Hot Air

Yer both wrong....

"Gases passing through an ideal gas turbine undergo three thermodynamic processes. These are isentropic compression, isobaric (constant pressure) combustion and isentropic expansion. Together these make up the Brayton cycle."

http://en.wikipedia.org/wiki/Gas_turbine

http://www.ueet.nasa.gov/StudentSite/engines.html

'Air powered Turbine' ... no combustion?
If all else fails, read the OP.

Yep. Compressed air running the turbine. No combustion.

Two choices: Compressed air at ambient temperature OR compressed air at some temperature that is higher than ambient. Same pressure from both.

Oh ok I get it, like air tools...I still don't get the purpose of the question....are you trying to gain thrust? or do you just want to ascertain if thrust is being created?

No thrust is being created.

Converting moving air to work. As it is phrased, it might be a wind turbine. Probably not, but it might be.

[edit] Oops, in #3 the wind turbine is shot down. So an air motor of some type.

For a specialized application, he's connecting the turbine shaft to an electrical generator.

So... assume we have two tanks of compressed air. One tank is at ambient. The other has been heated. The psi of each tank is the same.

The higher the molecular density, the more capacity for work, so I would guess the colder the better...

I agree with you on this. Isn't that why intercoolers were invented for turbo chargers? They are basically a compressed air powered turbine.

The real question is "how is the air being compressed?" Not by an electric motor driven by an air turbine powered generator by chance?

Do i sound cynical.

Jim

"The real question is "how is the air being compressed?" Not by an electric motor driven by an air turbine powered generator by chance?"

Gee! Won't that work?

Seriously, no. This is not another perpetual motion machine. As a part of the application, my buddy has access to an (essentially) infinite free heat source. (Geo thermal) So he's thinking of using that to heat the air, thus getting more bang with less total weight of air.

I'm not so sure this will work. The simple answer would be, "try it and see," but he doesn't have enough decent instrumentation to measure all the variables and draw conclusions.

Still not sure I've heard a definitive answer backed up with a rigorous "because."

Bill

You've told us the source of heat.

Now tell us the source of compressed air.

Well, that does it for me.

I cannot believe you are so incredibly stubborn!

Please read this.Electrons

It is another example of some one asking a question and refusing any logical attempt to have that question answered.

I have resisted a strong urge to call you ignorant, as well, cause maybe it isn't your fault you can't grasp the obvious.

Good bye!

If he doesn't have the instrumentation to see results, then there is no difference.

For an air motor with the desired output of the most work, you want to get the most mass flow that you can through the motor. Therefore you would want cold air since it has a higher mass density than hot air.

For your buddy's situation, the best answer is the KISS principle.

And a TANSTAFL to you too!

Disclaimer: OP still hasn't stated where his friend is getting the compressed air. Unless he has a free, unlimited source, energy has to be input to compress it and you basically have a gas turbine (Brayton Cycle) as shown by SolarEagle in post #1. The source of heat is theoretically immaterial.

I would like one of you 'anti-heaters' to run a little real-world test. Book a short flight on a turbo-prop. Before take-off, talk to the pilot (and co-pilot if you want a complete test), and tell him you have 'met' this anonymous technical internet group who have developed a disagreement on turbine behavior. The majority believe that if he reduces the heat input to the engine(s), he will get more power. So ask him, when he has reduced the throttle at cruise altitude, to cut off the fuel to prove your point. IMPORTANT: Warn him/her not to do this at full throttle as the extra power developed when he/she removes the heat may overstress the engine parts.

Please report back your results.

Thanks for a) making me dust off the old thermo books, and b) allowing me to have a little fun .

Thanks for helping bring things into light.

I strongly agree with you that when the system is taken as a whole, heating the air will always result in a positive change in the net work.

However, I would also like to point out the OP questioned specifically only the turbine power (not the system).

Here is what I know...

1) heating the air will result in an increase in the specific work of the turbine

2) heating the air will result in a change in the physical properties of the fluid (such as a lower density and a higher viscosity) that will tend to negatively affect the mass flow.

