unit - ii theory

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ENGINEERING THERMODYNAMICS

UNIT - II

4/24/2011

P.MOHANAVELU,SENIOR LECTURER,PANIMALAR ENGINEERINGCOLLEGE



UNIT - IISyllabus

Second law of thermodynamics – Kelvin’s and Clausius statements of second law. Reversibility andirreversibility. Carnot theorem, Carnot cycle, reversed carnot cycle, efficiency, COP. Thermodynamictemperature scale, Clausius inequality, concept of entropy, entropy of ideal gas, principle of increase of entropy

– availability.

2.1 LIMITATIONS OF FIRST LAW OF THERMODYNAMICS AND INTRODUCTION TO SECONDLAW OF THERMODYNAMICS

Energy can flow from a system in the form of heat or work . The first law of thermodynamics sets nolimit to the amount of the total energy of a system which can be caused to flow out as work. A limit is imposed,however, as a result of the principle enunciated in the second law of thermodynamics which states that heat willflow naturally from one energy reservoir to another at a lower temperature, but not in opposite direction withoutassistance. This is very important because a heat engine operates between two energy reservoirs at differenttemperatures.

Further the first law of thermodynamics establishes equivalence between the quantity of heat used and the mechanical work but does not specify the conditions under which conversion of heat into work is possible,neither the direction in which heat transfer can take place . This gap has been bridged by the second law of thermodynamics.2.2 HEAT RESERVOIR

Heat reservoir is the system having very large heat capacity i.e. it is a body capable of absorbing or rejecting finite amount of energy without any appreciable change in its’ temperature . Thus in general it may

be considered as a system in which any amount of energy may be dumped or extracted out and there shall be nochange in its temperature.

Heat reservoirs can be of two types depending upon nature of heat interaction i.e. heat rejection or heatabsorption from it. Heat reservoir which rejects heat from it is called source. While the heat reservoir which

absorbs heat is called sink. Sometimes these heat reservoirs may also be called Thermal Energy Reservoirs2.3 HEAT ENGINE

Heat engine may be precisely defined as “a device operating in cycle between high temperature source and low temperature sink and producing work” . Heat engine receives heat from source, transformssome portion of heat into work and rejects balance heat to sink. All the processes occurring in heat engineconstitute cycle.



2.4 HEAT PUMP AND REFRIGERATION

Heat pump refers to a device used for extracting heat from a low temperature surroundings and sending it to high temperature body, while operating in a cycle. In other words heat pump maintains a body or system at temperature higher than temperature of surroundings, while operating in cycle.

Refrigerator is a device similar to heat pump but with reverse objective. It maintains a body at temperature lower than that of surroundings while operating in a cycle.

η th and (C.O.P. )ref is always less than unity and ( C.O.P. )heat pump is always greater than unity2.2 SECOND LAW OF THERMODYNAMICS

The second law of thermodynamics has been enunciated meticulously by Clausius, Kelvin and Planck inslightly different words although both statements are basically identical. Each statement is based on anirreversible process . The first considers transformation of heat between two thermal reservoirs while thesecond considers the transformation of heat into work .2.2.1 Clausius Statement



“ It is impossible for a self acting machine working in a cyclic process unaided by any external agency, to convey heat from a body at a lower temperature to a body at a higher temperature ”. In otherwords, heat of, itself, cannot flow from a colder to a hotter body.2.2.2 Kelvin-Planck Statement

“ It is impossible to construct an engine, which while operating in a cycle produces no other effect except to extract heat from a single reservoir and do equivalent amount of work ” .

2.2.3 Equivalence of Clausius Statement to the Kelvin-Planck StatementAlthough the Clausius and Kelvin-Planck statements appear to be different, they are really equivalent in

the sense that a violation of either statement implies violation of oth er . Consider a higher temperature reservoir T 1 and low temperature reservoir T 2. Figure below shows a heat pump which requires no work and transfers anamount of Q2 from a low temperature to a higher temperature reservoir (in violation of the Clausius statement).Let an amount of heat Q1 (greater than Q2) be transferred from high temperature reservoir to heat engine whichdevelops a net work, W = Q 1 – Q2 and rejects Q2 to the low temperature reservoir. Since there is no heatinteraction with the low temperature, it can be eliminated. The combined system of the heat engine and heat

pump acts then like a heat engine exchanging heat with a single reservoir, which is the violation of the Kelvin-Planck statement.

