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Assignment - 2 Submitted By:- Jasreen Kaur (40516101416) BTech CT 4th Year Q1 ) 10 litres of a dilute aqueous solution of a hormone (concentration = 0.1 g/1) was contacted with 1 litre of an organic solvent at 20 °C. The solute concentration in the extract thus obtained was found to be 0.7 g/1. What would have been the solute concentration in the extract had the extraction been carried out at 4°C? State assumptions required. Ans :-
Raffinate Cr 10l
Feed 10l, C = 0.1 g/l
S ,1 Litre Organic Solvent
1l , 0.7 g/l ,Ce
Solute Balance
10*0.1 = 1*0.7 + 10*Cr 10Cr = 1- 0.7 Cr = 0.03
K = Ce/Cr = 0.7/0.03 = 23.334 K = Exp(Mur - Mue/RT)
( T = 20C = 293K and R = 8.314 )
LnK = (mUR - mUE/8.314*293) Ln(23.334) = ((Mur - Mue)/8.314*293) Mur - Mue = 7673.1212
If The Experiment was Done at 4C I.e. 273K K = Exp((Mur - Mue)/RT) K = Exp (7673.1212/8.314*277) Hence , K = 27..9942
Hence Ce/Cr = 27.9942
-1
10*0.1 = 1*Ce + 10*cR
- 2
From 1 & 2 1 = 27.9942 cR + 10Cr 1= 37.9942*Cr Hence Cr = 0.0263 Ce = 0.7368 The Concentration Of Solute in Solution at 4.C is 0.7368 Q2 ) A differential extractor is being used to extract an amino acid from an aqueous solution (concentration = 1 g/1) to an organic solvent. The extractor is 1 m long, the feed flow rate is 0.01 m 3/s, the extracting solvent flow rate is 0.005 m3/s, the average concentration difference across the stagnant film on the raffinate side is 0.1 g/1, and the raffinate concentration is 0.2 g/1. Calculate the mass transfer coefficient on the raffinate side if the interfacial area per unit height of the extractor is 100 m2/m. State assumptions made. Feed Flow Rate(F) = 0.01m^3/s , C=1g/l
Ans
ES Flow Rate(S) = 0.005m^3/s,C = 0.1 g/l
Extracted Liquid Er = ? To be Found Out
By General Material Balance F+S=R+E 0.01 + 0.005 = R + E R + E = 0.015 Now Doing Component Balance
Raffinate Here Conc Difference ΔCr = 0.1 g/l
(0.01)8(1) + (0.005)*(0) = (0.2*10^3)8R + E8(Ce) 10 = 200*R + E*Ce From Equilibrium Conditions Ce = K*Cr Ce = K*(0.2)*10^3 10 = 200*R + 200*K*E 10/200 = R + K*E Now Solving Equations 0.05 = R + K*E And 0.015 = R + E Hence We Obtain 0.035/(K-1) = E Hence R = 0.015 - E R = 0.015 - 0.035/(K-1) R = (0.015K - 0.015 - 0.035)/(K-1) R = (0.015K - 0.05)/(K-1) R = (0.015K - 0.05)/(K-1) Now For Difference in Concentration Across the Raffinate Portion We use Integration Integrated From Cr0 to Cr(dCr/Cr0 - Cr) = (Kr/R)^2 Hence After Integration We Obtain 2*R/100 = K Note here I have Left the Answer in terms of K because the Value of K is not Given.Also I found out that there was no example of Extraction where Temperature Change is Given,But this Question uses that.Hence I am Not Sure about the Authenticity of this Answer.