3) the turbine output is a function of the specific work and mass flow

That's it. It is conceivable that the turbine power could go either up or down, even though, as I said above, the net work of the system as whole will always go up.

Hold the phones guys, the combustion is to increase the pressure so the gases move faster through the vanes. It's not the hot air they want, it's the increased velocity. Going back to SolarEagle's diagram, the OP's question starts at stage 3 of the diagram, after the combustion, except his gas isn't hot.

So which gas has more the potential to do more work? Cold or hot?

(Hint: think about why there is such a thing as a reheat cycle)

The OP's question describes two states - one with hot gas, the other with gas at ambient - and asks for the difference in power output.

I agreed and have already stated my position on it.

Yup - I was replying to passington.

Wow.. and look at me jumping on you... eeep!! I'll pay more attention in the future!

True, I was just separating it from the diagram because the only difference in the OP's question was temperature. How the hot version became hot is outside of the question, the pressures are the same.

Well, you changed the problem with the 2 tanks analogy, unless the tanks have infinite volume.

OK, the basic equation for ideal work is: W = ∫p * dV, where

W = work done

p = pressure

V = specific volume (not absolute volume), or 1/density

So you can see that if you have the same mass flow rate, the specific volume V will be higher for the heated air since the density will be lower (@ constant pressure); therefore, more work will be done (assuming constant efficiency in the conversion process).

Imagine you have a compressor discharging to your turbine at some presssure, say 100 PSIG. Your turbine has a fixed nozzle, and the turbine load can absorb whatever power the turbine produces. Now start heating the air in the discharge line. The discharge pressure of the compressor will start rising because the nozzle cannot pass the same amount (mass) of lower density air through the same nozzle. You magically increase the nozzle size to keep the compressor discharge at 100 PSIG as you heat the air. This keeps the mass flow rate constant. You would see an increase in power production from your turbine.

You changed the problem with the 2 tanks. If you started with 2 equal volume tanks at equal temperature and pressure, and heated one, you would have to let some air out of it to prevent the pressure from rising. Then you wouldn't have equal amounts (mass) in both tanks, producing a different problem.

One thing others have touched on is that the temperature of the discharge air will drop as the pressure is reduced and as work is removed (regardless of the inlet temperature). Depending on conditions, if the air is not dry, you may have moisture condensation in the turbine. I worked at a plant that used diaphragm pumps powered by plant air. Even though the plant used a refrigerated dryer that reduces the moisture dew point to 35degF or so, at times the exhaust air would be fog due to the moisture. This wasn't a problem for the pumps, but some turbines could be damaged if too much moisture is generated. It can also adversely affect efficiency.

Hope this helps a fellow ME.

Are you sure that the V you're using is supposed to be specific volume and not ordinary volume? The dimensional analysis doesn't work out properly using W = pressure x 1/density.

I.e., Work = Newtons x meters;

Pressure x 1/density = (Newtons/meters-squared) x (Meters-cubed/kilograms) = (Newtons x meters)/Kg. Thus there is an extra 1/Kg term if you use specific volume.

Wikipedia just shows the formula as regular volume: http://en.wikipedia.org/wiki/Pressure-volume_work

Oops, you're absolutely right. This is usually what happens when I try to act smart → .

However, most of my explanation is still ok - I think, isn't it ? In other words, with the higher temperature case through the larger nozzle, the volume is greater, since V = mass / density, and the mass flow is constant.

I'm not sure yet. I'm thinking, given the same volume and pressure, the cooler tank has a higher mass of air, so the greater mass is capable of doing more total work from when the valves are opened until both tanks reach atmospheric pressure.

I think I still need to cogitate on it a bit more.

Hot air: less dense: less work.

That's why pilots perform density altitude calculations.

As heated air moves through the turbine, it will cool. This will result in less thrust.

It appears to me that you are correct.

"As heated air moves through the turbine, it will cool. This will result in less thrust."

HUH? As it cools, it will become *more* dense. Won't that lead to more thrust rather than less?

Heated air in the process of cooling reduces in volume and hence pressure inside the turbine.

