Understanding Steam at High Pressures: Enthalpy and Volume Explained

[Dogberry]

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

 

[Andrew Mason]

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 \Delta H = \int dH = \Delta U + <br /> \int VdP + \int PdV

If \int Pdv + \int VdP = 0 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. \int Pdv + \int VdP &gt; 0. As you compress saturated steam, some of the steam condenses.

AM