2.2.4 Corollary of second law of thermodynamicsWithout violating the first law, a machine can be imagined which would continuously absorb heat

from a single thermal reservoir and would convert this heat completely into work. The efficiency of such amachine would be 100 per cent. This machine is called the perpetual motion machine of the second kind (PMM2).2.3 CARNOT’S THEOREM

“It states that of all engines operating between a given constant temperature source and a givenconstant temperature sink, none has a higher efficiency than a reversible engine”.



(a) cyclic heat engines HE A and HE B operating between the same source and sink, of which HE B is

reversible HE A and HE B are the two engines operating between the given source at temperature T 1 and the given

sink at temperature T 2.Let HE A be any heat engine and HE B be any reversible heat engine. We have to prove that efficiency of

HEB is more than that of HEA. Let us assume that η A >η B. Let the rates of working of the engines be such thatQ1 A = Q1 B = Q1

Since η A > η B

W A > W B Now, let HE B be reversed. Since HE B is a reversible heat engine, the magnitudes of heat and work

transfer quantities will remain the same, but their directions will be reversed, as shown in Fig. (b). Since W A >W B, some part of W A (equal to W B) may be fed to drive the reversed heat engine E H B. Since Q1 A = Q1 B = Q1, theheat discharged by E H B may be supplied to HE A. The source may, therefore, be eliminated Fig.(c). The netresult is that HE A and E H B together constitute a heat engine which, operating in a cycle produces net work W A –W B while exchanging heat with a single reservoir at T 2. This violates the Kelvin-Planck statement of the secondlaw. Hence the assumption that η A > η B is wrong.

(b) HE B is reversed.



(c) HE A and HE B together violate the Kelvin-Planck statement.2.3.1 COROLLARY OF CARNOT’S THEOREM

‘ ‘The efficiency of all reversible heat engines operating between the same temperature levels is the same”.Refer Fig. 5.6. Let both the heat engines HE A and HE B be reversible . Let us assume η A > η B. Similar to

the procedure outlined in the Article 2.4, if HE B is reversed to run say, as a heat pump using some part of thework output ( W A) of engine HE A, we see that the combined system of heat pump HE B and engine HE A, becomes

a PMM 2. So η A cannot be greater than η B. Similarly, if we assume η B > η A and reverse the engine HE A, weobserve that η B cannot be greater than η A.

η A = η B.Since the efficiencies of all reversible engines operating between the same heat reservoirs are the same,

the efficiency of a reversible engine is independent of the nature or amount of the working substanceundergoing the cycle .2.3.2 EFFICIENCY OF THE REVERSIBLE HEAT ENGINEThe efficiency of a reversible heat engine in which heat is received solely at T 1 is found to be

From the above expression, it may be noted that as T 2 decreases and T 1 increases, the efficiency of the reversiblecycle increases.Since η is always less than unity, T 2 is always greater than zero and + ve.The C.O.P. of a refrigerator is given by

For a reversible refrigerator , using



Similarly, for a reversible heat pump

2.4 CARNOT CYCLEThe cycle was first suggested by a French engineer Sadi Carnot in 1824 which works on reversible cycle

and is known as Carnot cycle .

Any fluid may be used to operate the Carnot cycle which is performed in an engine cylinder the head of which is supposed alternatively to be perfect conductor or a perfect insulator of a heat. Heat is caused to flowinto the cylinder by the application of high temperature energy source to the cylinder head during expansion,and to flow from the cylinder by the application of a lower temperature energy source to the head duringcompression.

The assumptions made for describing the working of the Carnot engine are as follows:

(i) The piston moving in a cylinder does not develop any friction during motion.(ii) The walls of piston and cylinder are considered as perfect insulators of heat.

