Carnot efficiency does not consider heat transfer. Convective heat transfer (~A*h*delta(T)) is a major heat loss mechanism lowering the thermodynamic efficiency. In low temperature combustion, the delta(T) is smaller and hence less heat transfer from the combustion gasses to the wall.

Low temperature combustion brings benefits to engines in terms of both heat transfer (heat losses at the walls) and heat rejection to the exhaust. For natural aspirated engines, a higher temperature during the expansion stroke means more enthalpy loss to the exhaust path. For turbocharged engines, the question is far more complex.

in ideal case greater the delta T more will be efficiency....

Good question! 5 comments, sorry for the long reply.

1. Actually the Carnot cycle says that the efficiency is based on 1 - TL/TH, i.e. the ratio of the temperature at which heat is rejected (TL) to the temperature at which it is added (TH). TH by itself doesn't matter.

2. No real engine operates on the Carnot cycle. The Otto cycle is a somewhat reasonable approximation to what happens in your premixed-charge (e.g. gasoline, spark-ignition) engine. In the ideal cycle, only compression ratio affects efficiency because increasing compression ratio (r) decreases the ratio of TL/TH. The ideal efficiency can easily be inferred as follows:

a. The adiabatic compression law for an ideal gas is TL/TH = (VL/VH)^(k-1) = (1/r)^(k-1), where VL is the (larger) volume before compression (low temperature), VH is the (smaller) volume after compression, and k is the specific heat ratio of the gas.

b. Treating the Otto cycle as a large number of individual Carnot cycle "strips", the efficiency for each strip is 1 - TL/TH. Even though TL and TH are different for each strip in the Otto cycle, the RATIO of TL/TH is the same for each strip.

c. Combining efficiency = 1 - TL/TH with the adiabatic compression law, the classical formula for Otto-Cycle efficiency = 1 - 1/r^(k-1) is obtained.

3. Mr. Heyne is correct about the heat loss issue, however, if one halves the heat loss by halving the temperature, that means one also put in half the fuel and so can get only half the gross work output. So lowering combustion temperature alone does not increase efficiency, in fact, using half the fuel actually decreases the net efficiency. This is because the friction loss is almost the same regardless of whether you put in 1 unit of fuel or 0.5 units of fuel and the net efficiency (we call it "brake thermal efficiency") = (net work)/(heat in) = (gross work - friction loss)/(heat in).

4. So-called low-temperature engines (I suspect Mr. Mohamad is thinking about HCCI engines) are not more efficient because they operate at lower combustion temperatures, they are (at least, in principle) more efficient because they use higher compression ratio (r). The problem with higher compression ratio is of course knocking, which causes engine damage if not suppressed, but HCCI engines are essentially "controlled knocking" engines. They can operate at lower temperatures because one can burn mixtures having less fuel using compression ignition (with large r) than one can burn with conventional spark ignition. The lower temperatures are beneficial for reducing nitric oxide emissions. The challenge is controlling the knocking process.

5. There is far more I could say about this than this already too-long message. I hope it's ok to post links (it's to my personal, non-commercial site). If so, for more information see the home page of my air breathing engines course:

http://ronney.usc.edu/courses/ame-436/

and in particular lectures 8 - 10:

http://ronney.usc.edu/AME436S13/Lecture8files/

http://ronney.usc.edu/AME436S13/Lecture9files/

http://ronney.usc.edu/AME436S13/Lecture10files/

You may also find my spreadsheet model of engines useful:

http://ronney.usc.edu/AME436S13/AirCycles4Recips.xls

(There is a macro in this spreadsheet; don't worry, it's not a macro virus! If you're concerned, just open the file with macros disabled, the macro is needed for only one non-essential feature.)

I think another reason is combustion efficiency. Comb. efficiency is higher with high temperature of reactions, the combustion process in the first region (because the mixtures are not fully homogeneous, cannot start combustion at the same time) is more complete. Hence, the products include CO2 and H2O, which have high heat capacity. These products lead to reduce the temperature in expand stroke...

