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REL
28-11-2003, 08:27 AM
I am having a bit of trouble with the PH diagrams. I posted information on another forum, however, interest level is really low on such things there. I am working from the RSES manual 2 and Dossat's book.

I thought I had a pretty good handle on it, but I hit a road block when I started using Dossat's. I am really trying to self improve and just started off with the book like a green newbie page one.

Here is a copy of what I have posted and any help is appreciated. If you do not mind keeping your thoughts in IP instead of SI, I also would appreciate that as I do not yet have a handle on the conversions, working on that too :)

R134a
100 degree saturation temp, outlet of the condenser is point A with no subcooling.

This gives you
p = 138.83 psia
T = 100 F
h = 44.23 Btu/lbm
s = 0.89848 Btu/lbm Degree F
v = 0.339 cf/lbm

!

S is absolute entropy
s is specific entropy (lowercase)
sm specific per unit mass

Now ASHRAE list r134a @ 100 degrees at sf (f is subscript for liquid) .09142 and sg (g is subscript as well) at .21868 both are specific and subtraction results in 0.12726

Since sm = (1-X)(sf)+(x)(sg)
Where X is percent quality

So now I have to solve for X not sm since they gave me sm, my freakin brain hurts!
s = 0.89848 = (1-X)(.09142)+(x)(.21868)

Ok, Back at the book a bit indicates there is perhaps an error in the book, but my confidence is NOT high about that so I will explain.

If you change the temperature to 40 with 15% quality you have:

sm = (1-X)(sf)+(x)(sg)
sm = (1 - 0.15)(.05359)+(0.15)(0.22172)
sm = 0.788 Btu/lbm degrees F

To me, that states the entropy of R134a entering the evaporator with an 85/15 mix has an entropy of 0.788Btu's per pound of mass.

If that thinking is correct, then No number can replace X and come out with the result of the first problem as point A on the PH chart is right at the point of 100% liquid, with no subcooling.

REL
28-11-2003, 09:21 PM
Marc,

This is from chapter 10 "The theoretical saturated vapor-compression cycle" in Dossat's book, fifth edition. I am looking at the First step on the PH diagram, state point A which is at the outlet of the condenser.

The have marked the state point directly on the saturated liquid curve. They state the entropy and specifc volumes are taken from table 6.6 which is an ASHRAE table for the thermodynamic properties of R134a.

I developped "X" as percent quality because of lessons in chapter six indicating it. The second problem involving the 15% figure is part of an example in chapter six, not something I derived. I am still on the learning curve and do not want to cloud it with my own math.

REL
29-11-2003, 06:57 AM
Had to type it out as the scanner power supply has puked!

At the end of chapter six "Properties of vapors"

Quality is designated with the letter x. The quality of a liquid-vapor mixture is used to calculate the enthalpy, entropy, specific volume, and density properties of a mixture. Keeping in mind that any wet vapor (liquid-vapor mix) exist at its saturated state, it follows that its temperature will always be at its saturation temperature corresponding to the pressure of the mixture. Equations 6.1, 6.2 and 6.3 are used to calculate the specific volume, enthalpy and entropy of liquid vapor mixtures having a quality of x percent. The subscript m identifies mixture quantities, and subscripts f and g denote values for the saturated liquid and vapor phases respectively.

Vm = (1- x) (Vf) + (x) (Vg) 6.1
Hm = (1 ˇV x) (hf) + (x) hg) 6.2
Sm = (1 ˇV x) (sf) + (x)(sg) 6.3

(the V, H, and S are all lower case, but I am not aware of how to subscript with MS word)

Section 6.7 Optional analysis "calculating the properties of liquid vapor mixture."
(this is an exercise type of section with solutions provided to the user)

Determine the specific volume, enthalpy , and entropy of R-134a vapor having a saturation temperature of 40 (degree mark) F and a quality of 15%

Solution from table 1 , the specific volume (reciprocal of density) , enthalpy, and entropy of R-134a are :
Vf = 0.0125,
Vg = 0.9534,
Hf = 24.694,
Hg = 108.705,
Sf = 0.05359,
Sg= 0.22172