(iii) The cylinder head is arranged that it can be a perfect heat conductor or perfect heat insulator.(iv) The transfer of heat does not affect the temperature of source or sink.(v) Working medium is a perfect gas and has constant specific heat.(vi) Compression and expansion are reversible.

Following are the four stages of Carnot cycle :Stage 1. (Process 1-2). Hot energy source is applied. Heat added Q1 is taken in whilst the fluid

expands isothermally and reversibly at constant high temperature T 1.For the perfect gas as working fluid in isothermal process no change in internal energy occurs,Therefore U 2 = U 1And Q1 = W 1–2

Stage 2. (Process 2-3). The cylinder becomes a perfect insulator so that no heat flow takes place.The fluid expands adiabatically and reversibly whilst temperature falls from T 1 to T 2.

During adiabatic expansion say work W expn is available,Q2–3 = 0

From first law of thermodynamics;0 = ( U 3 – U 2) + W expn

Or W expn = (U 2 – U 3)



Stage 3. (Process 3-4). Cold energy source is applied. Heat rejected Q2 flows from the fluid whilst itis compressed isothermally and reversibly at constant lower temperature T 2.

From first law of thermodynamics applied on process 3–4, – Q2 = (U 4 – U 3) + (– W 3–4)

For perfect gas internal energy shall remain constant during isothermal process. Thus, U 3 = U 4 – Q2 = – W 3–4

Or Q2 = W 3–4

Stage 4. (Process 4-1). Cylinder head becomes a perfect insulator so that no heat flow occurs. Thecompression is continued adiabatically and reversibly during which temperature is raised from T 2 to T 1.

From first law applied on process 4–1For adiabatic process;

Q4–1 = 00 = ( U 1 – U 4) + (– W compr )W compr = (U 1 −U 4)

Efficiency of reversible heat engine can be given as;

Work done = W expn – W compr

And heat is supplied only during process 1–2, thereforeHeat supplied by the source = Q1=Q add

Substituting in the expression for efficiency

Also for a cycle

so Work done = Q 1 – Q 2

The Carnot cycle cannot be performed in practice because of the following reasons :(i) Frequent change of cylinder head i.e. of insulating type and diathermic type for adiabatic andisothermal processes is very difficult.(ii) Isothermal heat addition and isothermal-heat rejection are practically very difficult to be realized(iii) Reversible adiabatic expansion and compression are not possible.(iv) Even if near reversible isothermal heat addition and rejection is to be achieved then time durationfor heat interaction should be very large i.e. infinitesimal heat interaction occurring at dead slow speed.

Near reversible adiabatic processes can be achieved by making them to occur fast. In a piston-cylinder reciprocating engine arrangement such speed fluctuation in a single cycle is not possible.

2.5 THERMODYNAMIC TEMPERATURETake the case of reversible heat engine operating between two reservoirs. Its thermal efficiency is given

by the eqn.below,



The temperature of a reservoir remains uniform and fixed irrespective of heat transfer.This means that reservoir has only one property defining its state and the heat transfer from a reservoir is

some function of that property, temperature . Thus Q = φ ( K ), where K is the temperature of reservoir. Thechoice of the function is universally accepted to be such that the relation,

where T 1 and T 2 are the thermodynamic temperatures of the reservoirs.Zero thermodynamic temperature (that temperature to which T 2 tends, as the heat transfer Q2 tends to

zero) has never been attained and one form of third law of thermodynamics is the statement :‘‘The temperature of a system cannot be reduced to zero in a finite number of processes.” After establishing the concept of a zero thermodynamic temperature, a reference reservoir is chosen and

assigned a numerical value of temperature. Any other thermodynamic temperature may now be defined in termsof reference value and the heat transfers that would occur with reversible engine,

The determination of thermodynamic temperature cannot be made in this way as it is not possible to build a reversible engine. Temperatures are determined by the application of thermodynamic relations to other measurements.

