Rankine Cycle Question

http://en.wikipedia.org/wiki/Rankine_Cycle#The_four_processes_in_the_Rankine_cycle

The working gas is heated creating pressure, the pressure is used to create work, the expansion is after...

The answer is, molecules of the media will find its equilibrium state with its properties (pressure, temperature, volume)

Any energy either work or heat you will add or remove on it at a certain pressure or temperature or volume, it will find its equilibrium state.

Work and heat are just the consequent of differences of states. If 3 or more states are in equilibrium, there'll be no work or heat

I think people may be missing my main question/discussion.

SolarEagle, not to get off topic, but how can work be done without expansion (just pressure) since work is force times distance?

Noudge79, you have only highlighted the problem since it appears the system is losing heat by adding pressure. Equillibrium is easy to understand when the added pressure of compression increases heat or expansion reduces heat, but why does increased pressure reduce heat when a material expands; i.e. there is less heat in the system after it has expanded under pressure than if it had expanded without being under pressure?

The point of the Rankin cycle is that heat is being used up as work - molecular movement (heat) is being reduced when a material is expanding under pressure (doing work). In short, my question is why is molecular movement is being reduced by pressure when a material expands?

Heat has increased the molecules kinetic energy. The molecule has mass the heat has increase the molecules velocity. As the pressure is reduced the molecules give up this kinetic energy in expansion as they collide into each other. The kinetic energy is transfered mechanical devices in this expansion or back over to heat.

Yes, this is getting at my question.

Thanks.

Question, was it really molecular movement(kinetic energy) consequent to heat or the collission of molecules below stated by Noudge79?

What if molecules are all excited and yet, they find its way not to bump one another? Is there any heat created with that?

simple and obvious, molecular collission is a funtion of space in between molecules, and temperature is directly proportional to this collision. providing much space for molecules as they expand, decreases the rate of having them collide to one another.

As what we are thought way back our grade school years, gas has more space in between molecules than liquids and much more than solids.

Expansion is decrease in pressure, I did not somehow get expansion as increase of pressure, that's simply compression.

Just a note of clarification regarding the expansion/pressure issue. If the heated fluid is allowed to expand without any resistance to its expansion then no work would be done and no heat would be used up. But since the goal is to get work out of the system, resistance to the expansion is imposed, resulting in mechanical force times distance, i.e. work. The pressure I was refering to is not simply compression, but rather the pressure that is imposed by resisting expansion as the fluid is allowed to expand. Again, if there is no resistance imposed on the expansion, the amount of heat energy stored within the fluid should remain virtually the same as before the expansion occurs, with the drop in temperature being directly proportional to the increase in volume. However, if resistance to expansion is imposed, then some molecular movement is being transmitted to the resisting member (usually a piston or turbine blade) resulting in an overall loss of heat within the working fluid.

No useable work is done. Heat increased the molecules velocity they travel through a distance.

U-oh, now you are confusing me and concerning me, since I thought you, ozzb, and I were on the same page. What do you mean that no useable work is done? The whole point of the rankine cycle is to convert heat to work - work from the rankine cycle provides the great majority of the worlds electricity.

Did not the heat increase the velocity of the molecules(force)? Do they not travel a distance?

Yes and yes, but it is your first sentence, not the second one, that I don't understand. I agree that molecules move over a distance. I also see that some of this movement is being converted into work as a turbine or engine is being driven, thereby reducing the heat that remains in the working fluid. Again, what are you trying to say in "no useable work is being done"?

Why does the work have to be a turbine or an engine. Isn't the heat forcing the molecule to change velocity causing work to be done.

I think the moving molecule would be better defined in physics terms as energy with the potential to do work. Unless that movement it used to drive something else it is not doing work, as physics would define it. It is more like an inertia wheel or gyroscope, a body in motion that is staying in motion until it interacts with some other body. The gyroscope has the potential to do work. So, once you hook it up to a generator and put a load on the generator, the energy will be extracted from the gyroscope in the form of work on the generator. The turbine or engine is eliminated in this example because the gyroscope is already a mechanical device with mechanical movement, as opposed to molecular movement. With molecular movement, some form of mechanical device needs to be introduced in order to produce "work" as physics would define it. The molecular movement is self contained like the gyroscope, it is not really doing "work" until it acts on some other body.