Take an empty water bottle at room temperature, cap it tightly, and place it in the freezer. The air inside will reduce in volume and the pressure in the bottle will fall causing the collapse of the bottle.

This will happen inside the turbine....falling pressure = less thrust.

But if this causes lower pressure on the outlet side, surely that will give a higher pressure differential and result in more power output?

The outlet of the turbine is at atmospheric pressure. The warmer air cooling will not have a lower pressure than that at any time. You have simply started with air that cannot do as much work as colder air, and wasted additional energy in the process making it warm.

"You have simply started with air that cannot do as much work as colder air..."

Because?

Both the the warm and cool air are at the same starting pressure. The delta P is the same with both "types" of air. Why would warmer air be less capable of doing work?

Warmer air at the same pressure is less dense. So fewer molecules hit the blades per second.

Nuff said...

Apparently, not enough for the OP.

Evidently so...

KT4YE,

I am considering bestowing upon you the 1ST ever malar Award.

HUH? Not sure if this is an insult, compliment or...

Bill

Compare water with air. The more dense water at the same pressure will do more work than than air.

Jim

I see where you're coming from, but "The warmer air cooling will not have a lower pressure than that at any time." ... are you sure about that? Why wouldn't the cooler air (assuming it's cooler than ambient) lower the atmospheric pressure locally?

Jus' thinkin'.

(If there were such an effect, I suspect it would be masked by the lower density of the input, warmer air - but there may be some geometries (such as a long outlet pipe of the same diameter as the rotor) where it may be observable.

Jus' whitterin').

"So if warmer air is less dense (which is likely), then warm air would yield less thrust."

Yes, but less dense = lower pressure. However, in this problem, both the ambient air and the hot air are at the same pressure.

But I gave you a Good answer for the link to the hovercraft info.

Oops.

Didn't mean to repeat what you just said.

OK, I skimmed the posts and may have missed something. There has been talk about density - higher at lower temperatures. This would substantiate more driving force at lower temps.

Another factor is viscosity. Lower temperatures yield lower viscosities. This is a competing factor in the issue at hand. I do not have a definitive answer whether air at 50ºC will have more driving force than air at 20ºC or not. My gut tells me that colder air will have a greater driving force, that viscosity vs temperature has less of an effect than density vs temperature.

I will study this a bit. My first stop is here.

All you have to do is consider what does the work here. Collision of molecules of air with the working surface of the turbine is what does the work.

So, at the interface of turbine blade with air, warmer air, at a given pressure, will contain a certain number of molecules, depending on temperature of the air. Right? The number of molecules at 40°C will be greater than the number of molecules at 100°C. Right?

Now, consider the energy in a single molecule. More energy if hotter, less if cooler.

So, the question becomes, do the hot molecules exert more force on the turbine blades than cooler molecules?

No, because they are not confined, as in a balloon, and the initial velocity of molecules in a tube is the same in either case. It's the quantity that is different.

It's time for a toddy.

I believe you are right - that density is the main factor here - but, viscosity still plays a part. Are you arguing with me about cool air vs warm air? From what I can see between my post and your reply, we agree except maybe for the existence of a viscosity factor.

And no, not a toddy, but the 6th Guinness of the evening (it's later here)

No, not arguing about "viscosity",(not sure it's the correct term for a non-liquid fluid) but in the case of a gas, the mechanism that imparts the quality of resistance to flow is different, in that with true liquids molecular chains come into play and air has no chains. It's still about the number and magnitude of single molecular impacts that imparts the energy into the turbine.

It's still about the number and magnitude of single molecular impacts that imparts the energy into the turbine.
The sum and substance of the the question is answered here. GA

All fluids, gases and liquids, have viscosity. I am a ChemE and I know this.

If by "chains" you are talking about numbers of carbon atoms linked with each other with the attendant hydrogen atoms to form hydrocarbons, I understand the point you are trying to make.

In the case of water, there are no "chains" as is also true with air. The number of, the velocity of, and the mass of the molecules impinging on an impeller, whether gas or liquid, are directly proportional to the output of said impeller.

