Determine the entropy of steam at 1000 psia and a quality of

Determine the entropy of steam at 1000 psia and a quality of 50%

Solution

If we assume that the radiator does not expand or contract significantly with the temperature change, we have is a constant volume process. We can find the entropy change from the initial state, where we are given the temperature and pressure and the final state where we know the temperature and we also know that the specific volume will be the same as it is at the initial state. We can also use the information on the radiator volume to find the system mass.

At the initial state of 200 kPa and 150^oC, we find the following properties from Table A-6 on page 920: v₁ = 0.95986 m³/kg and s₁ = 7.2810 kJ/kgK. From Table A-4 on page 916, we see that, at the final temperature T₂ = 40^oC, the specific volume, v₂ = v₁ is between v_f(40^oC) = 0.001008 m³/kg and v_g(40^oC) = 19.515 m³/kg. Thus the final state is in the mixed region. We find the final quality from the following equation.

We can then use this quality to find the final entropy.

To find the total entropy change we have to find the system mass. We can find this from the radiator volume and the initial specific volume.

= –0.132 kJ/K

Note that we can have a negative entropy change and still satisfy the second law inequality that dS dQ/T because the heat transfer is negative. (We did not actually compute Q, but we know that heat has to leave the system to change the state from a vapor at 150^oC to a mixture of liquid and vapor at 40^oC.)