Notes for the Second Law of Thermodynamics

It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce an amount of work.

1.) Introduction

Not only do hotter liquids and gases contain more energy, but a bigger fraction of that energy can be converted by an engine to useful work. Discussion of the Second Law takes in the reversible engine, a theoretical cycle that would yield the maximum possible efficiency. To test the concept of the reversible engine we shall envisage it as operating between two reservoirs, a reservoir being a very large heat store that changes its own temperature only infinitesimally when heat is added or rejected ( $ Q = m c \Delta T $ and m c is huge ). Hotter reservoirs are called "sources" and colder reservoirs "sinks". A further concept is the reversible heat pump.

The essential background for these notes is given in the next section. For the purposes of context, the section offers an example of an engine cycle and relates the cycle efficiency to heat and work flows. The reverse of an engine cycle is a heat pump cycle - this too is explained. In subsequent sections two compatible versions of the Second Law are proposed; the important fact is that all engines must reject a certain amount of heat, without it being converted to work. Thereafter the most efficient cycle is a cycle comprising "reversible processes", and a temperature scale is constructed using the hypothetical reversible engine. This leads us to the concept of entropy. Entropy is property of the equilibrium state, that serves to indicate the extent of non-ideal irreversibility in a system and its surroundings. The previously presented equations for the entropy of ideal gases and pure liquids are proven. I conclude by showing how the engineer can read temperature-entropy charts so as to infer what is happening in a process.

2.) Machine Cycles and Cycle Efficiency

Cycle efficiency is the ratio of work output to heat input over a complete cycle of an engine.

A heat engine accepts heat from a hot reservoir, does useful work, and rejects heat to a cold reservoir. Flames and solar heat approximate hot reservoirs, or sources of heat. The atmosphere or an ocean can approximate a cold reservoir (or heat sink) - note the rejection of heat to power station cooling towers, automobile exhausts and radiators. Heat engines work in cycles, described in earlier notes . A closed system undergoes a cycle when it passes through a series of events which leave its final state equal to all aspects of its initial state. The working gas passes through a sequence of states, possibly mapped by p-versus-V, before returning to the start condition (e.g. the gas experiences states 1-2-3-4-1-2-3-4-1...). This is done by repeated expansion and compression of gas in a piston-cylinder, or by the use of compressors and turbines.

An illustrative example of a cycle was proposed by Sadi Carnot (1824). (We shall see later that this illustrates the hypothetic "reversible" heat engine.)

Figure 1: Carnot's theoretical engine. (a) Mechanism (b) pV diagram (c) Ts diagram. Press down the "engine" button to activate. To illustrate a heat pump, start with the piston at its lowest position (bottom dead centre) .

For the time being, focus on parts (a) and (b) of Figure 1; drive the cycle with the "Heat Engine" button. The following should be apparent for the four processes from 1-2-3-4-1.

The area underneath curves 1-2 and 2-3 on the pV diagram (Figure 1b) represents the work extracted over a complete cycle. The area under 3-4-1 represents the compression work added to the cycle. The net work per cycle is therefore area 1-2-3-4-1. With reference specifically to the Carnot cycle one notes:

where the operator || returns the modulus of a real number, for example $ |-5|=5$. The cycle's net transfers of energy $ W, Q_H, Q_L $ are marked on Figure 2(a), a simplified schematic of an engine cycle. Remember that the Figure shows the net work output, which is the difference between expansion work (out) and compression work (in). According to the First Law of thermodynamics the net work output and net heat input are equal.

The cycle efficiency is defined as the net work output of the cycle per unit of heat input.

Figure 2 Schematic representation of cycles(a) heat engine and (b) heat pump .

A heat pump is the functional reverse of a heat engine. The net addition of work to each cycle results in the extraction of heat from the cold reservoir and the transfer of heat to the hot reservoir. The domestic refrigerator provides a commonplace example; heat is extracted from the cabinet contents, mechanical work is applied to the compressor, and heat is rejected from coils at the back of the machine. A second example is the commercial air source heat pump; heat is extracted from the ambient atmosphere, typically at 5 to 20 $^oC$, and delivered to hot water at temperatures up to $60^oC$.

Figure 1 also shows the operation of Carnot's hypothetical heat pump - the reverse of the Carnot engine. (To operate keep the "Heat Pump" button depressed. The four cycle states have been renumbered in line with Rogers and Mayhew.) One notes:

This time the work addition to the cycle (the area under the pV curve 1-2-3) is greater than the work extracted (see pV curve 3-4-1) and there is net cyclic heat addition. Compared to the heat engine, the directions of heat transfer are reversed. Heat is transferred to the hot reservoir.

The net flows of energy to a heat pump are shown schematically on part (b) of Figure 2. The First Law can be written in similar form to that of the heat engine. Given that our interest is in the extracted heat, $ Q_H $,

Rather than efficiency, we tend to talk about "coefficient of performance", the quantity of heat added to the hot reservoir per unit of added work.

The COPs of the practical heat pumps used for space heating are in the range from 2.5 to 4.

3.) Kelvin-Planck Statement of the Second Law

"It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce an amount of work." (Rogers and Mayhew)

In the previous section I took some trouble to describe the concept of a cycle, and how it relates to a simple engine. A cycle exhibits identical start and end states. Carnot's hypothetical engine cycle necessitated not only the addition of heat, but also the rejection of some heat to a heat sink. The same is noted of all engine cycles proposed in leading text books (e.g. Keenan, Rogers and Mayhew, Cengel and Boles, Van Wylen) and all practical engines have some means of rejecting heat. A sceptic might propose thermal expansion of a metal bar as a process that does work with heat addition and no heat rejection; thermal expansion however is not a cyclic process because the temperature and length of the working material have changed. Similar reasoning inevitably deals with all other provocative challenges to the Kelvin-Planck statement.

