The Carnot Factor

Carnot allowed us to answer the following question Which part of heat Q, available at a temperature T T0, can at most be converted into useful work Provided the process is cyclic and conducted reversibly, the maximum amount of work available is given by 

The factor 1 - (T0/T) is often called the thermal efficiency, but we prefer to call it the Carnot factor For example, if heat is supplied at 600K and the temperature of the environment is 300 K, the Carnot factor is 1/2. We could also say that in this instance the quality q of every Joule of heat is 1/2 J/J, if we wish to express that at most half of the Joule of heat supplied can be made available for useful work with respect to our environment at T0  

So we define the quality of heat supplied at a temperature level T T0 as the maximum fraction available for useful work. Baehr [4] has called this part of heat the exergy of heat. The remaining part is unavailable for useful work and is called anergy. It is the minimal part of the original heat that will be transferred as heat Qo n to the environment. In Baehr s terminology, we could say that in this instance the ideal heat engine would achieve the following separation between useful and useless Joules  

Overall, the cycle takes up an amount of energy Q, produces an amount of work W0T, and releases an amount of heat Q - W0 r = Q, inm at the temperature T0 to the environment. 

This equation very clearly expresses the quality aspects of heat and must have tempted Sussmann [5] to a statement much inspired by Orwell in his 

---

In this chapter, we show that it is not so much energy that is consumed but its quality, that is, the extent to which it is available for work. The quality of heat is the well-known thermal efficiency, the Carnot factor. If quality is lost, work has been consumed and lost. Lost work can be expressed in the products of flow rates and driving forces of a process. Its relation to entropy generation is established, which will allow us later to arrive at a universal relation between lost work and the driving forces in a process. 

The enclosed area represents the amount of lost work in this plot of the Carnot factor against... 

From Figure 3.2, in which the Carnot factor has been plotted against the amount of heat transferred, we can conclude that the work lost is represented by the enclosed area between the temperature levels Thigh and Tlow and the points of entry and exit. 

The quality of heat is defined as its maximum potential to perform work with respect to a defined environment. Usually, this is the environment within which the process takes place. The Carnot factor quantitatively expresses which fraction of heat is at most available for work. Heat in free fall from a higher to a lower temperature incurs a loss in this quality. The quality has vanished at T0, the temperature of the prevailing environment. Lost work can be identified with entropy generation in a simple relation. This relation appears to have a universal value. 

Consider lkg/s of coal that is combusted with an adequate amount of air (approximately zero exergy contribution). The rate at which exergy flows into the system is therefore 23,583 kW. The combustion releases heat, namely, at a rate of 21,860 kW at a temperature T. Since we have created a heat source at temperature T, it is straightforward to compute the work potential (exergy) of this heat source. All we need to do is multiply the heat release rate (21,860 kW) by the Carnot factor 1 - (T0/T). This means that if the combustion takes place at temperature T = 1200 K for a fluidized bed reactor (Table 9.1), the efficiency of the combustion alone is combustion = (21,860/23,583) [1 - (T0/T)] = 0.93 [1 - (T0/T)] = 0.93 [1 - (298.15/1200)] = 0.7 This means that already 30% of the maximum work has been lost We summarize this simplified analysis in Figure 9.15. 

The heat is available at 1200 K, but there will be temperature differences in the heat exchanger, so more available work will be lost in the heat exchange process. What can we learn from this example If we examine the Carnot factor, the answer seems to be clear. If we increase the operating temperature of the combustor, we can increase the efficiency and lose less work in the process. For example, if we had chosen an operating temperature of 2000 K, as could be possible in the suspended bed, we would have obtained an efficiency of 0.79, which is quite considerable. However, any gain in efficiency could be offset by the increase in work necessary to pulverize the coal For the sake of simplicity, we have not included these in this analysis. From the point of view of efficiency of combustion,... 

