2
$\begingroup$

The Carnot efficiency limit shows the maximum efficiency of a heat engine as:

\begin{align} \eta & = 1-\frac{T_C}{T_H} \end{align}

I have often heard comments that $ T_H $ is the temperature limit of the materials used in the particular engine one is working with. Although this may be useful for someone designing a particular engine, I'm wondering what $ T_H $ stands for theoretically. As an example, if I am using gasoline or diesel for fuel, would the theoretical value of $ T_H $ correspond to the adiabatic flame temperature for those fuels? Again, I am not concerned at the present time if that temperature melts all the engine parts, I am interested in what theoretical efficiency limits I can achieve with particular fuels and compression ratios.

This leads me to a second question. If I use the adiabatic flame temperature for a particular fuel as my $ T_H $, I would like to use a $ T_H $ based on the adiabatic temperature of that fuel at different compression ratios. Does anyone know of a resource where I can find the adiabatic flame temperatures of, lets say gasoline or diesel, at different compression ratios? I am looking for a table with various temperatures so I don't have to do the math for each theoretical fuel or compression ratio.

Thanks for considering this question :)

Improve this question
$\endgroup$
10
  • $\begingroup$ The diesel engine is not a thermal engine in the sense defined by carnot, as such you cannot calculate the efficiency using the equation above $\endgroup$
    – user126422
    Commented Sep 28, 2016 at 20:23
  • 1
    $\begingroup$ $T_h$ is the temperature of the hot reservoir. Nothing to do with fuel. $\endgroup$
    – user108787
    Commented Sep 28, 2016 at 21:10
  • $\begingroup$ I should reiterate , "Although this may be useful for someone designing a particular engine". I am not talking about a particular engine, whether that be a diesel or otto, I am interested in understanding what the TH for a heat engine is in general.. This engine may or may not use diesel or gasoline or hydrogen. $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 0:02
  • 1
    $\begingroup$ Is this your question: If I had to calculate an upper bound on the efficiency of a real engine, which uses a particular fuel, by employing the formula $1-T_C/T_H$, what should I use for $T_H$? $\endgroup$
    – Deep
    Commented Sep 29, 2016 at 5:15
  • $\begingroup$ Yes, that is where I am going with this. The engine I am using is novel and does not, with the amount of investigation I have done so far, fit the exact upper efficiency limits of an otto or diesel cycle engine. Since that is the case, I am considering what the upper limits might be from a purely theoretical view. I didn't want to ask the question in a detailed way because I didn't want to burden anyone with the physics involved in trying to determine a theoretical efficiency limit for this particular engine (thus my vagueness) $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 5:40

6 Answers 6

Reset to default
2
$\begingroup$

In a Carnot cycle, a gas does work while its temperature lowers. If this is done irreversibly you get the maximum theoretical efficiency (constant entropy).

Real engines try to approach this but fail. But yes, when the thermal ratio (input/output) is greater you will get greater efficiency. Thus the drive for materials that can withstand high temperatures in the first stage of gas turbines , for example.

You can in principle improve the temperature at the input by increasing the fraction of oxygen in the air being combusted - if you don't have to heat nitrogen you can get a hotter flame, or if you like a higher temperature / pressure at the start of your Otto cycle.

Putting "real" numbers on this is the realm of engineering more than physics...

Improve this answer
$\endgroup$
2
  • $\begingroup$ "Putting "real" numbers on this is the realm of engineering more than physics..." Agreed, thank you for the reply, putting real numbers will be an exercise for the future for sure, for now, I am wondering what my limits might be for an engine I haven't quite wrapped my head around in terms of theoretical efficiency. For all I know it may follow an Otto or Diesel cycle but I can't assume that right now, I am hoping for better than that, given what I consider it's improved characteristics of expansion versus compression ratios and the way it decouples combustion from time constraints. $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 6:10
  • $\begingroup$ many of the Diesel and Otto theoretical values seem to incorporate the assumption (even though the assumption is admitted), that combustion happens under constant volume conditions and this seems very unsatisfactory to me, especially when one considers the inefficiencies related to time-loss or incomplete combustion, losses directly relating to non constant volume combustion for much of this portion of the Otto or Diesel cycle. This has me wondering what the theoretical limits might be for an engine that actually has constant volume combustion but also has a different cycle than most others. $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 6:22
1
$\begingroup$