The Second Law is axiomatic. It is self-evident, it has been observed in practice and has yet to be disproved. Referring back to Figures 1 and 2, we see that a cyclic engine not only accepts heat but must reject it too (see the first bullet point below). Any proposed cyclic engine connected to one reservoir only and producing work would be termed a “perpetual motion machine of the second kind”. The consequences of the Second Law are:

Example 4.010: In the sketches below, which cyclic engines are possible and which cyclic engines are impossible? Arrows point in the directions of energy transfer.

"It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce an amount of work."

Impossible, there is only one heat reservoir.
Impossible, there is only one heat reservoir. The two heat transfers are to/ from the single reservoir.
Possible, there are two heat reservoirs.
Possible, engine E2 "sees" a heat flow, as if it were connected to a hot reservoir. Engine E1 "sees" a heat loss, as if it were connected to a cold reservoir.
Possible, the net transfer of work is from the surroundings to the Engine. An example is frictional heating.

4.) Clausius Statement of the Second Law

Corollary 1: "It is impossible to construct a device that operates in a cycle and transfers heat from a cooler to a hotter body without work being done on the system by the surroundings." (Rogers and Mayhew)

The above is Clausius's statement of the Second Law. My own informal version of this statement is "when I unplug my domestic refrigerator, it stops working."

The Clausius and Kelvin-Plank Statements can be shown to conform to each other. The slide show in Figure 4 below shows this schematically. Slide 2 makes the provocative, incorrect assertion that an amount of heat, $Q_u=X \; kJ/cycle $, transfers from a cold reservoir to a hot reservoir. From slide 3 onwards one observes an adjacent heat engine cycle that accepts heat $ |Q_H| = |W| + |X| \; kJ/cycle $, produces work $|W| \; kJ/cycle $, and rejects to the cold reservoir heat $ |Q_L| = |X| \; kJ/cycle $. The net heat transfer to the cold reservoir is then $ |Q_L|-|Q_u|= |X| –|X| = 0 $. In slide 5 an equivalent thermal circuit omits the cold reservoir and transfers the engine's reject heat to the hot reservoir. Thereupon the (erroneous) suggestion is that a single (hot) reservoir produces a net amount of work $|W| \; kJ/cycle $. This contravenes the Kelvin-Planck statement.

Figure 4: Animation showing two statements of the Second Law to be compatible Press "previous" and "next" buttons to navigate.

5.) Reversible Processes

"When a fluid undergoes a [thermodynamically] reversible process, both the fluid and its surroundings can always be restored to their original states." (Rogers and Mayhew)

We must understand reversible processes as a prerequisite to studying reversible cycles. A reversible process is very similar to a quasi-equilibrium process, where process passes through a series of equilibrium states. The preconditions for a thermodynamically reversible process are as follows:

Additional preconditions concerning non-linear mechanical properties, electrical resistivity and chemical properties are not required in these notes.

DISCLAIMER: The following three examples concerning reversibility/ irreversibility are my own ideas that have yet to be formally peer-reviewed. I hope these slide shows and animations communicate ideas easily, so that the accepted text books will become easier to understand.

Example of solid-to-solid friction and irreversibility

Consider a simple mechanical analogy. A skier starts their journey in a U-shaped valley, from rest, on a slope at a height z=100 m above the valley floor. The system (human + skis) passes through state 1, state 2 and back to state 1.

Consider the net effect of the simple cycle 1-2-1. A frictionless version of 1-2-1 can happen with no work transfer to the skier and no heat transfer to the surroundings (snow). However, friction demands that some applied work is irreversibly converted to heat and transferred to the surroundings (i.e. the snow).

Example 4.020: With regard to the skier described above, (a) confirm that no mechanical work is required for the frictionless journey, (b) comment on the additional work required with friction. Explain why friction makes the descent thermodynamically irreversible? Make the additional assumption that the skis maintain instantaneous thermal equilibrium with surrounding snow (=at the same temperature). Also ignore aerodynamic drag.

During a cyclic process any friction causes the surroundings to transfer extra work irreversibly to the system, and the system to transfer extra heat irreversibly to the surroundings.

Problem Statement: Classification of processes - thermodynamically reversible or thermodynamically irreversible.
Diagram:

Figure Example 4.020: Simple mechanical problem demonstrating reversibility (a) frictionless process with no heat and work transfer (b) friction applies during descent, causing irreversible heat and work transfer.
Assumptions: See problem description. Also, a frictionless ascent (for the sake of simplicity).
Physical Laws: First Law, relates friction work to heat transfer. Energy balance relates potential energy, kinetic energy and friction work.
Calculation: For process 1-to-2 write the energy balance as,
$$ W_{12} + Q_{12} = KE_2+PE_2-KE_1-PE_1 = m(\frac{v_2^2-v_1^2}{2} +g (z_2-z_1)) $$
(a) Frictionless case: In the absence of friction there is no heat loss and $ Q_{12}= 0 $ and we are asked to confirm that $ W_{12}=0 $. If we put numbers into the above equation,
$$ W_{12} = m ( \frac{44.3^2-0^2}{2} +9.81(0-100)) = m \times 0 = 0 \quad (3s.f.) $$
This is a simple conversion of potential energy to kinetic energy. The reverse conversion happens on the ascent (with process 2-3 equivalent to a process 2-1), with zero work required. No work is transferred from the surroundings to the system, and in the absence of friction no heat is transferred from the system to the surroundings. The (frictionless) descent is reversible.

