Steam Tables 3

[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

 

[quark]

First thing is that your specific volumes are wrong by leaps and bounds. SV at first condition is 0.066 cu.mtr/kg and for second condition, it is 0.1386 cu.mtr/kg (presuming your pressures are absolute values). At higher pressure, the volume per unit mass will always be lesser than at lower pressure.

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.

 

[Dogberry]

Cheers - my units should have been dm^3/kg.

Does anyone know why this is true for the volume, i.e. a molecular description?

Thanks!

 

[Dogberry]

Just read up on latent heat and that makes sense.

Thank you.

 

[ione]

1. Compressibility factor account for behaviour of real gases which differ from ideal gases. The higher the pressure applied, the stronger the intermolecular interaction and so the lower the compressibility factor (for an ideal gas Z=1)

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.

 

[Dogberry]

That makes sense, however my friends at (that really reliable source) wikipedia state:

"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?

 

[quark]

You are misinterpreting the Wiki Statement.

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?

 

[ione]

Yeah it can happen that Z>1 and the reason is the one reported in your post (repulsive force prevailing), but for pressures and temperature reported in your OP it is not the case. Please take a look at the attached paper (figures 3.1 and 3.2).

 

[Dogberry]

Hey, thanks guys.

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

 

[vpl]

Dogberry,

Is this a school assignment?

Patricia Lougheed

******

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[ione]

Dogberry,

Take a look at Spirax web site:

http://www.spiraxsarco.com/resources/steam-tables.asp

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

 

[Dogberry]

Nope Patricia, I'm sorry to say I'm much better qualified than that. Just trying to understand exactly what I'm seeing in some results I've got.

 

[Dogberry]

Thanks ione. As you said, as the pressure increases (20 -> 40bar), Z decreases (Z<1). We can therefore see that the intermolecular attractive forces dominate and increase with the pressure in this range.

 

[ione]

Going deeper into this subject, despite the fact it could happen for some gases and with specific conditions (pressure, temperature) that Z is greater than unity, for superheated steam Z will always be lower than unity. Superheated steam is steam that, for a given pressure, is at a temperature higher than the saturation temperature for that specific pressure. Now to increase the repulsive forces between molecules, pressure has to increase. If you do this keeping the temperature constant, the superheat degree will progressively decrease, until the saturation line is reached and any further increase in pressure will produce some condensation effect thus entering the wet steam region.

 

[quark]

ione is putting you on right track. Nevertheless, I would like to give one clarification to my earlier post. I gave a sample calculation to show you that, by Wiki's statement, you should compare real and ideal values at one pressure and not real values at different pressures.

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.

 

[Dogberry]

Thank you to the both of you!