Paul clearly stated that low temperature combustion never lead to higher efficiency compared with high temperature efficiency. Carnot efficiency is the ceiling. As a rule of thumb most real engine efficiency is half of Carnot one. More realistic efficiency, which is bast of optimum power is 1 -sqrt(Tc/Th), which accounts for heat losses in an idealized approach..

Suggest not to limit the mind only on low temperature by the name of "Low temperature combustion". For gasoline low temperature combustion system, it normally a lean combustion system as well. The lean charge may contribute to a high efficiency more that the low temperature.

Hi Li,

Lean mixture reduces NOx formation, however, still can not break Carnot law. In lean mixture, we heat more N2 (N2 is about 78% of air). Can you give an example with high efficiency engine operates with lean mixture and compare the efficiency (W/Q) with Carnot efficiency for the same engine. If the engine efficiency near Carnot efficiency then we need to review our knowledge of 2nd Law of Thermodynamics.

My measure concerns is, we do not want to mislead the readers and new comers to the area. Thanks,

Hi Abdulmajeed

Thermodynamic cycle efficiency is only one of many factors which influence final or total efficiency of real internal combustion engine. If the question is clearly specified to discuss the relationship between combustion temp and thermodynamic cycle efficiency. Ok we can forget about the lean mixture. But if people talked about thermal efficiency of an engine, it's a general speaking and more like total fuel efficiency. In this case, I would like to suggest people to also think about other aspects than the theoretical thermodynamic cycle efficiency. For a low temperature gasoline combustion system, it is normally lean mixture. Then decrease of pumping loss and increase of specific heat capacity will play a big role. And low temperature doesn't only impact the Carnot efficiency in an international combustion engine, it will also decrease the decomposition of fuel elements, consequently improve engine efficiency.

I guess we probably need to also consider the entropy production. At low temperatures the level of entropy production is low which results to low level of exergy destruction. In that situation the ratio of heat output to heat input becomes appreciable in comparison to higher temperature where heat losses are often large. The Carnot cycle is an idealized case, in real systems the situation is often complex and exergy destruction levels at higher temperatures are quite appreciable hence lowering the efficiency of the system. Thanks

It depends on what you're doing to produce low-temperature combustion. Running lean will increase your combustion efficiency, since you should extract all the chemical potential energy from the fuel (produce no Unburned Hydrocarbons, UHC), but this may not improve the thermal efficiency because you will produce less power. However, if you use a process such as Exhaust Gas Recirculation (EGR), where you mix some of the exhaust gas with the charge (air entering the engine), you don't have to work as hard to create an ignitable mixture (be careful about knocking!). You also have extra thermal load from the injected CO2 and H2O in addition to the N2 in the air which means the same heat release (for a stoichiometric mixture) warms up the products to a lower temperature. This has the benefits like others said of reduced heat losses and reduced NOx production, which improve thermal efficiency.

Depending on the behavior of the substances, to low temperature is obtained major efficiency, but it is obtained minor power.

The engine cycle is an open cycle also its strokes are different from Carnot cycle strokes. The most important efficiency for an engine cycle is the thermal efficiency, which is the ratio of the produced mechanical work to the chemical value of the fuel consumed. When the combustion temperature is low(for any reason), the chemical reaction formed in combustion stroke becomes more complete and the burning products are more CO2 and H2O. On the other hand if this temperature is high in the products there exist H, H2, CO and other unburned hydrocarbons all having chemical value not converted to heat. This is the main reason for higher thermal efficiency of the low temperature engines.

Answer based on Thermodynamics and Heat Transfer: In any combustion chamber,

higher the temperature, higher heat losses and harder to control the heat leakage. and as you know the heat transfer (especially heat leakage) is an irreversible process (it means that you could connect that heat leakage to a heat engine and get work out of it). Therefore, when you do not need high intensive energy at high temperature, it is better to use low temperature combustion for higher efficiency.

