MAE 320 Thermodynamics HW 5 Assignment

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Russell Shields
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1 MAE 0 Therodynaics HW 5 Assignent The hoework is due Wednesday, October 9, 06. Each proble is worth the pots dicated. Copyg o the solution ro another is not acceptable.. Multiple choices questions. There ay be ore than one answer or each question. (6 pots) (a). The st law o therodynaics cannot be applied to: (A) Control olue syste as the ass can cross the boundary. (B) A steady operation stea turbe. (C) An unsteady operation stea turbe when the energy contaed with turbe changes with tie. (D) Diuser, which has dierent let and area. (E) None o aboe (b). Please circle the alid stateent(s) ab control olue syste: (A) Both the ass and energy can cross the boundary o a control olue. (B) Under the steady-low condition, the ass and energy contaed with the control olue syste rea constants. (C) Under the steady-low condition, the luid properties at an let or always rea constant (do not change with tie) but ay be dierent at let and. (D) Under the steady-low condition, the tensie properties o the substance contaed ust be hoogeneous (unior) eerywhere a control olue. (E) Gas turbes, throttlg ales, nozzles and diusers operated under steady condition can be treated as the steady-low control olue syste. (c). Please circle the alid stateents ab heat exchangers: (A) Heat exchangers are deices where two og luid streas are ixed together. (B) Heat exchangers are deices where two og luid streas exchange ass. (C) Heat exchangers are deices where two og luid streas exchange heat (theral energy) with ixg together. (D) Heat exchangers are deices where two og luid streas exchange echanical work with ixg together. (E) There is no heat transer between the two luids o an sulated heat exchanger. (d). Stea (superheated water apor) enters an adiabatic throttlg ale at MPa, 400 C and s at 00 kpa. Neglectg the change ketic energy, the teperature o stea g this throttlg ale is (A) below 50 C (B) between 50 to 00 C (C) between 00 to 400 C (D) oer 400 C (E) Cannot be decided based on the oration proided.

2 (e). Please circle the alid stateent (s) ab a steady-low deice: (A) The ass with the control olue always reas constant (B) The energy with the control olue always reas constant (C) The luid properties ay change with location oer the cross section o the lets or s. (D) The luid low at any pot o any let and is unior and steady and the luid properties do not change with tie or location oer the cross section o an let or. (E) None o the aboe (). For diuser, we generally consider (A) W 0 (B) ke 0 (the diuser is designed to conert the low work to ketic energy) (C) Q >0 (D) pe 0 (E) None o the aboe (g). The change ketic energy can usually be ignored or (A) Diusers (B) Nozzles (C) Stea turbes (D) Throttlg ales (E) Heat exchangers (h). Stea (H O) lows through a pipe with an ner diaeter o 0 c at a pressure o 00 kpa and teperature o 00 o C. The speciic low energy ( P ) o stea the pipe is ostly close to (A) 6. / (Note: you need d the speciic olue o stea at gien T and P) (B) / (C) 07. / (D) 07. (E) None o the aboe

3 . Concept and short answer questions (0 pots) (a) Please draw a scheatic diagra o a doestic hot water tank delierg hot water to bathroo and kitchen, describe the therodynaic process oled, and list the assuptions, equations, and paraeters needed or you to d the water teperature o your hot water tank. ( pots) (b)the general equation o st law o therodynaics or control olue syste can be expressed as equation (a): de Q CV W + ( h + ke + pe) ( h + ke + pe) (a) dt Please list the assuptions needed or this general equation to be sipliied to equation (b) W turbe ( h h ) (b) Equation (b) can be used to calculate the power put o turbe. (4 pots) Answer: The assuptions clude: (a) steady operation with one let and one : de CV 0, dt (b) heat transer can be ignored: Q 0 (c) the changes ketic energy and potential energy can be ignored, so the keand pe ters can be dropped. (d) there is only one let and one, With these assuptions, Equation (a) can be re-written as: 0 W + h h 0 (c) With the known turbe, equation (c) can be urther reised to: turbe ( h ) W h h h (d) (c) Please expla i there is heat transer present with an adiabatic heat exchanger. ( pots)

