ME 201 Thermodynamics

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1 8 ME 0 Thermodynamics Practice Problems or Property Evaluation For each process described below provide the temperature, pressure, and speciic volume at each state and the change in enthalpy, internal energy, and entropy.. Water as saturated vapor at 445 kpa goes isentropically to 70kPa. Substance Type: Compressible (clue words saturated and vapor) Process: Isentropic State State T = 47.49C T = 5. C P = 445 kpa P = 70 kpa v = m /kg v = m /kg h = kj/kg h = 57. kj/kg u = 556. kj/kg u = kj/kg s = kj/(kg K) phase: two phase x = 0.94 Italicized versions read rom steam tables s = kj/(kgk) phase: sat.vap. At state we know the pressure and that the substance is a saturated vapor. This is suicient inormation to ix the state. Then we can go to Table C.bSI and read the remaining properties or saturated vapor at 445 kpa. At state we know the pressure, but we also that we have an isentropic process, so that s = s = kj/(kg K) So with the pressure and entropy known at State, it is ixed. Next, we must determine the phase. We again go to Table C.bSI and ind that at 70 kpa s =.4740 kj/(kgk) and s g = 7.8 kj/(kgk) Since s lies between these values, so that we have a two phase mixture with s -s x = 0.94 s -s g Our temperature must be the saturation temperature at 70 kpa or 5.C. With the quality known, we can determine our other properties at state h u v = h = u = v + xh + xu + xv g g g = (0.94)(5.94) = 57. kj/kg = (0.94)(040.88) = kj/kg = (0.94)(.0) = m / kg

2 ME 0 Thermodynamics h = -69. kj/kg u = kj/kg s = 0 kj/(kg K). Air at 00 kpa and 98 K goes isotropically to 40 K. Substance Type: Ideal Gas (air) Process: Isotropic State State T = 98 K T = 40 K P = 00 kpa P = 4. kpa v = 0.85 m /kg v = 0.85 m /kg h = 0.00 kj/kg h = 4.4 kj/kg u = kj/kg u = 0.6 kj/kg = 6.7 kj/(kgk) = kj/(kgk) Italicized versions read rom air tables or calculated rom ideal gas equation. For state we know T and P, so the state is ixed. Using the ideal gas equation RT (0.86)(98) v = 0.85 m /kg P (00) Then rom the air tables, Table A.SI h = 0.00 kj/kg, u = kj/kg, = 6.7kJ/(kg K) At state, we know the temperature and that the process was isotropic, then v = v = m /kg which ixes our state. Then using the ideal gas law we solve or the pressure RT (0.86)(40) P =4.kPa v (0.85) Then rom the air tables, Table A.SI h = 4.4 kj/kg, u = 0.6 kj/kg, = kj/(kg K) h = 4.4 kj/kg u = 0.6 kj/kg P s = - - R ln = (0.86)ln(4./00) P = kj/(kg K)

3 ME 0 Thermodynamics. Helium at 00 kpa and 00 K goes isothermally to 00 kpa.. Substance Type: Ideal Gas (He) Process: Isothermal State State T = 00 K T = 00 K P = 00kPa P = 00 kpa v =. m /kg v = 6.4 m /kg Italicized calculated rom ideal gas equation. For state we know T and P, so the state is ixed. Using the ideal gas equation RT (.08)(00) v =. m /kg P (00) At state, we know the pressure and that the process was isothermal, then T = T = 00 K which ixes our state. Then using the ideal gas law we solve or the speciic volume RT (.08)(00) v = 6.4 m /kg P (00) We note that since we have an ideal gas and an isothermal process, our changes in enthalpy and internal energy will be zero. To calculate our change in s (since we don t have He tables) we need the speciic heat at constant pressure. Unortunately He is not on our c P table. This is because since He is a monatomic gas, its c P does not depend on temperature. We Google or the c P o Helium and ind 5.98 kj/(kgk). Then our changes are then given as h = 0 kj/kg u = 0 kj/kg T P s = c ln - R ln T P =.44 kj/(kg K) = (5.98) ln - (.08) ln Steam at 600C and 6 MPa goes isenthalpically to 0.0 MPa. Substance Type: Compressible (clue words steam) Process: Isenthalpic State State T = 600C T = 578. C P = 6 MPa P = 0.0 MPa = 0 kpa v = m /kg v = 9.9 m /kg h = kj/kg h = kj/kg u = 6.9kJ/kg u = 6.44 kj/kg s = 7.69 kj/(kgk) s = kj/(kgk) phase: sup.vap. phase: sup.vap. Italicized versions read rom steam tables,

