For each process described below provide the temperature, pressure, and specific volume at each state and the change in enthalpy, internal energy, and entropy.

1. Water as saturated vapor at 65 psia goes isentropically to 25 psia.

Solution:

At state 1 we know the pressure and that the substance is a saturated vapor. This is sufficient information to fix the state. Then we can go to Table A-5E and read the remaining properties for saturated vapor at 65 psia. At state 2 we know the pressure, but we also that we have an isentropic process, so that

s₂ = s₁ = 1.638 Btu/(lb_mR)

So with the pressure and entropy known at State 2, it is fixed. Next, we must determine the phase. We again go to Table A-5E and find that at 25 psia

s_f = 0.353 Btu/(lb_{mR) and sg} = 1.714 Btu/(lb_mR)

Since s₂ lies between these values, so that we have a two phase mixture with

Our temperature must be the saturation temperature at 25 psia or 240F. With the quality known, we can determine our other properties at state 2

Our changes are then given as

2. Air at 14 psia and 75F goes isotropically to 150F.

Solution:

For state 1 we know T and P, so the state is fixed. Using the ideal gas equation

Then from the air tables, Table A-17E

At state 2, we know the temperature and that the process was isotropic, then

which fixes our state. Then using the ideal gas law we solve for the pressure

Then from the air tables, Table A-17E

Our changes are then given as

3. Helium at 200 kPa and 300 K goes isothermally to 100 kPa.

Solution:

For state 1we know T and P, so the state is fixed. Using the ideal gas equation

At state 2, we know the pressure and that the process was isothermal, then

which fixes our state. Then using the ideal gas law we solve for the specific volume

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 we need the specific heat at constant pressure which we take from Table A-2(a) as 5.198 kJ/(kgK). Then our changes are then given as

4. Steam at 600C and 6 MPa goes isenthalpically to .01 MPa.

Solution:

At state 1 we know the pressure and the temperature. This is sufficient information to fix the state. To determine the fluid phase we go to the saturation pressure table, Table A-5, and find the boiling temperature at 6 MPa to be 275.64C which is less than 600C, so that we have superheated vapor. We can then go to the superheat vapor table, Table A-6, and read the remaining properties for at 6 MPa and 600C.

At state 2 we know the pressure, but we also that we have an isenthalpic process, so that

So with the pressure and enthalpy known at State 2, it is fixed. Next, we must determine the phase. We again go to Table A-5 and find that at 10 kPa

Since we must have a superheated vapor. We go to the superheat vapor table, Table A-6, and find that at 10 kPa and h of 3658, we interpolate to obtain

Our changes are then given by

5. Air at 1.6 MPa and 500 K goes isobarically to 200 K.

Solution:

For state 1we know T and P, so the state is fixed. Using the ideal gas equation

Then from the air tables, Table A-17

At state 2, we know the temperature and that the process was isobaric, then

which fixes our state. Then using the ideal gas law we solve for the specific volume

Then from the air tables, Table A-17

Our changes are then given as

6. Water at 10 psia and 190F goes isentropically to 100 psia.

Solution:

We note that at 10 psia the boiling temperature for water is 193.2F. So we have a subcooled liquid. We also note that our compressed liquid tables for water start at 500 psia. We will treat water under these conditions as an incompressible substance.

For state 1 both T and P are given so the state is fixed. To obtain a specific volume, we note that since specific volume only depends on temperature for an incompressible liquid, we can read it from the steam tables for saturated liquid at the given temperature. Then from Table A-4E, we have

At state 2, we know the pressure and recall that for an isentropic process for an incompressible liquid

Then looking up the specific volume

Finally we calculate our changes with

7. Refrigerant-134a at -25°C and 200 kPa goes isobarically to 10°C.

Solution:

At state 1 we know the pressure and the temperature. This is sufficient information to fix the state. To determine the fluid phase we will go to the saturation pressure table for refrigerant-134a, Table A-12, and find a boiling temperature of -10.09C. Since this is greater than our given temperature (-25C) we must have a subcooled liquid. We go to the compressed liquid table and find that it does not exist. We will have to treat state 1 as an incompressible substance, but because of state 2 , we may still need to use compressible substance tables.

To begin in treating state 1 as a incompressible liquid, we note that none of v, s, or u depend on pressure, so that there values should be equal to those of saturated liquid at

-25C. Interpolating from Table A-11

To determine the enthalpy which does have a pressure dependence, we consider a hypothetical process from saturation conditions to the compressed conditions, then

At state 2 we know the temperature, but we also that we have an isobaric process, so that

So with the pressure and temperature known at State 2, it is fixed. Next, we must determine the phase. We again go to Table A-12 and find that at 200 kPa the boiling temperature is -10.09C, which means now we have superheated vapor. The remaining properties are read from Table A-13

Our changes are then given by

8. Carbon dioxide at 2000°F and 1500 psia goes to 600F and 300 psia.

Solution:

For state 1we know T and P, so the state is fixed. Using the ideal gas equation

Our remaining properties can be read from the ideal gas tables, Table A-20E.

At state 2, we know the pressure and the temperature which fixes our state. Then using the ideal gas law we solve for the specific volume

Our remaining properties can be read from the ideal gas tables, Table A-20E.

Our changes are then given as

9. Refrigerant-134a goes from vapor to liquid at 70F.

Solution:

At state 1 we know the temperature and will assume we have saturated vapor. This is sufficient information to fix the state. We can then go to the saturation table,

Table A-11E and read the remaining properties for at 70F.

At state 2 we know the temperature and will assume we have saturated liquid. This is sufficient information to fix the state. We can then go to the saturation table,

Table A-11E and read the remaining properties for at 70F.

Our changes are then given by

10. A 1 m³ volume contains 0.2 kg of air at 250 K. Without changing the volume the pressure becomes 32 psia.

Solution:

For state 1we know T and with the volume and mass information we can calculate the specific volume

so the state is fixed. Using the ideal gas equation

Then from the air tables, Table A-17

At state 2, we know the temperature and that the process was isotropic, then

which fixes our state. Then using the ideal gas law we solve for the temperature

We go to the air tables, Table A-17, to look up the remaining properties and note that our temperature is too high for the table. Hence, we will need to use our constant specific heat approach. Our average temperature is 2041 K, which is also off the table. Let's just use the c_P at the highest temperature or 1.2161 kJ/(kg K)

Our changes are then given as