1
$\begingroup$

In an IC engine, the air fuel mixture is ignited in a cylinder resulting in temperature rise of the mixture. However, there is no heat flowing into the engine cylinder from some heat reservoir. Yet, in the P-V graph below, it says that heat $Q_H$ is flowing into the cylinder.

Question: From what heat reservoir is $Q_H$ coming?

PV graph in IC engine

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

The part of the cycle where the red arrow is pointing is the 'ignition'.

That's when the new fuel/air mixture taken into the engine is ignited and provides the energy input. So the heat comes from the ignition of the new fuel.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Normally heat energy is something that flows across boundaries. But here what's the boundary for heat flow? $\endgroup$
    – Sunil
    Oct 3, 2021 at 10:15
  • 1
    $\begingroup$ @ Sunil, yes it's different to many thermodynamic applications, but here the fuel is injected into the cylinder, so maybe that's the point where you could say it crosses the boundary from outside the engine to inside. The fuel has potential chemical energy in it, then when ignited it becomes heat energy. $\endgroup$
    – John Hunter
    Oct 3, 2021 at 10:17
  • $\begingroup$ So, the fuel air mixture after ignition becomes the heat reservoir? $\endgroup$
    – Sunil
    Oct 3, 2021 at 10:23
  • 1
    $\begingroup$ @ Sunil yes, you could describe it that way $\endgroup$
    – John Hunter
    Oct 3, 2021 at 10:24
  • 1
    $\begingroup$ @ Sunhil perhaps you could look at this question physics.stackexchange.com/questions/86924/… about the difference between heat and internal energy. Technically the ignited fuel suddenly has more internal energy (kinetic energy), then that energy (by heat flow) spreads to the rest of the mixture in the cylinder, the air and any spent fuel/air mixture that was not fully removed at the exhaust part of the cycle, that's the best I can do, hope you find a way to make sense of it all! $\endgroup$
    – John Hunter
    Oct 3, 2021 at 10:53
0
$\begingroup$

The diagram in the figure does not really represent the version of the Otto Cycle in an IC engine. It "simulates" this using what they call in the literature the "standard air Otto cycle." (Fundamentals of Engineering Thermodynamics, Moran et al; Introduction to Chemical Engineering Thermodynamics, Smith and Van Ness). In the IC version of the Otto cycle, there is no QH added in the first step, and energy is released to the gas by chemical reaction. In the standard air version of the Otto cycle, heat QH is added to an inert gas (air) to simulate the energy released by the chemical reaction.

The first law analysis of the IC version of the Otto cycle goes like this: In the first step, the system is adiabatic and isochoric, such that $$Q = W = \Delta U = 0$$In addition, for the ideal gas mixture that experiences the combustion reaction, the change in internal energy per unit mass is related to the internal energy change of reaction (per unit mass) by $$\Delta U=\Delta U_R+C_v\Delta T$$where $\Delta U_R$ is the internal energy change per unit mass in going from combustion reactants to combustion products at constant temperature. If we combine these two equations, we obtain: $$\Delta T=\frac{(-\Delta U_R)}{C_v}$$which, for an exothermic reaction is positive.

The first law analysis of the standard air version of the Otto cycle goes like this: In the first step, the system is isochoric, but receives heat $Q=Q_H$ from the surroundings, such that $$\Delta U=C_v\Delta T=Q_H$$So in this version of the process, neither the heat added nor the internal energy change is zero. From this equation, it follows that $$\Delta T=\frac{Q_H}{C_v}$$

Comparing these two versions of the Otto cycle, we see that the standard air version can be made to simulate the actual IC version of the cycle if we specifically choose $Q_H$ for the standard air version such that $$Q_H=-\Delta U_R$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.