Name:____________________________ 1Final Exam CHEN 4253-Design I Open Book, Open Notes Problem 1 ____________/25 Problem 2 ____________/25 Problem 3 ____________/25 Problem 4 ____________/25 Total ____________/100Name:____________________________ 2Problem 1. (25 Points) Determine the bare module cost (CBM) for an 8 ft diameter, 70 sieve tray distillation column with each tray 1 ft tall (2 in weir height) built in carbon steel that will operate at 100 psig. Do not consider the cost of the reboiler nor the condenser just the column and its ancillary items. Assume an operating temperature of 400 F, a weld efficiency of 95% and the strength for carbon steel of 15,000 lb/in2.Name:____________________________ 4Problem 2 (25 Points) Determine the bare module cost (CBM) for a heat exchanger that is heating 10,000 kg/hr No. 6 Fuel Oil at 20 psig from 80 C to 130 C on the tube side with 50 psig steam that condenses at 147 C on the shell side. The tubes are made of carbon steel as is the shell and the tube length is 20 ft (6m) (or longer). The oil has a heat capacity of 0.8 cal/gm/C and the steam has a heat of vaporization of 2260 kJ/kg.Name:____________________________ 6Problem 3 (25 points) A chemical plant is to be constructed in 2010 with operation scheduled to begin in 2013. In 2015, the plant is projected to operated at 85% capacity with Sales $11 Million Annual Cost of Manufacture (includes dep.) $ 6 Million a.) Calculate the return on investment in 2015 given that the total depreciable capital is $17 Million and the working capital is $2 Million. Assume straight-line depreciation at 8%/yr. b.) Calculate the cash flow in 2015 and discount it to present (2010) value assuming an effective interest rate of 15%. Assume MACRS depreciation with a 10 year class life and assume periodic interest.Name:____________________________ 8Problem 4 (25 Points) You are to design an optimal high-temperature gas phase reactor that maximizes the reaction selectivity for the synthesis of ethylene oxide and also manages the heat effects. Sketch the reactor configuration and give reasons for your choices. Reaction 1 CH2=CH2 + 3O2Æ2CO2 + 2H2O Reaction 2 CH2=CH2 + O2ÅÆ CH2-CH2 (ethylene oxide, desired) Reaction Data Reaction Heat of Reaction, ΔHºrxn Activation Energy for forward reaction, EA Order with respect to ethylene (CH2=CH2) for forward reaction Order with respect to oxygen (O2) for forward reaction Order with respect to ethylene oxide for backward reaction 1 -1411 kJ/mole 20 kJ/mole 1 2 No back reaction 2 +40 kJ/mole 25 kJ/mole 1 1 1 Reaction 1 has the following forward only reaction rate -r1 = k1[CH2=CH2]1[O2]2 where k1= 5x103 (L2/mole2/s) exp[-20 kJ/mole/(RgT)] and Rg is the gas constant with a value of 8.29 J/(mole*K). Reaction 2 has the following forward and backward reaction rates -r2 = k2[CH2=CH2][O2] – k-2[Ethylene oxide] where k2= 5x106 (L/mole/s) exp[-25 kJ/mole/(RgT)] and k-2= 5x109 (1/s) exp[-35 kJ/mole/( RgT)] The equilibrium constant for reaction 2 is given by: Keq-2=k2/k-2. The reactor is made of a material that can withstand 900 K and not higher. Useful equations: Please note that the reactor volume for a plug flow reactor is given by: ∫−−−=outAinAXXAAAordXFV and that for a stirred tank reactor is \ / OName:____________________________ 9outAinAoutAAorXXFV−−−−−=)( where the rate is evaluated at the outlet conditions. The heat balance equation for a reactor given by the equation rxnpAHQTTCXΔ−−−=)('1 where C`p is the heat capacity of the un reacted feed stream, T1 is the feed temperature and ΔHrxn is the heat of reaction at the temperature