elementary thermodynamic question
elementary thermodynamic question
(OP)
Could someone simply confirm my reasoning that to use a working fluid such kerosene rather than water in a engine like a steam turbine would be a lot more efficient since the latent heat of evaporation of kerosene is less than 1/2 that of water? I would need to travel 45mi to the nearest Univ. to ask an engineering professor this or to even look this up in a thermodynamics text since most public libraries wouldn't have such books. And, if I did look it up in a book I might have to read two chapters and still not get an answer to the specific question.





RE: elementary thermodynamic question
There are at least two pressures... one for vaporization and one for condensation. There will be a different latent heat.
You can find energy as area under the T-s diagram. Energy in under the upper curve... energy rejected under the lower curve... work related to area between the curves. Make the curve wider (by increasing latent heat of vaporization) doesn't seem to particularly drive toward higher efficiency since it tends to affect all 3 terms. Moving Thot up or Tcold down does have a straightforward tie to improving efficiency.
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(2B)+(2B)' ?
RE: elementary thermodynamic question
RE: elementary thermodynamic question
1. Fuels are better used by burning (i.e combustion)
2. Fluids used in closed loop cycles should be stable
3. Lower boiling point ---> Lower pressure --->(so what?)
4. All fuels have certain auto ignition temperatures
Now think a bit and tell us the other points.
Good luck,
RE: elementary thermodynamic question
RE: elementary thermodynamic question
I think I remember that Thermo text books use other theoretical fluids in the examples and the problems that have to be solved at the end of the chapter.
The problem comes not in the specific properties of the fluid, but rather with the specific problems of sealing the fluids in the machinery.
Refrigerant cycles are totally closed cycles, so they work well for what they are designed to do.
It is impractical, however to scale that up to the size needed for a working power plant. And, since shaft HP is needed to drive a generator (assuming that is what you intended to use a steam turbine for) in your question, then sealing the device becomes the paramount issue.
Water works well because it is easy to handle. If some steam leaks out (excepting BWR nuclear cycles) it isn't terribly bad. Since turbine seals on the lower than atmospheric pressure stages work on the basis of pulling some air into the seal from outside and then separating it from the internal working fluid - steam in the case of steam turbines, how would you propose to do that with kerosene or toluene or flourocarbons?
So it is not the fluid properties, it is the practicality of making a workable machine to utilize those properties that is the problem.
rmw
RE: elementary thermodynamic question
rmw:
Thank you for that trip through nostalgic lane. I too remember studying mercury as a working fluid in a power plant in 1958 under Dr. "Steam" Simmang at Texas A&M. However, as I recall, ours was not an academic, conceptual study. It involved a real, installed, and operating unit at Kearny, NJ.
I thoroughly agree with your basic, common sense evaluation of working fluids.
There were several thermodynamic advantages to using mercury as the working fluid – but when the engineering common sense started to get into 3rd gear, it was realized that there were grave disadvantages as well (trade offs):
1) Mercury vapor is poisonous, so that extreme caution must be exercised to prevent leakage in closed places;
2) Mercury is expensive, so that the cost of installing a mercury-vapor power plant is high;
3) The saturation pressure of mercury at low temperature is excessively low, approximately 0.4 psia at 402 oF. Thus, mercury would not be at all suitable for expansions to ordinary atmospheric temperatures.
4) As ordinarily obtained, mercury does not wet the surface of the metal that it contacts. This property introduces serious difficulties in transferring heat from the furnace without burning the tubes and shell. The addition of magnesium and titanium to mercury gives it the property of wetting steel.
The above is all explained in "Thermodynamics", Virgil Moring Faires, the Macmillan Co.,3rd Ed.
For more, up-to-date information on the subject of different working fluids in Thermodynamic cycles to to: http
I think you will enjoy the ample discussion of different working fluids there.
RE: elementary thermodynamic question
RE: elementary thermodynamic question
You had a good point there though it is a bit redundant in this specific case. Nevertheless, I liked your understanding of a bit (apparently) difficult concept of thermodynamics.
Only that, the OP is asking for crunching the area and not widening. The advantage with this (of lower latent heat fluid) is that heat input (Q1) is lower. Can you guess the trade off for this?
RE: elementary thermodynamic question
RE: elementary thermodynamic question
I see another issue, the lower the boiling point, the lower the pressure. The lower the pressure, the lower the work energy available.
TTFN
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RE: elementary thermodynamic question
Check out LNG liquifaction plants for the use of binary fluids in refrigeration, and see the work being done on geothermal power for the use of binary fluids in the power cycle. Alos, there is increased chatter on theuse of CO2 for a power cycle.
RE: elementary thermodynamic question
Before I was only thinking of the heat of vaporization as energy that was lost.
