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Carnot Cycle

Last updated date: 27th May 2024
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Reverse Carnot Cycle

To understand the reverse Carnot cycle, first, let us read the Carnot cycle. In 1824, French Physicist Sadi Carnot developed the Carnot cycle. It also highlights the concept of reversibility.

Carnot Cycle Theorem and Proof

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Consider two engines R and I operating between temperature T and T2 where T1 > T2 (Fig.). Tand T2 are acting as a heat source and heat reservoir, respectively. Also, consider engine I is more efficient than engine R. Engine R absorbs Q1 amount of heat from the source at T1 temperature and rejects Q2 amount of heat in the sink at temperature T2

In contrast, Engine I absorb Q1’ amount of heat from source at temperature T1  and rejects Q2’ amount of heat in the sink at temperature T2

Of these two engines, R is reversible, and I is irreversible.

Also, assume the two engines do the same quantity of work W in each cycle. Assume I is more efficient than R.

\[\frac{Q_{1}^{‘} - Q_{2}^{‘}}{Q_{1}^{‘}}\] > \[\frac{Q_{1} - Q_{2}}{Q_{1}}\] or \[\frac{W}{Q_{1}^{‘}}\] > \[\frac{W}{Q_{1}}\] or Q> Q1

And Q1’ – Q2’ = Q1- Q2 or Q2 - Q2’ = Q1 - Q1

Since Q> Q1 thus Q2 > Q2

Additionally, let's read the reverse Carnot cycle where the two engines grouped so that I drive R in the opposite direction. This whole system becomes a self-operating machine where I supply external work W and R absorbs this amount of work in its reverse cycle. Also, I in its cycle takes in heat Q1’ from the source and releases up heat Q2’ to the sink. R in its reverse cycle inputs heat Q2 from the sink and releases up heat Q1 to the source.

The net output of the complete cycle of the grouped engines is given by;

Increase of heat by the source at T1 = Q1 – Q1

Decrease of heat by the sink at T2 = Q2 – Q2

External work done on the system = 0

Thus, the coupled engines forming a self-operating machine unassisted by any external factor transfer heat continuously from between the two bodies maintained at different temperatures.

This deduction is conflicting with the second law of thermodynamics. According to the second law of thermodynamics, in any cyclic process of a self-operating machine, heat cannot be transferred from one body to another at a higher temperature. Hence the assumption is invalid. Thus, no engine can be more efficient than a perfectly reversible engine working between the same temperatures.

Also, consider two reversible engines R1 and R2 , where R2 is more efficient than R1. Continuing in the same fashion that R2 cannot be more efficient than R1. Hence all reversible engines operating between the same two temperatures have identical efficiency.

Thus, the efficiency of a perfectly reversible engine is independent of the working substance but varies on the magnitude of temperature present between the engines.

Phases of the Carnot Cycle

The working substance is in a quasi-static state in the Carnot cycle.

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Isothermal Expansion

The cylinder positioned on the source. The gas obtains the temperature of the source followed by the quasi-static expansion. When the gas expands, temperature decreases. Heat is allowed to pass in the cylinder via perfect conducting substance that touches the source. Hence phase one is slow isothermal expansion at a constant temperature.

Thus, the working substance expands from initial state A (P1, V1, T1) to state B (P2, V2, T1) at constant temperature T1 along with AB. W1 is the work done when Q1 is the heat absorbed at temperature T1

Q1 = W1 = \[\int_{V_{1}}^{V_{2}}\] PdV = RT1log\[_{e}\]\[\frac{V_{2}}{V_{1}}\] = area ABGEA

Adiabatic Expansion

The cylinder is placed on a non-conducting stand. Due to this, gas undergoes slow adiabatic expansion from the state B (P2, V2, T1) to the state C (P3, V3, T2), represented as BC, release its internal energy to perform work resulting in a decrease in temperature to T2 equivalent to sink. The amount of external work done W2 during adiabatic process 

W2 = \[\int_{V_{2}}^{V_{3}}\] P.dV = K\[\int_{V_{2}}^{V_{3}}\] \[\frac{dV}{V^{\gamma}}\]

(PV\[^{\gamma}\] = constant = K) = \[\frac{KV_{3}^{1-\gamma} - P_{2}V{2}}{1-\gamma}\] = \[\frac{P_{3}V_{3} - P_{2}V_{2}}{1 - \lambda }\](P\[_{2}\]V\[_{2}\]\[^{\gamma }\] = P\[_{3}\]V\[_{3}\]\[^{\gamma }\] = K) = \[\frac{RT_{2}-RT_{1}}{1-\gamma}\] = \[\frac{R(T_{1}-T_{2})}{\gamma - 1}\] = Area BCHGB

Isothermal Compression

The cylinder is kept on the sink at temperature T2. The piston is now gradually moved inside so that the work is done on the gas. Due to heat generation by compression temperature increase. The conducting substance touching the sink passes the heat to the sink, therefore maintaining its temperature T2.

In short, gas undergoes isothermal compression CD and its state changes from (P3, V3, T2) to the state  (P4, V4, T2)  while rejecting the heat Qto the sink. The work done on substance Wis 

Q2 = W3 = \[\int_{V_{3}}^{V_{4}}\] PDV = RT2 log\[_{e}\]\[\frac{V_{4}}{V_{3}}\]

= -RT2log\[_{e}\]\[\frac{V_{3}}{V_{4}}\] = areaCHFDC  

The cylinder is placed back in an insulating stand and the piston is gradually moved inside till the gas regains its original state A (P1, V1, T1). The work W4 performed from D (P4, V4, T2) to the final state A (P1, V1, T1) is 

W4 = \[\int_{V_{4}}^{V_{1}}\]P.dv = \[\frac{R(T_{1}-T_{2})}{\gamma - 1}\] = AreaDFEAD

FAQs on Carnot Cycle

1. What is the Carnot Theorem?

The two crucial outputs were derived from the second law of thermodynamics and hence inference were grouped together to form the basis of Carnot’s theorem. Therefore Carnot theorem is stated below

(a) Heat engines at different temperatures are less efficient than Carnot heat engines operating between reservoirs at the same temperature.

(b)  As a consequence of the Carnot theorem, it is also stated that Carnot engines between the pair of heat reservoirs are equally efficient and are independent of the working substance. 

Carnot cycle is based on the carnot theorem divided in four phases i.e.  isothermal and adiabatic expansion and compression

2. Refrigerators Act as a Reverse Heat Engine. Comment

True, in any heat engine heat flows from high temperature to low temperature. In contrast, in the refrigerator the heat flows from lower temperature to higher temperature in which a certain amount of work is done. Refrigerators  release the energy in the outer surrounding atmosphere from inside the fridge with the help of the electric motor. In simple terms, a motor is responsible for transferring the energy from inside of the fridge to the outside. Air conditioners also work on the same principle of reverse heat engine. Both refrigerators and air conditioners are assessed by a physical parameter called as coefficient of performance.