The SI unit of thermodynamic temperature is the Kelvin ( K ). The relation between thermodynamictemperature and Celsius scale, which is in common use is :

Thermodynamic temperature = Celsius temperature + 273.15°.The Kelvin unit of thermodynamic temperature is the fraction of thermodynamic temperature of

‘Triple point’ of water.2.6. CLAUSIUS INEQUALITYWhen a reversible engine uses more than two reservoirs the third or higher numbered reservoirs will not be

equal in temperature to the original two. Consideration of expression for efficiency of the engine indicates thatfor maximum efficiency, all the heat transfer should take place at maximum or minimum reservoir temperatures. Any intermediate reservoir used will, therefore, lower the efficiency of the heat engine. Practicalengine cycles often involve continuous changes of temperature during heat transfer. A relationship among

processes in which these sort of changes occur is necessary. The ideal approach to a cycle in which temperaturecontinually changes is to consider the system to be in communication with a large number of reservoirs in

procession. Each reservoir is considered to have a temperature differing by a small amount from the previousone. In such a model it is possible to imagine that each reservoir is replaced by a reversible heat engine incommunication with standard reservoirs at same temperature T 0. Fig. 5.4 shows one example to thissubstitution.



The system to which the heat transfer is effected is neither concerned with the source of energy itreceives nor with the method of transfer, save that it must be reversible. Associated with the small heat transfer dQ to the original system is a small work transfer dW and for this system the first law gives

Now consider the engine replacing the reservoirs and apply the second law to the new system in Figabove ( b). If the new system is not a perpetual motion machine of second kind, no positive work transfer is

possible with a single reservoir.

But by the definition of thermodynamic temperature scale

and by combination of above three equation

This is known as Clausius inequality .Let us now consider the case of a reversible engine for which

reverse the engine and for the reversible heat pump obtained it is possible to develop the expression,

The negative sign indicates that the heat transfers have all reversed in direction when the engine wasreversed. This means that for the same machine we have two relations which are only satisfied if in thereversible case,



For a reversible case, as the number of reservoirs used tends to infinity, the limiting value of thesummation will be

In words, the Clausius inequality may be expressed as follows :

“When a system performs a reversible cycle, then but when the cycle is not

reversible

2.7. ENTROPY“Entropy is a function of a quantity of heat which shows the possibility of conversion of that heat into

work. The increase in entropy is small when heat is added at a high temperature and is greater when heat addition is made at a lower temperature. Thus for maximum entropy, there is minimum availability for conversion into work and for minimum entropy there is maximum availability for conversion into work.” 2.7.1. Entropy—a Property of a System

Refer Figure below. Let us consider a system undergoing a reversible process from state 1 to state 2along path L and then from state 2 to the original state 1 along path M . Applying the Clausius theorem to thisreversible cyclic process, we have

(where the subscript R designates a reversible cycle)

Reversible cyclic process between two fixed end statesHence when the system passes through the cycle 1- L-2-M -1, we have

Now consider another reversible cycle in which the system changes from state 1 to state 2 along path L but returns from state 2 to the original state 1 along a different path N . For this reversible cyclic process, wehave



Subtracting equation (5.17) from equation (5.16), we have

As no restriction is imposed on paths L and M , except that they must be reversible, the Quantity is a function of the initial and final states of the system and is independent of the path of the process. Hence it represents a property of the system. This property is known as the “entropy”.2.7.2. Change of Entropy in a Reversible Process

Refer Figure of 2.7.1.Let S 1 = Entropy at the initial state 1, and

S 2 = Entropy at the final state 2.Then, the change in entropy of a system, as it undergoes a change from state 1 to 2, becomes

Lastly, if the two equilibrium states 1 and 2 are infinitesimal near to each other, the integral sign may beomitted and S 2 – S 1 becomes equal to dS .

where dS is an exact differential.Thus, from equation (5.19), we find that the change of entropy in a reversible process is equal toThis is the mathematical formulation of the second law of thermodynamics Equation (5.19) indicates that

when an inexact differential δ Q is divided by an integrating factor T during a reversible process, it becomes anexact differential .

The third law of thermodynamics states “ When a system is at zero absolute temperature,the entropyof system is zero ”.

It is clear from the above law that the absolute value of entropy corresponding to a given state of the

system could be determined by integrating between the state at absolute zero and the given state. Zeroentropy, however, means the absence of all molecular, atomic, electronic and nuclear disorders.