So no work was done to get your inertia wheel to some rpm? Magic?

How the inertia wheel got started is immaterial. Regardless if work went into the wheel to get it started, the point is that no work is coming out until it is hooked up to the generator (which is under an electrical load or friction load or whatever else might cause resistance to motion).

No analogy is perfect. In the case of the gyroscope, it was most likely started by a work input. In the case of the moving molecule it is being activated by a different type of energy input - heat.

What is being discussed is the work output, not the input (be it work or heat or chemical reaction or nuclear reaction, etc.). The analogy was given to make the distinction between potential energy and work output. Neither the moving molecule nor the gyroscope have a work output until they are transmitting forceful movement to another mechanical body.

Look at any diagram of the Rankine cycle and there is a clear distinction between what is called "heat" (or thermal energy) and what is called "work". It is just the way things are currently defined in the field of thermodynamics.

As you might noticed, looking at the steam tables and the procedure for the computation of work done. Losses are already accounted at final resulting Pressure and Temperature (Final state). Pure substance like water follow a property curve (P,T,v, s, H). If you can't find the property relationship on the tables then it must be some sort of "magic" or out of this world.

There is an concept called "Availability" and it's simpler to understand other than sorting out to molecular level(statistical thermodynamics). Losses also can be easily distinguised using the concept.

It's just the measurement of Available energy (final state will that of the environment-surrounding)

From that, you can already knew how much work or heat is lost to your source.

Go to a compressor. Turn it on. The pressure tank gets hot.

Turn the compressor off for awhile. The pressure tank gets cooler down to the ambient temperature. When the temperature grade approaches zero, it stops cooling off.

Let some of the pressure out of the tank...have it blow a turbine or something. The air coming out will be very cool. The tank will get cold.

Lets go back to the beginning. If you put the compressor tank in a hot place (over a fire for instance), the air will get hotter inside that tank. The pressure will also rise. The energy you get out of the nozzle when you release it will be greater because of the higher pressure.

So heating the air in the compressor tank can allow the enclosed air to do work. This is how a Stirling Engine will work. And this answers your question. However the Rankine cycle is a bit more complicated than that because it brings steam into it.

You can increase the amount of work you get by increasing the mass of the air. Water molecules will do this in the form of steam. You can push machinery around with this vapor just fine, (and they do with superheated steam all the time) however Mr William Rankine figured out that that the real trick to doing this is to have a really good steep temperature gradient. The steeper the temperature gradient, the better the system will work. He went so far as to actually condense the steam at one part of the cycle. This allowed him to use the latent heat in the super heated compressed water to flash into high pressure steam, and there were a lot of advantages to creating a flash boiler.

You can turn the water molecules back into liquid by compressing the steam. Compressing the steam laden air will raise the boiling point, and the steam will turn back into water. You can push that superheated water around like any other water, and not use a lot of energy to do this.

But it takes a lot of energy to compress the steam back into water. So the Rankine Cycle uses temperature to cool the steam, to condense it.

And you get work out of it.

Did this help? Or did I just manage to obfuscate the issue?

This is all good.

However, despite where we last ended up (discussing the difference between energy and work), the initial question was why the working fluid in the Rankine cycle loses heat when resistance to expansion is imposed in the form of work. Normally, when a fluid expands there is no heat loss only a spreading out of the heat over a greater volume - as restricted by the thermodynamic law "energy can neither be created nor destroyed." However, if there is an energy output (i.e. work) then some of the heat must be converted to work and the working fluid would then contain less heat per mass than it did before. Again, restricted by the same law.

In other words, because there is resistance to expansion (work being done) the heat within the working fluid is being reduced. Or said another way, expansion without resistance = no overall heat loss and expansion with resistance = some overall heat loss.

I just found it curious that resistance to expansion (pressurized expansion) would reduce heat when pressure typically increases temperature. But, I guess it isn't really pressure that increases temperature but rather a reduction of volume or compression(without allowing the heat to escape). From there we went into explaning the conversion of heat to work as molecular movement being passed on to a mechanical device (turbine, engine, etc.).

I think I am pretty well settled with what all is happening in the Rankine cycle. I think I'll let others mull around in whatever else they might find curious or confusing.

Here is the difficulty....

Your statement that "Normally, when a fluid expands there is no heat loss, only a spreading out of the heat over a greater volume" contains the seeds of disaster within it. A stumbling block on the road to understanding.

