Carnot factor

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 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 ... 

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. 

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