4 . Hot air steadily enters a gas turbe at 5 bar, 900 K and s at bar, 600 K and a elocity o 00 /s. The ass low rate o air is 0 /s. Ignorg the heat transer, changes potential energy, and the ketic energy let, calculate (4 pots) (a) The power put ro this turbe, kw. Solution: Analysis (a) This question is associated with st law application control olue syste; (b) the syste is control olue syste understudy operation; (c) There is only one let and one, and thus ; (d) The energy balance or this steady-low syste can be expressed the rate or as V 0 V dec Q + W + h + + gz Q + W + h + + gz () dt V Or 0 V dec Q W + h + + gz h + + gz dt The ketic energy at let is negligible: V n 0. Heat transer Q 0 or Q 0 Q 0 Changes potential energy eects ( pe 0) are also negligible. Equation () can be sipliied and re-written as V W h h () V Or W ( h n h ) () You can d the enthalpy o air at let and Table A-7: h hair ( 900K) 9. 9 h hair ( 600K) Load enthalpy and elocity to equation (), you can d: W ( h h V / ) 0 / s 9.9 / / 09.kW (00 ) s J / s 000J 4

5 W ( h h V / ) J Or 0 / s 9.9 / / (00 ) s / s 000J 09.kW Note: () the positie dicates the transer o work ro syste to enironent. () I the ketic energy at ( 00 ) was ignored, the power put o this turbe is s 59. kw. The error caused by this ignorance is ab.6%, which is ery sall. This explas why the changes ketic energy can be ignored when solg therodynaic questions associated with turbe, copressor, and an. (b) The cross-section area o the,. Solution: The ass low rate is a unction o olue low rate and speciic olue or density. ρ V A V A / () The speciic olue can be ound usg ideal gas equation: P ν RT RT () P Load speciic olue to equation (), you can get: RT A () VP The gas constant R0.87 kpa. /.K can be obtaed ro Table A-. Load known ariables to equation (), you can d the area o : RT (0 / s)(0.87kpa / K)(600K) 0. 7 VP (00 / s)(00kpa) A 5

6 4. Nitrogen at 80 kpa and 400 K enters an adiabatic diuser steadily at a low rate o 6000 /h at 0 /s and leaes at 00 kpa at 0 /s. (4 pots) (a) Describe the a unction o diuser ( pots) Answer: The a unction o diuser is the slow down the high speed low through the conersion o ketic energy to ternal energy by creasg the pressure o luid. (b) Fd the teperature o nitrogen (6 pots) Solution: Analysis (a) There is only one let and one, and thus. We take diuser as the syste, which is a control olue sce ass crosses the boundary. The energy balance or this steady-low syste can be expressed the rate or as V V 0 dec Q W + h + + gz h + + gz () dt For nozzle, you can ignore the heat transer Q 0, work transer W 0, and changes potential energy gz 0, Equation () can be sipliied: or, h ( h + V / h ) 0 V V h + () You can d the speciic enthalpy at let Table A-8,640 hn (400K) kol h hn (400K) MW N 8 kol Load h and elocity at let and to equation () h V V h / + ( 0 /s) ( 0 /s) / 000 /s 44.7 / The ole-based speciic enthalpy at : kh hn, hn MWN , kol Fro Table A-8, you can d teperature o N at with the known enthalpy usg terpolation: 6

7 T 40 K (c) The cross-section area at the o the diuser (6 pots) Solution: The equation or calculation o ass low rate: AV ( 40 40) K 44. K The area o diuser can be ound ass low rate can be calculated A V The speciic olue o air at the diuser can be ound usg ideal gas equation: RT P The are can be ound: A V ( kpa / K)( K) ( 00 kpa) / (6000 h h )(.6 600s 0 /s /)

