Answer to Question #186067 in Chemical Engineering for sheyi

Question #186067

CaCO3 precipitate is made by reacting CaO with an aqueous solution of Na2CO3, the by-product being NaOH. After decanting, slurry leaving the precipitation tank is 5% w/w CaCO3, 0.1% NaOH, and the balance water. 100 000 kg hr–1 of slurry is fed to a two-stage counter-current continuous washing system to be washed by 200 000 kg hr–1 of fresh water. Underflow from each unit contains 20% w/w solids. Calculate fraction of NaOH recovered and mass percentage NaOH of dried product. Would adding a third stage be worthwhile? 


1
Expert's answer
2021-04-29T07:57:55-0400



Overall Mass balance: 100 000 + 20,000=U + E


Now, overall solid balance, as 20% 20 led from both thickenes F. Xacoz + NX Cacoz = E. XCacoz + U. cacos


=> 100000×0-05 + 20000 x0 E×0 + Ux0.2 -


U= 25000 lb/hr. Putty U25OOD in B, E = 9500020/hr


Now, U': 1 (207. solid in both thickenes U/F)


So, U'= 25000 lb/hr, 20% solid (Cacoz) E' = U²+W-U (applying mass balance around 2nd


FE: 25000+20000-25000 E'= 20000 lb/hr


Applying Cantic (NaOH) belance about 1st thickener


same can be written for 2nd thickere, Now, if equilibrium in achieved in each of the thickeners.


the ratio of NaOH to wath will be same under flows & Oveffor,


ie for 1st thickenes. E.H Cassan's no soud (cacos) in O/F). E (1-NaOH) 1' (1-2 NADH)


Similarly, XH = 250 NOOH


Thus, from ) & (10), if we write for overall system (for both thickens)


хмаон = х маон


Now, overall


caustic balance gives


F. X NaOH EX NAOH + U2


100000 ×0.001 + WXD = 95000xx NaOH + 25000 × x NaOH


е лаон = 0.0008@охи маон % recovery of Naout in the extract = 0.0008x95000


0.0008 90H = 0.083%. (by wt.)


{ 2 per E F


X Naon X NaOH Naon 0.083%


1.e wt% NaOH in the dried Calog =


U. X soud + U. NADH


25000x 0.00083


25000 x 0.20 +25000x0 00 089


0.004149


1.e wt% NOOOH in dried Cacoz ( without watter)


= 0.4149%.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS