Steam Tables
Steam Tables
(OP)
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
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





RE: Steam Tables
Enthalpy of steam is sum of sensible and latent heats. You require higher latent heat at lower pressures and that is the reason the enthalpy values are higher at low pressures.
BTW, get SteamTab. This is very accurate, handy and free as well.
RE: Steam Tables
Does anyone know why this is true for the volume, i.e. a molecular description?
Thanks!
RE: Steam Tables
Thank you.
RE: Steam Tables
P1*V1/Z1 = P2*V2/Z2
V2 = P1/P2*Z2/Z1*V1
Being
P1 = pressure at state 1
P2 = pressure at state 2
Z1 = compressibility factor at state 1
Z2 = compressibility factor at state 2
V1 = specific volume at state 1
V2 = specific volume at state 2
2. The latent heat of vaporization (specific enthalpy of evaporation) plays the biggest role and the specific enthalpy of evaporation decreases as pressure increases. For saturated steam at 40 barg the specific enthalpy of evaporation is 1705.62 kJ/kg, whilst for saturated steam at 40 barg the specific enthalpy of evaporation is approx 1879.49 kJ/kg.
RE: Steam Tables
"At high pressures molecules are colliding more often. This allows repulsive forces between molecules to have a noticeable effect, making the volume of the real gas (Vreal) greater than the volume of an ideal gas (Videal) which causes Z to increase above one."
Suggesting the exact opposite. However, based on the steam tables and what you have just said, I take it wikipedia is wrong?
RE: Steam Tables
Suppose, the steam follows ideal gas laws and you don't know the specific volume of steam at 20 bar a. Then you should calculate it by the ideal gas law as,
V2 = P1V1 x T2/(T1 x P2),
Since temperature is same in both cases, the formula becomes,
V2 = P1V1/P2, So, it should be 0.066 x2 = 0.132 cu.mtr/kg.
Actual specific volume (Vreal) is 0.1386 cu.mtr/kg which is greater than the ideal condition specific volume of 0.132 (Videal) cu.mtr/kg.
Where is the confusion?
RE: Steam Tables
RE: Steam Tables
I guess the only problem is you don't know if the 0.066 is greater or less than ideal, so it's difficult to know whether repulsive or attractive forces are at work.
i.e. if 0.066 is greater than ideal and ideal is, say, 0.64, then at 20bar the repulsive forces are greater, but if 0.066 is less than ideal and ideal is, say 0.68, then the repulsive forces are less at 20bar and it is tending towards ideal.
Looking at those curves ione it seems that you need to know where you are on the curve to know whether the difference from ideal is expanding or contracting with changes in pressure.
Thanks again,
Ed
RE: Steam Tables
Is this a school assignment?
Patricia Lougheed
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RE: Steam Tables
Take a look at Spirax web site:
htt
Choose the option superheated steam region and then enter pressure and temperature. The site will give you all the info you need (including compressibility factor Z). At your pressures and temeprature Z<1 so repulsive forces do not prevail
RE: Steam Tables
RE: Steam Tables
RE: Steam Tables
RE: Steam Tables
In actual case, for both the pressures, Videal>Vreal, as already noted by ione.
Videal (@40bara) = 1000x8.314x(350+270)/(18x4000) = 71.59 liters/kg or 0.0716 cu.mtr/kg and at 20 bara it is just double that value, which is 0.1432 cu.mtr/kg.
RE: Steam Tables