As it is not practicable to get data at zero absolute temperature, the change in entropy is calculated either between two known states or by selecting some convenient point at which the entropy is given an arbitraryvalue of zero. For steam, the reference point at which the entropy is given an arbitrary value of zero is 0°C andfor refrigerants like ammonia, Freon-12, carbon dioxide etc. the reference point is – 40°C, at which the entropyit taken as zero.

Thus, in practice we can determine the change in entropy and not the absolute value of entropy.

2.8 ENTROPY AND IRREVERSIBILITY



We know that change in entropy in a reversible process is equal to .Let us now find the change inentropy in an irreversible process

Entropy change for an irreversible processConsider a closed system undergoing a change from state 1 to state 2 by a reversible process 1- L-2 and

returns from state 2 to the initial state 1 by an irreversible process 2- M -1 as shown in Figure above on thethermodynamic coordinates, pressure and volume.

Since entropy is a thermodynamic property, we can write

(Subscript I represents the irreversible process).

Now for a reversible process, from eqn below, we have

Substituting the value of in eqn above , we get

Again, since the processes 1- L-2 and 2- M -1 together form an irreversible cycle,Applying Clausius equality to this expression, we get

Now subtracting eqn. (5.23) from eqn. (5.22), we get

which for infinitesimal changes in states can be written as



Eqn. (5.24) states that the change in entropy in an irreversible process is greater than Combining eqns.(5.23) and (5.24), we can write the equation in the general form as

where equality sign stands for the reversible process and inequality sign stands for the irreversible process.It may be noted here that the effect of irreversibility is always to increase the entropy of the system.Let us now consider an isolated system. We know that in an isolated system, matter, work or heat cannot

cross the boundary of the system. Hence according to first law of thermodynamics, the internal energy of thesystem will remain constant.

Since for an isolated system, δ Q = 0, from eqn. (5.25), we get

Eqn. (5.26) states that the entropy of an isolated system either increases or remains constant.This is a corollary of the second law. It explains the principle of increase in entropy.

5.14. CHANGE IN ENTROPY OF THE UNIVERSE

We know that the entropy of an isolated system either increase or remains constant,

By including any system and its surrounding within a single boundary, as shown in Fig. 5.23, an isolatedsystem can be formed. The combination of the system and the surroundings within a single boundary issometimes called the Universe. Hence, applying the principle of increase in entropy, we get

(dS )universe _≥ 0where ( dS )universe = (dS ) system + (dS ) surroundings .

In the combined closed system consider that a quantity of heat δ Q is transferred from the system attemperature T to the surroundings at temperature T 0. Applying eqn. (5.24) to this process, we can write

(–ve sign indicates that heat is transferred from the system).Similarly, since an amount of heat δ Q is absorbed by the surroundings, for a reversible process, we can

write



Hence, the total change in entropy for the combined system

The same result can be obtained in the case of an open system.For both closed and open systems, we can write

Eqn. (5.27) states that the process involving the interaction of a system and the surroundings takes placeonly if the net entropy of the combined system increases or in the limit remains constant. Since all natural

processes are irreversible, the entropy is increasing continually.The entropy attains its maximum value when the system reaches a stable equilibrium state from a non-

equilibrium state. This is the state of maximum disorder and is one of maximum thermodynamic probability.TEMPERATURE-ENTROPY DIAGRAMIf entropy is plotted-horizontally and absolute temperature vertically the diagram so obtained is calledtemperature-entropy (T - s) diagram. Such a diagram is shown in Fig. 5.24. If working fluid receives a smallamount of heat dQ in an elementary portion ab of an operation AB when temperature is T , and if dQ is

represented by the shaded area of which T is the mean ordinate, thewidth of the figure must be .This iscalled ‘ increment of entropy ’ and is denoted by dS . The total heat received by the operation will be given by thearea under the curve AB and ( S B – S A) will

be corresponding increase of entropy.

From above we conclude that :

“ Entropy may also be defined as the thermal property of a substance which remains constant when substance is expanded or compressed adiabatically in a cylinder ”.



