thermo problem

[MPKG]

OK -- I am good at conceiving engineering problems -- but not always good at solving them.

If I have refrigerant (R22) at saturated vapor state in a fixed volume (rigid-walled tubing) which is closed tight and not open to atmospheric pressure and cool the tubing -- what happens to the temperature, pressure, entropy, and quality of the refrigerant inside the tubing?

My first thought was the qaulity, temperature, and entropy go down and the pressure stays the same but then what fills the space in the fixed volume when vapor is converted to liquid and takes up much less space?

I am thinking that everything goes down: pressure, temperature, entropy, and quality. So on a temperature-entropy diagram this process would be represented by a dashed line curving down and to the left. Would you agree?

 

[TD2K]

R-22 vapor still fills the vapor portion of the container along with some liquid.

You have a certain volume full of saturated R-22 at a given pressure and temperature. It has a certain density therefore you have a certain mass in the container. The properties of the R-22 are per the thermodynamic tables.

As you cool it down, you form vapor and liquid, essentially following both sides of the vapor/liquid envelope on the Mollier diagram. Let's say you cooled it 10 deg. The saturated liquid has a certain density. The saturated vapor has a certain density. You can work out how much of each you have to give you the same overall mass when starting out. The properties of each phase is just per the thermodynamic tables.

You never have 'just' liquid for the case you have described. The liquid always exerts its vapor pressure though as you reduce the temperature, you can have almost 'all' liquid.

 

[25362]

This constant mass, constant total volume cooling process is undefined. The move to the left and down on a pressure/enthalpy chart is right. You can even increase entropy a bit, if, for example, R-22 stays at the saturated dry vapour condition, at lower P,T values. Just look to a Mollier chart or a pressure/enthalpy chart, such as fig. 3-35 in Perry VI, and you'll see that the final conditions will be determined by the new enthalpy and by another parameter such as temperature or pressure.

 

[davefitz]

The device you described is called a "heat pipe" . The constant mass/ constant volume implies the pressure changes dramatically as temperature changes.

One problem with heat pipes is that they fail due to high-high pressures when overheated above their design rating.

In the case of R-22, as temperature drops, the pressure drops perhaps to vacuum, and the tube should be designed for full vacuum conditions.

 

[MPKG]

Thanks everyone for your responses.

25362, to better define the cooling process would it be OK to assume constant enthalpy -- kind of like what is done for a throttling process problem where going from states 1 to 2 you move down and to the right on a temperature-entropy diagram?

 

[TD2K]

If you are cooling or heating the tube, it's definitely not a constant enthalphy situation.

 

[superheat]

You will not have throttling in a fixed pipe. Throttling is done to control pressure and temp between evap and cond. It requires a compressor to pump the freon before the valve can throttle the flow.

Heat pipes rely on latent heat and 2 air streams to boil and condense freon to transfer heat.

A pipe full of freon would just be a thermal storage system.

 

[poetix99]

25363: Constant mass, constant volume cooling process is not undefined. The average density and the temperature, so long as the system is a mixture of liquid and vapor, will uniquely determine the state of a single species system. Otherwise the pressure (or some other property) must also be measured.

Davefitz: A heat pipe is a subset of constant volume, constant mass systems. The system described, depending upon the geometry of the container, and where the heat exchange takes place, is not necessarily a heat pipe.

As TD2K said, this is definitely NOT a constant enthalpy process that you seem to have described.

MPKG: If you do intend to have designed a heat pipe, there are many other considerations that must be factored into effective design of heat pipes. You should consult a specialized text for this. I do not have any references to offer to you.

 

[25362]

to poetix99, you just said it: when the final average density and temperature for a liquid/vapour mixture, and/or any other property for a single phase, is measured, the process is defined. Otherwise, it isn't.

Looking to a Mollier chart, starting with a saturated dry gas, one may stay as such or condense it, partially or totally, on cooling. As long as the necessary and sufficient final state properties are not determined the process path is undefined.

 

[poetix99]

25362: In the case of a fixed volume (a closed vessel) and a fixed mass (whatever I've sealed into the closed vessel; presumably I know the mass that I've placed inside) the average density is known, and is equal to the initial density. The density is invariant throughout the "process".

Let's assume that the temperature can be measured.

What's indeterminate? One cannot freely choose states on a Mollier chart with the closed system we are considering; you must recognize that certain states can't be attained.

BTW, in the initial case of saturated vapor you've last described, with the fixed volume as MPKG had originally stated, you WILL NOT "totally condense" mass that had started as saturated vapor; there is too much volume in the system. There will be a vapor fraction - even if the condensate changes to solid. (However, if we were to wander into three-phase states in this conversation, we would indeed need another property, but I don't think you were referring to that circumstance.)

 

[235zzzo]

Hi MPGK!

You have to keep in mind, unfortunely, that the thermodynamic property, called entropy, s = 0 (ideal process) or s >0 (real process)and never s < 0 (impossible process) as a consequence of the Second Law.
Cheers.
zzzo

 

[jb48]

Seems to me that this a state function, and hence the final state can be determined independent of the path.