At this point, it is useful to point out that simply computing the efficiency based on the Carnot factor is an exercise that should be performed with care. The Carnot factor, C, is given by C = 1 - (T0/T). As shown in the example, we can compute the group -CArH/Exin to get the efficiency Strictly speaking, this is not always true, but why The answer is very subtle. Carnot s analysis holds only for infinite heat reservoirs, and if the heat (which can be viewed as the useful product) is transferred, the temperature of the reservoir (the product mixture) also changes So the correct way of computing the efficiency based on the heat is to take into account the fact that the temperature T is not constant, but is variable and will decrease to T0 ... 

This is the definition of the physical exergy of the effluent stream The computation of the terms will yield the physical component of the stream, and the combination with the chemical and mixing components will allow for the computation of the efficiency. The question now remains Why did the computation of the efficiency based on the Carnot factor give the correct number The answer is that since the temperature of the effluent gases is fixed, it mimics an infinite heat reservoir, and therefore n[AH - TqAS] simplifies to nAH[ 1 - T0(AS/AH)] = nAH[ 1 - (T0/T)], since AG = AH - TAS = 0 at equilibrium. 

According to this definition, the quality of mechanical or electrical energy is equal to unity and that of thermal energy at a temperature, T, is equal to the Carnot factor, 1 - TQ/T. For chemical reactions, the exergy ratio (a-) represents that fraction of the delivered energy that could be converted to thermodynamic work by a reversible process and has a value most often (but not always) between zero and unity. 

Using data for a fluid of your choice, other than steam, calculate values for the Carnot factor using the Clapeyron equation. 

This theoretical efficiency is much greater (by a factor of about 2) than that of a thermal combustion engine, producing the reversible work, WT, according to the Carnot s theorem ... 

Physical Interpretation of the Absence of the Carnot Efficiency Factor in Electrochemical Energy Conversion... 

The above equation is a generalization of the Carnot relation. The ratio between the exergy and the heat Ex/q is called the exergy factor. When T < T0, there is a lack of energy in the system the value of Ex/q greatly increases for... 

Fig. 2 Efficiency as a function of temperature of the energy conversion of a fuel cell and a conversion process limited by the Carnot s factor.

Note that (25) is just the Carnot coefficient of performance Tc/AT multiplied by a factor that depends upon Zyx and the hot and cold side temperatures Tn and Tc. In addition, the maximum value of the product ZyxTis unity. 

Consider a system programmed to travel the upper portion of the Figure 5.4 Carnot cycle. Let the initial and final states be i and iii, respectively. (a) What is the value of the merit factor Take the pressure resolution to be 1% of the range, (b) What is the value of bits Take the pressure and volume resolution to be 10% of the respective ranges. 

This ratio should be called the conversion factor and not the efficiency of the cycle. The reversible Carnot cycle is as efficient as any cycle can possibly be, when operating between the given pair of temperatures, and should thus be regarded as 100% efficient. Any irreversible cycle has a lower conversion factor and may thus be said to be inefficient relative to the reversible one. 

Though Sadi Carnot used the caloric theory of heat to reach his conclusions, his later scientific notes reveal his realization that the caloric theory was not supported by experiments. In fact, Camot understood the mechanical equivalence of heat and even estimated the conversion factor to be approximately 3.7 joules per calorie (the more accurate value being 4.18 J/cal) [1-3]. Unfortunately, Sadi Carnot s brother, Hippolyte Camot, who was in possession of Sadi s scientific notes from the time of his death in 1832, did not make them known to the scientific community until 1878 [3]. That was the year in which Joule published his last paper. By then the equivalence between heat and work and the law of conservation of energy were well known through the work of Joule, Helmholtz, Mayer and others. (It was also in 1878 that Gibbs published his famous work On the Equilibrium of Heterogeneous Substances). 

An example of a process that can deliver work by absorbing heat from a hot reservoir and rejecting heat to a cold reservoir is the Carnot engine. This is an idealized model consisting of a sequence of processes, each of which is assumed to be reversible. A reversible process is one that can be reversed by an infinitesimal change in the external conditions. For instance, in order to compress a gas reversibly, the external pressure at any moment should be P + AP, where P is the gas pressure at that moment and AP is a small pressure inaement. The reversible compression can be changed to a reversible expansion by changing the external pressure to P - AP. A reversible process consists of steps in which the system is at equilibrium. In a reversible process there are no losses due to friction or other factors. 

---