Note. ‘ s’ stands for specific entropy whereas ‘ S ’ means total entropy ( i.e., S = ms ).CHARACTERISTICS OF ENTROPY

The characteristics of entropy in a summarised form are given below :1. It increases when heat is supplied irrespective of the fact whether temperature changes or not.2. It decrease when heat is removed whether temperature changes or not.3. It remains unchanged in all adiabatic frictionless processes.

4. It increases if temperature of heat is lowered without work being done as in a throttling process.5.17. ENTROPY CHANGES FOR A CLOSED SYSTEM5.17.1. General Case for Change of Entropy of a Gas

Let 1 kg of gas at a pressure p1, volume v1, absolute temperature T 1 and entropy s1, be heated such that itsfinal pressure, volume, absolute temperature and entropy are p2, v2, T 2 and s2 respectively. Then by law of conservation of energy,

dQ = du + dW where, dQ = Small change of heat,

du = Small internal energy, anddW = Small change of work done ( pdv).

Now dQ = cvdT + pdvDividing both sides by T , we get

Integrating both sides, we get

This expression can be reproduced in the following way :According to the gas equation, we have

Substituting the value of in eqn. (5.28), we get



Again, from gas equation

Putting the value of in eqn. (5.28), we get

5.17.2. Heating a Gas at Constant VolumeRefer Fig. 5.25. Let 1 kg of gas be heated atconstant volume and let the change in entropy andabsolute temperature be from s1 to s2 and T 1 to T 2respectively.Differentiating to find small increment of heat dQ

corresponding to small rise in temperature dT .dQ = cvdT

Dividing both sides by T , we get




5.17.3. Heating a Gas at Constant PressureRefer Fig. 5.26. Let 1 kg of gas be heated at constant pressure, so that its absolute temperature

changes from T 1 to T 2 and entropy s1 to s2.

Differentiating to find small increase in heat, dQ of this gas when the temperature rise is dT .

Dividing both sides by T , we get




5.17.4. Isothermal Process

An isothermal expansion 1-2 at constant temperature T is shown in Fig. 5.27.Entropy changes from s1 to s2 when gas absorbs heat during expansion. The heat taken by the gas isgiven by the area under the line 1-2 which also represents the work done during expansion.

In other words, Q = W .

5.17.5. Adiabatic Process (Reversible)

During an adiabatic process as heat is neither supplied nor rejected,



This shows that there is no change in entropy and hence it is known as isentropic process.Fig. 5.28 represents an adiabatic process. It is a vertical line (1-2) and therefore area under this line is nil

; hence heat supplied or rejected and entropy change is zero.

5.17.6. Polytropic ProcessRefer Fig. 5.29.The expression for ‘entropy change’ in polytropic process ( pvn = constant) can be obtained from eqn.

(5.28)



Substituting the value of in eqn. (5.28), we get



AVAILABLE AND UNAVAILABLE ENERGY‘Available energy’ is the maximum portion of energy which could be converted into useful work by

ideal processes which reduce the system to a dead state (a state in equilibrium with the earth and itsatmosphere). Because there can be only one value for maximum work which the system alone could do whiledescending to its dead state, it follows immediately that ‘Available energy’ is a property .

A system which has a pressure difference from that of surroundings, work can be obtained from anexpansion process, and if the system has a different temperature, heat can be transferred to a cycle and work can

be obtained. But when the temperature and pressure becomes equal to that of the earth, transfer of energyceases, and although the system contains internal energy, this energy is unavailable energy.

The theoretical maximum amount of work which can be obtained from a system at any state p 1 and T when operating with a reservoir at the constant pressure and temperature p 0 and T 0 is called ‘availability’.AVAILABLE ENERGY REFERRED TO A CYCLE

The available energy (A.E.) or the available part of the energy supplied is the maximum work outputobtainable from a certain heat input in a cyclic heat engine (Fig. 6.1). The minimum energy that has to berejected to the sink by the second law is called the unavailable energy(U.E.), or the unavailable part of theenergy supplied.

For the given values of the source temperature T 1 and sink temperature T 2, the reversible efficiency,

For a given T 1, η rev. will increase with the decrease of T 2. The lowest practicable temperature of heat rejection isthe temperature of the surroundings, T 0.



