7 replies to this topic

[nikosv]

As a result of laboratory filtration of the CaCo3 suspention at the constant pressure ΔP1=1,5 bar it was obtained (t=20^o)
T[min] V[dm³]
10 25
20 41

Filtration (t=20^o) will be made in the large filtre if filtration area will be 10 times larger than in the laboratory filter and the pressure will be ΔP2=5 bar. Cake is incompressible.

Calculate
1) the volume of the filtrate obtainece after 1 hour in the large filter.
2) filtration rate after 1 hour.



V²+2CV=KT

2,5²+2C*25=K*600 (sec)
4,1²+2C*41=K*1200 (sec)

C=2,3944 (m³???)
K=0,03037 (m⁶/s ???)

-----
(K2/K1)=(10A₁/A₁)2*(ΔP₂/ΔP₁)

(C2/C1)=(10A₁/A₁)

so ..
K2=10,12323

C2=23,944

so the volume of the filtration in the large filter must be ..
V²+2C₂V=K₂T (T=3600sec=1 hour)

V²+47,888V-36443,638=0

αx2 + βx + γ = 0


so i find x₁,x₂=(+)168,45 m3,(-)-432,68 m3
So the volume of the filtration must be 168,45 (because this is the possitive one)

to calculate the filtration rate

V²+2C2V=K2T

=> 2V*(dV/dT)+2C2*(dV/dT)=K2

=> (dV/dT)=(K2/2*(V+C2)

So ..

(dV/dT)=0,02630859


IS THIS CORECT ??
CAN YOU HELP ME PLEASE ??

Edited by nikosv, 21 March 2010 - 06:40 PM.

[kkala]

Units need special care in such calculations to avoid confusion.
Laboratory data (25 cm3 at 10 min & 41 cm3 at 20 min) is adequate to specify C=23.94 cm3 and K=182.2 cm6, by solving the system of two equations (linear to C and K).
Industrial data will be represented by: K2=182.2*10^2*(5/1.5)=60733 cm6 = 6.073E-8 m6; C2=23.94*10=239.4 cm3 = 2.394E-4 m3 (according to the formulas used).
Then V2= -C2+SQRT(C2^2+K2*T), etc.

Edited by kkala, 21 March 2010 - 05:13 PM.

[nikosv]

Units need special care in such calculations to avoid confusion.
Laboratory data (25 cm3 at 10 min & 41 cm3 at 20 min) is adequate to specify C=23.94 cm3 and K=182.2 cm6, by solving the system of two equations (linear to C and K).
Industrial data will be represented by: K2=182.2*10^2*(5/1.5)=60733 cm6 = 6.073E-8 m6; C2=23.94*10=239.4 cm3 = 2.394E-4 m3 (according to the formulas used).
Then V2= -C2+SQRT(C2^2+K2*T), etc.

Thank you Kosta for your reply ..
it was realy helpfull ..
i did some changes to the post ..
with the new results ..
can you check the way of solving this if its correct ?? [i cant understand why you used this eqn: V2= -C2+SQRT(C2^2+K2*T)]

thank you again (ευχαριστω .. !!)

[breizh]

Hi NIKOS ,

I found this website very helpful
http://www.filtratio.../simulation.htm


Breizh

[kkala]

i did some changes to the post ..with the new results ..can you check the way of solving this if its correct ?? [i cant understand why you used this eqn: V2= -C2+SQRT(C2^2+K2*T)]