Heat is measured by a thermometer. Since the temperature of expanding air measureably decreases you must be using the word "heat" incorrectly. Let us replace the word "heat" (which is measured by a thermometer) with the word "energy".

Your statement will then be modified to say "Normally when a fluid expands there is no energy loss, only a spreading out of the energy over a greater volume".

It is quite true that there is no heat (energy) loss in a chamber which simply expands however, the temperature drop and the resultant temperature gradient is VERY significant. If this chamber is a cylinder head, then when the chamber expands, it is being forced to expand by the energy of the gasses inside it.

The act of performing work will cause the fluid (air or steam) to lose energy. We can calculate how MUCH energy it looses by checking the temperature of the fluid, but as we have seen, temperature depends upon pressure.

The more energy it loses as it drives that piston down, or that turbine around, the "cooler" it will get and the more efficient it will be.

Really.

Yes, we are on the same page. I was using the word heat as heat energy, not temperature. I realize temperature drops significantly when a volume expands. Confusion of terms...

The greater question lies in what the temperature difference is between the volume of fluid that expands without being under pressure (doing no work) and the volume of fluid that expands while being under pressure (doing work)? This has been my question all along.

If the temperature is the same, then the first law of thermodynamics is incorrect because energy is being created in the form of work. If we cling to the "first law" then the temperature differential must be proportional to the amount of work done (keeping other extraneous factors - poor insulation, friction, etc. - out of the equation).

I didn't want to get too distracted by discussions of terms, I just wanted to get at this one basic question: is there less (heat) energy in the expanded fluid because it has expanded under pressure (with resistance imposed) and why?

Also, I am not interested in the "because this is what the first law of thermodynamics states" sort of answers. I'm looking for descriptions of what is physically happening.

For example, the velocity of each molecule is being reduced more when they are pushing against a piston or turbine blade that is resisting their expansion than if that piston or turbine blade was allowed to move at a pace that would keep up with the expansion rate that would happen if the fluid enclosure were to suddenly disappear.

The confusing part of the whole question comes in when one realizes that compressing the fluid in a well-insulated non-expanding enclosure does NOT reduce the molecular movement, which by the same reasoning should also physically restrict molecular movement.

I don't know how many different ways I have to state the same question. (I think it has been about a dozen times now.) I just hope that someone can truly understand what I am asking.

The energy in each molecule is a certain thing. It will stay the same unless you actually physically heat it. This energy would be independent of how much it is compressed.

The energy of the molecules being crowded together is quite another thing. In reality, they are hard to distinguish, because one can trade off between the two energies. And we do, every time we fire up the old diesel engine! But it is important to realize that they are two quite different entities.

Lets look at the second one...the energy it took to compress the air. You have added energy into the system by compressing it. You get work from the system by decompressing it. Doesn't matter if it is pushing a piston or being dissipated in the air.

Confusion must result because one way to compress the air is by heating it. But then the system will stabilize at a hotter range unless you have a cooling tower. The heat energy you used to compress the air does not just go away, it must be cooled someplace. A radiator or a cooling tower, or just thrown away. (like in a diesel engine)

If you take a cylinder (just a closed metal tube will do), and stand it up in front of you. Drop a pison down into the cylinder. It will bounce up and down until friction slows it to nothing. If you heated the air in the bottom of the cylinder, make it hotter and hotter, the air will expand, the cylinder will rise a bit, but it will still bounce the same.. nothing different will happen until you create an energy gradient. You do this by creating a "cool end". A cooling tower, or basically, a place for the heat energy to go. In stirling engines, they create a whole extra cylinder for that!

So you must build in an energy gradient. This animation shows the temperature versus pressure gradient at work. This one is a little more complicated....

Bill

Now this, my friend, is helpful!

I never thought of heat input as a way of compressing the air, which puts everything in perspective. The decompressed air may have the same amount of energy in it, but you still need to input some form of work (be it bleed off, recompression, and reheat or simply mechanical compression) to get back to where you started. The best way, of course, being to drain the heat energy out in a heat sink, make the container increasingly smaller, and then reintroduce more heat.

Many thanks.

Look at the molecular action on either side of the piston during the expansion stroke. If you figure out from this consideration then you will be well on your way to understanding molecular level gas kinetics.