Prepared by:

Pradeep Kumar Gupta

Assistant Professor

Department of Mechanical Engineering

Thermal reservoir

Thermal reservoir is that part of the system which has a large heatcapacity i.e., it is a body which is capable of absorbing or rejectingany amount of heat without affecting its temperature.

Heat source: The reservoir which is at high temperature andsupplies heat is known as heat source or source. Such as furnace,combustion chamber etc.

Heat sink: The reservoir which is at low temperature and receivesheat is known as heat sink or sink. Such as atmosphere, sea, riveretc.

Second law of thermodynamics

Second law of thermodynamics may be stated in various ways given as under:

(a) Clausius statement

(b) Kelvin Plank statement

(a) Clausius Statement: According to Clausius statement, “it is impossible for self acting device (or machine), while operating in a cyclic process, to transfer heat from a reservoir at a lower temperature to a reservoir at a higher temperature without any external work being done on it.

Or

In other words, “heat cannot naturally flow from a colder body to a hotter body.”

(b) Kelvin- Planck statement: According to Kelvin- Planck statement, “it is impossible to construct a device (or heat engine) working on a cyclic process, whose only aim is to convert heat energy from a single thermal reservoir into an equivalent amount of work.”

Heat engine

A heat engine is a system that converts heat (thermal energy) orchemical energy of working substance (any type of fuel) tomechanical energy, which can then be used to do mechanical work.

Some facts about heat engines:

1. Heat engine receives heat form a higher temperature body(source: such as nuclear reactor, solar energy etc.)

2. Heat engine converts part of this heat into useful work.

3. Heat engine rejects the remaining waste heat to the lowtemperature body (sink: such as rivers, atmosphere etc.).

4. Heat engine operate in a cycle.

Efficiency of heat engine

Efficiency of a heat engine may be defined as the ratio of net work done by theengine to the heat supplied (or in simple words it is defined as the ratio ofoutput to input). If as per the fig. WHE is the work done by the engine and Q1 isthe heat supplied.

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒

𝐻𝑒𝑎𝑡 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑Since,𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝐻𝑒𝑎𝑡 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑 − ℎ𝑒𝑎𝑡 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑

𝑊𝐻𝐸= 𝑄1 − 𝑄2

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝐻𝑒𝑎𝑡 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑 − ℎ𝑒𝑎𝑡 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑

𝐻𝑒𝑎𝑡 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑

𝜂 =𝑄1 − 𝑄2𝑄1

𝜂 = 1 −𝑄2𝑄1

Coefficient of performance: It is defined as the ratio of heatextracted from the cold body to the amount of work done on therefrigerant.

𝐶. 𝑂. 𝑃. =𝑄

𝑊Where,

Q= amount of heat extracted from the cold body

W= amount of work done on the refrigerant

Refrigerator: Refrigerator is a device which is used to maintain the temperaturelower to its surroundings.

Or

Refrigerator is a reversed heat engine or a heat pump which extract heat from a coldbody and delivers it to a hot body.

Refrigerator is a reversed heat engine which either cool or maintain the temperatureof the body (T1) lower than the atmospheric temperature (Ta). This is obtained byextracting heat (Q1) from a cold body and delivering it to a hot body (Q2). In doing so,work Wr is required to be done on the system. According to the first law ofthermodynamics,

𝑊𝑟 = 𝑄2 − 𝑄1

The performance of a refrigerator is defined by the ratio of heat extracted from thecold body (Q1) to the amount of work done on the system (Wr). It is known ascoefficient of performance. Mathematically it is written as,

(𝐶. 𝑂. 𝑃. )𝑟=𝑄1𝑊𝑟=

𝑄1𝑄2 − 𝑄1

Heat Pump: Heat pump is also a reversed heat engine which extracts heat (Q1) from a cold body and delivers it to a hot body.

According to the first law of thermodynamics,𝑊𝑝 = 𝑄2 − 𝑄1

The performance of a pump is defined by the ratio of heat delivered to the hot body (Q2) to the amount of work done on the system (Wp). It is known as coefficient of performance. Mathematically it is written as,

(𝐶. 𝑂. 𝑃. )𝑝 𝑜𝑟 𝐸𝑃𝑅 =𝑄2

𝑊𝑝=

𝑄2

𝑄2−𝑄1

Where EPR is known as Engine performance ratio.

