Introductory Chemical Thermodynamics Equations - Final Exam | CHE 21100, Exams of Chemistry

Download Introductory Chemical Thermodynamics Equations - Final Exam | CHE 21100 and more Exams Chemistry in PDF only on Docsity! CHE211 CHEMICAL ENGINEERING THERMODYNAMICS Final Exam Fall Semester 2008 Time Allowed: 120 minutes Total Marks: 100 marks Open Book Exam Answer all questions Formula sheet and Thermodynamic data supplied. Any calculators allowed. SHOW ALL WORKING ON THE EXAM PAPER. STUDENT NAME: ______________________________ PUID: ______________________________ QUESTION 1 (a) During an isentropic, steady flow expansion, it is possible to prove that: VdPdH  . Derive an expression for work done per kg of fluid for an isentropic, steady flow expansion from P1 to P2 for a fluid which obeys the following truncated Virial Equation of State: 2 22 )( )( 1 RT PBC RT BP Z   where B and C are the second and third virial coefficients. [Hint: start by writing down the first law for a steady flow, isentropic expansion (turbine).] (10 marks) (b) Outback Australia is sunny and hot for most of the year. A thermal solar power plant claims to generate 5kW of electricity for every 50kW of solar energy captured by a specially designed solar pond. The pond can be considered a constant temperature heat reservoir at 60°C. Although the atmospheric temperature is very high, by taking advantage of evaporative cooling, we can exchange heat in the condenser of our power plant at 20°C. Evaluate this claim by comparing the efficiency of this power plant to a Carnot cycle using the same temperature reservoirs. (10 marks) (c) Steam at 5000 kPa and 500°C is expanded through a simple adiabatic throttling valve. At the inlet conditions, the enthalpy and entropy of the steam are 3433.7 kJ/kg and 6.9770 kJ/kg.K respectively. a. What does the first law tell us about the enthalpy of the exit steam? b. What does the second law tell us about the entropy of the exit steam? State clearly any assumptions you make. (10 marks) (d) We can write a fairly general expression for vapor-liquid equilibrium for a mixture as follows, provided we assume the gas phase can be treated as an ideal solution:         RT PPV PxPy sat i l isat iiiii )( exp A binary mixture of ethanol and water has an azeotrope at 0.89 mole fraction ethanol at atmospheric pressure and a temperature of roughly 90°C. Simplify the above expression as much as is reasonable for use in solving VLE problems for the ethanol/water mixture at atmospheric pressure. State clearly the assumptions you make to simplify the equation and justify each assumption qualitatively with one sentence only. (10 marks) [Hint: the math is trivial. The key is to clearly state and justify your assumptions.] Total = 40 marks QUESTION 2 (a) I am exploring the idea of separating a binary mixture by multiple flash separation. The mixture is throttled to a low pressure where it separates into a two phase mixture. The more volatile component will concentrate in the vapor phase. A flash tank is used to separate the vapor from the liquid. I repeat the process by recompressing the vapor phase and then flashing it into a second tank and so forth. Consider one flash separation stage only. I have a mixture of 80 mole% benzene and 20 mole% Toluene. This mixture is flashed to a pressure of 1 Bar. The measured temperature in the flash tank is 85°C. Under these conditions calculate the composition of the liquid and vapor phases and the mole ratio of vapor to liquid. The vapor pressures of the two component are given by: Benzene: 572.217/ 81.2726 7819.13ln        CtkPa P sat  Toluene: 625.217/ 96.3056 9320.13ln        CtkPa P sat  You may assume that this mixture obeys Raoult’s Law. (14 marks) (b) Unlike the benzene-toluene system, the benzene-cyclohexane binary system forms an azeotrope at 0.525 mole fraction benzene at a temperature of 77.6°C and a total pressure of 1.013 Bar (atmospheric