Something that might interest you would be supercritical fluids. These are compounds that, when raised above their critical temp and pressure, have viscosities on the order of gases, but densities on the order of liquids. Having high density and low viscosity make them ideal as cleaning agents. When I was with PNNL, we were doing research on cleaning critical components with CO₂ in it's supercritical state (T > 31ºC and P > 1071 psia). Nice thing about it is that, once you've dissolved the oils, etc, you merely need to lower the pressure and they fall out of solution and the CO₂ can then be reused.

Sorry to go on so - just remeniscing on some interesting work I did in the past.

Thanks, any more good stories?

Jim

Hi Jim,

That's my younger brother's name - we call him Jimmy.

I don't know how to take your post. Did I offend you? Do you think that what I said wasn't true?

I know I went on a bit and I can understand how that could bore (but it was interesting to me!)

Best Regards,

Mike

Interesting, sounds similar to non-Newtonian fluids

Right. Chains was a poor choice of words. I'll have to ponder this and see if I can come up with a better choice of words.

Anyway, I think we agree that hot air won't push as hard as cool air against the turbine blades. I think we are flogging another corpse here.

Yes, point taken, but I still enjoyed the dialogue. I am striving to improve my communication skills, but it can be tedious as I never learned to type.

Now that winters over, typing is a lot easier because I can now type without my mittens on :)

No problem Lyn! We're on the same page. It really is worthwhile to talk about these various problems/topics, learning new things every day.

Take care!

Like you said, one stage. No condenser to utilize the temp drop.

If there is a free supply of heat, consider going for more stages, like a condenser. Even if the heat is not enough to boil water, it may be enough to boil something else to drive the turbine or some other heat engine.

My experiments with simple steam engines showed an almost doubling of power using a condenser.

My answer is... it depends!!

The power from the turbine can be viewed as the product of the specific work and the mass flow. For the same pressure and a higher inlet temperature the specific work for the turbine will increase. This means that you will get more energy per unit mass of air flow. However, the mass flow is dependent on many factors such as density, viscosity, velocity, piping size, materials, fittings, valves, etc. and will likely decrease.

So which one will dominate? It all depends on how your friend's system is set up.

The OP is designing a simple one-stage turbine where air is the working fluid. He doesn't specify whether this is an axial turbine, like a fan, or a radial turbine. The friend maintains that "heated air -- at the same pressure -- will yield higher output power." That's confusing, and may be the reason this thread is a muddle that lacks resolution.

Diffusion of momentum through the boundary layers against the blades rotates the radial or axial turbine and drives a generator. You have abundant pressure from the geothermal source so the temperature contribution to enthalpy (energy content of material), from additional attempts to couple energy into the material by molecular collisions from a heated gas seems small enough to disregard.

It's the momentum that counts in any collision, like with a molecule on a turbine blade. So mass is important and cold air being denser might appear to be an advantage in a pressurized turbine. What's the point of heating the pressurized feed if not to increase momentum of the molecules?

Heat transfer into air is slow, and a useful turbine has high mass flow, so with so many molecules flowing past the heat exchange surfaces, additional heating would be inadvisable.

Separating the hot molecules (high velocity) from the cold ones can be done in the Vortex Tube. Thermal separation, as if by Maxwell's Demon, is accomplished in a continuous process in axial counterflow through a tube from a tangential feed to a conical impedance at the hot end, and a high velocity rebound jet through the self-organized counterflow reactor. The pressure at the cold end, where the pressure is high enough for spot cooling, is less than the feed pressure, and so is the pressure out of the hot end. A net pressure drop through the process shows some work has been done in there to overcome the entropy and establish the persistent axial counterflow.

Another way the working fluid can flow through the turbine is simultaneously radially inward and outward (radial counterflow). In the shear layer between counter-rotating boundary layers, persistent radial counterflow between counter-rotating radial turbines is assisted by extracting low enthalpy fluid axially. The steam separates as in a radial tree of vortex tubes. I'm talking about a radial turbine along these lines. http://www.freepatentsonline.com/7987677.pdf

Without pressure difference air flow will not take place. If there is no pressure difference air will be stagnant. Even on heating if same pressure is maintained, turbine can not generate power.