# Heat of a Diesel Cycle

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1. Apr 6, 2015

### cloud809

1. Ideal gas cycle - diesel engine cycle. You begin at some volume V1. Perform an adiabatic compression to V2. Perform an expansion to V3 at constant pressure. Next, perform an adiabatic expansion to V4, Last, it moves back to V1 at a fixed volume, while decreasing temperature back to original starting point.

NOTE: Ignore "specific volume," and just consider it to be generic "volume" for my case.

Find Qin and Qout in terms of V1, V2, V3, and T1.

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2. 1st law: ΔU = W + Q, where W = -∫PdV

We're told that the 1st law is the assumed starting point for this problem.

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3. For Qin, I began that ΔU = 0 for 2→3, so -W2→3=Qin. This yielded

Qin=- (-∫23PdV)

or Qin = P(V3-V2).

Am I right to assume that ΔU=0 for this 2→3 phase? I can't find how to incorporate V1 and T1 if my method was ok.

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Then, for Qout, I noted that W = 0 for 4→1, so ΔU4→1=Qout.

So, I found that ΔU = Nk(f/2)BΔT, or Qout = NkB(f/2)(T1-T4). And again, I'm running into the same issue of not filling the parameters of the original question.

2. Apr 6, 2015

### cloud809

One thing I just realized is that for 2→3, I shouldn't have said ΔU = 0. Instead, I should have set up Qin as:

Qin = ΔU - W, where

ΔU = NkB(f/2)ΔT

ΔU = NkB(f/2)[(PV3-PV2)/NkB]

and W is still W=-P(V3-v2)

so, once simplified (Qin = ΔU - W), Qin=P[V3((f/2)+1)-V2((f/2)+1)].

This still doesn't put it in terms of V1, V2, V3, T1 however.

3. Apr 6, 2015

### TSny

Try to express P in terms of volume and temperature. Don't forget the adiabatic law $TV^{\gamma-1} =$ const.

4. Apr 6, 2015

### cloud809

I've considered something similar and figured it out I believe.
I used P1V11+f/2=P2V21+f/2, solved for P2, and substituted that into my Qin equation. And since there was still a P1 in there, I used the PV=NkBT for P1 to put it in terms of T1. Now my Qin satisfies all 4 variables (V1, V2, V3, T1).

Now to spend a bit more time and figure out Qout...

5. Apr 6, 2015

Sounds good!