Units need special care in such calculations to avoid confusion. This concerns also me, since K=182.2 cm6/min in the previous post (not 182.2 cm6). Considering your new calculation, answer is revised as follows .
Laboratory data (V=25 dm3 at T=10 min & V=41 dm3 at T=20 min) is adequate to specify C=23.94 dm3 and K=182.2 dm6/min by solving the system of two equations (linear to C and K).
I believe V is not flow rate but total volume from T=0 (flow rate is decreasing with time, not V). You can use 25*(25+2C)/10=K=41*(41+2C)/20, so C=23.94 dm3 and K=182.2 dm6/min .
Industrial data will be represented by: K2=182.2*10^2*(5/1.5)=60733 dm6/min = 3.644 m6/h; C2=23.94*10=239.4 dm3 = 0.2934 m3 (according to the formulas used).
Then V2= -C2+SQRT(C2^2+K2*T), etc.
V= -C+SQRT(C^2+K*T) is the positive root of V^2+2*C*V-K*T=0. Inserting K2, C2 as above and T=1 h, one can find
V=0.2934 m3 + SQRT(0.2934^2+3.644*1 m6)=2.225 m3. dV/dT can give the flow rate at any time T .
Hopefully, Nikos, above is helpful. I dare say many mistakes are due to unit conversions (even by experienced engineers); skill can be got by exercise and writing the units beside any formula used.

Edited by kkala, 22 March 2010 - 02:46 AM.

[nikosv]

Hi NIKOS ,

I found this website very helpful
http://www.filtratio.../simulation.htm


Breizh

i have checked this web site ..
but i was not helpfull at all .. because the data that are given to analyse are just volume,time,pressure ..

thanks ..

@kostas i ll check it again later and i ll reply ..
thanks for you time and your help all of you ..

Edited by nikosv, 22 March 2010 - 10:54 AM.

[nikosv]

i did some changes to the post ..with the new results ..can you check the way of solving this if its correct ?? [i cant understand why you used this eqn: V2= -C2+SQRT(C2^2+K2*T)]

Units need special care in such calculations to avoid confusion. This concerns also me, since K=182.2 cm6/min in the previous post (not 182.2 cm6). Considering your new calculation, answer is revised as follows .
Laboratory data (V=25 dm3 at T=10 min & V=41 dm3 at T=20 min) is adequate to specify C=23.94 dm3 and K=182.2 dm6/min by solving the system of two equations (linear to C and K).
I believe V is not flow rate but total volume from T=0 (flow rate is decreasing with time, not V). You can use 25*(25+2C)/10=K=41*(41+2C)/20, so C=23.94 dm3 and K=182.2 dm6/min .
Industrial data will be represented by: K2=182.2*10^2*(5/1.5)=60733 dm6/min = 3.644 m6/h; C2=23.94*10=239.4 dm3 = 0.2934 m3 (according to the formulas used).
Then V2= -C2+SQRT(C2^2+K2*T), etc.
V= -C+SQRT(C^2+K*T) is the positive root of V^2+2*C*V-K*T=0. Inserting K2, C2 as above and T=1 h, one can find
V=0.2934 m3 + SQRT(0.2934^2+3.644*1 m6)=2.225 m3. dV/dT can give the flow rate at any time T .
Hopefully, Nikos, above is helpful. I dare say many mistakes are due to unit conversions (even by experienced engineers); skill can be got by exercise and writing the units beside any formula used.

i understand the eqn "V= -C+SQRT(C^2+K*T)"

in my exersise .. i calculate all the units in the S.I. (m,sec,..etc)
so i tranform dm to m , min and hour into sec .. ect ..
this is my final edit for this exersise .. i hope that its correct through the end .. !!
..

thanks a lot for your help ..
i hope you ll help me again in any problem i ll have .. !!

[kkala]

Nikos, key for correct results is the constants C and K from the laboratory test. In the SI units, C=23.94 dm3 = 0.02394 m3 and K=182.2 dm6/min = 3.0367 m6/s are different to the values used in your calculation.
In SI units lab test gives: V=0.025 m3 at T=600 s and V=0.041 m3 at T=1200 s.
According to them, values of constants C and K of V^2+2*V*C=K*T are found to be exactly as above (after solving the relevant system).
E.g. system equations are 0.025^2+2*0.025*C=K*600, as well as 0.041^2+2*0.041*C=K*1200 in SI, (C in m3, K in m6/s).
Attention to this point may be worthwhile.