In practice, it is standard to use the fuel conversion efficiency as a measure of an engine efficiency. This efficiency is defined as the ratio of the work per cycle and the product of the mass fuel by its lower heating value. Also, it is possible to define a thermal conversion efficiency as the ratio of the fuel conversion efficiency and the combustion efficiency. For hydrocarbons, the value of the fuel conversion efficiency closely approximates the availability conversion efficiency, which is defined as the ratio of the work per cycle and the maximum useful work that can be obtained by the system engine-atmosphere. It must be noted that, in contrast with the efficiency of a thermodynamic heat engine, the availability conversion efficiency is limited to 1 by the second law (and not to the efficiency of the Carnot's cycle). Then, it is not appropriate to use the Carnot's cycle as a reference measure in order to evaluate the efficiency of an internal combustion engine.

Thermal efficency for all engine cycles could be presented as

eta= 1-heat removed/heat added.

By this way thermal efficiency is higher, when heat removed is lower (heat removed is nearly 0 it is the best solution). So engine should operate in such manner, that fumes temperature is nearly equal ambient temperature.

This is possible when the temperature increase in the combustor is low. Small temperature after the combustion proceses cause small temperature after decompresion process. By this way engine outlet temperature is low and heat removed of the engine is low. Thermal efficiency is high, but engine work is low.

You can check when you draw Otto, Diesel or Bryton cycles for small and high combustion temperature increase.

Actually for the ideal Otto and Brayton cycles, the efficiency depends only on the compression ratio, not the amount of heat added. In those cycles, as one adds more heat, the temperature of heat addition increases (obviously) but so does the temperature at which heat is rejected. The ratio of TL/TH is the same for each part of the cycle for the Otto and Brayton cycles, and thus the efficiency (E) is independent of the heat addition, which is obvious from the formulas E = 1 - 1/r^(k-1) for Otto or E = 1 - 1/r^*((k-1)/k) for Brayton. (r is the volume ratio for Otto, and the pressure ratio for Brayton.) This is because the constant-pressure or constant-volume curves on the T-s diagram diverge for increasing s (heat addition). Refer to the attached diagram for details.

I think we are going out of the imposed Q: The Q can any cyclic engine violet Carnot efficiency? Or decreasing the source temperature increases the efficiency of operating for the same engine?

Regardless of technology used, the answer is NO.

However, comparing one type of engine with another type of engine, in this case the used technology make difference.

I still think that the rate and level of exergy (availability) destruction plays a significant role in this question. We recall that the Q is how we can explain why low temperature combustion engines achieve higher thermal efficiency. In a low temperature system, the change in entropy which is equal to the entropy production is small compared to a high temperature phenomenon, hence the exergy (availability) destruction which lowers the work output is small. This implies that the maximum useful work extractable from a given energy input will be high. If we then calculate the efficiency for such low temperature system it will be high, since the level of exergy destruction (the product of delta S and the environment temperature) is less. Exergy or available work is a second law tool that helps to pinpoint areas in systems where greater energy losses occur. Very useful in real systems where irreversibility plays a significant role. In the situation under discussion, the level of irreversibility is naturally small being a low temperature operation. We recognize that the high level of exergy destruction in high temperature combustion systems is partly what gave birth to the idea of co-generation in power plant technology. In such hybrid systems, the combined efficiencies are higher.

Hi Taib,

I think you answered yourself!

You say that when the combustion process undergoes low temperature, that leads to reduction in heat losses and all the heat produced from the combustion process will be converted into kinetic energy(useful work).This of course will increase the thermal efficiency.

Please see my answer above.

Dear All,

Some of us, forget that Exergy, Ex, of heat is no more than heat (quantity), Q, scaled by Carnot factor (1-Tc/Th), i.e., Ex=Q (1-Tc/Th). In other words, the quality of heat related with the temperature of the source. Therefore, higher temperature of the source means higher exergy. Low temperature source has low exergy, of course low temperature certainly means less heat losses. But, do not forget that the exergy of that source is also low. The bottom line, low temperature does not give high efficiency compared with high temperature heat source. As mentioned, for internal combustion engines, the efficiency is related with compression ratio, which is implicitly related to temperature of gas and higher compression ratio, has knock problem......

