Otto & Diesel Cycles: Heat Addition/Rejection Processes

[s@ikiran]

why are heat addition and rejection processes in an otto cycle constant volume processes?similarily why does heat addition occurs at constant pressure in a diesel cycle?

 

[SteamKing]

why are heat addition and rejection processes in an otto cycle constant volume processes?similarily why does heat addition occurs at constant pressure in a diesel cycle?

These are idealizations of what actually occurs in real engines.

For the otto cycle engine, combustion of the fuel and the exhaust of the burned fuel occur relatively quickly, while the piston is momentarily stopped at the top and bottom of the stroke, respectively.

https://en.wikipedia.org/wiki/Otto_cycle

For the diesel engine, the engine is designed so that the injection and combustion of fuel occurs just as the piston has reached top dead center. As the piston starts to move downward in the power stroke, fuel is injected and auto-ignited by the compressed air in the cylinder. For a brief time, the additional pressure created by the gases from the burned fuel offsets the loss of pressure in the cylinder as the piston travels downward.

https://en.wikipedia.org/wiki/Diesel_engine

 

[Harikrishnan R]

In otto Cycle Combustion of fuel (high Calorific Energy) is happened with in less time than diesel cycle, so the pressure is suddenly increased in otto cycle (Constant volume Process), than diesel engines. But heat rejection process is same for both cycles because there is no combustion of fuel...

 

[s@ikiran]

thank you for the answer

 

[s@ikiran]

These are idealizations of what actually occurs in real engines.

For the otto cycle engine, combustion of the fuel and the exhaust of the burned fuel occur relatively quickly, while the piston is momentarily stopped at the top and bottom of the stroke, respectively.

https://en.wikipedia.org/wiki/Otto_cycle

For the diesel engine, the engine is designed so that the injection and combustion of fuel occurs just as the piston has reached top dead center. As the piston starts to move downward in the power stroke, fuel is injected and auto-ignited by the compressed air in the cylinder. For a brief time, the additional pressure created by the gases from the burned fuel offsets the loss of pressure in the cylinder as the piston travels downward.

https://en.wikipedia.org/wiki/Diesel_engine

thank you bro i got your point
now i need another question to be answered

in a carnot cycle the work done is taken by the relation (work done =heat supplied -heat removed)
in a otto cycle the work done can be found by the same relation and it is also given as (work output=workdone by the system-work done on the system)

in otto cycle the work done in compression and expasion are considered whereas in carnot cycle it is not. why what is the reason behind it?

 

[SteamKing]

thank you bro i got your point
now i need another question to be answered

in a carnot cycle the work done is taken by the relation (work done =heat supplied -heat removed)
in a otto cycle the work done can be found by the same relation and it is also given as (work output=workdone by the system-work done on the system)

in otto cycle the work done in compression and expasion are considered whereas in carnot cycle it is not. why what is the reason behind it?

It's not clear what you mean here.

Work and heat are equivalent.

In all cycles, Carnot included, the amount of work which can be provided is based on the difference between the highest temperature obtained in the cycle and the lowest temperature obtained.

https://en.wikipedia.org/wiki/Carnot_cycle

The Carnot cycle extracts all available heat between these two temperature extremes, hence it is considered to be 100% efficient. All other cycles, like the Otto and the Diesel, are not able to extract this much available heat and turn it into work, even though they may operate between the same temperature extremes as the Carnot cycle, hence their efficiencies are less than 100%.

 

[s@ikiran]

It's not clear what you mean here.

Work and heat are equivalent.

In all cycles, Carnot included, the amount of work which can be provided is based on the difference between the highest temperature obtained in the cycle and the lowest temperature obtained.

https://en.wikipedia.org/wiki/Carnot_cycle

The Carnot cycle extracts all available heat between these two temperature extremes, hence it is considered to be 100% efficient. All other cycles, like the Otto and the Diesel, are not able to extract this much available heat and turn it into work, even though they may operate between the same temperature extremes as the Carnot cycle, hence their efficiencies are less than 100%.

ok i will make it clear
in carnot cycle heat supplied and heat removed during isothermal processes are considered while finding work
why isn't the work done during adiabatic compression and expansion considered?

 

[SteamKing]

ok i will make it clear
in carnot cycle heat supplied and heat removed during isothermal processes are considered while finding work
why isn't the work done during adiabatic compression and expansion considered?

In the Diesel and the Otto cycles, the work done compressing the gases inside the cylinder is not available as output from the engine; this is energy which is lost, as far as useful work is concerned. Similarly, the expansion of the exhaust gases also removes energy from the system, which might otherwise be converted to work.

In both the Otto and Diesel cycles, we are trying to model what happens in an actual mechanical device. The Carnot cycle is based entirely on theoretical considerations: there is nothing mechanical which actually comes close to working as the Carnot system describes.

 

[s@ikiran]

are the net work done and work output both equal?

 

[SteamKing]

are the net work done and work output both equal?

They should be.

 

[s@ikiran]

i saw in a book that for an otto cycle
work done=heat supplied-heat rejected i.e.,w=(cv(t3-t2)-cv(t4-t1))(constant volume process)
net work done=work done by system during adiabatic expansion-work done on system during adiabatic compression i.e.,w=((p3v3-p4v4)/(ϒ-1)-(p2v2-p1v1)/(ϒ-1))
now are these both equal.i mean if we are asked to find work can we use any of the above equations or are there any exceptions?

 

[SteamKing]

i saw in a book that for an otto cycle
work done=heat supplied-heat rejected i.e.,w=(cv(t3-t2)-cv(t4-t1))(constant volume process)
net work done=work done by system during adiabatic expansion-work done on system during adiabatic compression i.e.,w=((p3v3-p4v4)/(ϒ-1)-(p2v2-p1v1)/(ϒ-1))
now are these both equal.i mean if we are asked to find work can we use any of the above equations or are there any exceptions?

It's going to depend on the information given in a particular problem.

 

[Randy Beikmann]

s@ikiran, you said,
"in carnot cycle heat supplied and heat removed during isothermal processes are considered while finding work
why isn't the work done during adiabatic compression and expansion considered?"

This is not true. In every thermodynamic cycle, the work done in ALL processes must be accounted for, as well as the heat transferred. In the Carnot Cycle, work is done during all four processes, positive during expansion, and negative during compression. Heat, however, is only transferred in two: the isothermal expansion and compression processes. The net work is the sum for all four processes. Divide that by the net heat addition, and you get the Carnot efficiency 1-TL/TH.

 

[s@ikiran]

s@ikiran, you said,
"in carnot cycle heat supplied and heat removed during isothermal processes are considered while finding work
why isn't the work done during adiabatic compression and expansion considered?"

This is not true. In every thermodynamic cycle, the work done in ALL processes must be accounted for, as well as the heat transferred. In the Carnot Cycle, work is done during all four processes, positive during expansion, and negative during compression. Heat, however, is only transferred in two: the isothermal expansion and compression processes. The net work is the sum for all four processes. Divide that by the net heat addition, and you get the Carnot efficiency 1-TL/TH.

hi sir,
thank you for your reply
now do you mean that work done in adiabatic expansion and adiabatic compression are equal for a carnot cycle and hence they cancel each other?

i am still a btech 3rd year mechanical student
the thermal part has been started a few months ago
i still don't have clear idea about all the concepts and so my questions may seem a bit ridiculous sometimes but i am really interested in this thermal part and hence i want to learn it completely so i request you to take my questions positively

 

[Randy Beikmann]

s@ikiran, the main thing to learn is that in a cycle, all processes must be included in the analysis, for heat and work. But for heat, you are mainly interested in how much you need to put IN, to power the cycle.

In the Otto cycle, 1) the isentropic compression consumes work, 2) the constant-volume heat addition does not, 3) the isentropic expansion produces work, and 4) the constant-volume heat rejection does not. For heat, 1) the isentropic compression transfers no heat, 2) the constant-volume heat addition transfers heat in (of course), 3) the isentropic expansion transfers no heat, and 4) the constant-volume heat rejection transfers heat out (of course).

In the Carnot cycle, 1) the isothermal compression consumes work, 2) the isentropic compression consumes work, 3) the isothermal expansion produces work, and 4) the isentropic expansion produces work. For heat, 1) the isothermal compression transfers heat out, 2) the isentropic compression transfers no heat, 3) the isothermal expansion transfers heat in, and 4) the isentropic expansion transfers no heat.

So in the Otto cycle, only two processes involve work, and two involve heat transfer. In the Carnot cycle, all four processes involve work, and two involve heat transfer.

In both cycles, you must algebraically add the work produced in all four processes (some may be zero) to find the net work output. Divide that by the heat added (at constant volume in the Otto cycle, isothermally in the Carnot cycle), and you get the efficiency.

 

[Randy Beikmann]

s@ikiran, I do need to apologize for misspeaking in my first post. I just realized it. As I said in the second post, efficiency does NOT have to do with net heat addition, only the heat added at high temperature (as from burning a fuel). So regardless of how the cycle operates, efficiency is the net work divided by the heat transferred in: Wnet/Qh.

Again, I'm sorry if I led you off course.

 

[s@ikiran]

s@ikiran, I do need to apologize for misspeaking in my first post. I just realized it. As I said in the second post, efficiency does NOT have to do with net heat addition, only the heat added at high temperature (as from burning a fuel). So regardless of how the cycle operates, efficiency is the net work divided by the heat transferred in: Wnet/Qh.

Again, I'm sorry if I led you off course.

no sir there's no need to apologize
there's nothing like you led me off course but it's just thought provoking

consider an otto cycle:
In the Otto cycle, 1) the isentropic compression consumes work, 2) the constant-volume heat addition does not, 3) the isentropic expansion produces work, and 4) the constant-volume heat rejection does not. For heat, 1) the isentropic compression transfers no heat, 2) the constant-volume heat addition transfers heat in (of course), 3) the isentropic expansion transfers no heat, and 4) the constant-volume heat rejection transfers heat out (of course).
you must algebraically add the work produced in all four processes (some may be zero) to find the net work output.
as you said above in order to get work output we must add all the work being done in all the processes
i.e.,(work output=Wexpansion-Wcompression)→eqn 1
this gives an expression in terms of pressure and volume.
from first law of thermodynamics,
we can obtain work also by the equation (work output=heat supplied-heat rejected)→eqn 2
this gives an expression in terms of specific heats and temperature differences.
now are the values obained and the significance of both equations equal?
in case these both are equal,can we use eqn 1 as work output while finding efficiency?

in common usage efficiency means ratio of work output to useful input work right?
work and heat both are forms of energy and while considering thermal efficiency,why do we take heat addition only as input work?
work input is also given during compression process right. if it is so then why aren't we considering this compressor work as input work ?
why can't efficiency be η=Wnet/(Qs+Wcompressor)

 

[Randy Beikmann]

"are the values obtained and the significance of both equations [heat balance and work balance] equal?"
In these ideal cycles, the net heat added (based on specific heats) must equal the net work produced. In a real engine, it's better to use measured P vs. V data, because heat is lost through the cylinder walls, piston rings leak, etc.

"in common usage efficiency means ratio of work output to useful input work right?"
Energy efficiency is defined as "net work you get out" divided by "energy you put in." So there are two main types of energy efficiency: 1) thermal efficiency, and 2) mechanical efficiency.
Thermal efficiency is "net work"/"heat added." You want to know how completely the engine converted the energy you paid for (in the fuel) to work. This is used for heat engines like we are talking about.
Mechanical efficiency is "work output"/"work input." This is used to judge how well a device passes mechanical energy along. For example, if you have a gearbox and put 500 joules of work into the input shaft, and it transmits 450 joules to the output shaft, its mechanical efficiency is 90%.

I think this addresses your question "why aren't we considering this compressor work as input work ?" In an engine, the compression work is accomplished by the engine itself, not by some other device. The kinetic energy of the rotating parts is reduced during compression - the engine slows down. During the expansion, the engine speeds up. The net output work is the expansion work minus the compression work. So only the heat from a fuel is added to the engine, and that's what efficiency is based on. Thermal efficiency.

By the way, this whole area is explained much more completely in "Physics for Gearheads," Chapters 12-15. All these terms are defined and reinforced in the discussion and in the figures. ;-)

 

[s@ikiran]

"are the values obtained and the significance of both equations [heat balance and work balance] equal?"
In these ideal cycles, the net heat added (based on specific heats) must equal the net work produced. In a real engine, it's better to use measured P vs. V data, because heat is lost through the cylinder walls, piston rings leak, etc.

"in common usage efficiency means ratio of work output to useful input work right?"
Energy efficiency is defined as "net work you get out" divided by "energy you put in." So there are two main types of energy efficiency: 1) thermal efficiency, and 2) mechanical efficiency.
Thermal efficiency is "net work"/"heat added." You want to know how completely the engine converted the energy you paid for (in the fuel) to work. This is used for heat engines like we are talking about.
Mechanical efficiency is "work output"/"work input." This is used to judge how well a device passes mechanical energy along. For example, if you have a gearbox and put 500 joules of work into the input shaft, and it transmits 450 joules to the output shaft, its mechanical efficiency is 90%.

I think this addresses your question "why aren't we considering this compressor work as input work ?" In an engine, the compression work is accomplished by the engine itself, not by some other device. The kinetic energy of the rotating parts is reduced during compression - the engine slows down. During the expansion, the engine speeds up. The net output work is the expansion work minus the compression work. So only the heat from a fuel is added to the engine, and that's what efficiency is based on. Thermal efficiency.

By the way, this whole area is explained much more completely in "Physics for Gearheads," Chapters 12-15. All these terms are defined and reinforced in the discussion and in the figures. ;-)

thank you so much sir.this piece of information you gave me is really useful and now i understand this concept very clearly.
hope you can help me out with your valuable information whenever i need it.once again thank you very much and i will surely try to keep in touch with you.

 

[jlefevre76]

Sometimes a diesel engine can be better characterized by a dual cycle. It just depends on how the combustion occurs.

As you can see, a dual cycle has an isometric heat addition process, then an isobaric heat addition process. This is still an idealization, but better approximates the process in some engines. Technically everything you do in thermodynamics is an idealization of the actual processes. Another example is that the expansion and compression processes are never truly isentropic, but polytropic as some heat is transferred during the expansions and compressions.