Applying equation 6.1 the specific volume of the mixture is:

Vm = (1.0 ˇV0.15) (0.0125) + (0.15) (0.9534)
Vm = 0.1536 ft3/lbm (3 is super script for cubic feet)

Applying equation 6.2 the enthalpy of the mixture is

Hm = (1.0 ˇV 0.15) (24.694) + (0.15) (108.705)
Hm = 37.30 Btu/lbm

Applying equation 6.3 the specific entropy of the mixture is:

Sm = (1 - 0.15)(.05359)+(0.15)(0.22172)
Sm = 0.788 Btu/lbm (degree mark) F

(Again, Capitol letters used inappropriately)


Chapter 10 The theoretical saturated vapor-compression cycle.

10.3.1 Locating point A of the saturated cycle on the PH diagram. (R-134a is the refrigerant on the illustrated diagram at 100 degrees F)
State point A is located near the bottom of the condenser where the refrigerant is a saturated liquid at the condensing temperature and pressure. Since the refrigerant at point A is always a liquid in this theoretical cycle, the point is located somewhere along the saturated liquid line of the ph diagram. If either the pressure, temperature, or enthalpy of state point A is known, its location on the diagram is found by following the known variableˇ¦s property line to where it intersects the saturated liquid line. Point A is drawn at the intersection of these two lines.

The properties of the refrigerant at point A are:

p = 138.83 psia
T = 100 F
h = 44.23 Btu/lbm
s = 0.89848 Btu/lbm (degree mark) F
v = 0.339 ft3/lbm

The entropy and specific volume are read from table 6-2. In practice, pressure and temperature are readily measurable with service gauges and a thermometer. Therefore, point A can always be easily plotted on a ph diagram.

Now back to me here and not the book. Table 6-2 does not list "s = 0.89848 Btu/lbm (degree mark) F " any where at all for R-134a It does list the Sf and Sg values as Sf = 0.09142 and Sg = 0.21868

The same information may be obtained from http://www.suva.com.mx/pdf/TPSuva134a.pdf page 3. There is a slight discrepancy in that table and the one in the book from ASHRAE 1997. I have the ASHRAE handbook and there is no discrepancy between it and Dossat's book. Perhaps DuPont is off a little, who knows.

Anyway, thanks for the read here, it certainly is not a topic a lot of folks take interest in at all. I am well aware that a service tech would not have to fully understand such things to be effective. I myself subscribe to the theory that the more I do understand about it, the more effective I can be servicing.

frank
29-11-2003, 08:41 AM
To subscript in MS Word - highlight the text you want to subscript, right click, choose format font, check the subscript box and exit

REL
01-12-2003, 09:46 PM
Thanks for the assistance Marc. I have discovered the problem now, it is a typo. It is unfortunate that they give the answer within the problem instead of stating how to derive the answer first.

I did learn from your mathmatical process. Since x is percent of quality and you can not have 634% quality it made me review further.

It seems when Horan "updated" revised or whatever to the ASHRAE tables in Dossats book, he made a typo. It should have been 0.089848.

I was looking to solve for the answer instead of solving the equation. Ironicly, I made the exact same error in my calculation at 15% quality. It should have been 0.0788 Btu/lbm degrees F.

Thanks for your help,
Rich

Andy
01-12-2003, 11:31 PM
Hi:)
I learned using steam tables and I never remember this amount of confusion:confused: :eek:
Anybody still use these:confused:
They served me alright at the time, and I am ashamed to say I have never used dryness fractions since.:D
What tended to put me off was the questions we were given about working out the dryness fraction at a chill room compressor after the suction line had run tru a freezer.
I suspect dryness fractions would be more suited to people working on steam turbines:rolleyes:
Regards. Andy:)

Andy
02-12-2003, 09:25 AM
Hi Marc:)
hmm AKV''s are a newish thing, but the problem of minimal suction superheat at the compressor inlet, even when there was superheat at the cooler outlet is not. This usually happens in systems where capacity reduction is used, such as central pack systems. Suction accumulators should be fitted to these types of applications without exception:mad:
But then if designers did this there would be no work for technicians then:D
I can see where dryness fractions would aid in the sizing of these accumulators.
Kind Regards. Andy:)

REL
03-12-2003, 01:23 AM
Originally posted by Marc O'Brien
You probably asked the same questions on alt.hvac, lol, what did you expect? That is now hvac-talk No.2 :)

I was "in the box" and could not get out.