In a Carnot engine, $T_H$ is the temperature of the hot reservoir and $T_C$ is the temperature of the cold reservoir. In a Carnot engine, you can transfer energy out of the hot reservoir into an element of your engine, up until it is as hot as the hot reservoir. You can transfer energy into the cold reservoir from an element of your engine, up until it is as cold as the cold reservoir. The reservoirs are assumed to have unbounded heat capacity, so they never change temperature.

Carnot has no concept of fuel. The source of the raised temperature of $T_H$ is not specified in that abstract engine. However, it would be trivial to show that $T_H$ cannot exceed the temperature of your burning fuel because the burning fuel is the hottest material in the system. The actual $T_H$ that you can use depends on what approaches you use to map your real-life system (with fuels and real heat capacities) into the abstract Carnot engine (with no fuel and unbounded heat capacities in the reservoirs).

Improve this answer
$\endgroup$
5
  • $\begingroup$ "The actual TH that you can use depends on what approaches you use to map your real-life system (with fuels and real heat capacities) into the abstract Carnot engine (with no fuel and unbounded heat capacities in the reservoirs)." That is where this will become interesting but for now, your reply, "However, it would be trivial to show that TH cannot exceed the temperature of your burning fuel because the burning fuel is the hottest material in the system", has me thinking about how close the engine design I have will mirror a Carnot heat engine. $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 6:00
  • $\begingroup$ Cort, I am back in the office again. Thanks for the concise response to my question. I am not sure about the rules of this forum since I just signed up last week. Would it be frowned upon for me to get more specific about the details of the heat engine I am referencing; I don't want it to become engineering. I am interested to see what type of species others consider it to most resemble. There are some distinct differences between the four stroke operation of this and the typical otto or diesel family in a few terms, one being pressure volume relationship in certain portions of the cycle. $\endgroup$
    – randyj
    Commented Oct 2, 2016 at 10:00
  • $\begingroup$ Specifically, unlike the Otto, the isentropic expansion ratio will not be equal to the compression ratio so contrary to this statement from Wiki... "The volume ratio V{4}/V{3} is called the "isentropic expansion ratio". (For the Otto cycle is the same as the compression ratio V{1}/V{2}." ... these ratios will be different, probably closer to a 3 to 1 volume ratio. $\endgroup$
    – randyj
    Commented Oct 2, 2016 at 12:40
  • $\begingroup$ Also from the Wiki. article on the Otto engine, $\endgroup$
    – randyj
    Commented Oct 2, 2016 at 12:42
  • $\begingroup$ The temperature relationship between points 1,2,3, and 4 will be disproportional. Specifically T4\T1 will not equal T3\T2 $\endgroup$
    – randyj
    Commented Oct 2, 2016 at 12:45
1
$\begingroup$

I am interested in what theoretical efficiency limits I can achieve with particular fuels and compression ratios.

As user115350 has stated, the efficiency of a diesel engine is

$$\eta = 1-\frac{1}{r^{\gamma-1}}\left(\frac{\alpha^\gamma -1}{\gamma (\alpha -1)}\right)$$

As you probably know, diesel engines have compression rations of around 20:1 and efficiencies of about 40%. In fact, the only limit on the compression ratio is the strength of the material from which the engine is made.

As I said in my comment, $T_h $ is not fuel related, it is used in connection with "ideal engines". A Carnot engine is the most efficient engine, but it has a very slow cycle of operation, so much so that if you ran a car with it, people could easily walk past you.

This leads me to a second question. If I use the adiabatic flame temperature for a particular fuel as my $T_h$, I would like to use a $T_h$ based on the adiabatic temperature of that fuel at different compression ratios.

So in light of what I have said above, this is based on a wrong assumption about $T_h$.

