Repost - Please, need to get done by tomorrow, stuck big time

[ChronicQuantumAddict]

Please help with the following question, completely lost:
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 ratio of specific heats, Cp/Cv).

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.

I was able to graph the thing. Now i need to know how to calculate the heat in between each step, i am clueless as how to do this. i know once i get that, i can easily find the total work, and i know the expression for the efficiency and everything else.

My main problem is how do u get the heat of each step, and i do know that some steps have no heat exchange. please help, this is due tomorrow, and i may have to post it on the board in class.

Thanks very much

 

[Pengwuino]

Do not double post and read the FAQ

 

[Moonbear]

You've already been helped in your original thread.