(r).
The average rate during the final 0.54 min of drying is 8(ft3/min)/(ft2)(r).
Fig. 5: Air rate as a function of time
The total air rate = 0.14(4.9) + 0.54(8) = 5 (ft3/min)/(ft2)(r).
Since there are 1.82 min/r
the air rate is 5/1.82 = 2.75 (ft3/min)/ft2
Since the required surface are for 5000 Ibm/h is 84.7 ft2
air rate is = (2.75 ) ( 84.7) = 233(ft3/min) = 6.6 (m3/min)
Remarks:
This type of filters are recommended to be used in wide rang of industrial filtration processes because:
1- It can be easily operated in continuous mode. 2- Its cost is low compared with the other filters. 3- can be designed in deferent sizes.
References:
1- Avinash Gupta, Hand book of Chemical Engineer calculation 2- J. M. COULSON & J. F. RICHARDSON ,Chemical Engineering Series , V2
3- Wikipedia website
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