Knowing X represented quality, I limited the possibilities to values less than 1 while trying to solve the equation. As I knew the equation was correct, and the value could not exceed 1, I questioned my own possible application of the wrong equation for the value.

I have spent more than a few years in this trade without a full grasp on a lot of the science behind it. I saw your thoughts on training a person to assist you and the process you would take them through. I have also observed your postings for over 4 years and knew Dossat's book would be of value. Your thoughts about Modern and Refrigeration Tech were on target with my own thinking.

I did pose the question on HVAC-TALK as there are some good people there, and some are interested in learning. There is also a lot of ego stroking and odd people. At the same time, you have highly skilled people like Prof-Sporlan visiting to answer some of the tougher questions. Now days alt.hvac has developed into mostly whining and who can tell the nastiest story about who. There are still some good people there, but there is not much information getting exchanged.

This site has far more in depth discussions about our craft. There is a lot of good talented and educated people contributing to it. It made far more sense to put it here where Andy, the other Andy, Dan, yourself, Gary or a host of other people would answer it with something beyond "You do not need that. Why are you trying to figure that out" types of replies.

Thanks again for the assistance,
Rich

Prof Sporlan
04-12-2003, 02:02 AM
Entropy = J/K = Joule per Kelvin

Is it a complex concept?
Somewhat, perhaps... since it cannot be measured, and it has a broad definition ranging from a measure of unavailable energy to the amount of chaos.


Just looking at it's definition or formula, unit, whatever, J/K:
There are numerous way to calculate. One way:
s = - (<i>d</i>f / <i>d</i>T)<sub>v</sub>
where:

s = entropy, kJ/kg-K
f = Helmholtz energy, kJ/kg
T = temperature, K

With a thermodynamic process, one is less concerned with an absolute value of entropy, but with the change of entropy, which is of course is positive for all real world closed system processes.

Which brings up an interesting question. Where on the scale of min to max entropy does the universe currently reside? If the universe will someday reach maximum entropy, heat death if you will, and started at minimum entropy, the big bang perhaps, where on this scale do we currently reside?


Just got confirmation, the Prof does indeed visit hvac-talk and is on his way there this very minute...

Prof, give em that white mans magic, that German touch !!
LOL. A short lesson on the Second Law should create some turmoil :)

RogGoetsch
04-12-2003, 09:00 AM
Originally posted by Prof Sporlan

Which brings up an interesting question. Where on the scale of min to max entropy does the universe currently reside? If the universe will someday reach maximum entropy, heat death if you will, and started at minimum entropy, the big bang perhaps, where on this scale do we currently reside?

LOL. A short lesson on the Second Law should create some turmoil :)

Let me try this one out here:

I was on a white-water river a few years back and I was admiring how the kayakers could ride upriver on eddy currents along the banks even as we in rafts went hurtling down in midstream. It reminded me of life in the universe because I think of the concept of entropy and the behavior of energy/matter through time as being like a great river, inexorably descending from order to chaos. In the midst of this stream, some of the matter begins to organize itself into more complex structures, a process we call "life".

Does life follow the laws of entropy? Of course the matter of which living systems are composed continues its merry way down the river no matter how much we may try to retard the flow, but this force that organizes the matter, this "life" force, what of it? I tend to view life as evidence of invisible fingers rippling the waters of energy/matter in non-entropic directions. Am I wrong? Is life mathematically explanable with something on the order of eddy currents?

frank
05-12-2003, 08:20 AM
It's easy enough to understand that if work is done on a gas to compress it and almost all of that work manifests as specific heat, adding to the gases temperature, then the process may theoretically be 100% reversible. But dq/T ??

