ME 201 Thermodynamics

Transcription

1 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: Isenthalpic T K T 100 K P 1 1 MPa P MPa h kj/kg h 98.1 kj/kg φ kj/(kg K) φ 8.01 kj/(kg K) For state 1 we know T and P, so the state is fixed. Then from the air tables, Table A-SI h kj/kg, φ kj/(kg K) At state, we know the pressure and that the process was isenthalpic, then h 0 h h1 h h1 98.1kJ / kg which fixes our state. Obviously, T T K φ 8.01 kj/(kg K) Our entropy change is given as P s φ - φ1 - R ln -(0.87) ln kj/(kg K) P 1 1. Steam at.5 MPa and 400 C undergoes a polytropic process with polytropic exponent 1. to a final pressure of 0. MPa. Determine the internal energy change. Substance Type: Phase Change Substance (steam) Process: Polytropic with n1. T C T 178. C P 1.5 MPa P 0. MPa v m /kg v m /kg u kj/kg u 60.8 kj/kg We know both our temperature and pressure at state 1, so our state is fixed. Then we need to determine the fluid phase. From the saturation pressure table we find T sat (at.5 MPa).94 C 1

2 Since this is less than T 1, we have superheated vapor. Going to the superheat tables at.5 MPa and interpolating, we find v m /kg and u kj/kg We now use our polytropic process condition for which Pv n constant or P 1 v n n 1 P v Solving for v 1/n 1/1..5 v v1 (0.15) m /kg P 0. To determine the fluid phase at state, we go to the saturation pressure table at 0. MPa to find: v f m /kg and v g m /kg Since v is greater than v g, we must have superheated vapor. From the superheat tables at 0. MPa, we find T 178. C and u 60.8 kj/kg, Finally, u kj/kg. H at.5 MPa and 400 K undergoes a polytrophic process with polytropic exponent 0.8 to a final pressure of 0. MPa. Determine the internal energy change. Substance Type: Ideal Gas (H ) Process: Polytropic with n1.6c T K T 75.1 K P 1.5 MPa P 0. MPa v m /kg v m /kg u kj/kg u kj/kg For state 1 we know T and P, so the state is fixed. Then from the ideal gas tables, Table A.6SI u kj/kg The ideal gas law can be used to calculate the specific volume, or RT1 (.016)(400) v1 0.6 m /kg 500 At state, we know the pressure and that the process was polytropiv, then Pv n constant or P 1 v n n 1 P v Solving for v v 1/n 1/0.8 v1 P.5 (0.6) m /kg

3 which fixes our state. The temperature at state can be calculated from the ideal gas law vp (7.5814)(00) T 75.1 K R.016 Returning to the H tables we find u kj/kg Our entropy change is given as u u u kj/kg 4. Steam at 5 MPa and 700 C goes isometrically to 1.75 MPa. Determine the enthalpy change. Substance Type: Phase Change Substance (steam) Process: isometric (constant v) T C T 178. C P 1 5 MPa P 1.75 MPa v m /kg v m /kg h kj/kg h kj/kg Phase: phase with x We know both our temperature and pressure at state 1, so our state is fixed. Then we need to determine the fluid phase. Since P 1 > MPa we have superheated vapor at state 1. Then from the superheat tables v m /kg and h kj/kg We have an isometric process so that v v m /kg To determine the fluid phase at state, we go to the saturation pressure table at 1.75 MPa to find: v f m /kg and v g m /kg Since v is between these two values we must have a phase mixture with quality x v v v v f g f Our enthalpy at is then given by h x h g + (1-x )h f (0.165)(800.70) + ( )(961.70) kj/kg Finally, h kj/kg

4 5. Air at 500 K and 0.8 MPa goes to 00 K isentropically. Determine the internal energy change. Substance Type: Ideal Gas (air) Process: Isentropic T K T 00 K P MPa P u kj/kg u 8.67 kj/kg φ kj/(kg K) φ For state 1 we know T and P, so the state is fixed. Then from the air tables, Table A-SI u kj/kg, φ kj/(kg K) At state, we know the temperature and since we only want the internal energy change, we can look up u directly, u 8.67 kj/kg Then u u kj / kg u 1 6. Water at MPa and 150 C goes to 140 C isentropically. Determine the enthalpy change. Substance Type: Phase Change Substance (water) Process: isisentropic (constant s) T C T 140 C P 1 MPa P MPa s m /kg s m /kg h kj/kg h kj/kg Phase: phase with x We know both our temperature and pressure at state 1, so our state is fixed. Then we need to determine the fluid phase. Since T sat (at MPa) 1.7 C We have subcooled liquid. Going to the compressed liquid table we find that our pressure is below 5 MPa. This indicates that we can use an incompressible substance model for the subcooled liquid. For an incompressible liquid we have s c P,avg T ln T 1 4