8 5. A copressor copresses the saturated R-4a apor at bar to 4 bar and 50 o C at a low rate o 0. /s. Ignorg the heat transer and the changes potential energy and ketic energy, detere (4 pots) Note to TA: Please pay attention to ass low rate when gradg this hoework. The ass low rate this question is 0. /s. I student used 0.0 /s, then this hoework should be graded as zero as it was the solution or an old hoework. (a) The olue low rate o R-4a at the. Analysis: The olue low rate is a unction o ass low rate and speciic olue V With the known ass low rate, 0. / s, you need d the speciic olue at gien teperature and pressure at prior to the calculation o olue low rate. To d the speciic olue, you need d the phase o R-4a and then the speciic olue with the proper property table. Solution: The phase o a substance can be ound by coparg the gien teperature (pressure) with saturation teperature at gien pressure (teperature): Table A-, T T 8.9 C < T 50 C 4bar 400kPa The R-4a the copressor as superheated apor. The speciic olue o R-4a at can be ound Table A- Superheated Vapor 4bar,50 0.4MPa,50 C The olue low rate can be calculated: V s s (b) The power needed or this copressor ( kw) Solution: the st law o therodynaics o a steady operation control olue syste can be expressed as: Q W + h + ke + pe h + ke + pe ( ) ( ) 0 With the ignorance o heat transer, the changes potential and ketic energy, and the assuption o one let and one, this equation can be sipliied to W copressor + ( h h ) 0 W h h copressor ( ) The speciic enthalpy o R-4a can be ound property Table. State : Saturated R-4a apor at bar00 kpa h bar 00kPa (Table A-: Pressure Table) State : Superheated R-4a Vapor, which has been ound (a) 8

9 h 4bar,50 C h@ 0.4MPa,50 C (Table A-: Superheated R-4a) The power consuption o copressor can be ound: W copressor ( h h ) 0. ( ) kW s s Note: The negatie sign dicates that copressor consues power its operation, which is consistent with our coon knowledge o copressor. 9

6. A condenser (heat exchanger) is designed to cool / o water apor at MPa and 500 o C to a saturated liquid at MPa. The coolg water is puped ro a local rier at 5 o C.

10 6. A condenser (heat exchanger) is designed to cool / o water apor at MPa and 500 o C to a saturated liquid at MPa. The coolg water is puped ro a local rier at 5 o C. Ater passg through the condenser, the teperature o the coolg water is restricted by the enironental code not to exceed 5 o C. Considerg the coolg water is copressible liquid, detere the iu ass low rate o coolg water. (6 pots) T 5 o C w? P MPa s/ T 4 <5 o C w? P MPa T 500 o C s/ Note to TA: Please pay attention to the axiu teperature o the coolg water when gradg this hoework. The the axiu teperature o the coolg water is 5 C. I student used 0 C, then this hoework should be graded as zero as it was the solution or an old hoework. Solution: st Law o therodynaics under steady operatg process can be expressed as: ( h + ke + pe) ( h + ke + pe) 0 Q W + (a) For heat exchanger, the heat transer between heat exchanger and enironent, work transer, changes potential energy and ketic energy can be ignored: The st law equation or this steady-low process can be re-arranged as: h h 0 or h h (b) There are two lets and two s as shown Figure, the orer equation can be urther re-written as: sh + wh sh + wh4 (c) This equation can be urther re-written to d the ass low rate o coolg water: s ( h h ) w (d) ( h4 h ) The change enthalpy o liquid water (copressible substance) can be ound usg equation: h h C ( T ) w (e) T Load equation (e) to (d), you can get; s ( h h ) s ( h h ) w () ( h h ) C ( T T ) 4 In equation (), the ass low rate o stea, the enthalpy o stea at let, and, let teperature o coolg water are either known oration or paraeters can be ound with known w 4

11 oration. Accordgly, the iu ass low rate o coolg water is obtaed with axiu teperature o coolg water, T 4 5 C At let : water, P MPa 000kPa, T 500 C, Sce 000kPa C < T, the phase o water at let is superheated apor. Fro Table A-6, h 479. / ; Exit : water, P MPa 000kPa, saturated liquid Fro table A-5, h h 76.5 / Fro Table A-, C w 4. / K Hence: ( h h ) s / ( ) /, iuw 77.5 ( ) o o C T T (4. / C) (5 5) C w 4,axiu /