4 ME 0 Thermodynamics At state we know the pressure and the temperature. This is suicient inormation to ix the state. To determine the luid phase we go to the saturation pressure table, Table C.bSI, and ind the boiling temperature at 6 MPa to be 78.C which is less than 600C, so that we have superheated vapor. We can then go to the superheat vapor table, Table C.cSI, and read the remaining properties or at 6 MPa and 600C. h = kj/kg, u = 6.9 kj/kg, s = 7.69 kj/(kg K), v = m /kg At state we know the pressure, but we also that we have an isenthalpic process, so that h = h = kj/kg So with the pressure and enthalpy known at State, it is ixed. Next, we must determine the phase. We again go to Table C.bSI and ind that at 0 kpa h =9.75 kj/kg and h = kj/kg g Since h > hg, we must have a superheated vapor. We go to the superheat vapor table, Table C.cSI6, and ind that at 0 kpa and h o kj/kg, we interpolate to obtain T = 578. C, u = 6.44 kj/kg, s =0.099 kj/(kg K), v Our changes are then given by h = 0 kj/kg u = -.95 kj/kg s =.966 kj/(kg K) = 9.9 m 5. Air at.6 MPa and 500 K goes isobarically to 00 K. Substance Type: Ideal Gas (air) Process: Isobaric State State T = 500 K T = 00 K P =.6 MPa = 600 kpa P = 600 kpa v = m /kg v = m /kg h = 05.8kJ/kg h = kJ/kg u = 47. kj/kg u = kj/kg = 7.77 kj/(kgk) = 6.0 kj/(kgk) Italicized versions read rom air tables or calculated rom ideal gas equation. For state we know T and P, so the state is ixed. Using the ideal gas equation RT (0.86)(500) v = m /kg P (600) Then rom the air tables, Table A.SI h = 05.8 kj/kg, u =47.kJ/kg, = 7.77 kj/(kg K) At state, we know the temperature and that the process was isobaric, then = P =600 kpa P /kg 4

5 ME 0 Thermodynamics which ixes our state. Then using the ideal gas law we solve or the speciic volume RT (0.86)(00) v = m /kg P (600) Then rom the air tables, Table A.SI h = kj/kg, u = kj/kg, = 6.0 kj/(kg K) h = -0.8 kj/kg u = kj/kg P s = - - R ln = (0.88)ln() P = kj/(kg K) 6. Water at 70 kpa and 88 C goes isentropically to 690 kpa. We note that at 70 kpa the boiling temperature or water is C. So we have a subcooled liquid. We also note that our compressed liquid tables or water start at 5 MPa. We will treat water under these conditions as an incompressible substance. Substance Type: Incompressible Substance (liquid water, no phase change) Process: Isentropic State State T = 88C T = 88C P = 70 kpa P = 690 kpa v = m /kg v = m /kg Italicized values are determined rom incompressible subtance equations. Bold values are calculated For state both T and P are given so the state is ixed. To obtain a speciic volume, we note that since speciic volume only depends on temperature or an incompressible liquid, we can read it rom the incompressible substance table or water (Table B.SI). Then rom we have v = m /kg At state, we know the pressure and recall that or an isentropic process or an incompressible liquid T s = c ln = 0 T T = T = 88C So the speciic volume does not change v = m /kg 5

6 ME 0 Thermodynamics Finally we calculate our changes with T s = c ln = 0 kj/(kg K) T u = c (T - T ) = 0 kj/kg h = c (T - T ) + v avg (P - P ) = v avg (P = ( )(690-70) = 0.64 kj/kg 7. Rerigerant- at -5 C and 00 kpa goes isobarically to 0 C. Substance Type: Compressible (clue words rerigerant) Process: Isobaricpic State State T = -5C T = 0C P = 00 kpa P = 00 kpa v = m /kg v = m /kg h =.7 kj/kg h = 5. kj/kg u =. kj/kg u = 9.97 kj/kg s = kj/(kgk) s =.05 kj/(kgk) phase: sub.liq. phase: sup.vap. Italicized versions read rom steam tables At state we know the pressure and the temperature. This is suicient inormation to ix the state. To determine the luid phase we will go to the saturation pressure table or rerigerant-, Table C.cSI, and ind a boiling temperature o -.C. Since this is greater than our given temperature (-5C) we must have a subcooled liquid. We go to the compressed liquid table and ind that it does not exist. We will have to treat state as an incompressible substance, but because o state, we may still need to use compressible substance tables. To begin in treating state as a incompressible liquid, we note that none o v, s, or u depend on pressure, so that there values should be equal to those o saturated liquid at -5C. Interpolating rom Table C.SI u =. kj/kg, s = kj/(kg K), v = m /kg To determine the enthalpy which does have a pressure dependence, we consider a hypothetical process rom saturation conditions to the compressed conditions, then = h (@T ) + v (@T ) [P - P (@T )] h sat =.5 + ( )( ) =.7 kj/kg At state we know the temperature, but we also that we have an isobaric process, so that P = P = 00 kpa So with the pressure and temperature known at State, it is ixed. Next, we must determine the phase. We again go to Table C.bSI and ind that at 00 kpa the boiling temperature is -.C, - P ) 6