5 7. Methanol at 0 C and 100 kpa is heated to 50 C and 00 kpa. Determine the enthalpy change Substance Type: Incompressible (methanol) Process: Isentropic T 1 0 C T 50 C P kpa P 00 kpa The enthalpy change for an incompressible substance is given by c (T - T ) v (P - P ) h P, avg 1 + avg 1 where c P,avg and v avg will be evaluated at T 1 + T C 9 K Going to Table B.5SI we find (after interpolating) c P,avg.51 kj/(kg K) and ρ avg kg/m Then 1 h (.51)(50-0) + (00-100) 75.4 kj/kg Air at 00 K and 10 kpa goes isometrically to 0 kpa. Determine the enthalpy change. Substance Type: Ideal Gas (air) Process: Isometric (const. v) T 1 00 K T P 1 10 kpa P 0 kpa h kj/kg h kj/kg v m /kg v 5.74 m /kg For state 1 we know T and P, so the state is fixed. Then from the air tables, Table A-SI h kj/kg We can calculate the specific volume from the ideal gas law, or RT1 (0.87)(00) v m /kg 10 At state, we know the pressure and that the volume stays constant. Hence v v m /kg which then fixes our state. We can calculate the temperature form the ideal gas law to obtain v1 (0)(5.74) T1 400K R 0.87 Then from the air tables h kj/kg And h h ( 96) kJ / kg h 1 5

6 9. Saturated liquid water at 170 C goes to 0 kpa isenthalpically. Determine the entropy change. Substance Type: Phase Change Substance (water) Process: isenthalpic (constant h) T C T P kpa P 0 kpa s 1.09 kj/(kg K) s m /kg h kj/kg Phase: sat.liq. h kj/kg Phase: phase with x We know our temperature and that we have saturated liquid at state 1, so our state is fixed. Going to the saturation temperature table we find P 1 P sat kpa h 1 h f kj/kg s 1 s f.09 kj/(kg K) For state, we use the isenthalpic condition to write h h kj/kg To determine the fluid phase at state, we go to the saturation pressure table at 0 kpapa to find: h f 51.1 kj/kg and h g kj/kg Since h is between these two values we must have a phase mixture with quality x h h h h f g f Our entropy at is then given by s x s g + (1-x )s f (0.198)(7.9090) + ( )(0.81).40 kj/(kg K) Finally, s kj/(kg K) 10. CO at 400 K and 00 kpa goes isentropically to 500 K. Determine the volume change. Substance Type: Ideal Gas (CO) Process: Isentropic (const. s) T K T 500 K P 1 00 kpa P 0 kpa φ kj/(kg K) φ kj/(kg K) v m /kg v 5.74 m /kg For state 1 we know T and P, so the state is fixed. Then from the air tables, Table A-SI φ kj/(kg K) We can calculate the specific volume from the ideal gas law, or RT1 (0.968)(400) v m /kg 00 At state, we know the temperature and that the entropy change is zero. Hence 6

7 P s R ln P φ φ 1 which allows us to solve for P φ - φ P exp (00)exp R Our final specific volume is RT (0.968)(500) v 0.9 m /kg P 66.9 Finally v v v m / kg 66.9 kpa 11. Aluminum at 00K and 100 kpa is heated isobarically to 500K. Determine the internal energy change. Substance Type: Incompressible (aluminum) Process: Isentropic T 1 00 K T 500 KC P kpa P 100 kpa The enthalpy change for an incompressible substance is given by c (T - T ) u P,avg 1 where c P,avg will be evaluated at T 1 + T K Going to Table B.10SI we find c P,avg J/(kg K) Then u (406.98)(500-00) kj/kg 7