How Is Heat Calculated in a Diesel Cycle Using Volumes and Temperatures?

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Homework Help Overview

The discussion revolves around calculating heat in a diesel cycle, specifically focusing on the relationships between volumes and temperatures during the cycle's phases. The original poster outlines the steps of the ideal gas cycle and seeks to express heat transfer (Qin and Qout) in terms of specific variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the first law of thermodynamics and considers the changes in internal energy and work done during different phases of the cycle. They question the assumption of ΔU being zero during the 2→3 phase and seek to incorporate all relevant variables into their equations.

Discussion Status

Participants are actively exploring different approaches to express Qin and Qout in terms of the specified variables. Some have provided guidance on using the adiabatic law and substituting pressure in terms of volume and temperature. There is a recognition of progress made by the original poster in expressing Qin with all required variables, while Qout remains to be fully addressed.

Contextual Notes

Participants note the importance of adhering to the assumptions of the ideal gas law and the specific conditions of the diesel cycle. The discussion also highlights the challenge of incorporating all necessary parameters into the equations while following the constraints of the problem.

cloud809
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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.

300px-DieselCycle_PV.svg.png

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.
 
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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.
 
cloud809 said:
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.
Try to express P in terms of volume and temperature. Don't forget the adiabatic law ##TV^{\gamma-1} =## const.
 
Thanks for the reply TSny.
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...
 
Sounds good!
 

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