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Thermodynamic Processes

Posted by steamerandy 
Thermodynamic Processes
October 12, 2011 03:02PM
I have some questions about steam processes. This started in another thread, "Waht about this valve"

The question is what is the reasion that steam expands in different ways.

Case 1: Isentropic process. An isentropic process is a reversible adiabatic process. During this process, the entropy is constant. An adiabatic process is one during which no heat is transfered to or from the working substance. In the isentropic process, the working substance loses or gains energy as work, not as heat transfered.

Case 2: Throttling process. In a throttling process the working substance flows adiabatically from a reagion of high pressure to a region of low presssure with out doing work. This case for an ideal gas has constant temperature as well as constant enthalpy.

The isentropic case can be a steady or non-flow process while case 2 is a flow process,

Case 3: isothermal process. The isothermal process is a non-flow change of state of a working substance during which work is done and the temperature remains constant.

Now for an ideal gas going through case 2 or case 3 the resualt would be the identical. In case 2 we have no heat transfer but in case 3 there is heat transfer to maintain temperature.

In case 3 the work, W, is equal to the heat transfered, Q:

W = Q

In case 2 the work, W is equal to the transfered heat, Q as well. They both = 0

W = Q = 0

Actually the logic and math explanations of both case 2 and 3 make sense in and of themselves. But the change of states are identical for both casses. An example of case 2 where there two containers conected by a valve, one containing a gas at some pressure and the other a vacume. The valve is opened and the pressure allowed to equalize.

What is the ideal process of an air compressor compression stroke. The actueal process with out heat transfer would be isentropic. But sense the end result is that the output air be at room temperature the ideal process is isothermal. Thus cooling fins and water jackiting to cool the compressor cylander to remove the heat of compression. The intake stroke of an air compressor is ideally isenthalip. But there is work being done to create the partial vacume to draw air into the cylander.

This is more a point of interest. I know when to apply them and how to calculate them.

Andy



Edited 4 time(s). Last edit at 10/12/2011 03:06PM by steamerandy.
Re: Thermodynamic Processes
October 12, 2011 04:33PM
Quote
Andy
But sense the end result is that the output air be at room temperature the ideal process is isothermal. Thus cooling fins and water jacketing to cool the compressor cylander to remove the heat of compression.

"Thus cooling fins and water jackiting to cool the compressor cylinder to remove the heat of compression."

Quote
Andy
What is the ideal process of an air compressor compression stroke

Its actually the reverse in actual practice, for example, with air you must consider depressurization, and cryo potential. A turbine using air does infact get cold towards the exhaust area. However the use of fins or heating jacket is the same "too heat"

With an IC engine they are cooling fins, since the heat byproducts of combustion must be managed.

So when using air as an ideal gas in an expander, ambient temps of around 80°f can be absorbed by so-called cooling fins.


Jeremy
Re: Thermodynamic Processes
October 12, 2011 10:14PM
Jeremy, I don't think you understand the point. Compression of air in a cylander is isentropic. The temperature will increase. But the ideal cycle in the case of an air compressor is not modeling the actual processs. It is modeling what is wonted to happen. So to make the compression come closer the the isothermal (constant temperature) process heat must be extracted from the air as it is compressed.

And yes air is sometimes heated ahead of an air engine to avoid cold exhaust temperatures. But mainly to avoid condensation. And is done ahead of the engine. Hot air is not wonted in the tank as on cooling, which will happen in the tank or in the lines, pressure will be reduced.

However you missed the point of this thread entiarly. It's not about air compression. It's about thermodynamic processes. In specific the isenthalipic and throttling processes following the same path from start to finish but are different in that one converts heat to work or viseversa while the other doesn't. It's a hard thing to explain. It comes from the conservation of energy law. But it is a bit strange in that they both follow the same path. How ever I know of no actual isenthalpic natural process. The air compressor is an example of it's application only. But it is not the actual or natural process. Throttling on the other hand has many applications.

In general all thermodynamic cycles are not really modeling the actual machine. They are simplified and ment to be used as a tool for compairson. With one exception. The Carnot cycle was developed to find the limits of efficiency. The Carnot cycle gives the highest efficiency of any engine or compressor operator between a given set of temperatures. No other engine cycle can be more efficienct then the Carnot cycle. How ever it can not be implemented in reality. The Carnot cycle is made up of four reversable processes. Two isentropic and two isothermal. All heat is transfered in by an isothermal process at a constant high temperature and rejected at a constant low temperature isothermal process. These are seperated by the isentropic processes. As all processes of the Carnot cycle are reversable the cycle can equally convert work to heat or heat to work. Because of the processes that make up the Carnot cycle, it efficiency becomes independent of the working substance and is only dependent on the temperatures it is operatoring between. So the Carnot cycle efficency is simply:

(TH - TL) / TH

Which brings us to the 95% efficiency clame you made on the LSR thread the other day. In generaly the otto cycle can at best atain 60% of the Carnot cycle. Now if we take 32F as the Cannot low temperature what must the high temperature be to get 95% efficiency. That would mean that for the Carnot cycle to atain 95% efficienct it would have to be operatoring between 9373.73F and 32F. And sense the otto cycle is only around 60% as efficient as the Carnot cycle you would have a Carnot cycle operatoring at grater then 158% efficiency which is imposable.

