meebal 2013 exam 2 final version - omega chi epsilon at...

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)

C6H14 + 9.5O2 → 6CO2 + 7H2O

C6H14 + 6.5O2 → 6CO + 7H2O

Stream 1 (100 mol/s) containing hexane (C6H14) is burned with excess air (Stream 2). The dry basis molar composition of the product gas (Stream 3) is 7% CO2, 2% CO, 0.265% C6H14, and the remainder is O2 and N2. 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=8.10765, B=1750.286, C=235.00 (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 N2 in Stream 3 (mol/s) e. (5 pts) Percent excess air fed in Stream 2 (%)

MEEBAL Exam 2 November 2013 2. (25 points)

100 mol/s of C3H8 (Stream 1) is completely burned in an adiabatic reactor with 25% excess air (Stream 2) in the following reaction:

C3H8 (g) + 5O2 (g) à 3CO2 (g) + 4H2O (v)

No CO is formed. C3H8 (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 C3H8 (g), O2 (g), N2 (g), CO2 (g) and H2O (v) at 25 °C and 1 atm for the energy balance.

Heat capacities:

CO2: Cp = 0.050 kJ/mol·°C

H2O (v): Cp = 0.039 kJ/mol·°C

O2: Cp = 0.033 kJ/mol·°C

N2: Cp = 0.031 kJ/mol·°C

ΔĤ°rxn = -2044 kJ/mol

Calculate:

a. (5 pts) Molar flow rate of O2 in Stream 3 (mol/s) b. (5 pts) Mole fraction of H2O 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)


Methanol (CH3OH) is produced in the following reaction:

CO + 2H2 → CH3OH

A fresh feed of CO (g) and H2 (g) at 25 °C, 5 atm is fed to a continuous vapor phase reactor in stoichiometric proportion at a rate of 18 m3/h (Stream 1). The product stream [CH3OH (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 H2 (g), CO (g) and CH3OH (v) at 25 oC for the energy balance, and assume pressure contributions to the enthalpy are negligible.

Standard heats of formation at 25 °C:

ΔHf (CH3OH) = -201.2 kJ/mol

ΔHf (CO) = -110.52 kJ/mol

Specific enthalpies:

HCO (127 °C) = 2.99 kJ/mol

HH2 (127 °C) = 2.943 kJ/mol

HH2O (35 °C) = 104 kJ/kg

HH2O (Saturated Stream at 9 bar) = 2772.1 kJ/kg

Heat capacity constants (Cp=a+bT+cT2):

CpCH3OH (kJ/(mol oC)): 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)


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 = 6.84471, B = 1060.793, C = 231.541 (T in °C, P in mm Hg) cp, pentane (l) = 155.4x10-3

(kJ/mol·°C) cp, pentane (v) = 114.8x10-3 + 34.09x10-5 T (kJ/mol·°C) Tb, pentane = 36.07 °C ΔĤv, pentane (Tb) = 25.77 kJ/mol Antoine equation constants for hexane: A = 6.88555, B = 1175.817, C = 224.867 (T in °C, P in mm Hg) cp, hexane (l) = 216.3x10-3

(kJ/mol·°C) cp,hexane (v) = 137.44x10-3 + 40.85x10-5 T - 23.92x10-8 T2 (kJ/mol·°C) Tb,hexane = 68.74 °C ΔĤv, hexane (Tb) = 28.85 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 (oC)

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meebal 2013 exam 2 final version - omega chi epsilon at...

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