pressure). At this temperature, the vapor pressure of pure cyclohexane is 0.980 Bar and that for benzene can be calculated from the expression given above. Sketch the P-x-y diagram for the benzene-cyclohexane system making use of your knowledge of the liquid and vapor compositions and total pressure at xbenzene =0.0, 0.525 and 1.0. Label the regions that are liquid, vapor, and liquid-vapor mixture. (6 marks) Total = 20 marks WORKING FOR QUESTION 2 HERE WORKING FOR QUESTION 2 HERE WORKING FOR QUESTION 3 HERE QUESTION 4 Consider a simple, gas phase decomposition reaction: A(g) →2B(g) This reaction takes place at 1000K and 10 Bar. Initially there are 5 moles of A and no moles of B present. The standard Gibbs energy and enthalpy of formation at 25°C A and B are: molJH molJG molJH molJG Bf Bf Af Af /600 /100 /200 /500 0 , 0 , 0 , 0 ,     (a) Calculate the value of the equilibrium constant K at 1000K assuming the temperature effect can be predicted from the van’t Hoff equation. (5 marks) (b) Calculate the final mole fraction of component A assuming (i) the system reaches equilibrium, and (ii) the gas mixture behaves as an ideal gas. (10 marks) (c) I want to increase the conversion of A in this reaction. Propose one way of doing this by varying the reaction conditions. Explain why your suggestion will work in one or two sentences. (5 marks) [Hint: If you have trouble calculating K in part (a), assume a reasonable value for K in solving parts (b) and (c).] WORKING FOR QUESTION 4 HERE SdTVdPdG SdTPdVdA VdPTdSdH PdVTdSdU     fRTddG ln Pf    10 0 10 lnln 1ln      P dP ZZ P satsat satl satsat sat P RT PPV P f f          )( exp              i ii i ii nTPi i dxSdTVdPdG dnnSdTnVdPnGd n nG j    )( )( ,, iiiii fxPy  ˆ           i i iiii id i i iiii id i ii id i id i i ii id i id i xxRSxS xxRTGxG VxVVV HxHHH ln ln ; ; P P x y K sat i i i i  22 11 12 xy xy  iv i i i f f RT G K                 0 0 ˆ exp          TT H K K 11 ln 0   )(ˆ 0 TK P P y iv i ii           Constants and Conversions                  1 2 ln 1 015.273 81.9 1013251 101 11 314.8 2 1 2 5 2 11 x x dx x CK msg PaAtm Pabar NmPa KJmolR x x  Table E.1; Values of Z° y= T, 0.30 0,35 0.40 0.45 030 OSS 0.60 0.65 0.70 0,78 0.90 O85 0.90 0.93 0.95 0.97 0.98 Ow 100 101 1.02 LOS Lo Lis 1.20 130 140 1.30 1.60 1.70 80 1.90 0.0100 0.0029 O06 0.0024 0.0022 0.0021 0.9804 0.9849 O.9881 0.9904 0.9922 0.9935 0.9956 0.9954 0.9959 0.9961 0.9963 0.9965 0.9966 0.9967 0.9968 0.9969 0.9971 0.9975 0.9978 0.9981 0.9985 0.9983 0.4991 0.9993 0.9994 0.9995 0.9996 0,9997 0.9998 0.9999 1.0000 1.0000 1.0000 1.0001 1.0001 0.1000 =9<.2000 04000 6.6000 0.0290 0.0261 0.0239 0.0221 0.0207 @.0195 O.0186 0.0178 0.8958 0.9165 0.9319 0.0436 0.0579 0.0522 0.0877 0.0442 0.0813 0.0390 0.0372 0.0356 O.0344 0.0336 0.8539 0.8810 0.9015 OSS 0.9174 0.9227 0.9253 0.9277 0.9300 0.9322 0.9343 0.9401 0.9485 0.9554 O.9611 0.9702 0.9768 0.9818 0.9856 0.9886 0.9910 OU158 01757 O1043 OIS6S 0.0953 0.1429 0.0882 0.1322 0.0825 0.4250 0.0778 01166 0.074 0.1109 00710 01063 0.0687 0.1027 0.0670 0.100) O.0661 0.0985 0.0661 0.0983 0.7800 = 0.1006 0.3059 0.6635 0.8206 0.6067 0.8336 0.7240 0.8398 0.7360 O84SS 0.7471 O8509 0.7574 0.8561 0.7671 O.8610 0.7761 O8743 0.8002 0.8930 0.8323 0.9081 0.8576 0.9205 0.8779 0.9396 0.9083 0.9534 0.9298 0.9636 0.9456 09714 06,9575 O9775 0.9667 0.9823 0.9739 09861 0.9796 0.9892 0.9842 09937 0.9910 09969 0.9957 0991 G.9990 10007 1.0013 10018 1.0030 1.0035 1.0055 LOO$3 1.0066 0.2000 O2HS o.1904 0.2762 OL047 0.1553 0.1476 0.1415 0.1366 0.1330 0.1307 0.1391 0.4321 0.1339 01410 O.S580 0.5887 0.6138 0.6355 0.6542 0.6710. 0.7130 0.7649 0.8330 08764 0.9062 0.9278 0.9439 0.9563 0.9659 0.9735 0, 9886 0.9948 0.9990 1.0021 1.0043 1007S 1.0090 0.2892 0.2604 0.2379 02200 0.2056 0.1939 0.1842 0.1765 0.1703 0.1656 0.1626 0.1614 0.1630 0.1664 O.1NS O.1779 0.1884 0.1959 0.2901 OAR 05146 0.6026 0.6830 0.7443 0.7838 0.8438 0.8827 0.9103 0.9308 0.9463 0.9583 09734 1.0031 10087 1O0T Lous