(b) Friction case: Friction leads to a loss of kinetic energy and corresponding heat loss to the surroundings, so that the skier does not reach the original velocity of $v_2 = 44.3 m/s $. If no work is done during the descent.
$$ Q_{12} = KE_2+PE_2-KE_1-PE_1 = m(\frac{v_{2,f}^2-v_1^2}{2} +g (z_2-z_1)) $$
where subscript f indicates the velocity affected by friction. If the "with friction" and "without friction" processes are compared,
$$ Q_{12} = m(\frac{v_{f,2}^2-(44.3^2 m^2 s^{-2}) }{2}) $$

To return the skier to their original altitude work must be done during the ascent. For the sake of simplicity, let us assume that the ascent is frictionless. The First Law of thermodynamics for a cyclic process tells us that the net transfer of heat-plus-work during 1-2-1 is zero. Hence,
$$ W_{21} = -Q_{12} $$
(This assumes that there is no friction during the ascent.)
Discussion: For the sakes of simplicity and illustration, take work transfer as coming from the surroundings, rather than originating from the muscles of the skier. The Kelvin-Planck statement tells us that the heat transferred back to the surroundings, $ |Q_{12}| $, cannot be converted back to an equal amount of work, $ |W_{12}| = X$. This is the subject of Figure 5 below.)

Example 4.020 is straightforward because the skier only does sufficient work to overcome friction. In this case the First Law for a cyclic process tells us that frictional work is equal to frictional heat transfer. The Second Law tells us that this particular energy transfer cannot be reversed, as shown below.

Figure 5. Net frictional work and heat transfer in (a) the descent-ascent cycle (b) an attempt to revert the surroundings to their original state fails owing to the Kelvin-Planck statement.

Example of an adiabatic process, friction and reversibility

Fluid friction causes irreversibility; the impact of friction is worsened by any departure from quasi-equilibrium.

Friction is present in all fluid flows. (For practical purposes it can sometimes be ignored in gas flows.) The precise mechanisms of turbulence and viscosity do not need to be understood for this discussion; just two things should be appreciated. Firstly, rather than at an interface, friction causes energy conversion and restraining forces within the fluid volume. Secondly, the restraining forces increase with the fluid velocity; slowly moving fluids suffer less friction.

The effects of friction are more easily appreciated when a process departs radically from quasi-equilibrium. All practical processes deviate to some extent from thermodynamic equilibrium, although assumed equilibrium thermodynamics often suffice for engineering purposes. Let us, however, consider an extreme departure. Figure 6 considers an adiabatic expansion between two volumes, $V_1$ and $V_2$. (The term "adiabatic" means that no heat will be transferred to or from the surroundings.) The Figure's "reversible" option animates a quasi-equilibrium, frictionless process from 1-to-2. One notes internal energy conversion, $U_2-U_1=W_b$ where $W_b$ indicates the moving boundary work. This energy conversion reduces gas temperature from $T_H$ to $T_L$. (The frictionless adiabatic curve intersects both isotherms on part (b)). An adiabatic, frictionless compression then reverses the process, returning gas to state 1 and taking an amount of work $ | W_b| $ from the surroundings. Process 1-2 is reversible because process 1-2-1 leaves both the system and its surroundings unchanged; the net work transfer is zero and the net heat transfer is zero (these were, after all, adiabatic processes). The ultimate result of the cycle is zero change to the internal energy in the surroundings, and zero change to the rotational kinetic energy of the flywheel. Friction makes matters more complicated.

Pressing Figure 6's "irreversible" button brings about a (hypothetical) catastrophic failure. At the crank's 12-o'clock position the connecting rod fails (and is erased from the animation) such that expanding gas moves the piston very quickly to the bottom of the cylinder. I mark the centre of mass (C.O.M.) of the gas with a green circle. Following the sudden expansion a pressure wave moves the C.O.M. through an underdamped oscillation, rather like an underdamped spring-mass . Let us neglect the mass and kinetic energy of the piston, and any sliding friction between piston and cylinder, and assume that the initial gas pressure is very high so that the expansion minimal does work against the surrounding atmosphere. The the initial energy conversion is principally from internal energy to kinetic energy (hence the moving C.O.M.) and then, by frictional damping, back to internal energy. By the time the flywheel crank reaches its six-o'clock position friction damping has returned all kinetic energy to internal energy (restoring temperature to approximately $T_H$ ). The net effect is that between states 1 and 2 there is minimal work ( $ W_{12} \approx 0 $ ), and no heat transfer ( $ Q_{12}=0 $).

When the flywheel reaches 6-O'clock the connecting rod is reinstated and the piston cylinder is returned to state 1 isothermally at $T_H$ and reversibly. Moving boundary work must be applied to repressurise the gas, and by the First Law there must be a (nearly) equal heat rejection to the surroundings.

The net effect of the cycle, 1-2-1, is that work $W_{12}$ has been transferred to the system from the surroundings in exchange for heat $ -Q_{12}$. Like the previous example, the Kelvin-Planck statement tells us that this energy transfer cannot be undone (see Figure 5).

