M10 has a whole section devoted to demonstrating that a fixed adiabatic volume cannot undergo condensation. As I'll show, this is an unsurprising result. But the section arises from reviews of an earlier paper suggesting hurricanes were driven by condensation-induced contraction. Reviewers objected that latent heat release would outweigh the tendency to contract to to loss of water vapor volume. So there was no contraction to provide energy.

On the Air Vent threads, various experiments to study condensation/contraction were proposed involving air in bags and bottles, etc. I have been trying to explain how they lack the adiabatic expansion that causes cloud condensation. Bodies of air rise, and as they do so they expand, doing work and cooling. If they are saturated with water vapor, some of this condenses. Because the cooling is adiabatic, the latent heat is retained and slows down the cooling and condensation process. The result is that the air parcel becomes warmer than its surroundings (though cooler than it was originally) and so expands relative to nearby air.

Below the jump, I'll describe a stationary volume-change process with condensation, and calculate the expansion effect.

Consider a volume of saturated air held adiabatically in a chamber with a piston which can force an expansion. For definiteness, let's say that V=25 m

^{3}at T=300°K, containing n=1000 moles of air+water at P=100 kPa. Latent heat (of vap) L=45 kJ/mol. There are n

_{w}=36 moles of water vapor present.

As the piston moves, there are three governing equations:

**Conservation of energy**

1) n c

_{v}dT + L dn_{w}+ P dV = 0**Ideal Gas Law:**

2) P dV + V dP = n R dT + dn

_{w}RT**Clausius-Clapeyron Equation**

3) RT

^{2}dn_{w}= L n_{w}dT (see update)[Update - a small error here. I've identified dn

_{w}/n

_{w }with dp

_{w}/p

_{w}But in fact

p

_{w =p(}n

_{w}/n)

_{ }.and p is not constant. So it should be

3) RT

^{2}(dn

_{w}/n

_{w}+ dp/p)= L dT

It makes a small difference - full update to come (7.47 am 9 Nov)

It's here (140 pm) - I've marked new in brown, and obsolete in grey]

I prefer to put these in non-dimensional form, using a notation Dy =1/y dy

*(y=T,P etc)*. That means that Dy is a proportional change in y. So:

*Divide (1) by PV or nRT*

4) c

_{v}/R DT + (L*n_{w})/(R*T*n) Dn_{w}+ DV = 0*Divide (2) by PV or nRT*

5) - DT - n

_{w}/n Dn

_{w}+ DP + DV = 0

*Divide (3) by R*T*

^{2}*n_{w}
6) - L/(R*T) DT + Dn

_{w}+ DP = 0#### Solving

There are three equations in the four variables (DT, Dn_{w}, DP, DV). However, one of these, DP or DV, is specified (the amount the piston is moved).

If you set DV=0, then there is just one solution DT= Dn

_{w}=DP=0. In other words, no condensation - nothing can happen. That's what Sec 2 of M10 was about.

Fixing DP is most natural for atmospheric comparison, since P is a measure of altitude. Dividing the equations by DP gives the derivatives ( DT/DP, Dn

_{w}/DP, DV/DP). Remember these are proportional derivatives. If you change P by p%, then you change V by (DV/DP)*p %.

I did a first solution with n

_{w}=0. This is in effect the dry air case. The R code is below.

With the diatomic c

_{v}/R = 5/2, it gave, as expected,

DT/DP=2/7=0.2857, DV/DP=5/7= -0.714

That means that if P decreases by 1%, then the volume V increases by 0.71% and T decreases by 0.28%

Then I did the solution for the above case at 300 °K, 100 kPa. The answers were:

_{w}/DP=1.14, DV/DP=-0.896

DT/DP=0.106, Dn

_{w}/DP=0.916, DV/DP=-0.861Added: Plot of % variation (dry and wet) of temp, volume and wv with varying P. Note that with expansion, P diminishes. |

First observation - wet DT/DP is much smaller. Condensation is stabilizing the temperature, as you'd expect. And a 1% drop in P causes 0.916

But the answer I was looking for is the third. A 1% drop in P increases the volume by 0.861 0.896% - more than it does for dry air (0.714%). The condensation causes the volume to expand relative to non-condensing air.

