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Filtration - Plate Filter Press


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#1 Starlord25

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Posted 23 September 2016 - 09:20 AM

Greetings all.
Can someone please provide some help regarding the following question?

A plate and frame filter press is to be built to handle 0.1m3 solids per m3 filtrate. ruv is 5x10^10
Designed to operate at 500kN m-2 constant pressure with cycle times below:

140seconds to dismantle press
100seconds to clean EACH plate
140seconds to reassemble press

Plate area is 0.5m2. Frame thickness 25mm

By expressing cake volume in terms of 'n' plates and filtrate volume (V) in terms of 'n' plates, calculate filtration time.

#2 Starlord25

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Posted 23 September 2016 - 09:23 AM

All help is appreciated regarding this one as I have now spent 2 weeks attempting different calculations but can't get very far.

#3 breizh

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Posted 23 September 2016 - 10:07 PM

http://www.solidliqu...rs/pressure.htm

Hi,

You may find direction through this link . Why don't you submit your work for people to comment ?

 

My thoughts

 

note : a great website for filtration , may not be spot on for your query .

 

http://www.bhs-filtr...om/technical/#1

 

Good luck.

 

Breizh

Attached Files


Edited by breizh, 24 September 2016 - 01:51 AM.


#4 Starlord25

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Posted 26 September 2016 - 12:13 PM

Thanks for the response.

I haven't got any real workings to post regarding this one unfortunately.
The previous 2 parts to this problem, I have completed.
This part seems to elude me entirely.
:(

#5 Starlord25

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Posted 26 September 2016 - 12:14 PM

Not any meaningful workings anyway.
I will try and post what I do have later tonight.

#6 Starlord25

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Posted 27 September 2016 - 02:38 PM

Ok
The only related workings I have are volume of cake in terms of n

Volume of cake = 0.0125n

But how do I express filtrate volume (V) in terms of n?
I think if I can get my head round that, I can have a decent go.

Please help! :)

#7 Starlord25

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Posted 29 September 2016 - 12:14 PM

Ok.
Below is my latest attempt:

Volume of cake = 0.0125n

Volume of filtrate = 0.125n (as 0.1m3 solids per m3 filtrate)

Substitution into rate equation:

t= (ruv V^2)/(2A^2 p)

t = ((5x10^10)x0.125n^2)/(2(0.5n)^2x500000)

t = 12500seconds

#8 Starlord25

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Posted 29 September 2016 - 12:17 PM

Following on from this though to find number of plates at optimum conditions:

Time out of service = filtration time

t = 280 + 100n

280 + 100n = 12500

n = 122.2

Hence number of plates = 123 rounded up.

Seems unreasonably large amount?

#9 breizh

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Posted 29 September 2016 - 08:56 PM

hi ,

To me t is related to the filtration time , not the cycle time .

I may be wrong .

 

Number of plates is often big , discontinuous process  and hours are necessary to remove the cake manually . These days you can find automatic system like the one describe on the link 1.

 

Good luck.

 

Breizh


Edited by breizh, 30 September 2016 - 09:43 PM.


#10 Starlord25

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Posted 30 September 2016 - 06:32 AM

You are correct, however under optimum conditions, the filtration time should equal the time out of service (in my understanding).
This is why I have related t to time out of service in my equations.

#11 Starlord25

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Posted 06 October 2016 - 06:27 AM

Just to give any readers some feedback, I've just found out my answers are correct.
Despite the doubt :)

Thanks to everyone who gave up their time to provide some input. :)

#12 sgkim

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Posted 07 October 2016 - 05:57 AM

Starlord25, 

 

Please follow the sequence for determining the number of the plates of the filter:

 

1. Filtration load:   slurry volume to be treated per hour:     Q   [m^3 slurry/h]

2. Filtration cycle Time:      t      [hour per cycle (h/cy)]

   (t is complete cycle time required for precoating/ feeding / blowing / washing / rinsing/ sqeezing/ dismantling / cake discharging / media cleaing / reassembling)   

3. Bulk cake volume fraction in slurry:  f  [m^3 cake /m^3 slurry] (= 0.1 m^3 cake/m^3 slurry)

4. Hourly cake volume to be filtered:   Vh = f * Q   [m^3 cake /h]  

5. Cake volume per plate:  v    [m^3 cake/plate]  (= 0.025 m^3 cake/plate]

6. Volume of cake per cycle:  Vc = Vh * t  [m^3 cake /cy]

7. Number of plates of filter:  n = Vc/v   [plates/cy]

8. Number of Cycles per hour:  C = 1/t  [cy/h]

9. Slurry quantity per cycle  qc  = Q/C  [m^3 slurry/cy] 

10. Slurry feeding time per cycle:   ts  [min/cy]

11. Slurry feed rate: Fs= qc/ts  [m^3 slurry/min] 

12. Wash, rinsing liquid volume required per cake volume: w, r  [m^3 wash-liquid/m^3 cake],[m^3 rinsing-liquid/m^3 cake]

13. Total wash, rinsing liquid required per cycle:  W = w*Vc,  R = r*Vc   [m^3 wash-liquid/cy],[m^3 rinsing-liquid/cy]

14. Washing, rinsing time per cycle:   tw, tr [min/cy]

15. Washing, rinsing liquid flow rate:  Fw= W/tw, Fr=R/tr   [m^3 wash-liquid/min],[m^3 rinsing-liquid/min]

....

similar fashion can be queued.

 

The filtration cycle time, t, is dependent mostly on the character of the particles in the slurry.  Filtration time for every cycle shall be determined considering the charactor of the particles and the mechanical factors involved.  

 

If the number of plates, n, exceeds that commercially available, then select the larger frame model or increase the number of units for parallel processing.

 

And, the "time" can not be calculted but to be called upon by yourself.

 

Hoping the above may help,  

 

Stefano


Edited by sgkim, 14 October 2016 - 05:35 AM.





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