Question regarding Heat transfer in Carnot Engine

Question

A Carnot Engine is a theoretical engine unlike a Sterling Engine which can be made practically.

Some of the drawbacks of Carnot Engine are,

1)The Heat Transfer occurs only during isothermal process(compression and expansion),this is because the working material (ie) gas or fuel used, if it's at a different temperature(technically the temperature of the working material is infinitesimally smaller)than the hot reservoir ,then some amount of heat get's used up to attain thermal equilibrium (ie) it's irreversibly lost .

Now my question is ,if $Q_H$ is the heat supplied by the $Heat$ $Reservoir$ ,then the heat is used to do some work of say $dW=pdV$ and if entire heat $Q_H$ gets used as Work,then how does it transfer $Q_c$ amount of heat to the cold reservoir?

And it seems so odd to say that once the Isothermal processes are over ,it expands /contracts adiabatically ,are we physically removing the sources here ? Is the only use of Adiabatic process in "Imaginary Carnot Cycle" to produce a proper cycle?

Answer 1

First, it is not quite right to say that some other kind of heat engine is practical while the Carnot engine is not. Rather, many kinds of heat engine can be made to work in actual practice, and they will follow a cycle which one can try to model theoretically. The full details of a real system can never be captured completely in a simple model, but it is useful to introduce simple or idealized cycles and study them theoretically. There is such an idealization for the Stirling engine, for example.

However it is true that the Carnot engine is used in the study of thermodynamics more for its value as a theoretical tool than as an engine one might build. One can build such an engine, understanding that the physical engine will never match the simple idealized one exactly.

Now let's unpack your questions about the idealized Carnot engine. When applied to a simple mechanical system such as a gas, the Carnot cycle consists of isothermal heat input, then adiabatic expansion, then isothermal heat output, then adiabatic contraction.

First, the amount of work done during a cycle is given by $$ W = Q_h - Q_c $$ where $Q_h$ is the heat coming in at the heat input stage and $Q_c$ is the heat going out at the heat output stage. Your question proposes that $Q_c$ might be zero. This is ruled out by the Second Law of thermodynamics. The Kelvin statement, for example, rules it out directly. But by studying the entropy we can take this a bit further. If the whole process is reversible then the entropy coming in must equal the entropy going out, so we find: $$ \frac{Q_h}{T_h} = \frac{Q_c}{T_c} $$ where $T_h$ and $T_c$ are the temperatures at the hot and cold isothermal stage, respectively. This is a quantitative statement which makes it clear that $Q_c \ne 0$. Now you might want to ask if we can have $T_c = 0$ (or very close to $0$). This is ruled out by a more subtle point: as you approach absolute zero the heat capacities goes to zero so it is no longer possible to transfer heat without changing the temperatures of the reservoirs. But anyway it is not possible to have a system with $T_c$ exactly zero.

Finally, you ask about the adiabatic stages and is this physically realistic (what makes the heat transfer stop?). This can be made as physically realistic as the other parts of the cycle. One might, for example, introduce a mechanism which changes the thermal conductivity of the region between the working fluid and the heat source or heat drain. There are some clever ways of doing this using superconductivity, but in any case it is physically possible. Another way is simply to move the working fluid from one part of a physical machine to another, without contracting or expanding. You could have three regions: one in contact with a hot reservoir, one with good thermal isolation, and one in contact with a cold reservoir. Expansion and contraction can happen in each region. The fluid is moved between regions as required.

Answer 2

Is the only use of Adiabatic process in "Imaginary Carnot Cycle" to produce a proper cycle?

Yes; adiabatic processes are required in the Carnot cycle to ensure reversibility.

In contrast, if one finished the isothermal expansion (contraction) at the hot (cold) reservoir temperature and then transferred the system to contact the cold (hot) reservoir, then a thermal gradient would arise at the interface. Energy flowing down a gradient generates entropy, which is unacceptable in this context.

Therefore, we need to add some additional expansion (contraction), which occurs adiabatically, to bring the system to the temperature of the new reservoir. In this way, no thermal gradient ever arises.

Answer 3

Now my question is ,if $Q_H$ is the heat supplied by the $Heat$ $Reservoir$ ,then the heat is used to do some work of say $dW=pdV$ and if entire heat $Q_H$ gets used as Work, then how does it transfer $Q_c$ amount of heat to the cold reservoir?

I have the impression you think the heat $Q_C$ has to come from the heat $Q_H$. But that's not the case. The heat $Q_C$ comes from energy transferred to the system from the surroundings in the form of compression work done on the system by the surroundings. The compression work causes heat transfer by raising the system temperature infinitesimally greater than than the temperature of the cold reservoir.

Some of the drawbacks of Carnot Engine are 1)The Heat Transfer occurs only during isothermal process(compression and expansion)

Not only is it not a drawback, it is an advantage that all the heat transfers occur during the isothermal processes. The efficiency $\eta$ of the Carnot cycle, or any heat engine cycle for that matter, is the net work done divided by the gross heat added, or

$$\eta=\frac{W_{NET}}{Q_H}=\frac{Q_{H}-Q_{C}}{Q_H}=1-\frac{Q_C}{Q_H}$$

Figures 1 and 2 below are T-S diagrams for the Carnot heat engine cycle and an arbitrary reversible cycle both operating in the same temperature range.

In FIG 1 all the heat received in the Carnot cycle is from a single hot reservoir $T_H$ while all the heat rejected is to a single cold reservoir temperature $T_C$ of the reversible isothermal processes. The area enclosed in the cycle is the net work done.

In the arbitrary cycle of FIG 2, operating in the same range of temperatures as the Carnot cycle, but with multiple reservoirs instead of two, less total heat $Q_H$ is received and more heat $Q_C$ rejected, for a lower thermal efficiency.

...if it's at a different temperature(technically the temperature of the working material is infinitesimally smaller)than the hot reservoir ,then some amount of heat get's used up to attain thermal equilibrium (ie) it's irreversibly lost

But there is no finite temperature difference between the system and hot reservoir in the Carnot cycle, the difference being infinitesimal as you parenthetically acknowledged. That's what makes the cycle theoretically reversible instead of irreversible. That's the advantage of the Carnot (and any reversible) cycle.

Even if there were a finite temperature difference making the cycle irreversible, no heat from the hot reservoir would be irreversibly "lost" from the hot reservoir. What would be "lost" is the opportunity to do maximum net work due to the entropy generated by heat transfer over the finite temperature difference. To complete the cycle, that generated entropy would have to be transferred to the cold reservoir in the form of more heat $Q_C$ than for the reversible process. Since the net work equals $Q_{H}-Q_C$ less net work is done.

And it seems so odd to say that once the Isothermal processes are over ,it expands /contracts adiabatically ,are we physically removing the sources here ?

The sources are not physically removed following the isothermal processes. The system is thermally insulated from the sources following the isothermal processes.

Is the only use of Adiabatic process in "Imaginary Carnot Cycle" to produce a proper cycle?

They don't just complete the cycle, but complete it in such way to maximize thermal efficiency. Linking them by non isentropic (non constant S) processes lowers the efficiency, as shown for the arbitrary cycle of FIG 2.

Hope this helps.