1. (25 points) C 6 H O 2 6CO 2 + 7H 2 O C 6 H O 2 6CO + 7H 2 O

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1 MEEBAL Exam 2 November 2013 Show all work in your blue book. Points will be deducted if steps leading to answers are not shown. No work outside blue books (such as writing on the flow sheets) will be considered. No outgoing text messages are allowed during the exam. You must use given stream numbering in the problems (no points given if different numbering system is used). Report all answers with three significant digits. All pressures are absolute unless otherwise specified. You are allowed to use two pages of notes and a calculator (no textbooks, computers or tablets such as ipads are allowed). You must pass in your test sheet with your blue book for your exam to be graded (put your name on the exam sheet). 1. (25 points) C 6 H O 2 6CO 2 + 7H 2 O C 6 H O 2 6CO + 7H 2 O Stream 1 (100 mol/s) containing hexane (C 6 H 14 ) is burned with excess air (Stream 2). The dry basis molar composition of the product gas (Stream 3) is 7% CO 2, 2% CO, 0.265% C 6 H 14, and the remainder is O 2 and N 2. The stack gas in Stream 3 leaves at a pressure of 760 mm Hg. Assume the gases are ideal. Antoine s equation constants for water: A= , B= , C= (T in C, P in mm Hg) Calculate: a. (5 pts) Fractional conversion of hexane b. (5 pts) Mole fraction of water in Stream 3 (wet basis) c. (5 pts) Dew point of stack gas ( C) d. (5 pts) Molar flow rate of N 2 in Stream 3 (mol/s) e. (5 pts) Percent excess air fed in Stream 2 (%)

2 MEEBAL Exam 2 November (25 points) 100 mol/s of C 3 H 8 (Stream 1) is completely burned in an adiabatic reactor with 25% excess air (Stream 2) in the following reaction: C 3 H 8 (g) + 5O 2 (g) à 3CO 2 (g) + 4H 2 O (v) No CO is formed. C 3 H 8 (Stream 1) enters the reactor at 25 C, and the air (Stream 2) is preheated to 200 C. All products of the combustion reaction leave the reactor as gas (Stream 3). Assume a reference of C 3 H 8 (g), O 2 (g), N 2 (g), CO 2 (g) and H 2 O (v) at 25 C and 1 atm for the energy balance. Heat capacities: CO 2 : C p = kj/mol C H 2 O (v): C p = kj/mol C O 2 : C p = kj/mol C N 2 : C p = kj/mol C ΔĤ rxn = kj/mol Calculate: a. (5 pts) Molar flow rate of O 2 in Stream 3 (mol/s) b. (5 pts) Mole fraction of H 2 O in Stream 3 c. (5 pts) Temperature of Stream 3 for the adiabatic reactor ( C) d. (5 pts) Temperature of Stream 2 for the adiabatic reactor if the temperatures of Streams 1 and 3 are 25 and 1500 C, respectively ( C) e. (5 pts) Q for a non-adiabatic reactor if the temperatures of Streams 1, 2 and 3 are 25, 200 and 1000 C, respectively (kw)

3 MEEBAL Exam 2 November (25 points) Methanol (CH 3 OH) is produced in the following reaction: CO + 2H 2 CH 3 OH A fresh feed of CO (g) and H 2 (g) at 25 C, 5 atm is fed to a continuous vapor phase reactor in stoichiometric proportion at a rate of 18 m 3 /h (Stream 1). The product stream [CH 3 OH (v)] emerges at 127 C. Heat is removed from the reactor at a rate of 20 kw, which is used to heat water (Stream 3) from 35 C to saturated steam at 9 bar. Assume a reference state of H 2 (g), CO (g) and CH 3 OH (v) at 25 o C for the energy balance, and assume pressure contributions to the enthalpy are negligible. Standard heats of formation at 25 C: ΔH f (CH 3 OH) = kj/mol ΔH f (CO) = kj/mol Specific enthalpies: H CO (127 C) = 2.99 kj/mol H H2 (127 C) = kj/mol H H2O (35 C) = 104 kj/kg H H2O (Saturated Stream at 9 bar) = kj/kg Heat capacity constants (Cp=a+bT+cT 2 ): Cp CH3OH (kj/(mol o C)): a = 42.93x10-3, b = 8.301x10-5, c = -1.87x10-8 Calculate: a. (5 pts) Molar flow rate of CO in Stream 1 (mol/h) b. (5 pts) Specific enthalpy of methanol in Stream 2 using the reference conditions specified in the problem statement (kj/mol) c. (5 pts) Fractional conversion of CO d. (5 pts) Mole fraction of each component in Stream 2 e. (5 pts) Mass flow rate of water in Stream 3 (kg/h)

4 MEEBAL Exam 2 November (25 points) A mixture of liquid pentane (30 mol/s) and liquid hexane (70 mol/s) at 25 C and 6 atm are fed into an evaporator (Stream 1). In the evaporator, some of the feed liquid is vaporized. The temperature of the evaporator is maintained at 60 C by the addition of heat. The pressure of the evaporator is 1 atm. Assume Stream 2 is an ideal gas. Assume a reference of pentane and hexane (both as liquids) at 25 C and 6 atm for the energy balance. Assume the enthalpies are independent of pressure. Antoine equation constants for pentane: A = , B = , C = (T in C, P in mm Hg) c p, pentane (l) = 155.4x10-3 (kj/mol C) c p, pentane (v) = 114.8x x10-5 T (kj/mol C) T b, pentane = C ΔĤ v, pentane (T b ) = kj/mol Antoine equation constants for hexane: A = , B = , C = (T in C, P in mm Hg) c p, hexane (l) = 216.3x10-3 (kj/mol C) c p,hexane (v) = x x10-5 T x10-8 T 2 (kj/mol C) T b,hexane = C ΔĤ v, hexane (T b ) = kj/mol Calculate: a. (5 pts) Mole fraction of pentane in Stream 2 b. (5 pts) Volumetric flow rate of Stream 2 (L/s) c. (5 pts) Specific enthalpy of pentane in Stream 2 using the reference conditions specified in the problem statement (kj/mol) d. (5 pts) Heat added to evaporator (kw) e. (5 pts) Temperature of Stream 1 to achieve the same concentrations of hexane and pentane in Streams 2 and 3 if Q = 0 and the evaporator temperature is 60 C ( o C)

Fundamentals of Combustion

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