 

[chicopee]

W/ YOUR SEALED RIGID TUBE CONDITION, THE FOLLOWING HAPPEN WHEN HEAT IS REMOVED FROM THE FREON:
1. SPECIFIC VOLUME REMAINS CONSTANT
2. PRESSURE AND TEMP. DECREASE
3. QUALITY DECREASES
4. ENTROPY DECREASES
5. ENTHALPY DECREASES

THE KEY TO YOUR PROBLEM IS THE CONSTANT SPECIFIC VOLUME. SKETCH A P-v DIAGRAM AND ASSUME AN INITIAL SATURATED VAPOR STATE FROM WHICH NOTATE PRESSURE, TEMP, SPECIFIC VOLUME, QUALITY, ENTROPY AND ENTHALPY FROM A R-22 PROPERTY TABLE. AS HEAT IS REMOVED, ASSUME A FINAL STATE SATURATED PRESSURE FROM WHICH YOU CAN DETERMINE SATURATED TEMPERATURE; SINCE YOU ALSO KNOW THE SPECIFIC VOLUME, YOU CAN CALCULATE THE QUALITY, ENTROPY AND ENTHALPY.
INTERPOLATION MAY BE NECESSARY TO GET THE INITIAL AND FINAL PROPERTIES MENTIONED.

 

[ivymike]

You have to keep in mind, unfortunely, that the thermodynamic property, called entropy, s = 0 (ideal process) or s >0 (real process)and never s < 0 (impossible process) as a consequence of the Second Law.
Cheers.
zzzo

it's been a long time since I took thermo, but I thought that the second law was only applied to closed systems. Since heat is being removed from this system, it's not closed. If you &quot;draw a bigger box,&quot; such that no energy enters or exits the box, then you can apply the second law. What I think you'll find is that even though s decreases for the closed volume discussed above (as h is removed), s will increase by an equal or greater amount in whatever ends up catching the h from the volume. If you know the initial state of this system, and you know how much h is removed from the chamber, I believe that you know the final state (assuming you have the right tables and can do the math). As one of my fav. thermo profs used to say, &quot;it's all in the h's.&quot;

 

[ivymike]

To highlight just how long it's been since I took thermo, a vague recollection has just come to me that &quot;u&quot; is the prefered energy variable for a system like this (not &quot;h&quot. Either way we're talking about the internal energy of the working fluid (and possibly the container)...

 

[25362]

To chicopee, sorry, but your conclusions don't seem right to me: first of all, this freon was given at the saturated dry vapour condition as a start. As such, assuming the rigid container doesn't suffer any appreciable change, cooling could:

decrease entropy,
increase entropy,
leave entropy unchanged,
specific volume may change if there is condensation,
P and T decrease,
quality drops,
quality stays constant.

A diagram &quot;enthalpy/log pressure&quot; as shown in Perry VI, fig. 3-35, is more explicit than anything else.

 

[235zzzo]

Hi!

So we come to a &quot;main question&quot;: How to decrease the Entropy, or d(s) < 0. Is it possible?

I invite my co-fellows to specific processes, either open or close systems, where entropy, s &quot;goes down&quot; physically speaking, I mean.

In my coming holidays I'll take a thermodynamics manual to read again, with pleasure. As a matter I believe the Second Law is applied to all systems. It has an universal nature, hasn't it? Otherwise it is not a Law !?

Just be careful to define the boundary conditions. That's very important. Or some confusion can arise.
Let's keep the discussion open, please.
Cheers.
zzzo

 

[25362]

Lets keep the discussion open as zzzo says.
One way to define entropy is by relating it to the unavailability to do work (the quality of energy). During an irreversible process in which the entropy of a system increases by dS, energy E=T[sub]min[/sub]dS, becomes unavailable to do work, where T[sub]min[/sub] is the lowest temperature available to the system.

However, entropy can decrease (order increases).

Running a refrigerator decreases the entropy of its contents. But we must do work for heat to flow in the direction it doesn't normally flow. If we enlarge this closed system to include with whatever else supplies the refrigerator with work, we would find the entropy of this new closed system doesn't decrease. The entropy decrease at the refrigerator is offset by increases elsewhere in the system. If irreversible processes occur anywhere in the system, then there is a net entropy increase.

By way of reverse logic: any system whose entropy seems to decrease cannot be a &quot;closed&quot; system. If we enlarge the boundaries to encompass whatever may exchange energy with the original system, the entropy of the larger system will not decrease.

Life on Earth is remarkable in the sense that it is full with examples of entropy decreasing processes: the building of a house, the growth of a living thing, this printed page, etc., in an universe governed by a tendency toward disorder.

However, Earth is not a closed system. It gets high-quality energy from the Sun. When considering the Earth-Sun system, the entropy decrease associated with life and civilization is more than compensated by the increase in entropy of the Sun's nuclear fusion processes.
Ultimately, we have the 2nd Law of Thermodynamics:
The entropy of the universe can never decrease.

 

[25362]

Poetix99, TD2K and Chicopee all are right and I am wrong. I somehow &quot;forgot&quot; (shame on me) we are dealing with a constant average density (mass and volume) of the fluid in a &quot;closed&quot; container of invariable dimensions.

Therefore, finding only one paramater, P, or T is enough to determine how much liquid and how much vapour (i.e., quality) are there upon cooling.

On a two-phase system with one component the phase rule would give us one degree of freedom as sufficient to fix all the intensive properties of the system.

Yes MPKG, on a P,T chart, the cooling curve would move down and to the left, following a line of constant density, thus pressure, temperature, quality, entropy (kJ/kg.K) as well as the average enthalpy (kJ/kg) would decrease on cooling as repeatedly said by the above-mentioned experts.