Now if you were talking about engine efficiency which is actually thermal efficiency divided by the ideal cycle thermal efficiency that is a different matter. Harry's engine is getting better then 30% messured thermal efficiency.



Edited 1 time(s). Last edit at 10/12/2011 10:23PM by steamerandy.
Re: Thermodynamic Processes
October 13, 2011 07:18PM
Quote
Andy
However you missed the point of this thread entiarly. It's not about air compression. It's about thermodynamic processes. In specific the isenthalipic and throttling processes following the same path from start to finish but are different in that one converts heat to work or viseversa while the other doesn't. It's a hard thing to explain.

sigh

ya, and it may also be called "poly-tropic" : )


Jeremy
Re: Thermodynamic Processes
October 13, 2011 07:47PM
[en.wikipedia.org]
htp://en.wikipedia.org/wiki/Polytropic_process
Jeremy
Re: Thermodynamic Processes
October 13, 2011 08:10PM
Hi jeremy

Ideal gas formula such as polytropic are not are not applicable to a real gas
when Cv is not constant. Cv varies and ideal gas formula can only be
used for very small changes. The closer to the saturation line the larger the
error using ideal gas formula. You need to be around 400F above
saturation before steam acts some what like an ideal gas. But really enthalpy
does not vary with temperature in a liner relation hardly any ware in the
range of temperatures and pressures we are interested in for steam engines. And
if you are close to the critical point things are vary strange.
Re: Thermodynamic Processes
October 14, 2011 01:00AM
In an engine things get even more strange. Earlier today I came accross some old SACA articles, then saw this thread.

Here is a link to the second part of a three part study. [www.steamautomobile.com] It starts on page 20.

In it they test a 1 3/4" bore by 1 3/4" stroke V4 uniflow with an auxilary exhaust valve. They had the equipment capable of producing an indicator diagram when at speed. During the first part of the compression stroke the polynominal was 1.65, during the later part it was 1.07, they state that the theoretical nominal under those conditions was 1.32.

Their theory was that the high 1.65 was when the steam was cooler then the cylinder walls and absorbing heat, the 1.07 was when the steam was hotter and the cylinder walls and head were absorbing heat.

Note that the test engine has .292 ci per sq in of the cylinder head, piston head and cylinder walls. The Stanley 30 hp engine 4 1/2" by 6 1/2" has .836 ci per sq in. The Duke of Gloucester with its 18" by 28" has 3.4 ci per sq in. I honestly believe that the ratio of engine volume to surface area has a huge effect on the effective compression and expansion ratios of an engine in regards to the polynominal coefficient, I really don't see how it couldn't, even when the increased strokes per minute of a small engine is considered.

More food for thought.

Caleb Ramsby
Re: Thermodynamic Processes
October 14, 2011 04:12AM
Here are some plots of the isentropic k exponent calculated as k = Cp/Cv These plots were generated with a constant entropy over the range. The k factor is varing and you would not be able to get a constant entropy using a constant k factor. You simply do not get accurate results using ideal gas formula with steam. They are being ploted from the high pressure to the low against pressure. My dT/dd|s function is not working in the saturated region. The plots have one point in the saturated as it was programed to stop when qualty was less then 1. The top chart is a plot of the entropy of for that run. You can see it varies a bit. In plot 1 the entropy varries from 1.5335657589052 to 1.5335657589054 A variance of 0.0000000000002. Note the last plot was with a relative high supper temperature of 1200F at the start. The others were at lower temperatures.

Andy



Edited 3 time(s). Last edit at 10/20/2011 04:42PM by steamerandy.


Re: Thermodynamic Processes
October 16, 2011 10:20AM
To avoid the problem of low volume to surface area ratio, use the most forgiving materials on the head, the piston top and the cylinder walls. If you go to low specific heat, low conductivity an low specific gravity, you are best off. Aluminum is horrible. Cast iron is good. Etc. Ceramic coatings are even better.

Karl Petersen
ben
Re: Thermodynamic Processes
October 16, 2011 12:13PM
For fun,, to accent Karl's comment,,,Look up the strength/hardnsss of Al,,,7075-T6,,,at temp just above a steaming hot cuppa TEA,,,I was suprised to see a pushrod looking like a hot noodle in an engine,,,no other dammage tho,,,Ben
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