8/18/2019 Sample Calculation -Constant Pressure filtration

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Weight of (Filtrate + Bucket2) = 3.2 kg

Weight of Bucket2 = 0.550 g

Weight of Filtrate = (3.2-0.550) g

  = 2.65 kg

Densit of Water at !oo" #e"$erature = %%&.&& kg'"3

olu"e of Filtrate = (2.65'%%&.&&) "3

  =2.656*0-3 "3

u"ulati,e olu"e Flo = (2.06+2.66) *0-3 "3

  = 5./6*0-3 "3

u"ulati,e #i"e = (30+30) sec

  = 60 sec.

t', = (60-30) ' 2.656*0-3

  = **2%5.5* s'"3

14 7#89:

1a"$le calculation for o;ser,ation nu";er 2 at 20 $sig

8/18/2019 Sample Calculation -Constant Pressure filtration

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Fro" fig-/ (gra$h for 20-$sig $ressure)

8t is foun that the constants of filtration are<

1lo$e of cur,e a = *&**.2 s'"6

8nterce$t at a>is ; = *0326 s'"3

Don ti"e = / "in 25 sec

  = /×60+25 sec

  = 265 sec

#he o$ti"u" ,olu"e of filtrate $er ccle

o$t = √(2t'a)

  = √(2×265'*&**.2)

  = 0.55& "3

#he o$ti"u" filtration ti"e

to$t = a,o$t2'2+;,o$t

  = *&**.2 (0.55&)2 0.5+*03260.55&

  = 60*&.03 s

 Theoretical fltration time, t =a

2×V 

2+b×V 

  =1711.2

2×0.005462

2+10326×0.005462

  = 56.43 sec

8/18/2019 Sample Calculation -Constant Pressure filtration

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,erage flo rate of filtrate 

Qavg =V 

t +t d 

=0.005462

56.43+265

  = 1.7   × 10-5 m3

/sec

At room temerat!re o" #4 $eg cel

%ol!&ilit' o" (a()3 in *ater = 16.6 +g/

%o, concentration o" (a()3, (=16.6

16.6+1

  = 0.43# +g (a()3/+g sol!tion

late height= .5 inch

late *i$th= .5 inch

Area o" late, A =.5   ×9.5 in#

  = 0.#5 in#

  =0.05#3 m# 

%eci"ic ca+e resistance, 2=a A

2∆P

Cμ  

=1711.2

×0.05823

2

×137857.14

0.9432×0.00089  

=.53   ×  10 m g-1

ass o" the ca+e, * =CV 

 A

  =0.9432×0.005462

0.05823

+g

8/18/2019 Sample Calculation -Constant Pressure filtration

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  =0.05 g

(a+e esistance, c = 2*

  = .53   ×  10   × 0.05

  = .4   ×  107 m

esistance o" the fltering me$i!m, m =bA∆P

 μ  

=10326×0.05823×137857.14

0.00089   m

  = .31   ×10 10 m