ME 00 Thermodynamics II Exam 1 September 7, 01 8:00 p.m. 9:00 p.m. Name: Solution Section (Circle One): Sojka Naik 11:0 a.m. 1:0 p.m. Instructions: This is a closed book/notes exam. You may use a calculator. You must start from the most basic form of the governing equations and simplify as necessary to solve each problem. You must show all work for full credit. Please keep your eyes on your own exam. If you are caught cheating you will get a zero for the exam and your name will be turned over to the Dean of Students. Please feel free to use the tables/equation sheet provided with the exam. If you need to interpolate any property data, feel free to use the closest value to save time. State clearly the source of property data. Question Total Score 1 40 40 0 Total 100
A vertical piston-cylinder device initially contains kg of R-14a at 800 kpa and 80C. At this state, the piston is touching on a pair of stops at the top. The mass of the piston is such that a pressure of 500 kpa is required to move it. A valve at the bottom of the cylinder is opened and R-14a exits from the cylinder. The piston is observed to move and the valve is closed when 1 kg of R-14a is removed from the cylinder and 1 kg of R-14a at 500 kpa and 0C remains inside the cylinder. R-14a Valve Calculate the heat transfer (kj) during the process. (40 points) Assumptions 1. Neglect KE and PE effects. Uniform flow of R-14a leaving from the cylinder at average condition between the initial and final states of R-14a Solution Considering mass balance for the control volume around the cylinder: dmcv mi m e Integrating: mcv m m1 me me m1m 1kg dt inlets exits Considering energy balance for the control volume around the cylinder: decv d i e U CV KE CV PE CV Q CV W CV m i h i gz i m e h e gz e dt dt inlets exits Integrating: UCVQW mehemu mu 1 1 QW mehe Qm u mu W m h 1 1 e e State 1: Table A-11: T sat (8 bar) = 1.C T 1 > T sat SHV Using Table A-1: v 1 = 0.064 m /kg, u 1 = 89.89 kj/kg, h 1 = 16 kj/kg State : Table A-11: T sat (5 bar) = 15.74C T > T sat SHV Using Table A-1: v = 0.04188 m /kg, u = 9.4 kj/kg, h = 60.4 kj/kg
Use this as extra space for Problem 1. Assuming uniform flow of R-14a out of the cylinder at average condition between the initial h1 h kj and final states: he 88.17 kg m Initial volume of R-14a: V1 mv 1 1 kg 0.064 0.0658 m kg m Final volume of R-14a: V mv 1 kg 0.04188 0.04188 m kg Boundary work due to piston movement: 1 W PdV P V V 1 500 kpa 0.04188 0.0658 m 11.7 kj Heat transfer during the process: kj kj kj Qmu mu 1 1W mehe1 kg 9.4 kg 89.89 11.7 kj 1 kg 88.17 kg kg kg Q 6.9 kj ; negative sign indicates heat transfer from the cylinder to surrounding air
1. Air enters a nozzle steadily at 00 kpa and 87C with a velocity of 50 m/s. The nozzle is poorly insulated and heat loss of 6.6 kj/kg occurs to the environment. Air exits the nozzle at 95 kpa with a velocity of 00 m/s. Assume that the air surrounding (environment) the nozzle is at 100 kpa and 17C. (a) Calculate the temperature (K) of air at the nozzle exit. (b) Determine the exergy destruction (kj/kg) for the nozzle. (c) Calculate flow exergy (kj/kg) of the air exiting the nozzle. (15 points) (15 points) (10 points) Assumptions 1. Steady state. One-dimensional flow. Neglect changes in PE 4. Passive device i.e. no mechanical work 5. Air behaves as an ideal gas Solution (a) Considering energy balance for the control volume around the nozzle: de nozzle 1 1 qnozzle wnozzle h1h gz1zh qnozzle h1 dt Using Table A-: h 1 = h(60 K) = 60.58 kj/kg kj kj 50 00 kj m J h 6.6 60.58 ; 1 1 kg kg 1000 kg s kg kj h 10. kg Using Table A-, the temperature of air the nozzle exit: T 10 K (b) Considering entropy balance for the control volume around the nozzle (including some part of the surrounding air): ds nozzle nozzle nozzle q s1s q nozzle nozzle s s1 dt T T 0 0
Use this as extra space for Problem. q nozzle 0 0 P nozzle s s1 Rair ln T0 P1 R kj Rair 0.87 M air kg-k Using Table A-: s 0 1 = s 0 (60 K) = 1.8854 kj/kg-k, s 0 = s 0 (10 K) = 1.7498 kj/kg-k kj 6.6 kg kj kj 95 kpa kj nozzle 1.7498 1.8854 0.87 ln 0.04 17 7K kg-k kg-k 00 kpa kg-k Exergy (available energy) destroyed due to entropy generation (irreversibility): kj ed T 0 90 K 0.04 kj ed 58.7 kg-k kg (c) Flow exergy of the air exiting the nozzle: ef, h h0 T0 s s0 gz 0 0 P ef, h h0 T0 s s0 Rair ln P0 Using Table A-: h 0 = h(90 K) = 90.16 kj/kg, s 0 0 = s 0 (90 K) = 1.6680 kj/kg-k 00 kj kj kj 95 kpa kj e f, 10.90.16 90 K 1.7498 1.6680 0.87 ln kg kg-k kg-k 100 kpa 1000 kg kj e f, 41.4 kg
. Answer the following questions. Show calculations (part a) or justify your answers with one sentence or an equation as necessary (parts b through e). (a) A heat engine receives heat from a source at 17C at the rate of 600 kw and rejects heat to the ambient reservoir at 7C at the rate of 19 kw. What is the exergetic efficiency (%) of the heat engine? (8 points) (b) Exergy can never be negative. Thermal efficiency of the given heat engine: W 600 19 kw out Q in Q out IHE, 68% Q Q 600 kw in in Maximum possible thermal efficiency of the heat engine in the given temperature limits: TC 00 K ImaxHE,, 1 1 80% T 1500 K H Exergetic efficiency of the given heat engine: IHE, II, HE 85% ImaxHE,, True False Insufficient Information Exergy is useful work potential; cannot be negative ( points) (c) Exergy change can never be negative. ( points) True False Insufficient Information Change in exergy is negative when system moves towards the state of environment (d) Exergy destruction can never be negative. ( points) True False Insufficient Information e d T 0 ; entropy generation () is positive for irreversible process and zero for reversible reversible process
(e) Consider two closed systems in an environment of P 0 and T 0. System A has P A = P 0 and T A = T 0 while system B has P B = P 0 and T B < T 0. Ignoring KE and PE effects, which system has higher exergy? ( points) System A System B Insufficient Information System A has no exergy since it is at the condition of the environment