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 1 100 K T 100 K P 1 1 MPa P MPa h 1 98.1 kj/kg h 98.1 kj/kg φ 1 8.01 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 h1 98.1 kj/kg, φ 1 8.01 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 1 100 K φ 8.01 kj/(kg K) Our entropy change is given as P s φ - φ1 - R ln -(0.87) ln 0.979 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 1 400 C T 178. C P 1.5 MPa P 0. MPa v 1 0.15 m /kg v 1.081 m /kg u 1 98.8 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
Since this is less than T 1, we have superheated vapor. Going to the superheat tables at.5 MPa and interpolating, we find v 1 0.15 m /kg and u 1 98.8 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) 1.081 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 0.001061 m /kg and v g 0.8855 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 60.8-98.8-18.00 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 1 400 K T 75.1 K P 1.5 MPa P 0. MPa v 1 0.6 m /kg v 7.5814 m /kg u 1 1048.61 kj/kg u 4706.0 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 1 1048.61 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) 0. 7.5814 m /kg
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 4706.0 kj/kg Our entropy change is given as u u u 1 4706.0-1048.61 657.9 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 1 700 C T 178. C P 1 5 MPa P 1.75 MPa v 1 0.0165 m /kg v 0.0165 m /kg h 1 767.59 kj/kg h 11.74 kj/kg Phase: phase with x 0.165 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 1 0.0165 m /kg and h 1 767.59 kj/kg We have an isometric process so that v v 1 0.0165 m /kg To determine the fluid phase at state, we go to the saturation pressure table at 1.75 MPa to find: v f 0.001165 m /kg and v g 0.115 m /kg Since v is between these two values we must have a phase mixture with quality x v v v v 0.0165 0.001165 0.115 0.001165 f g f 0.165 Our enthalpy at is then given by h x h g + (1-x )h f (0.165)(800.70) + (1-0.165)(961.70) 11.74 kj/kg Finally, h 11.74-98.8-176.08 kj/kg
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 1 500 K T 00 K P 1 0.8 MPa P u 1 147.1 kj/kg u 8.67 kj/kg φ 1 7.77 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 u1 147.1 kj/kg, φ 1 7.77 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 8.67 147.1-18.54 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 1 150 C T 140 C P 1 MPa P MPa s 1 0.0165 m /kg s 0.0165 m /kg h 1 767.59 kj/kg h 11.74 kj/kg Phase: phase with x 0.165 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
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 1 100 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 0 + 50 0 C 9 K Going to Table B.5SI we find (after interpolating) c P,avg.51 kj/(kg K) and ρ avg 788.5 kg/m Then 1 h (.51)(50-0) + (00-100) 75.4 kj/kg 788.5 8. 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 1-96.00 kj/kg h 10.11 kj/kg v 1 5.74 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 1-96.00 kj/kg We can calculate the specific volume from the ideal gas law, or RT1 (0.87)(00) v1 5.74 m /kg 10 At state, we know the pressure and that the volume stays constant. Hence v v 1 5.74 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 10.11 kj/kg And h h 10.11 ( 96) -199.11kJ / kg h 1 5
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 1 170 C T P 1 791.870 kpa P 0 kpa s 1.09 kj/(kg K) s 0.0165 m /kg h 1 718.764 kj/kg Phase: sat.liq. h 718.764 kj/kg Phase: phase with x 0.165 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 791.870 kpa h 1 h f 718.764 kj/kg s 1 s f.09 kj/(kg K) For state, we use the isenthalpic condition to write h h 1 718.764 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 609.96 kj/kg Since h is between these two values we must have a phase mixture with quality x h h h h 718.764 51.1 609.96 51.1 f g f 0.198 Our entropy at is then given by s x s g + (1-x )s f (0.198)(7.9090) + (1-0.198)(0.81).40 kj/(kg K) Finally, s.40 -.09 0.1948 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 1 400 K T 500 K P 1 00 kpa P 0 kpa φ 1 7.67 kj/(kg K) φ 7.5980 kj/(kg K) v 1 0.957 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 1 7.67 kj/(kg K) We can calculate the specific volume from the ideal gas law, or RT1 (0.968)(400) v1 0.957 m /kg 00 At state, we know the temperature and that the entropy change is zero. Hence 6
P s 0-1 - R ln P φ φ 1 which allows us to solve for P φ - φ1 7.5980-7.67 P exp (00)exp R 0.968 Our final specific volume is RT (0.968)(500) v 0.9 m /kg P 66.9 Finally v v v1 0.9 0.957-0.1718 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 1 100 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 00 + 500 400 K Going to Table B.10SI we find c P,avg 406.98 J/(kg K) Then u (406.98)(500-00) 81.40 kj/kg 7