CHEN 200 Final with Solutions, Spring 2011

CHEN200 Introduction to Chemical Engineering FINAL - SOLUTIONS 1 9th August 2010 P1. [20 marks, 2 marks for part (a), 4 marks for (b), 6 marks for (c) and 8 marks for (d)] (a) Convert density in lbm/ft3A to kg/m3 = 85.3/(2.2*2.832*10-2)= 1369.1 kg/m3 Volume of pond = 50*15*2 = 1,500 m3. Therefore the mass of the pond is= 2.05356*106 kg Weight = 9.81*2.05356*106 = 20.15*106 N (b) (i) (ii) (c) If US gallons were used the answer would be 5770 planes. (d) (i) 200 crystal/min.mm; 10 crystals/min.mm2. (ii) CHEN200 Introduction to Chemical Engineering FINAL - SOLUTIONS 2 (iii) P2. [20 marks, 3 marks each for parts (a) and (b), 6 marks for (c) and 8 marks for (d)] (a) (i) 2600*1.934*10-2=50.284 psi, (ii) (275/33.9)*101.325=822kPa, (iii) (3.00*1.01325*105)/(1002)= 30.40 N/cm2, (iv) 1atm – (20*10/760) = 0.737 atm, (v) 25.0*51.71= 1292.75 mm Hg (gauge), (vi) (25.0+14.696)*51.71= 2052.68 mm Hg (absolute). (b) C (i) T = -10 oF to oR, oC, K, T =-10oF+459.7=449.7oR/1.8=249K−273=-23°C (ii) ΔT = 85 oC = 85K, = 85 oC*1.8= 153oF= 153oR, (iii) ΔT = 150 oR =150oF/1.8 =83.3oC=83.3K (c) (i) ��� � 14.32� � �� � 10.45����� � 459.4� (ii) 0.316 lb-mole of gas and 2.65 lbm of CO (iii) 2735.5oF . The ideal gas equation is an approximation of the real behavior of gases where several simplifying assumptions are made (no interactions between the molecules, no interaction with the surface, etc.). The ideal gas equation does not apply at high temperatures and pressures when the conditions approach the critical point. (d) Assumptions: Standard conditions, Density of water is 1,000 kg/m3, Basis is 1 m3 Abu Dhabi Kharg Island MW PPM (mg/L) Weight F Mol F PPM (mg/L) Weight F Mol F Chloride (Cl) 35.453 24,300 0.55 0.50 23,000 0.56 0.50 Sulfate (SO4) 96.064 3,420 0.08 0.03 3,100 0.07 0.02 Bicarbonate (HCO3) 61 200 0.005 0.002 171 0.004 0.002 Carbonate (CO3) 60 0 0 0 24 0.001 0.0003 Sodium (Na) 23 13,900 0.32 0.44 13,044 0.32 0.44 Magnesium (Mg) 54.94 1,560 0.04 0.02 1,500 0.04 0.02 Calcium (Ca) 40.08 600 0.014 0.01 520 0.013 0.01 SG 1.029 1.031 No. of moles 1,372 1,292 TDS 43,980 41,359 Weight % 4.27% 4.01% CHEN200 Introduction to Chemical Engineering FINAL - SOLUTIONS 3 P3. [25 marks, 8 marks each for parts (a) and (b), 6 marks for (c) and 3 marks for (d)] 2C2H4 O2 2C2H4O C2H4 3O2 2CO2 � 2H2O (a) Basis: 100 mol fed to reactor/h ⇒ 25 mol O2/h, 75 mol C2H4/h Reactor 5 unknowns (n1 - n5) -3 atomic balances -1 - % yield -1 - % conversion 0 D.F. Strategy: 1. Solve balances around reactor to find n1- n5 2. Solve balances around mixing point to find nO2, nC2H4 (1) % Conversion ⇒ n1 = .800 * 75 (2) % yield: (0.200)(75) mol C2H4 � �� ��� ����� ������ ���� = n3 (production rate of C2H4O) (3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4 (4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5 (5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5 (6) O2 balance (mix pt): nO2 = 25 – n2 (7) C2H4 balance (mix pt): nC2H4 = 75 – n1 Overall conversion of C2H4: 100% (b) Calculate the quantities specified in part (b). n1 = 60.0 mol C2H4/h , n2 = 13.75 mol O2 /h, n3 = 13.5 mol C2H4O/h n4 = 3.00 mol CO2/h, n5 = 3.00 mol H2O/h, nO2 = 11.25 mol O2/h nC2H4 = 15.0 mol C2H4/h, 100% conversion of C2H4 (c) CHEN200 Introduction to Chemical Engineering FINAL - SOLUTIONS 4 P4. [15 marks, 6 marks parts (a) and 9 marks for part (b)] (a) (i) Pfinal = 243 mm Hg . Since liquid is still present, the pressure and temperature must lie on the vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the (ii) Assuming ideal gas behavior for the vapor, (b) Ethyl acetate has a vapour pressure of 118.3 mm Hg at 29.5oC and a normal boiling point of 77.0oC. Estimate the vapour pressure at 45oC using (i) the Antoine equation: (ii) (iii) CHEN200 Introduction to Chemical Engineering FINAL - SOLUTIONS 5 P5. [20 marks, 8 marks part (a) and 12 marks for part (b)] C2H4(g) ½ O2(g) C2H4O(g) C2H4 � 3O2 � 2CO2 � 2H2O Basis: 2 mol C2H4 fed to reactor (a) (b) References for enthalpy calculations :C (s) , H2 (g) , O2 (g) at 25 oC CHEN200 Introduction to Chemical Engineering FINAL - SOLUTIONS 6 Bonus Problem [10 marks]