12 7. A 4 L (olue) o pressure cooker a kitchen operates at 00 kpa with an area o 0. With well controlled heat transer ro range, liquid water is aporized to saturated apor, which s the pressure cooker at a constant ass low rate. Initially, the pressure cooker contas saturated water. Ater two hours o cookg, o saturated water apor let the pressure cooker. Durg this process, the heat loss ro the pressure cooker to the kitchen is 00. The changes potential and ketic energy can be ignored. (6 pots) Note to TA: Please pay attention to the heat loss ro cooker to kitchen, which is 00 this question. I student used 00, please ark this hoework as zero as the student ay hae copied ro past year solution. a) Please expla whether this pressure cooker is operatg under steady-low process or not; Answer: This pressure cooker is not operatg under steady operation as the ass o water contaed the pressure cooker is decreasg with tie. b) Fd the elocity o saturated apor, /s Solution: Notes: water s this pressure cooker as saturated stea. The elocity o stea g the pressure cooker can be ound usg the equation calculatg the ass low rate o stea: VA ρv ρva (a) This equation can be re-arranged (re-written) to calculate the elocity: V (b) A Note: With the known area, you still need d the ass low rate, speciic olue o saturated apor (c) t 600s s hour hour g kPa Load the cross-section area, ass low rate and speciic olue to equation (b), you can d the elocity: s V 4. 6 A s c) The speciic olue, quality, speciic ternal energy o water at State and State Solution: State : L V L 0. 00

13 At 00 kpa, , g The phase o water pressure tank at state can be ound by coparg the speciic olue o water with the speciic olue o saturated liquid water and saturated water apor g. Sce x u State : < < g, so the water is saturated ixture g u + xu g L V L At 00 kpa, ( ) , g The phase o water pressure tank at state can be ound by coparg the speciic olue o water with the speciic olue o saturated liquid water and saturated water apor g. Sce x < < g, so the water is saturated ixture g 0.00 With the known quality, you can d the speciic ternal energy o the saturated ixture. u u + xu g d) The heat transerred ro the range to the pressure cooker durg this one-hour process, Solution: the st law o therodynaics or a transient operation control olue syste can be expressed as: Q Q + W W + h + V + gz h + V + gz ECV, ECV, U CV, There is no work transer and ass transer to the pressure cooker. The changes the potential energy can be ignored. This equation can be sipliied to: U CV,

14 Q Q h + V E CV, E CV, U CV, U CV, u u Note: please expla i or i u u ( u u) can be applied this pressure cooker question. The heat transer to the syste can be ound: Q u u + Q + h + V J 000J 8. The concern to global issues is always debatable due to its direct and direct ipact on our counity, town, and state, etc but sees general not aectg the lie o college students. Howeer, this ay not be true West Virgia. The tuition ee WVU has been raised the past years due to the budget cut and expected reenue loss West Virgia, and other unexplaable reasons raised by our State Goernent, and Uniersity Adistrator. Please write an essay to expla the possible connection between WVU tuition ee creases with global issues, and describe how a echanical engeer with therodynaic knowledge can help solg global issues your uture career. The essay ust be at least 000 words, typed usg Ties New Roan typeace and a ont size o pot. It would be appreciated i you can clude a title your essay and list at least three reerences cited. (0 pots) Suggested Book Probles to Reiew: These probles will not be collected, but are suggested as a reiew. Chapter 5: 5, 0,, 4, 7,,, 4, 47, 57, 6, 68, 78, 87, 90, 97, 06, 7,, 4, 6 4

Number of extra papers used if any

Number of extra papers used if any Last Nae: First Nae: Thero no. ME 00 Therodynaics 1 Fall 018 Exa Circle your structor s last nae Diision 1 (7:0): Naik Diision (1:0): Wassgren Diision 6 (11:0): Sojka Diision (9:0): Choi Diision 4 (8:0):