7 ME 0 Thermodynamics which means now we have superheated vapor. The remaining properties are read rom Table C.cSI h = 5. kj/kg, s =.05 kj/(kg K) v = m /kg, u Our changes are then given by h =.86 kj/kg u = kj/kg s =.085 kj/(kg K) = 9.97 kj/kg 8. Carbon dioxide at 70 K and 0 MPa goes to590 K and MPa. Substance Type: Ideal Gas (CO ) State State T =70 K T = 590 K P = 0 MPa = 0,000 kpa P = MPa = 000 kpa v = m /kg v = m /kg h = 9.0 kj/kg h = 8.46 kj/kg u = kj/kg u = 7.0 kj/kg = 6.58 kj/(kgk) = kj/(kgk) Italicized calculated rom ideal gas equation. For state we know T and P, so the state is ixed. Using the ideal gas equation RT (8.4/44)(70) v = m /kg P (0000) Our remaining properties can be read rom the ideal gas tables, Table A.7SI. h = 9.0 kj/kg, u = kj/kg, = 6.58 kj/(kg K) At state, we know the pressure and the temperature which ixes our state. Then using the ideal gas law we solve or the speciic volume RT (8.4/44)(590) v = m /kg P (000) Our remaining properties can be read rom the ideal gas tables, Table A.7SI. h = 8.46 kj/kg, u = 7.0 kj/kg, = kj(kg K) h = kj/kg u = kj/kg P s = - - R ln kj/(kg K) P 7

8 ME 0 Thermodynamics 9. Rerigerant- goes rom vapor to liquid at C. Substance Type: Compressible (clue words rerigerant & vapor) State State T = C T = C P = MPa P = MPa v =0.09m /kg v = m/kg h = kj/kg h = kj/kg u = kj/kg u = 56. kj/kg s = kj/(kgk) s = 0.4 kj/(kgk) phase: sat.vap. phase: sat.liq. Italicized versions read rom rerigerant tables At state we know the temperature and will assume we have saturated vapor. This is suicient inormation to ix the state. We can then go to the saturation table, Table C.aSI and read the remaining properties or at 70F. h =96.55 kj/kg, s = kj/(kg K), u =79.08 kj/kg, v = 0.09 m /kg, P = MPa At state we know the temperature and will assume we have saturated liquid. This is suicient inormation to ix the state. We can then go to the saturation table, Table C.aSI and read the remaining properties or at C. h = kj/kg, s = 0.4kJ/(kg K), u = 56. kj/kg, v = m /kg, P Our changes are then given by h = kj/kg u = -.74 kj/kg s = kj/(kg K) = MPa 0. A m volume contains 0. kg o air at 50 K. Without changing the volume the pressure becomes 0 kpa. Substance Type: Ideal Gas (air) Process: Isotropic State State T = 50 K T = 8 K P = 4.5 kpa P = 0 kpa v = 5.0 m /kg v = 5.0 m /kg h = -48. kj/kg h = u = kj/kg u = = kj/(kgk) = Italicized versions read rom air tables or calculated rom ideal gas equation. 8

9 ME 0 Thermodynamics For state we know T and with the volume and mass inormation we can calculate the speciic volume V.0 v = 5 m /kg m 0. so the state is ixed. Using the ideal gas equation RT (0.86)(50) P =4.5 kpa v (5) Then rom the air tables, Table A.SI h = -48. kj/kg, u = kj/kg, = kj/(kg K) At state, we know the temperature and that the process was isotropic, then v = v = 5.0 m /kg which ixes our state. Then using the ideal gas law we solve or the temperature vp (5)(0) T = 8 K R (0.87) We go to the air tables, Table A.SI, to look up the remaining properties and note that our temperature is too high or the table. Hence, we will need to use our constant speciic heat approach. Our average temperature is 04 K, so we go to Table A.SI and ind c =.559 kj/(kg K) and c V,avg = kj/(kg K) h = c (T - T ) = kj/kg u = (c s = c - R)(T - T ) = 47.7 kj/kg T P ln - R ln =.674 kj/(kg K) T P 9

ME 201 Thermodynamics

ME 201 Thermodynamics Spring 01 ME 01 Thermodynamics Property Evaluation Practice Problems II Solutions 1. Air at 100 K and 1 MPa goes to MPa isenthapically. Determine the entropy change. Substance Type: Ideal Gas (air) Process: