Otto Cycle - Thermodynamics - Efficiency

[ChronicQuantumAddict]

Please help with the following question, completely lost: :yuck:

The behavior of a four stroke gasoline engine can be approximated by the so-called Otto cycle. The process is as follows (6 steps):

1) Isobaric Intake (from V=0) at atmospheric pressure up to volume V1 (maximal volume), Temperature T1

2) Adiabatic compression to volume V2, Temperature T2

3) Isochoric increase of Temperature during ignition to T3

4) Adiabatic expansion to V1 (max volume), Temperature T4

5) Isochoric decrease of temperature (to T1)

6) and Isobaric exhaust to V=0 at atmospheric pressure.

Sketch the process in a P-V diagram. Assume the working substance is an ideal gas and calculate the efficiency of the process. Express the result in terms of the temperatures T1, T2, T3, and T4. If the compression ratio is given by r = V1/V2, express the efficieny with r. For an Engine with r = 8, and gamma = 1.5 what do u obtain? (Note gamma is the ration of specific heats, Cp/Cv).

Please help, i don't know how a graph can be sketched on here, because i have an idea of the graph, but if someone could post this diagram along with a clue or the procedures on how to obtain the answer it would be much appreciated. Thanks

 

[siddharth]

Efficiency is going to be (Net work done by the system)/(Total Heat supplied to the system).
So, you have to express the net work done and total heat supplied in terms of the known quantities.
Try calculating the heat supplied to or lost from the system in each step. Since change in the net Internal energy is zero for the whole cyclic process, if you can find the heat supplied to or lost from the system in each step, you can easily find the work done for the whole process.

 

[ChronicQuantumAddict]

hmm, can u be a little bit more specific, how would i get the heat supplied? and thanks

 

[ChronicQuantumAddict]

ok, i have drawn the steps in the diagram, but i am still confused on how to obtain the heats in each step. i know i need to use W_net = W1 + W2 + W3 + ... = (Closed Integral) P*dV to obtain the net work, and then with that, i can calculate the efficieny. but i am still stuck, any other pointers? thanks

 

[ChronicQuantumAddict]

same here, due tomorrow, please help...

 

[ChronicQuantumAddict]

well I am quite sorry for reposting, but i did not understand the help, and no one was replying to the post, so i reposted to get it back up to the top so people can see it. Can someone please help me a little more?

 

[OptimusPrime]

Your question is very unclear. Please make it clearer. Thanks!

 

[ChronicQuantumAddict]

i can't make it clearer, that it the question, i have found the expression from the book to be:

Efficiency = 1 - [(T4-T1)/(T3-T2)] = 1 - [1/(r^(gamma-1))]

where gamma = ratio of heat capacities, and r = compression ratio. But i do not know how to obtain this expression from the heat and work. Can someone help me to see how this is done?

At the moment i am trying to work backwards from this expression to try and obtain the method to derive it. any help is much appreciated, its due in 2 hours

 

[OptimusPrime]

Well, why did you post it here at the last minute? Come on, what do you expect?

 

[ChronicQuantumAddict]

Well, why did you post it here at the last minute? Come on, what do you expect?

Actually, i posted it here several days ago, but i was able to solve the problem, thanks everyone who helped.

 

[abdulhanan]

first watch the image.
http://en.wikipedia.org/wiki/File:P-V_otto.pngefficiency=output/input.
output=heat supplied(h.s)-heat rejected(h.r)
input = heat supplied.

effi = (h.s-h.r)/h.s=1-(h.r/h.s)

h.s = mCv(T3-T2)
h.r = -mCv(T1-T4) = mCv(T4-T1)

effi. = 1-((T4-T1)/(T3-T2))
sorry for late reply.
i saw this site just today.

 

[abdulhanan]

first watch the image.
http://en.wikipedia.org/wiki/File:P-V_otto.png


efficiency=output/input.
output=heat supplied(h.s)-heat rejected(h.r)
input = heat supplied.

effi = (h.s-h.r)/h.s=1-(h.r/h.s)

h.s = mCv(T3-T2)
h.r = -mCv(T1-T4) = mCv(T4-T1)

effi. = 1-((T4-T1)/(T3-T2))



sorry for late reply.
i saw this site just today.