ME104_FinalExam_Solutions_Spring2009_NoRestriction

ME104_FinalExam_Solutions_Spring2009_NoRestriction -...

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NAME___ SOLUTIONS _______ INSTRUCTOR :(circle one) Kalnitsky 8am Levy 9am Smith 10am Oztekin 10am Macpherson 11am LEHIGH UNIVERSITY DEPARTMENT OF MECHANICAL ENGINEERING AND MECHANICS ME 104 — THERMODYNAMICS I Final Exam (Closed Books and Notes) SPRING 2009 ______________________________________________________________________________ INSTRUCTIONS: 1. No unauthorized books, notes, or other similar materials are permitted. 2. Use thermodynamic tables provided as necessary. 3. Information stored in the memory of a calculator can not be used in the exam and receives no credit. 4. Show all your assumptions clearly. 5. Show ALL the steps in the solution of problems, both analytical and numerical, ON PAPER to receive credit. 6. LEGIBILITY is essential or the exam can not be graded properly. ______________________________________________________________________________ Problem No. Points Grade 1 25 2 25 3 25 4 25 5 25 6 25 Total 150 The below set of thermodynamic equations may be useful. C P = R + C v C P / C v = k T dS = dU + P dV T dS = dH - V dP Δ s = C v dT / T + R ln ( v 2 / v 1 ) Δ s = C v,ave ln ( T 2 / T 1 ) + R ln ( v 2 / v 1 ) Δ s = C P dT / T - R ln ( P 2 / P 1 ) Δ s = C P,ave ln ( T 2 / T 1 ) - R ln ( P 2 / P 1 ) Δ s = s 2 0 - s 1 0 - R ln ( P 2 / P 1 ) P 2 / P 1 = P r2 / P r1 v 2 / v 1 = v r2 / v r1 P 2 / P 1 = ( v 1 / v 2 ) k T 2 / T 1 = ( P 2 / P 1 ) (k – 1)/k T 2 / T 1 = ( v 1 / v 2 ) k – 1 η C = w s / w a η T = w a / w s
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ME104 FINAL EXAMINATION NAME____ SOLUTIONS __________ Problem 1 Water enters a well-insulated double tube heat exchanger at a rate of 3 kg/min as a saturated liquid at 50 kPa and leaves at 250 ° C. The heating of the water is accomplished by heat transfer from a hot stream of air that enters the heat exchanger at 1000 ˚C and leaves at 450 ° C. Assume ideal gas behavior for the air with constant specific heats. Neglect the pressure drops of the fluids in the heat exchanger. Also, neglect changes in kinetic and potential energies. a) Calculate the mass flow rate of air. b) Calculate the rate of heat transfer to water. c) Calculate the rate of entropy generation in the heat exchanger. First law for the total heat exchanger () ( ) 0 h h m h h m W Q win wout water ain aout air = + = For air with constant specifice heats @ 1.1 kJ/kg-C: () () () kg kJ 605 C 450 1000 C kg kJ 1 . 1 T T cp h h aout ain air aout ain = = = () K kg kJ 6223 . 0 273 1000 273 450 ln K kg kJ 1 . 1 P P ln R T T ln cp s s s ain aout ain aout air ain aout air = + + = = = Δ For water-steam: K kg kJ 0912 1 s and kg kJ 54 340 h win win = = . . @ sat. water and P=0.050 MPa K kg kJ 3568 . 8 s and kg kJ 2 . 2976 h wout wout = = @ P=0.050 MPa and T=250C For mass flow rate of air, from first law energy balance: () ( ) () () () () () min . . . min
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This note was uploaded on 07/12/2010 for the course ME 104 taught by Professor Gomatam during the Spring '08 term at Lehigh University .

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ME104_FinalExam_Solutions_Spring2009_NoRestriction -...

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