ME 200 Thermodynamics 1 Fall 2017 Exam 3

Transcription

1 ME 200 hermodynamics 1 Fall 2017 Exam Circle your structor s last name Division 1: Naik Division : Wassgren Division 6: Braun Division 2: Sojka Division 4: Goldenste Division 7: Buckius Division 8: Meyer INSRUCIONS his is a closed book and closed notes exam. Equation sheets and all needed tables are provided. Significant credit for each problem is given if you identify your system and its boundary, draw the relevant energy flows on a diagram i.e. Energy Flow Diagram (EFD), start your analysis with the basic equations, list all relevant assumptions, and have appropriate units and use three significant figures. here is no need to re-write the given and fd. Do not hesitate to ask if you do not comprehend a problem statement. For your own benefit, please write clearly and legibly. You must show your work to receive credit for your answers. Do not write on the back of any page because it will not be scanned so will not be graded. If you need extra paper raise your hand and a proctor will supply it. IMPORAN NOE he use of PDAs, Blackberry-type devices, cell phones, laptop computers, smart watches or any other sources of communication (wireless or otherwise) is strictly prohibited durg examations. Dog so is cheatg. If you brg a smart watch, cell phone, or other communication device to the examation, it must be turned off prior to the start of the exam, placed your backpack, and the backpack must be stored below your seat. It shall be reactivated only after you leave the examation room for the fal time. Otherwise it is a form of cheatg and will be treated as such. SECOND IMPORAN NOE he only calculators allowed for use on this exam are those of the I-0X series. No others.

2 1. [20 pots] Circle the correct answer (no partial credit) for each. (a) Entropy of air treated as an ideal gas depends only on temperature. (rue or False) (b) Enthalpy of air treated as an ideal gas depends only on temperature. (rue or False) (c) Enthalpy of water treated as an compressible substance depends only on temperature. (rue or False) (d) Heat transfer is always zero durg an isothermal process. (rue or False) (e) he entropy change of a substance undergog an ternally reversible process is always zero. (rue or False) (f) Entropy of a fluid undergog an adiabatic, steady-state throttlg process usg a flow restriction (e.g. valve) device (Increases, Decreases, Remas the Same) (g) Entropy of water treated as an compressible substance undergog an isothermal process (Increases, Decreases, Remas the Same) (h) Entropy of a pure substance undergog a phase change from saturated vapor to saturated liquid at constant pressure (Increases, Decreases, Remas the Same) (i) Change entropy of a fluid havg undergone a complete cycle a reversible Carnot heat enge is (Positive, Negative, Zero) (j) Change entropy of a fluid havg undergone a complete cycle an irreversible heat enge is (Positive, Negative, Zero)

3 2. [40 pots] A piston-cylder device contas 0.25 of air itially at a temperature of 27C and an absolute pressure of 1 bar (State 1). he air undergoes a compression process, where PV 1. = constant, until the volume is 20% of the itial volume and the absolute pressure is 8.1 bar (State 2). Durg the compression process, 44.5 of work is done on the air. he cylder is fitted with a coolg water jacket all around its er wall. he coolg water jacket contas 1.75 of liquid water. he water is itially at a temperature of 25C and an absolute pressure of 1 bar (State ) at the start of the air compression process. Heat transfer occurs only between air the cylder and water side the coolg jacket sce the water jacket is perfectly sulated on its side. Molecular weight of air: /kmol Specific heat of liquid water: 4.18 /-K Use the closest value ideal gas table; do not terpolate. (a) What is the fal temperature (C) of water durg the compression process? (b) Calculate the entropy change (/K) for the air. (c) Fd the entropy change (/K) for the water. (d) Determe the entropy generation for the entire process (both air and water). Identify appropriate system or systems on the sketch provided, show mass/energy teractions (EFD), list any assumptions and basic equations, and provide your solution. here is no need to re-write the given and fd.