#### Interpretation

This study was just of air in a confined space. But it can be applied to saturated air rising in parcels in the atmosphere. The dry lapse rate is the temperature gradient which is neutral for air moving without condensation (more here). So the moist air is cooling more slowly than the lapse rate as it rises, and is expanding relative to air on its level.#### Update - more algebra

Equations 4-6 can be rewritten in even fewer non-dimensional parameters. I'll take the ratio c

_{v}/R = 5/2.
Then with b=L/(R*T), and c=(L*n

_{w})/(n*R*T)
7) 2.5 DT + c Dn

_{w}+ DV = 0
8) - DT - (c/b) Dn

_{w}+ DP + DV= 0
9) -b DT + Dn

_{w}+ DP = 0
Eqns 7-9 are linear homogeneous equations in the four variables:
(DT, Dn

_{w}, DP, DV) The answer can be expressed in terms of a vector a=( a_{1},a_{2},a_{3},a_{4}) and an arbitrary parameter h:
DT=a

_{1}h, Dn_{w}=a_{2}h, DP=a_{3}h, DV=a_{4}h.
Finding a is a matter of linear algebra (school algebra or determinants). I'll just state the result and verify:

a

_{1}= 1 + c + c/b, a_{2}= b - 3.5, a_{3}= 3.5+c+b*c, a_{4}= -(2.5+b*c+2.5*c/b-c)
Substitute in (7) with h=1 (h will cancel):

2.5*a

_{1}+ c * a_{2}+ a_{4}= 2.5+2.5*c+2.5*c/b + c * b-3.5*c - 2.5 - b*c+-2.5*c/b+c = 0
And (8): -a

_{1}-(c/b)* a_{2}+ a_{3}+ a_{4}= 1- c-c/b - c+3.5*c/b + 3.5 + c + b*c - 2.5 - b*c-2.5*c/b+c = 0
And (9): -b*a

_{1}+ a_{2}= -b-c-b*c + b- 3.5 + 3.5+c+b*c = 0
You can use this result to get any desired derivative, eg DV/DP = a

_{4}/a_{3}.
Remember again that D is proportional, so dV/dP = (V/P) a

Update. I see that Eq 3.74 of Rodrigo Caballero's PhysMet lecture notes gives a version of DT/DP which is equivalent to (1+c)/(3.5+bc). He says this is an approximation (3.72). I think that is equivalent to omitting the last term in my Eq 2, which would indeed give this expression. However, I believe the inclusion of that term is more accurate.

_{4}/a_{3}.Update. I see that Eq 3.74 of Rodrigo Caballero's PhysMet lecture notes gives a version of DT/DP which is equivalent to (1+c)/(3.5+bc). He says this is an approximation (3.72). I think that is equivalent to omitting the last term in my Eq 2, which would indeed give this expression. However, I believe the inclusion of that term is more accurate.

**Appendix - R code**

T=300

N=1000

Nw=36 #0

P=100000

R=8.3

Cv=R*2.5

L=45000

LRT=L/R/T

m=matrix(c(Cv/R,-1,-LRT, LRT*Nw/N, -Nw/N,1, 1,1,0, 0,1,1),3,4);

M=m[,c(1,2,4)]; rh= -m[,3];

ddV=solve(M,rh); # D/DV volume derivatives

M=m[,c(1,2,3)]; rh= -m[,4];

ddP=solve(M,rh); # D/DP pressure derivatives

print(ddV); print(ddP);

Minor point: L can only be considered constant over a narrow temperature range. It does vary with temperature. See for example here.

ReplyDeleteDeWitt,

ReplyDeleteYes, I've calculated derivatives at 300K only. If you want to track over a big change, these need to be integrated, including variable L.

Nick,

ReplyDeleteCondensation during expansion can be the basis for a heat pump. The expansion process is analysed in my paper: N.G. Barton “An Evaporation Heat Engine and Condensation Heat Pump”, ANZIAM J, Vol 49 (2008), 503-524. My analysis uses the set of variables: V, T, P_air, P_vap, rho_air, rho_vap and is an alternative

to your approach. My analysis is not limited to one temperature, and the latent heat is taken as a tabulated function of T. The ideal gas law is

still used. Your basic conclusion is correct of course: there are competing effects of latent heat (that acts to warm the air and increase the volume) and condensation (which acts to decrease the volume). Latent heat is dominant, so the moist adiabatic lapse rate is slower than the dry rate.

Thanks, Noel,

ReplyDeleteYes, I appreciate that you have done a full cycle analysis, which is necessary for analysing an engine. Here I'm just trying to demonstrate the dominance of the latent heat. As I said at the Air Vent, I think your engine is a very good demonstration of this, on the evaporation side rather than condensation.

Nick,

ReplyDeleteI commented on Lucia's blog basically the same point as you stated. The condensation releases 540 cal/g of removed water and the water vapor volume loss gives a much smaller total volume loss than the air heated by this excess energy gives an increase in volume. I used a 1% by weight water vapor content and showed it gave a net 6% volume increase. Lucia came back with some irrelevant comment, and the blog comments went on ignoring this simple fact. Obviously the cooling from adiabatic expansion when rising is the basic cause, and the dew point drops faster than the pressure, so at some point the water vapor condenses out.

Leonard,

ReplyDeleteYes, a lot of people have found a LH/mass factor of about 6. Dr Rosenfeld, in his review of an earlier AM paper, estimated (S12437) 223/40=5.6. Jim D got about 5.7.

The latent heat is released. It won't just vanish.