Does anyone know of a resource where I can find the adiabatic flame temperatures of, lets say gasoline or diesel, at different compression ratios? I am looking for a table with various temperatures so I don't have to do the math for each theoretical fuel or compression ratio.

This page Adiabatic Flame Temperatures gives details of various fuels, but all at constant pressure. Unfortunately, it is not easy to find any more information than that.

Improve this answer
$\endgroup$
4
  • $\begingroup$ We may be trying to put this question in a preconceived box. Let me reiterate, I am not talking about any particular engine nor am I mentioning length of times for work or heat transfer. Getting back to this hypothetical heat engine with a TH and a TC, I understand the TC to be ambient (or whatever surrounding cold sink it is) but what about the TH or high heat sink, if you are using fuel for this engine, would not the fuel be the high heat sink? also from this, would not the adiabatic flame temperature of that fuel be the highest possible TH for the heat engine using this fuel? $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 0:23
  • 1
    $\begingroup$ @randyj As CountTo10 says, TH is the reservoir temperature and is the temperature at which heat is transferred to the working fluid. In theory, you have to decompose the cycle into a (uncountable) heap of concatenated inifnitessimal Carnot cycles and apply the TH concept to each one separately in something like the Diesel cycle - even then hard to define. In general, TH winds up limited by engineering: superheated steam, for example, must be kept below about 700C (IIRC) in a steam turbine otherwise most steels used in turbines will yield - even though the flame temperatures are much higher. $\endgroup$
    – Selene Routley
    Commented Sep 29, 2016 at 0:34
  • $\begingroup$ For theoretical purposes, TH is the highest reservoir temperature you have, correct? If my experiment was on the moon and my TC was on the surface of the dark side and I wanted the TH to be on the bright side but found I could create an even higher TH by a combustion reservoir that was 100 degrees C above the surface temperature of the bright side of the moon, my combustion reservoir would legitimately be the TH I should use if I was interested in efficiency, correct? $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 1:03
  • $\begingroup$ By "interested in efficiency" I mean the efficiency potential of the heat engine itself, neglecting the energy needed to heat the combustion reservoir $\endgroup$
    – randyj
    Commented Sep 29, 2016 at 1:07
1
$\begingroup$

Carnot efficiency does not care about fuel but the adiabatic flame temperature is the theoretical maximum of $T_H$. Adiabatic temperature is not a measured property, but a calculated one. Thus it is not listed in a table. It is the theoretical temperature limit of combustion if

  1. Combustion is complete
  2. No energy is lost to the surroundings.

Say you know temperature and pressure of your intake, you can find the enthalpy of the uncompressed gas. Then you need to find enthalpy change in compression. If compression is an isothermal process, internal energy change equals work done to the system. So enthalpy change is pressure change times volume. Pressure change can be found from compression ratio and volume is your engine capacity. If combustion is also adiabatic, enthalpy does not change. Then you can find the adiabatic flame temperature provided data are available.

Of course in the real world where process is not adiabatic, temperature is non-uniform and combustion is not instant or complete, the material design process is much more complicated. Adiabatic flame temperature is an over-specification as far as engine temperature is concerned. You would find real life piston and cylinder material melting much lower than adiabatic flame temperature.

Improve this answer
$\endgroup$
0
$\begingroup$

Sorry everyone, I am new to this site and to the app. I would like to rate some of the excellent responses but I am concerned I may do harm. I'll wait till I'm back in the office and on my pc :)

Improve this answer
$\endgroup$
0
$\begingroup$

You can read wikipedia about diesel cycle. The cycle is not Carnot cycle, you can not use $\eta = 1- \frac{T_C}{T_H} $, which is only applied to Carnot cycle. For diesel cycle, the theoretical efficiency can be calculated using below. $$\eta = 1-\frac{1}{r^{\gamma-1}}\left(\frac{\alpha^\gamma -1}{\gamma (\alpha -1)}\right)$$ where $r$ is compression ratio. As you can by increasing compression ratio, you can improve the engine's efficiency.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.