Whilst such large deviations from equilibrium would rarely be observed in practice, most processes will go through a sequence of steps each of which deviates somewhat from equilibrium. Under steady-flow, irreversibility is dealt with by an empirical parameter, the isentropic efficiency for turbines , or the separately defined isentropic efficiency for compressors.

Figure 6: Reversible and irreversible adiabatic processes (a) machine (b) p V diagram. Keep "reversible" or "irreversible" buttons depressed. C.O.M. indicates centre of mass. Gas space is coded blue for $T_L$ and red for $T_H$.

Example of isothermal process, temperature difference and reversibility

In a reversible process, any heat transferred to the fluid is by virtue of infinitesimally small temperature differences.

Figure 7 sets up a reversible isothermal process. In the expansion stoke (1-2) the working gas is at temperature $T_i$ and accepts heat from a reservoir at the same temperature. In the compression stroke (2-1) the working gas retains temperature $T_i$ and rejects heat to the reservoir, also at $T_i$. Both the system and the surroundings (reservoir and flywheel) are returned to their original condition.

Real world processes are irreversible. They take place over a finite duration of time in equipment with a finite heat transfer area. Were the working gas and reservoir to be maintained at identical temperature, Newton's rule of heat transfer would demand either infinite process duration or an infinite heat transfer area. In the real-world this is unrealistic. The irreversible example in Figure 7 maintains the working gas at temperature $T_i$ and (for a frictionless process) the moving boundary work for forward and reverse processes are identical. However for forward process 1-2 heat is added from a hot reservoir at $T_H > T_i $ and for reverse process 2-1 heat is rejected to a cold reservoir at $T_L < T_i $. The net effect of reversing the cycle is to transfer energy from the hot reservoir to the cold reservoir. The Clausius Statement of the Second Law tells us that this heat transfer cannot be undone.

Figure 7: Reversible and irreverible isothermal processes. (a) Machine (b) pV Diagram. The gas in the piston-cylinder is at intermediate temperature, $T_i$, colour coded purple. The reversible process returns heat to the same reservoir. The irreversible process shifts heat from a hotter reservoir (red) to a colder reservoir (blue).

6.) Reversible Cycles

Corollary 2: "It is impossible to construct an engine operating between only two reservoirs which will have a higher efficiency than a reversible engine operating between the same two reservoirs." (Rogers and Mayhew)

A reversible engine cycle comprises a set of reversible processes. Complementary reading of Keenan, Wylie, and/ or Rogers and Mayhew is highly recommended.

Proof of the second corollary

All processes in a reversible cycle are themselves reversible. For instance, in the Carnot Heat Engine in Figure 1 the option of reversing to heat pump operation simply multiplies each of the two heat flows and four work flows by (-1). The net transfers of energy, shown in Figure 2, are also reversed. Reversible processes occur under quasi-equilibrium, with zero friction and infinitesimally small temperature driving forces whenever heat is transferred. It seems reasonably that for a given cyclic heat transfer, $Q_H$, a reversible engine will produce more work output than any irreversible one.

Figure 8 shows a proof. One starts with reversible and irreversible engines connected to the same heat reservoirs and receiving the same quantity of heat, $Q_H$. The trick in slide 2 is to reverse the reversible engine so that it becomes a reversible heat pump. The hypothetical machine formed by merging the pump and engine is connected to a single reservoir, and from the Kelvin Planck statement cannot produce a net amount of work output. Hence $W_{rev} > W_{irr}$.

7.) Thermodynamic Temperature Scale and Carnot Efficiency

Corollary 3: "All reversible engines operating between the same two reservoirs have the same efficiency." (Rogers and Mayhew)

Corollary 3 follows naturally from the proof given for Corollary 2. The "Irreversible Engine" could be replaced with a second reversible engine, which cannot perform better or worse than the first reversible engine. I like to think about "types" of reversible engine: Carnot, Sterling and Erikson all proposed cycles that, theoretically, could operate reversibly. So, for example, a theoretical Carnot engine and a theoretical Sterling engine operating between the same thermal reservoirs would exhibit identical efficiencies. (In practice any of the aforementioned engines is far from reversible.)

Corollary 4: A scale of temperature can be defined which is independent of any particular thermometric substance, and which provides an absolute zero of temperature.

Consider first a material temperature scale, namely the Celcius scale. A section of a thermometer's capillary tube is divided into 100 equal degrees, spanning the range from the ice point to the boiling point of water. When the thermometer bulb is located in a sample of a fluid, the indicated temperature is taken as being in direct proportion to the expansion of the instrument's working liquid (generally alcohol). A thermodynamic scale is based on thermodynamic principles, with no reliance on any thermometric fluid. A set of thought experiments will demonstrate that (1) an alternative thermodynamic scale of temperature can be divided into degrees by thermodynamic principles; (2) there is a minimum possible temperature, termed "absolute zero"; (3) the efficiency of a reversible heat engine is a simple function of reservoir temperature.

