Heat engine efficiency limit 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 :)
 A: 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...
A: 
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.
A: 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).
A: 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


*

*Combustion is complete

*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.
A: 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. 
A: 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 :)
