Understanding Steam at High Pressures: Enthalpy and Volume Explained

  • Context: Graduate 
  • Thread starter Thread starter Dogberry
  • Start date Start date
  • Tags Tags
    Steam
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
1 reply · 3K views
Dogberry
Messages
1
Reaction score
0
Hi,

Can someone please explain this to me:

Looking at the steam tables, if I have steam at 40bar and 350C then the specific volume is ~66.4 m^3/kg and enthalpy is 3095 kJ/kg.

If I halve the pressure to 20bar (still at 350C), thus 'doubling' the specific volume to 138.56 m^3/kg, the enthalpy becomes 3138.6 kJ/kg.

Two questions:

1. The volume is not exactly double, it's more than that, and while understand this is not an ideal gas, my understanding of the compressibility factor led me to believe the volume should be greater than expected at higher pressures. This is the other way round.

2. From my understanding of enthalpy: h = U + PV, because the volume has 'doubled' and the pressure has halved, the enthalpy should have stayed approximately constant. It doesn't. In fact, to achieve the same enthalpy, I would need to drop the temperature to 330C (@20bar). I don't understand why, where is this drop captured in the equation? Surely lower temperature = lower enthalpy?

Any help will be greatly appreciated and thank you for your time!

Ed
 
Physics news on Phys.org


Dogberry said:
Hi,

Can someone please explain this to me:

Looking at the steam tables, if I have steam at 40bar and 350C then the specific volume is ~66.4 m^3/kg and enthalpy is 3095 kJ/kg.

If I halve the pressure to 20bar (still at 350C), thus 'doubling' the specific volume to 138.56 m^3/kg, the enthalpy becomes 3138.6 kJ/kg.

Two questions:

1. The volume is not exactly double, it's more than that, and while understand this is not an ideal gas, my understanding of the compressibility factor led me to believe the volume should be greater than expected at higher pressures. This is the other way round.

2. From my understanding of enthalpy: h = U + PV, because the volume has 'doubled' and the pressure has halved, the enthalpy should have stayed approximately constant. It doesn't. In fact, to achieve the same enthalpy, I would need to drop the temperature to 330C (@20bar). I don't understand why, where is this drop captured in the equation? Surely lower temperature = lower enthalpy?

Any help will be greatly appreciated and thank you for your time!
Start with dH = dU + d(PV). So [itex]\Delta H = \int dH = \Delta U + <br /> \int VdP + \int PdV[/itex]

If [itex]\int Pdv + \int VdP = 0[/itex] there is no change in enthalpy if temperature is constant.
But for steam, this is not the case because it is not an ideal gas. [itex]\int Pdv + \int VdP > 0[/itex]. As you compress saturated steam, some of the steam condenses.

AM