What would a PV curve for a Quasiturbine engine look like?

  • Thread starter Thread starter taijin
  • Start date Start date
  • Tags Tags
    Curve Engine
AI Thread Summary
A discussion on the PV curve for a Quasiturbine engine highlights the challenge of visualizing its performance characteristics. Participants suggest exploring related resources, including articles and diagrams on Otto cycles, as a basis for understanding the Quasiturbine's PV curve. One user emphasizes the importance of considering temperature and entropy diagrams alongside the PV curve. The consensus is to adapt existing Otto cycle diagrams to better represent the Quasiturbine engine's unique features. Overall, leveraging existing resources and modifying them appears to be the recommended approach for the report.
taijin
Messages
1
Reaction score
0
Im working on a report for my physics class on engines and I wanted to do a report on the quasiturbine engine but I can't figure out what a PV curve for a Quasiturbine engine would look like?
http://quasiturbine.promci.qc.ca/
 
Physics news on Phys.org
Last edited by a moderator:
I'm not familiar with this engine, but just a thought: sometimes it's helpful to look at the temperature/entropy diagram as well.
 
disregard my last reply. I was just reading about this engine, and you probably want the P/V diagram, check out factbug.org

It says this engine operates around an otto cycle. I'm certain you can google a graphic of an otto cycle pv diagram. Then, perhaps modify it for your purposes?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top