exam1_sol_2006

exam1_sol_2006 - ENGRD 221 Thermodynamics (Prof. N....

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ENGRD 221 –Thermodynamics (Prof. N. Zabaras) Prelim I 10/12/2006 Page 1 of 9 Thursday October 12, 7:30 pm – 9:30 pm Closed books and notes. Wireless or any other form of internet/phone connection is strongly prohibited during the exam. Answer all questions. Make sure your answers are legible. Circle your final answer. The TAs and instructor will not respond to any questions during the exam. If you think that something is wrong with one of the problems below, please state your concern in your exam books. Problem 1 (15 points) Multiple choice questions. Answer all questions – explanations are not needed and will not be graded, only provide your answer. (1) (3 points) A 2 kW electric resistance heater submerged in 5-kg water is turned on and kept on for 10 minutes. During the process, 300 kJ of heat is lost from the water. Assuming the specific heat of water to be 4.18 kJ/kg. o C, the temperature rise of the water is (a) 0.4 o C (b) 43.1 o C (c) 57.4 o C (d) 71.8 o C (e) 180.0 o C Answer : Applying the first law of thermodynamics for this system Q – W = Δ U; Q = -300 kJ, W = -2kW x 600 sec 1 kJ 1 kW s = -1200 kJ Δ U = mC Δ T (m = mass of water = 5 kg, c = specific heat = 4.18 kJ/kg. o C) Therefore, Δ T = 43.1 o C (2) (3 points) An ideal gas process for which P v k = constant, where k=c p /c v is an (select the correct answer or answers) (a) incompressible process (b) isobaric process (c) adiabatic process (d) isentropic process (e) isothermal process Answer: The correct answers are (c,d) (3) (3 points) When a closed system with internal irreversibilities undergoes a process from state 1 to state 2, the entropy change, S 2 – S 1 , and entropy production, σ , can be a) negative and positive, respectively.
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ENGRD 221 –Thermodynamics (Prof. N. Zabaras) Prelim I 10/12/2006 Page 2 of 9 b) negative and zero, respectively. c) positive and negative, respectively. d) positive and zero, respectively. e) all of the above. Answer: (a) negative and positive respectively; σ is always positive for this process. (4) (3 points) Air (considered an ideal gas) is compressed from room conditions to a specified pressure in a reversible manner by two compressors: one isothermal and the other adiabatic. If the entropy change of air is Δ s isothermal during the reversible isothermal compression and Δ s adiabatic during the reversible adiabatic compression, the correct statement regarding the entropy change of air per unit mass is a) Δ s isothermal = Δ s adiabatic = 0 b) Δ s isothermal = Δ s adiabatic > 0 c) Δ s adiabatic > 0 d) Δ s isothermal < 0 e) Δ s isothermal = 0 Answer: (d) Δ s isothermal < 0; For the reversible adiabatic compression process, Δ s adiabatic = 0. However, for the reversible isothermal compression process for an ideal gas, Δ s isothermal = -Rln(p 2 /p 1 ) < 0,
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This note was uploaded on 07/09/2009 for the course M&AE 326 taught by Professor Psiaki during the Spring '08 term at Cornell.

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exam1_sol_2006 - ENGRD 221 Thermodynamics (Prof. N....

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