The reason I suggested kerosene was that it boils at about 157 deg C and could be turned back into liquid easily at nearly the usual ambient temp + its low latent heat which doesn't seem to matter much.
I knew someone who had worked at an oil refinery and once talked about some of this stuff going through the pipes at such high temperature and pressure that if the pipe sprang a leak it would turn into a blow torch.
RE: elementary thermodynamic question
I think I agree with quark. There is an example problem in my physics book calculating the entropy of changing 1kg of water to steam at 100 degC and it would take 1/2 that for kerosene at its boiling point (some additional energy needed to raise it to that temp). Of course, the less energy you put into something the less there is available to get out.
RE: elementary thermodynamic question
It seems to me that a refrigerant like R-134 would work well in applications like geothermal, where you might not get as high of a temperature as you do in a combustion process. The Chena Geothermal Plant in Alaska uses R-134.
http://www.chenapower.com/geothermal-power/
RE: elementary thermodynamic question
Or the boiler pipes are made of that unobtanium-impossibleakathrough alloy.
RE: elementary thermodynamic question
What I originally had in mind was a small reflective solar collector electric generating system which should provide sufficient electric power for a small house that would stand alone away from other objects-not an industrial sized unit. There would still be some hazards to take precautions for though. I recently learned that the sun provides 1kw/sqm of power at the Earth's surface when its shining, of course. With 4 sqm of suitable mirrors and assuming 30% to 40% efficiency of the system and (they say) 95% efficiency for the generator this should provide sufficient power many think the average homeowner should be using. When I hear some speak of their electric bills, however, I can't imagine how they waste so much.
RE: elementary thermodynamic question
> 1120 W/m^2 is applicable for about 2 hours a day, at best, when the sun is directly overhead. At other times, the sun's angle and increased path through the atmosphere reduces the amount of energy available.
> 30%*95%*4kW = 1.14kW, which, if you could generate that 24/7, would be a decent amount of energy. However, you'll find that you can receive less than about 5kWh/m^2 per day, so your 4m^2 of collector will only result in about 5.7 kWh of energy per day, which is about 1/4 of what most houses need.
TTFN
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RE: elementary thermodynamic question
The real problem is if one wants to use much electric heat. I already know many people share my point of view. Actually, I read some steam engines have achieved upwards of 60% efficiency, but that may not be easy to do.
RE: elementary thermodynamic question
Best rule of thumb for solar PV and heat energy is: 9:00 to 3:00 (local solar time) is the effective time limits on energy received regardless of receiver type or solar tracking system, and this only if there is no shadows blocking the receiver. (At my house, for example, my neighbors' trees block all direct sunlight all day except for 1:30 to 2:15.)
This 6-hour "Local solar exposure" window is only an average though: We (Northern hemisphere) will get that only on March 22 and Sept 22. More exposure time in the summer, less in the winter. On the face of it, that would help your proposal, since AC use is greater in the summer. But the thermal lag in houses really means that - though exposure will not change, the AC won't be running nearly as hard (if at all) between the first two hours when the panels are receiving energy) but must keep running from 3:00 pm through about 6:00 pm - when almost no energy is coming in the array, but the house and its contents are still very hot.
The 1000 per sq meter drops with any dirt, dust, pollen, leaves, or smudges on the array upper surface. It drops with the atmospheric conditions too: Haze, clouds (obviously), rain, higher humidity, block energy from the array.
Latitude MUST be included: The loss of energy increases as you go north proportional the thicker atmosphere: at its limit, only the equator gets the "advertised" theoretical value. Everywhere else is less proportional to the cos of the latitude -> and this loss must be multiplied by the loss factor of the rotation of the earth (the 9:00 - 3:00 factor approximated above.)
The 1000 per sq meter is reception is theoretical for a sq meter panel only if that panel is continuously aimed directly at the sun: If you use a flat panel aimed "south" -even if that panel is daily changed to track the changooing elevation angle of the sun's path, that cosine loss of the fixed panel must be factored in as well: Tracking units are "expensive" - often, in my costing, the tracking unit, its power supply and controller, its mounts and their foundations or roof-top mounts to the house are as expensive as the solar PV array itself. If the flat panel is kept at a "generic angle" and not rotated to track the sun - usually the house's latitude - then you have both a sun's seasonal elevation cosine loss factor and a sun's daily rotation loss factor to be multiplied in...
Check therefore, your theorectical reception against a house at latitude 24 (south FL), latitude 32 (interstate 20 or the mid-south), latitude 40 and latitude 45. Recongize that most of the US and all of Europe lives further than 40 north. And not all houses have an effective south-facing roof surface. (My own does not, for example, so my solar panels were very ugly when I had them installed.)
Your idea has merits - but please keep calculating.
RE: elementary thermodynamic question
Roof-mounted close-coupled thermosiphon solar water heaters are also considered an indirect saving on the electricity bill.