Consider the train of reversible heat engines and reservoirs on part (a) Figure 9. At the top of the train is a "reference" reservoir at known condition, for example one could choose the melting point of copper, and to which we shall ascribe temperature $T_{ref} $. At the "bottom" of the train is a reservoir at temperature close to absolute zero, $T_o \rightarrow 0$. (The Kelvin-Planck statement forces some heat rejection from engine 1. To make this vanishingly small $T_o \rightarrow 0 $ ). Otherwise each reservoir, i, accepts heat $ Q_i $ from neighbouring engine i+1 and rejects equal heat $Q_i $ to neighbouring engine i. Let us choose reservoir temperatures such that all engines produce equal amounts of work, $ \sigma $ per engine per cycle. Allocate in turn an additional one degree of temperature to each internal reservoir; hence $T_o \rightarrow 0 , T_1 = 1 , T_2 = 2 , ... $. In a general form,

By applying the First Law to a generic scale engine (Figure 9b) we get $ |Q_{i+1}|=|Q_i|+|\sigma| $ and applying this to successive engines from the "absolute zero" reservoir and upwards, $Q_1=\sigma, \; Q_2=2\sigma, \; Q_3=3\sigma ... $. In a general form,

Figure 9: Temperature Scale (a) The scale proper, here comprising seven engines producing equal work $ \sigma $ and six intermediate reservoirs at temperatures 1,2,...6 degrees (b) Generic engine within the scale (c) Measurement engine showing (d) Rejected heat proportional to temperature (e) 'Standard' heat engine.

The (red) boundary on part (a) includes engines that operate between $T_{ref}$ and some as yet unknown temperature $T$ see part (c) on Figure 9. In conformity with the third corollary my combination of several - specifically four - reversible engines into a single reversible engine has no impact on work production. Given that along the temperature scale (part a) Q/T is constant then the rejected heat Q (part 9.c) indicates sample temperature,

(see Figure 9(d)). The Kelvin-Planck statement necessitates some heat rejection to the sample and $ |Q|>0 $. Figure 9d thus illustrates how the temperature scale forces an absolute zero, $T>0$.

Figure 9e shows a "standard" heat engine. Using the nomenclature on part e, let us manipulate and rewrite the previous equation as,

Then manipulation and combination of the above two equations gives the important Carnot efficiency .

No real world engine will achieve this efficiency. In practice, however, it is a useful sanity check for any calculation, or for any claim made by inventors. The Carnot efficiency formula offers designers and inventors two very strong hints - operate engines with hotter heat sources or colder heat sinks.

As a simple example, let us return to the temperature scale on Figure 9a. Consider a reversible engine operating between the reservoir temperature, $ T_{ref} \equiv T_H = 7 $ and a temperature $ T_L = 3 $. Both temperatures are contained within the red box, within which I count 4 horizontal arrows corresponding to useful work. Below $ T_{ref} \equiv T_H $ I count 7 horizontal arrows corresponding to heat input, $ |Q_H| $. The cycle efficiency is then $ work/heat \; input = 4 \sigma / 7 \sigma = 57per\;cent $. Or, using the Carnot formula,

Recall that my arbitrary seven-division scale used the melting point of copper as a reference. This gives a hot reservoir temperature of 1333 K and a cold reservoir temperature of 762 K. Irreversibility will reduce work output and hence increase rejected heat, $ |Q_L| $. In the rest of these notes we shall make use of the following for both reversible and irreversible engines.

8.) Engines Connected to Many Thermal Reservoirs

Corollary 5: "The efficiency of any reversible engine operating between more than two reservoirs must be less than that of a reversible engine operating between two reservoirs which have temperatures equal to the highest and lowest temperatures of the fluid in the original engine.." (Rogers and Mayhew)

This seems to me to be fairly obvious, but a little rigorous thought helps us (ultimately) to develop the concept of entropy.

Figure 10(a) shows a reversible engine cycle connected to two hot reservoirs, providing $Q_{H1}$ and $ Q_{H2}$, and one cold reservoir. Let us start with a physical argument. In part (b) the engine is decomposed into two separate reversible engines, B1 and B2, and four reservoirs (albeit with two at identical temperatures, $T_{L1}=T_{L2}=T_L$ ). So long as engines A1, B1 and B2 are reversible the overall conversion of heat to work will be the same, following the third corollary . The engine B2, fed from an intermediate reservoir with $T_{H2} \lt T_{H1} $ , must produce less work per cycle, following the above arguments concerning temperature scale (or from the Carnot Efficiency). The total work per cycle is less than if all heat had been provided by a reservoir at temperature $T_{H1}$.

In mathematical terms, recall that for a single reversible engine with two reservoirs the thermodynamic temperature scale dictates,

$$ \frac{|Q_{H1}|}{T_{H1}} - \frac{|Q_{L1}|}{T_{L1}} + \frac{|Q_{H2}|}{T_{H2}} - \frac{|Q_{L2}|}{T_{L2}} = 0 \qquad two \; reversible \; cycles $$

Noting that $Q_L = Q_{L1}+Q_{L2} $ and $T_{L1}=T_{L2} \equiv T_L$ one can manipulate the above to get,

$$ \frac{|Q_{H1}|}{T_{H1}} + \frac{|Q_{H2}|}{T_{H2}}- \frac{|Q_{L}|}{T_L}= 0 \qquad reversible \; cycle \; with \; three \; reservoirs $$

$$ |Q_L| = T_L \times (\frac{|Q_{H1}|}{T_{H1}}+\frac{|Q_{H2}|}{T_{H2}}) \qquad reversible \; cycle \; with \; three reservoirs $$

At $T_{H2}=T_{H1} $, the problem reverts to the two-reservoir problem and yields heat rejection $|Q_L|$ for a two reservoir problem with heat input $ Q_H \equiv Q_{H1}+Q_{H2} $. Thereafter, any reduction in $T_{H2}$ acts to increase heat rejection and reduce work output..