Consider a finite process l-m, in which heat is supplied reversibly to a heat engine (Fig. 6.2). Taking anelementary cycle, if dQ 1 is the heat received by the engine reversibly at T 1,

For the heat engine receiving heat for the whole process l-m, and rejecting heat at T 0



Thus unavailable energy is the product of the lowest temperature of heat rejection, and the change of entropy of the system during the process of supplying heat (Fig. 6.3).DECREASE IN AVAILABLE ENERGY WHEN HEAT IS TRANSFERRED THROUGHA FINITE TEMPERATURE DIFFERENCE

When transfer of heat takes place through a finite temperature difference, there is a decrease in theavailability of energy so transferred. Consider a reversible heat engine operating between temperatures T 1 andT 0 (Fig. 6.4). Then

Assume that heat Q1 is transferred through a finite temperature difference from the reservoir or source atT 1 to the engine absorbing heat at T 1′, lower than T 1 (Fig. 6.4). The availability of Q1 as received by the engine atT 1′ can be found by allowing the engine to operate reversibly in a cycle between T 1′ and T 0 receiving Q1 andrejecting Q2′.



The loss of available energy due to irreversible heat transfer through finite temperature difference between the source and the working fluid during the heat addition process is given as



Decrease in available energy, A.EThus the decrease in A.E. is the product of the lowest feasible temperature of heat rejection and the

additional entropy change in the system while receiving heat irreversibly, compared to the case of reversibleheat transfer from the same source. The greater is the temperature difference (T 1 – T 1′), the greater is the heat rejection Q 2′ and the greater will be the unavailable part of the energy supplied (Fig. 6.5).

Energy is said to be degraded each time it flows through a finite temperature difference. That is, why thesecond law of thermodynamics is sometimes called the law of the degradation of energy , and energy is said to‘run down hill’ .AVAILABILITY IN NON-FLOW SYSTEMS

Let us consider a system consisting of a fluid in a cylinder behind a piston, the fluid expandingreversibly from initial condition of p1 and T 1 to final atmospheric conditions of p0 and T 0.Imagine also that the system works in conjunction with a reversible heat engine which receives heat reversiblyfrom the fluid in the cylinder such that the working substance of the heat engine follows the cycle O1LO asshown in Fig. 6.6, where s1 = s L and T 0 = T L (the only possible way in which this could occur would be if aninfinite number of reversible heat engines were arranged in parallel, each operating on a Carnot cycle, each one

receiving heat at a different constant temperature and each one rejecting heat atT 0

). The work done by theengine is given by

The heat supplied to the engine is equal to the heat rejected by the fluid in the cylinder.

Therefore, for the fluid in the cylinder undergoing the process 1 to 0, we have

The work done by the fluid on the piston is less than the total work done by the fluid, since there is nowork done on the atmosphere which is at constant pressure p0



Work done on atmosphere

Hence, maximum work available

Note. When a fluid undergoes a complete cycle then the net work done on the atmosphere is zero

The property, a = u + p0v – T 0 s (per unit mass) is called the non-flow availability functionIRREVERSIBILITY

The actual work which a system does is always less than the idealized reversible work, and thedifference between the two is called the irreversibility of the process .

Thus, Irreversibility, I = W max – W This is also sometimes referred to as ‘ degradation ’ or ‘ dissipation ’.For a non-flow process between the equilibrium states, when the system exchanges heat only with

environment, irreversibility ( per unit mass ),

The same expression for irreversibility applies to both flow and non-flow processes.The quantity T 0 (Δ s system + Δ s surr .) represents (per unit mass) an increase in unavailable energy (or

energy).EFFECTIVENESS

Effectiveness is defined as the ratio of actual useful work to the maximum useful work .The useful output of a system is given by the increase of availability of the surroundings.

For a compression or heating process the effectiveness is given by

The effectiveness of an actual process is always less than unity . Thus effectiveness of a process is themeasure of the extent to which advantage has been taken of an opportunity to obtain useful work.

unit - ii theory

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