That would be the CARNOT cycle, the perfect cycle that all other systems try to emulate, however, it is impossible to produce a system without losses. :)

It's the same with perpetual motion.

Prof Sporlan
06-12-2003, 03:20 AM
I'm not getting in fuzzy feelings looking at the entropy related expressions? I need to mess with a bit of calculus rather, I guess.

Prof ?

LOL! You seem to have a need to be "touchy-feely" with your fine analytic capabilities… :) This can be a bit of a problem for a property that cannot be measured. But it is not insurmountable. :)

The ds = Q / T equation assumes an isothermal process. Consider the following isothermal process…

You have a static universe consisting of 2 large planets, lets call them A and B.

Planets A and B are uniformly at temperatures 400K and 300K respectively. And assume that heat radiation transfers 1000 kJ/kg from planet A to planet B.

Using the above equation, the entropy change in planet A: -1000 kJ/kg / 400 = -2.50 kJ/kg-K.

The entropy change in planet B: 1000 kJ/kg / 300 = 3.33 kJ/kg-K.

Add these entropies together: 3.33 – 2.50 = 0.83 kJ/kg-K.

The Second Law holds true! :)

Prof Sporlan
06-12-2003, 05:41 PM
For a process in which temperature changes, the entropy change must be found by integration, i.e.:

delta s = int(ds) = int(dQ / T)

In the two planet example, we assume the planets are so large that the small heat transfer will not change the average temperatures of the planets. Therefore, we consider this process isothermal.

frank
07-12-2003, 03:01 PM
Spot on Marc

We had a 70's theme dinner party with a few friends and it was a hugely funny night. No doubt you can see that they took the P**S out of me! :D

Dan
09-12-2003, 01:42 AM
The entropy change in planet B: 1000 kJ/kg / 300 = 3.33 kJ/kg-K.

Add these entropies together: 3.33 – 2.50 = 0.83 kJ/kg-K.

The Second Law holds true!

I give up. What happened to the 0.83 kJ/kg-K and how does this assist the proof of the second law of thermodynamics?

Is this energy lost to this particular exchange but yet available to another? Is there some fraction or difference yet to be extracted from this energy in another exchange? I am still puzzled by the concept of entropy. Sometimes I hear it is less useable energy, other times I hear it as more dispersed energy; the latter strikes me as reversible and the former not.

frank
09-12-2003, 08:52 PM
Dan

If you consider an electric motor being driven by 1kw of energy input, you know that you are not going to get 1kw of output due to inefficiencies in the motor as well as friction losses, heat losses etc. this lost energy is entropy.

Prof Sporlan
10-12-2003, 03:32 AM
The First Law says energy is conserved, i.e., you can’t lose it. More specifically, energy, or its equivalent in mass, cannot be created or destroyed.

The Second Law says when you change energy from one state to another, you "lose" some in the process. But in reality, this energy isn’t lost; it is only in a form that is not available to perform useful work.

Consider the aforementioned 2 planet example. Ignoring the assumption of it being an isothermal process and the entropy calculations, what is happening here? We have energy in the form of radiant heat being transferred from plant A to planet B. Over time, planet A will cool, and planet B will warm up. Assuming we do not have anything else going on here, the temperatures of the planets will approach each other.

So energy is being conserved. But why is entropy increasing? Where is this unavailable energy?

Consider that for a process to have no change in entropy, the process must be reversible. As plant A cools down, is there any way for it to return to 400K?

Prof Sporlan
11-12-2003, 04:41 AM
the process only becomes reversible once the two planets are at equal temperatures
Assuming we have radiant heat as the only means of converting energy, then everything stops when the planets are at equal temperatures. We are then at max entropy and nothing else can happen.

But this example is intentionally over simplistic to show how the First and Second Laws work. The mass of the planets have an energy equivalent, and further energy conversions would be possible, which would lead to further increases in entropy.