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

AI Thread Summary
The discussion focuses on calculating heat in a diesel cycle using the first law of thermodynamics and the ideal gas law. The process begins with adiabatic compression and includes phases of constant pressure expansion and further adiabatic expansion. The user derives expressions for Qin and Qout, initially struggling to incorporate all variables, particularly V1 and T1. After adjustments, they successfully express Qin in terms of V1, V2, V3, and T1 by substituting pressure with the ideal gas law. The conversation concludes with the user planning to apply similar methods to determine Qout.
cloud809
Messages
3
Reaction score
0
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.


--

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

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


--

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.

--

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.
 
Physics news on Phys.org
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!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top