(𝐶. 𝑂. 𝑃. )𝑝 𝑜𝑟 𝐸𝑃𝑅 =𝑄1

𝑄2 − 𝑄1+ 1

(𝐶. 𝑂. 𝑃. )𝑝= (𝐶. 𝑂. 𝑃. )𝑟+1

Carnot cycleThis cycle was given by Nicolas Leonard Sadi Carnot, a French engineer.

Let us consider some elements for making analysis of Carnot’s cycle:

i. A working substance is assumed to be a perfect gas.

ii. Two heat reservoirs at different temperatures one is at high temperature and other is atlower temperature.

iii. Cylinder walls are perfectly insulated.

(a) P-v diagram (b) T-s diagram

Isothermal expansion (Process 1-2): The working substance (air) is expanded isothermally (i.e. atconstant temperature T1= T2) as shown by the curve 1-2 on P-v and T-s diagrams. During the processthe pressure decreases from P1 to P2 and volume increases from v1 to v2. Heat supplied on the airduring the isothermal expansion process is given by

Q1-2= Area 1-2 -s2-s1

Q1-2= T1(s2-s1)

Adiabatic expansion (Process 2-3): During this process, the air is allowed to expand adiabatically.The reversible adiabatic expansion process is represented by the curve 2-3 on P-v and T-s diagram.The temperature of working substance falls from T2 to T3 at constant entropy. During adiabaticprocess no heat is absorbed or rejected by the air.

Isothermal compression (Process 3-4): The air is compressed isothermally (i.e. at constanttemperature T3= T4) as shown by the curve 3-4 on P-v and T-s diagrams. During the process thepressure increases from P3 to P4 and volume decreases from v3 to v4. Heat rejected by the air duringthe isothermal expansion process is given by

Q3-4= Area 4-3-s2 –s1

Q3-4= T3(s2-s1)

Adiabatic compression (Process 4-1): During this process, the working substance is allowed tocompress adiabatically. The reversible adiabatic compression process is represented by the curve 4-1on P-v and T-s diagram. The temperature of working substance increases from T4 to T1 at constantentropy. During adiabatic process no heat is absorbed or rejected by the working substance.

The net work done= Heat supplied – Heat rejected

Work done= Q1-2 – Q3-4

Work done= T1(s2-s1) - T3(s2-s1)= (T1-T3) (s2-s1)

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒

𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑

𝜂 =𝑇1 − 𝑇3 𝑠2 − 𝑠1𝑇1 𝑠2 − 𝑠1

𝜂 =𝑇1 − 𝑇3𝑇1

= 1 −𝑇3𝑇1

Carnot’s Theorem

It states that all heat engines operating between a given constanttemperature source and a given constant temperature sink, nonehas a higher efficiency than a reversible engine.

Let two engines E1 and E2 operate between the given sourceat temperature T1 and the given sink at temperature T2 as shown infigure.

Let E1 be the heat engine and E2 be a reversible heat engine. It has toprove that the efficiency of E2 is more than that of E1. Let us assumethat it is not true and 𝜂1 > 𝜂2. Let the rates of working of theengines be such that

𝑄1′ = 𝑄1

′′ = 𝑄1Since 𝜂1 > 𝜂2

𝑊1𝑄1>𝑊2𝑄2

So that,𝑊1 > 𝑊2

Now, let E2 be reversed. Since E2 is assumed as a reversible heat engine, the magnitudes of heat and workquantities will remain the same, their directions will be reversed as shown in figure.

Since𝑊1 > 𝑊2 , some part of W1 may be fed to drive the reversed heat engine ∃2.

Since 𝑄1′ = 𝑄1′′ = 𝑄1, the heat discharged by ∃2 may be supplied to E1. The source may, therefore, beeliminated. The net result E1 and ∃2 together constitute a heat engine which, operating in a cycle,produces net work W1-W2 while exchanging with a single reservoir at T2. This violates the statement ofsecond law(Kelvin- Planck). Thus the assumption that η_1>η_2 is wrong.

Therefore,

𝜂2 ≥ 𝜂1

Corollary of Carnot’s theorem

The efficiency of all reversible heat engines operating between the sametemperatures is equal.