4 Extra Space for Problem 2 Assumptions Quasi-equilibrium Ignore KE change Ignore PE change Neglect friction No other work except boundary work Air is an ideal gas Liquid water is an compressible liquid Specific heat of liquid water remas constant with temperature change Basic Equations de dt Integratg: QW U KE PE ds Q j m s m dt system j j, boundary s Integratg: Q S boundary generation Solution (a) Initial volume of air the cylder: R m air K mairrair M 1 air -K V m P1 P1 100 kpa Fal volume of air the cylder: V2 0.2V m PV 2 2 Fal temperature of air the cylder: 2 mairrair 810 kpa m Fal temperature of air the cylder: K K u system u and u u K Q W m h ke pe m h ke pe 490 K

5 Extra Space for Problem 2 Considerg air side the cylder as the system, heat transfer durg the compression process: Q12 W12 U2 U1 mair u2 u Considerg liquid water side the coolg jacket as the system durg the polytropic Q W U U m u u m C compression process: water 4 water water 4 Heat transfer from air occurs only to the liquid water the coolg jacket sce the side of the coolg jacket is perfectly sulated Q4 Q12 10 emperature change of liquid water durg the compression process: Q K m waterc water K Fal temperature of water: C (b) Entropy change for the air: 0 0 P bar Sair mair s2 s1 Rair ln ln P1 1bar -K S air K (c) Entropy change for the water: K Swater mwatercwater ln ln -K K S water K (d) Considerg air the cylder and liquid water side the coolg jacket as the combed system, entropy generation for the entire process: Q S Sair Swater K boundary

6 . [40 pots] A solar-powered power plant uses the sun s radiation to boil water. At peak operatg conditions, the rate of radiation heat transfer to the boiler is 420 MW. he wor fluid is water/, with data at each state provided the table below; all the pressure values are absolute. Q boiler m 2 m W turbe m 4 m Q 1 boundary 5 C 278 K environment m 4 State P, bar, C h, /, m / s, /-K P, bar sat, C h f, / h g, / s f, /_K s g, /_K (a) Calculate the mass flow rate (/s) through the boiler. (b) Compute the isentropic efficiency (%) of the adiabatic turbe. (c) Fd the entropy generation (kw/k) for the adiabatic turbe. (d) Determe the total entropy generation (kw/k) for the assumg heat transfer occurs to an environment of temperature 5C. (e) Show the cycle on -s diagram relative to the vapor dome and the relevant les of constant pressure. Label the axes and four states and dicate the process directions with arrows. Critical temperature and pressure of water are 74C and 221 bar, respectively.

7 Extra Space for Problem Assumptions Quasi-equilibrium Steady state, steady flow One-dimensional, uniform flow Ignore KE change Ignore PE change Boiler: W CV 0 urbe and Pump: Q 0 Basic Equations dm dt de dt system system m CV m Q W m h ke pe m h ke pe ds dt system Q m s m s j generation j j, boundary Solution (a) Mass balance for the boiler: m2 m m Q m h h Energy balance for the boiler: boiler Mass flow rate of through the boiler: m Q boiler s h h m s (b) Mass balance for the turbe: m m4 m Energy balance for the actual turbe: wactual h h For the isentropic process through the turbe: s s4s K s f,10c < s 4s < s g,10c saturated liquid-vapor mixture s4 s s f,10 C For saturated liquid-vapor mixture: x4s s s g,10 C f,10 C

8 Extra Space for Problem x h h h s f,10 C 4 s 4s h4 s h h g,10 C f,10 C Energy balance for the isentropic turbe: wisentropic h h4s wactual Isentropic efficiency of the turbe: turbe turbe 88.4% w isentropic 1589 (c) Mass balance for the turbe: m m4 m Entropy balance for the actual turbe: turbe m s4 s s -K turbe kw 75.7 K (d) Mass balance for the : m4 m1 m Energy balance for the : Q m h1h s Q 257,780 s Entropy balance for the condernser: Q 257,780 kw m s1s K s -K boundary kw 16. K (e) -s diagram ( C) = 600 C P 2 = P = 40 bar P 4 = P 1 = bar 2 = 10.1 C 2 2s 1 = 4 = 10 C 1 s 1 = s 2s = s 4 s = s 4s = s4 = s 2 = s (/-K)