L. Introduction to Chemical Engineering ETH Zürich. Page 1 of 7 FS Prof. Marco Mazzotti Written Examination,

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1 Introduction to Chemical Engineering ETH Zürich Page 1 of 7 FS L Introduction to Chemical Engineering Prof. Marco Mazzotti Written Examination, , 15:0016:00 Student Name: ETH Student Number: Student Signature: Instructions This exam contains 7 pages (including this cover page) and 4 questions. All questions have to be answered on this form. The forms have to be handed in to the instructors not later than the time specied at the beginning of the test. All questions shall be answered without using any helping material (that is pocket calculators, text books, lecture notes, or copies of slides). All cellphones have to be switched o during the examination. Do not use red as a color for writing or drawing. Attempts to deceive will be punished according to the ETH disciplinary code: Among other punishments, this examination may be declared to be failed. You don't need to reach the maximum of 40 points to obtain the maximum grade. Anweisungen Diese Prüfung besteht aus 4 Fragen auf 7 Seiten (inklusive dieser Titelseite). Alle Fragen sollen auf der Angabe beantwortet werden. Nach dem Ende der Prüfung müssen die Angaben dem Prüfungssteller abgegeben werden. Alle Fragen müssen ohne Zuhilfenahme von Hilfsmaterial (Taschenrechner, Bücher, Notizen oder Kopien der Präsentationen) gelöst werden. Alle Mobiltelefone müssen während der Prüfung ausgeschaltet sein. Bitte verwenden Sie keine rote Farbe zum Schreiben oder Zeichnen. Versuchter Betrug wird entsprechend der ETH Disziplinarordnung geahndet: Neben anderen Strafen kann versuchter Betrug zum Nicht-Bestehen der Prüfung führen. Sie müssen nicht die Gesamtanzahl von 40 Punkten erreichen, um die Höchstnote zu erhalten. Question: Total Points: Score:

2 Introduction to Chemical Engineering Page 2 of 7 Problem 1 Thermodynamic equilibrium a) (i) No mechanical equilibrium (P A P B ) (ii) T A = T B =T (iii) Using the chemical potential µ w,a (T A, P A, x A ) = µ w,b (T B, P B, x B ) µ w,a (T, P A, x A ) = µ w,b (T, P B, x B ) µ w,a (T, P A ) = µ w,b (T, P B, x B ) b) Using the isofugacity condition and the provided equation for the fugacity of a liquid: P V P A fw,a L (T, P A ) = fw,a L (T, P A ) = Pw V z 1 exp P dp + v RT dp fw,b L (T, P B, x B ) = x w,b fw,b L (T, P B ) = Pw V exp 0 P V fw,a L (T, P A ) = x w,b fw,b L (T, P B ) P A v x w,b = exp RT dp log(x w,b ) = P B 0 v RT (P A P B ) z 1 P dp + P V (T ) P B P V (T ) v RT dp Knowing P B = P A + ρg h: logx w,b = v RT (ρg h) Please notice that the molar volume v is the one of water, which does not depend on the presence of other components. c) Osmotic pressure d) Calculations are reported in the following logx w,b = v RT (ρg h) 1 x w,b = v RT (ρg h) x s,b = mol m 3 ( 1000 kg m 3 8 J mol 1 K 1 10 m s m ) = K

3 Introduction to Chemical Engineering Page 3 of 7 Problem 2 Flash I a) To construct the x y diagram one can choose at least ve points from the T xy diagram provided. The table reports some points useful for the construction of the composition diagram (please, note that it is necessary to round to the closest value to easily reconstruct the xy diagram). T [ C] x y The equilibrium curve is reported in blue in Figure 1. b) By using the mass balance for the ash one can easily obtain the operating line. Its intersection with the equilibrium curve obtained in a) will give the operating point. α = V/F = 0.5 z = (1 α)x + αy y = α 1 α x + 1 α z (1) which gives the following operating line (in red in Figure 1): y = 1.2 x The corresponding composition of the liquid and vapor phase is x = 0.4 y = 0.8 The operating temperature can be retrieved from the T xy diagram and is 87.5 C. c) The stream to be processed is the liquid one, the only one allowing to satisfy the specications. In fact, since the vapour product of the rst ash has a molar fraction y already higher than 0.7, it cannot be further processed to produce a new vapour phase with lower concentration. From the T xy diagram one can easily read the composition of the streams and the temperature of the ash chamber. x = y = 0.7 T = 92.5 (2) Using the operating line for the second ash, one gets the liquid and vapor rates.

4 Introduction to Chemical Engineering Page 4 of 7 Figure / Abbildung 1 m = α = α α = 5/17 Hence V = 25 mol/s and L = 60 mol/s. Alternatively the lever arm rule can be used, using the T xy diagram. α = 5/17

5 Introduction to Chemical Engineering Page 5 of 7 Problem 3 Chemical reactor design a) I) II) c A = (1mol/L) exp [( s 1) t ] c A = (1mol/L) exp [( s 1) (800 s) ] = (1mol/L) exp ( 1.6) = mol/l b) I) II) c A = c in A exp ( kq ) V = (1mol/L) exp [( L 1) V ] (0.1mol/L) = (1mol/L) exp ( kq ) V k Q V = 2.3 ( 0.5 L s 1 ) V = s 1 = 575 L c) I) II) c A = cin A 1 + k Q V = 1mol/L 1 + ( L 1) V = 1 + k Q V k Q V = 9 ( 0.5 L s 1 ) V = s 1 = 2250 L

6 Introduction to Chemical Engineering Page 6 of 7 Problem 4 Schröder-van Laar equation a) The Schröder-van Laar equation relates the solubility of compound i, x i, at temperature T to the solution properties, γi, the melting temperature of the solute, T m i and its melting enthalpy, h m i. b) The condition required is the equilibrium between the solute in the solid and liquid phase. Hence: f S i (T, P ) = f L i (T, P, x) f S i (T, P ) = f L i (T, P )x i γ(t, P, x) (3) c) The assumption of ideal solution leads to γ i = 1. Rewriting the Schröder-van Laar equation: ( h x m s = exp s R ( h x m w = exp w R ( 1 T m s ( 1 T m w 1 )) T 1 )) T (4) The presence of the eutectic can be justied by the presence of a temperature where the two terms on the right-hand side of the equations are the same. Hence an intersection in the two solubility lines can be identied. The qualitative trend of the phase diagram is shown in Figure 2. d) Solubility would be the same in all solvents. Figure / Abbildung 2

7 Introduction to Chemical Engineering Page 7 of 7 e) Rearranging the equations, and considering a constant α: ln(x i γ i ) = hm s R lnx i lnγ i = α lnx i A i (1 x i ) 2 = α A i (1 x i ) 2 = lnx i + α ( 1 T 1 T m s ) = α The left-hand side depends on the solvent, while the right-hand side is only solute dependent. The intersection of the curves will give the solubility. Solvent I is the one where the sugar is most soluble (lower x i ), followed by II and III. In the following graphics, for instance, α has been taken equal to 1. F1 = A i (1 x ) 2 F2 = lnx + α α = 1 with A i = +1, A i = 0, A i = 1 (5)

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