Figure 10 A heat engine connected to three reservoirs (a) schematic showing one cycle of engine A (b) decomposition into two separate engine cycles, B1 and B2, bounded by a red rectangle. For engine cycle B2 a reduced $T_i$ corresponds to increased heat loss through $Q_{L2}$ and reduced efficiency.

Corollary 6: The following Clausius Inequality applies to engine and heat pump cycles,

where T defines reservoir temperatures and Q defines reservoir-to-machine heat transfer. The cyclic integral is zero if the cycle is reversible and negative otherwise. The left hand side tells us one of three things about a proposed cycle:

The Clausius Inequality can be seen as an extension of the two-reservoir and three-reservoir problems. Note the frequent occurrence in all these problems of the Q/T quotient. Let us define the direction of cyclic heat flows with the IUPAC sign convention such that heat flows into the engine have positive sign and heat flows out of the engine have negative sign. Thus $ Q_H = +|Q_H| $ and $Q_L = -1 \times |Q_L| $. For the simple Carnot cycle replace H with 1 and L with 2 so that,

$$ \frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0 \qquad reversible \; cycle \; with \; two \; reservoirs $$

Likewise for the three reservoir problem setting $H1 \rightarrow 1; \; H2 \rightarrow 2; \; L \rightarrow 3 $ yields,

$$ \frac{Q_1}{T_1} + \frac{Q_2}{T_2} + \frac{Q_3}{T_3} = 0 \qquad reversible \; cycle \; with \; three \; reservoirs $$

I argue that a reversible cycle comprising an arbitrary number of reservoirs can be decomposed into many sets of two reservoir cycles (in the same way as the three reservoir problem). So the more generic form of the above becomes,

$$ \sum \frac{Q_i}{T_i} = 0 \qquad reversible \; cycle \; with \; several \; reservoirs $$

Let us now tackle the general form that is the Clausius Inequality. As an illustrative example, consider the radiator water in an automotive engine, typically entering the engine block at 333K and leaving at 363K. In approximate terms one could argue that the engine exchanges heat with 30 cold reservoirs, at separate temperatures of $ 363, 364, 365 ... 362K $. A similar argument applies if one takes the example of a coal-fired power cycle, wherein heat is transferred to the steam cycle from combustion gases over a range of temperatures, not untypically $1300, 1299, 1298, ... 420K$. (Here I have used increments of one degree, but could have chosen even smaller values, 0.1K, 0.01K etc.) To indicate vanishingly small heat transfer from each reservoir $Q \rightarrow dQ $ and,

$$ \int_{cycle} \frac{dQ}{T} = 0 \qquad reversible \; cycle \; with \; very \; many \; reservoirs$$

Finally, any irreversibility results in less conversion of heat to work, and thus more rejected heat (i.e. more $ |Q_L| $ or more "negative Q"). Thus the more general case is,

It is important to remember that dQ refers to heat passing from reservoir to engine and T refers to reservoir temperature . The next example shows the compatibity between the Clausius inequality and other aspects of the Second Law.

Example 4.030: Check that the Clausius Inequality is in agreement with the Kelvin-Planck Statement and the five corollaries listed in Sections 3 to 8.

The following forms of the inequality will be useful to us.

$$ \int_{cycle} \frac{dQ}{T} \leq 0 $$ $$ \int_{cycle} \frac{d |Q_H| }{T_H} - \int_{cycle} \frac{d |Q_L| }{T_L} \leq 0 \qquad heat \; engine \; only $$ $$ \int_{cycle} \frac{d |Q_L| }{T_L} - \int_{cycle} \frac{d |Q_H| }{T_H} \leq 0 \qquad heat \; pump \; only $$

The Kelvin Planck statement : Consider a single reservoir connected to an engine, as shown in Figure 3 . The reservoir is a hot reservoir and we find

$$ \int_{cycle} \frac{dQ}{T} = \frac{|Q_H|}{T_H} \ge 0 \qquad impossible $$

The Clausius Statement of the Second Law (Corollary 1): Consider a machine located between the hot and cold reservoirs, receiving heat $|Q|$ from the cold reservoir and adding heat $|Q|$ to the hot reservoir. Q has negative sign between machine and hot reservoir and positive sign between machine and cold reservoir. So for the "heat pump" form of the Clausius inequality,

$$ \int_{cycle} \frac{d |Q_L| }{T_L} -\int_{cycle} \frac{d |Q_H| }{T_H} = \frac{|Q|}{T_L}-\frac{|Q|}{T_H} = |Q|\frac{T_H-T_L}{T_H \; T_L} \geq 0 $$

So long as $T_H \ge T_L $ the $LHS \ge 0 $ and the above is impossible

Irreversibility and friction : In the simplest example, the process loses heat throughout the cycle and the surroundings act as a cold reservoir. Let $Q_{fric}$ be the heat loss associated with friction for combined forward and reverse processes. Let $T_{sur}$ be the temperature of the surroundings, $$ \int_{cycle} \frac{dQ }{T } \equiv -\frac{|Q_{fric}|}{T_{sur}} \le 0 $$ confirms that the process is irreversible.

With regards to friction and adiabatic processes there is likewise a net heat loss to the surroundings and the above equation applies. With regards to finite temperature differences consider the equation derived above for transfer of a quantity of heat $|Q|$ between reservoirs,

$$ \int_{cycle} \frac{d |Q_H| }{T_H} - \int_{cycle} \frac{d |Q_L| }{T_L} = |Q|\frac{T_L-T_H}{T_H \; T_L} $$

So long as $ T_L \le T_H $ the LHS is negative and the situation is irreversible.