Let, both the heat engines E1 and E2 be reversible and 𝜂1 > 𝜂2. If E2 isreversed i.e. it works as a heat pump using some part of work output ofengine E1. We obtained that combined system of heat engine E1 and heatpump E2, becomes a perpetual motion machine (PMM-II). So, 𝜂1 cannotbe greater than 𝜂2. Similarly if we assume that 𝜂2 > 𝜂1and reverse theheat engine E1, we observe that 𝜂2 cannot be greater than 𝜂1. Therefore

𝜂1 = 𝜂2

Since, efficiencies of all reversible heat engines operating between thethermal reservoirs are same. The efficiency of reversible heat engine isindependent pf the nature or amount of the working substance takingpart in the cycle.

Equivalence of Kelvin- Planck statement and Clausius statement

Though Kelvin- Planck statement and Clausius statement of second law ofthermodynamics appear two different interpretations of the same basic fact,but both these statements are equivalent in all aspects. For establishingequivalence of the two statements, it has to be proved that violation of Kelvin-Planck statement implies the violation of Clausius statement and vice-versa.This is explained as under:

Let us consider a system as shown by Fig. 4.4 (a). In this system, a heat engine of 100% thermalefficiency (i.e. PMM-2) is violating the Kelvin –Planck statement as it converts the heat energy(Q1) from a single high temperature body at T1, into an equivalent amount of work (W= Q1).This work output can be used to drive the heat pump which receives an amount of heat Q2 froma low temperature body at T2 and rejects heat Q1+Q2 to a high temperature body at T1. If thecombination of heat engine and a heat pump is taken as a single system (Fig.4.4(a)), then theresult is a device that operates in a cycle and has no effect other than the transfer of heat Q2from a low temperature body to a high temperature body, thus it is a violation of Clausiusstatement. Hence it proves that violation of Kelvin-Planck statement leads to violation ofClausius statement.

Now let us consider another system as shown by fig. 4.4 (b). In this system, a heat pump (i.e.PMM-2) is violating the Clausius statement as it transfers heat from a low temperature body atT2 to a high temperature body at T1without any disbursement of work. Let a heat engineoperating between the same heat bodies that receives an amount of heat Q1 from a hightemperature body at temperature T1 and does work (WHE= Q1-Q2) and rejects an amount of heatQ2 to a low temperature body at temperature T2. If the combination of heat pump and heatengine is taken as a single system (Fig.4.4(b)), then the result is a device that operates in a cyclewhose entire effect is to remove heat at the rate of (Q1-Q2) and convert it completely into anequivalent amount of work, thus it is violation of Kelvin- Planck statement. Hence it proves thatviolation of Clausius statement leads a violation of Kelvin-Planck statement.

Thus from the above we can say that both the statements of second law of thermodynamicsare complimentary to each other.

Perpetual motion machine of the first kind (PMM-1)

A device or a machine which violates the first law of thermodynamics (i.e.the energy can neither be created nor be destroyed it can only betransformed from one form to another) is termed as perpetual motionmachine of the first kind (PMM-1).

Perpetual motion machine of second kind (PMM-2)

A heat engine that violates the second law of thermodynamics (i.e. a heatengine which converts entire of the heat energy in mechanical work) istermed as perpetual motion machine of second kind (PMM-2).

Concept of reversibility

A process is reversible if the system passes through a continuous series of equilibrium states.

Process 1-2 followed by 1-a-2 and process 2-1 followed by 2-a-1, is areversible process.

Process 1-2 followed by 1-a-2 and process 2-1 followed by 2-b-1, is anirreversible process.

Clausius inequality

It was first given by R.J.E. Clausius (1822-1888), a German Physicistand is expressed as,

𝛿𝑄

𝑇≤ 0

According to above equation, “the cyclic integral of δQ/T is alwaysless than or equal to zero. This inequality is holds good for allreversible and irreversible cycles.”

Entropy

Entropy may be defined as, “a thermodynamic quantity representingthe unavailability of a system's thermal energy for conversion intomechanical work, often defined as the degree of randomness ordisorder in the system.”

Or

“The cyclic integral of the quantity𝛿𝑄

𝑇for a reversible cycle being

equal to zero indicate that𝛿𝑄

𝑇is a point function and is therefore

property of a system. This property is termed as entropy. It isexpressed by ‘S’. Mathematically it is written as:

12𝑑𝑆 = 1

2 𝛿𝑄

𝑇in kJ/K

𝑑𝑆 = (𝑆2 − 𝑆1) = 12 𝛿𝑄

𝑇in kJ/K