Some one mentioned that combustion heat converts into kinetic energy! the exergy of KE equals to KE (1/2 mV^2) by second low it is impossible to convert the heat to KE. However, you can say that exergy of heat (Q(1-Tc/Th)) can be converted to KE theoretically.

Thank you for all answers. I think to wrap up, the low temperature combustion itself does not increase thermal efficiency, rather the reduction of heat loss. Most LTC engines, be it HCCI, PCCI or RCCI, operate at higher compression ratio(as required for compression ignition which leads to thermal efficiency increase. Another factor is the fact that fuel formulations for LTC engines must posses high autoignition properties, resulting in late combustion (after TDC). This results in delayed pressure rise compared to SI which means negative work due to early ignition(before TDC) is reduced.

Actually any combustion that occurs before or after TDC reduces efficiency. The highest efficiency is obtained when all of the combustion occurs at the minimum volume, i.e. TDC - which is what the ideal Otto Cycle assumes.

Dear Mohamad

Based on Thermodynamics and Heat Transfer:

LTC (Low Temperature Combustion) is more efficient because they use higher compression ratio (r). Combining efficiency = 1 - TL/TH with the adiabatic compression law, the classical formula for Otto-Cycle efficiency = 1 - 1/r^(k-1) is obtained. The problem with higher compression ratio is of course knocking, which causes engine damage. But HCCI engines are essentially "controlled knocking" engines. They can operate at lower temperatures because one can burn mixtures having less fuel using compression ignition (with large r) than one can burn with conventional spark ignition. The lower temperatures are beneficial for reducing nitric oxide emissions. The challenge is controlling the knocking process. Furthermore, in LTC condition, temperature gradient and heat losses are decreased. So, in LTC Diesel engine Efficiency is better in comparison conventional Diesel engine.

Professor A.A. Ranjbar

Dear all

It seems to me that some new answers are not related to the initial question. The question was:"How do we explain why low temperature combustion engines achieve higher thermal efficiency?"

Please notice that every parameter has its own effects on the engine efficiency. For instance, as the compression ratio increases the cycle maximum pressure rises and there would be a better expansion stroke (work stroke). These are two reasons for the efficiency enhancement with compression ratio increase. So please don't mix up. Also please consider the meaning of the efficiency.

I'd like to remind again that when the combustion temperature is low, the chemical reaction in combustion stroke becomes more favorable, because in the combustion products less H, H2, CO and other unburned hydrocarbons will be present and instead more CO2 and H2O will be present among the combustion products. In fact compared with high temperature engines in the low temperature engines more sensible internal energy is released due to combustion. This is why the low temperature engines have higher thermal efficiency.

Dear all respected colleagues:

Lets have all answers be fully considered. Just God knows !

Professor A.A. Ranjbar

Professor Ronney. I agree with you that theoretically, efficiency is highest when combustion happen at constant volume (extremely short period of time). However from practical experience (my own and few other works), a longer distributed burning during power stroke with relatively lower peak pressure a few degrees after TDC (5-15 deg CA ATDC) are favorable to engine conversion efficiency and SFC. This idea has been discussed elaborately in a combustion seminar recently and it leads to more interest in reactivity-controlled compression ignition (RCCI) engine.

The main difference from conventional Diesel engines and HCCI ("low temperature") that determine this increase in efficiency is that in the first one there is a premixed combustion (constant volume combustion: high efficiency) followed by a diffusion combustion (constant pressure combustion: low efficiency) while in the HCCI there is only an highly premixed combustion (homogeneous-> constant volume combustion: high efficiency);a bit lower compression ratio is adopted in order to reduce the knocking problem and to retard the time of the ignition to increase further the mixing quality; the high mixing quality (homogeneous) determines very low emissions and a further increase of efficiency

In this case we should reduce the amount of heat rejected to surroundings, i.e. we should reduce the ratio of heat rejected to heat supplied to the engine. This can be done be reducing the average temperature of the exhaust gases. However, the exhaust temperature must be always greater than the dew point. We may need larger surface areas to insure the desired rate of the heat transfer for closed systems.