The second Corollary : For two reservoirs only, the Clausius Inequality becomes,

$$ \frac{ |Q_H| }{T_H} - \frac{ |Q_L| }{T_H} \leq 0 \qquad cycle $$

Term $Q_L$ has its minimum when LHS =0, the condition for a reversible cycle. The corresponding maximum value of $ |W| )$ is the greatest possible value, achieved for a reversible cycle.

$$ |Q_{L,min}| = |Q_H| \frac {T_L}{T_H} \qquad reversible $$

The Third corollary : For fixed reservoir temperatures, the above equation shows the minimum fraction of heat rejected to the cold reservoir to be fixed. The inequality has no terms that allow for the type of engine, only whether or not it is reversible.

The Fourth Corollary : With regards to the Carnot efficiency, one notes the expression for $Q_{L,min}$ above. From the First Law of Thermodynamics , the corresponding maximum work is, $$ |W_{max}| = |Q_H|-|Q_{L,min}| = |Q_H|(1-\frac {T_L}{T_H}) $$

The maximum, Carnot efficiency is

$$\eta_{Carnot} = \frac{|W_{max}|}{|Q_H|} = 1-\frac {T_L}{T_H} $$

Similary if one considers the Thermodynamic temperature scale,

$$ T = T_{ref} \frac{|Q|}{|Q_{ref}|} \qquad reversible \; cycle $$

This can be rearranged as

$$ \frac{|Q_{ref}|}{ T_{ref}} - \frac{|Q|}{T} = 0 $$

This equivalent to the "equals condition" of the Clausius Inequality.

9.) Entropy - a Thermodynamic Property of a Fluid

"Corollary 7: There exists a property of equilibrium state of a closed system such that a change in its value is equal to $$ S_2-S_1 = \int_1^2 \frac{dQ_{rev}}{T} $$ for any reversible process undergone by the system between state 1 and 2." (Rogers and Mayhew)

The “closed system” is often a fluid, and entropy is often a fluid property. For example, consider the air/ gas sample enclosed in the piston-cylinder forming part of a Carnot engine, Figure 1 . In the cycle this fluid is subject to a change in its properties, $p, V, T, U, H$. But it is also subject to heat addition and heat rejection, and therefore a change in a sixth property, entropy S. It is difficult to find an intuitive definition of entropy: Corollary 7 is probably as good as it gets. It is important to recall, from Corollary 4 onwards, that $Q/T$ is important and constant in simple reversible systems.

Proof of the seventh corollary.

Figure 11: Two reversible cycles showing that entropy is a property. The two cyclic integrals of $ dQ_{rev}/T $ are decomposed as $ \int_A * + \int_C * = \int_B * + \int_C * = 0 $. The identical values for paths A and B indicate that $ \int dQ_{rev}/T $ is a thermodynamic property.

Consider in Figure 11 two reversible cycles comprising forward paths, A and B, and the same return path, C. The progress of the two cycles thus formed is mapped by plotting two properties from the set $p, V, T, U, H$. The "equals form" of the Clausius Inequality applies to both reversible cycles, AC and BC.

$$ \int_{cycle} \frac{dQ_{rev}}{T} = \int_{1,2,A} \frac{dQ_{rev}}{T} + \int_{2,1C} \frac{ dQ_{rev}}{T} = 0 $$

$$ \int_{cycle} \frac{dQ_{rev}}{T} = \int_{1,2,B} \frac{dQ_{rev}}{T} + \int_{2,1C} \frac{ dQ_{rev}}{T} = 0 $$

$$ \int_{1,2,A} \frac{dQ_{rev}}{T} = \int_{1,2,B} \frac{dQ_{rev}}{T} = -\int_{2,1,C} \frac{dQ_{rev}}{T} $$

That is, the group $ \int_{1,2} dQ_{rev} \ T $ is independent of path (A or B) and hence is a fluid property, termed entropy, S.

We should treat sign conventions carefully. For the Clausius inequality "positive heat" is added to an engine whereas for entropy "positive heat" is added to a fluid which might be working fluid in the engine but might also be the contents of a thermal reservoir.

Corollary 8 The entropy of any closed system which is thermally isolated from the surroundings either increases or, if the process undergone by the system is reversible, remains constant.

In a thermally isolated system work may cross the system boundary, but no heat crosses the system boundary. Unlike energy, entropy does not follow principles of conservation.

Proof of the eighth corollary

Consider the two adiabatic forward (1-2) processes shown on Figure 6. One is reversible, one is irreversible.

Throughout the adiabatic, reversible, forward process $dQ_{rev}=0$. By definition, entropy is constant,

$$ S=constant \implies S_2 - S_1 = 0 \qquad adiabatic \; reversible \; process $$

For the irreversible, forward process , state-1-to-state-2, "adiabatic" means $dQ=0$ is retained throughout the process but this time friction acts such that the process is almost isothermal. We do not know immediately the value of $dQ_{rev} $ required for the calculation of entropy. On Figure 6 we propose a near isothermal reversible return process , state-2-to-state-1, with cooling. Apply the Clausius Inequality thus,

On the right hand side the first term is equal to zero (because the process is adiabatic) and the second relates to reversible cooling. Thus the net impact of the irreversibility in the first part of the cycle is to give the cyclic integral a negative value,

Reversing end and start states gives the entropy change during the forward, irreversible, adiabatic process. Recall that the term "adiabatic" is synonymous with "thermally isolated",

Reflection on the two reservoir problem.

Consider the sketch on Figure 9(e) . Treat the "thermally isolated system" as comprising two reservoirs plus engine and bounded by the picture's frame. For the thermodynamic temperature scale (Fig 9.d),

Calculate the entropy changes in the hot and cold reservoirs, for reversible processes

$$ \Delta S_H + \Delta S_L = (-1) \times ( \frac{|Q_H|}{T_H} - \frac{|Q_L|}{T_L}) = 0 \qquad reversible $$

An irreversible process exhibits greater heat rejection from the engine, $|Q_L|$, and hence greater heat transfer to cold reservoir and a net production of entropy .

Example 4.030 showed the Clausius Inequality to conform with the Second Law and other corollaries. Taking a similar view of the eight corollary one finds the following.

10.) Calculation of Entropy

Pure liquids and ideal gases

Analytical expressions exist for the entropies of pure liquids and ideal gases.

To obtain an expression for reversible heat transfer, start with the Non Flow energy equation, written for vanishingly small changes.

Internal energy is a property, and its independence on path us to achieve dU by means of a reversible path,

Let us start with a pure liquid. There is zero reversible moving boundary work, $dW=0 $, and the independence of liquid specific heat capacity, $ c $, on process renders unnecessary the qualification of $ c $ as isochoric.

$$ S_2-S_1 = \int_1^2 dS = mc \int_1^2 \frac{dT}{T} = m c \; ln (\frac{T_2}{T_1}) $$

For the purpose of publishing tables, it is more convenient to talk in terms of specific entropy, $ s=S/m$. Consider a datum at temperature $T_o$ (where $s=0$).

$$ s = \frac{S}{m} = c \; ln ( \frac{T }{T_0} ) \qquad pure \; liquid \; specific \; entropy \; at \; T$$

Now consider the instance of an ideal gas. Start with the differential and reversible form of the NFEE.

To eliminate pressure, the Ideal Gas Law, is substituted in the form $p=mRT/V$

$$ s = c_v ln ( \frac{T}{T_o} ) + R \; ln ( \frac{v}{v_o} )$$ $$ s = c_p ln ( \frac{T}{T_o} ) - R \; ln ( \frac{p}{p_o} )$$

At a reference (or datum) condition where $s_o=0$, $T_o$ is the temperature, $v_o$ is the specific volume, and $p_o$ is the pressure.

An interactive T-s diagram for air is available here or ( when I have updated file names) here

Using the above relationships where appropriate, try sketching following process as T-vs-s:

Check your answers against the interactive chart. Which has the steepest gradient, a constant volume process or a constant pressure process?

Entropy of steam and refrigerants

Non ideal gases such as steam do not offer analytical expressions for entropy. One can however integrate known values of enthalpy or internal energy. In relation to a datum condition 0,

The isobar shape for steam is rather different to air, because steam undergoes a phase change. The T-s interactive chart for steam is at here .

Example 4.040: Water at 6 bar is heated from $0^oC$ to the saturation temperature of $159^oC$. Further heating converts the water from saturated liquid to saturated vapour. The enthalpies at the three states of interest are $0, 670, 2757 \; kJ/kg $. Estimate the corresponding entropies - the datum state is at $0^oC, 1.013bar$.

Take the liquid form as a pure liquid, with entropy independent of pressure. At $0^oC$ there is no entropy change if water is pressurised from $ 1.013 $ to $ 6 $ bar. If the temperature is raised to $159^0C \equiv 432 K $ the average specific heat capacity is,

$$ c = \frac {h_f(159^o C) - h_f (0^oC)}{159-0} = \frac {670-0}{159-0} = 4.21 kJ/kg$$

The specific entropy is,

$$ s_f(159^oC) = \int_{273}^{432} \frac{dh}{T} = c \int_{273}^{432} \frac {dT}{T} = c \; ln ( \frac{432}{273} ) = 1.932 kJ/kg K $$

With regards to entropy change at constant pressure,

$$ s_g(159^oC)-s_f(159^oC) = \int_f^g \frac{dh}{T} = \frac{h_g(159^oC)-h_f(159^oC)}{T}=\frac{2757-670}{432}=4.831 kJ/kgK $$ $$ s_g(159^oC) = 1.932+4.831=6.763 \; kJ/ kg \, K $$

Discussion: Strictly speaking, I should have used the internal energy of the liquid to obtain entropy. When I consult tabulated data, I find a very small difference in the third significant figure, $u_f(159^oC)=669kJ/kg$ versus $h_f(159^oC)=670kJ/kg$. The tabulated values for entropy are $0, 1.931 \; and \; 6.761 \; kJ/ kg \, K $.

11.) Temperature-Entropy Diagrams

The temperature-entropy diagram is often used to track processes and cycles. An example is given in Figure 1(c) , the Carnot cycle. The horizonal lines represent constant temperature (isothermal) processes and the vertical lines represent constant entropy (isentropic) processes.

Compressors and turbines inevitably operate irreversibly. It is customary to show a reversible process as a solid line, and an irreversible process as a dotted line. In large compressors and turbines, with minimal heat transfer to/ from the equipment, the entropy of the working fluid is always increased. Examples of such T-s diagrams are in my notes on open systems. .

As identified in Example 4.030, the shape of the temperature/ entropy isobar can be explored in the interactive charts available at the bottom of the contents page . For steam one must be aware of the wet steam region, where phase change takes place. Other T-s plots of steam isobars are given in Figure 5 of earlier notes.

Topic 4a. The Second Law of Thermodynamics and its Corollaries