Degradation and Quality of Energy

6. Degradation and Quality of Energy

Consider a Carnot cycle operating between thermal energy reservoirs at the high and low temperatures T H and T L , respectively. The term (1– T 0 /T H ) represents the quality of the energy or the work potential per unit energy that can be extracted in the form of heat from a thermal energy reservoir at a temperature T H . Therefore, the heat Q H extracted from the ther-

mal energy reservoir at T H has the potential to perform work equal to Q H × quality = Q H

(1– T 0 /T H ) where quality of energy at T H is

given by (1– T 0 /T H ). It is seen that at a specified temperature, the quality equals the efficiency of a Carnot engine that is operated between TERs at temperatures T

and T 0 . Assume that it is possible to re- move 100 kJ of heat from hot gases that are at 1000 K (and which constitute a thermal energy reservoir) using a Carnot engine operating between 1000 K and the ambient temperature of 300 K (cf. Figure 22a ). Us- ing the engine under these conditions, it is possible to produce a work output of

Figure 21: The heating of a coffee pot using a Carnot heat pump.

Figure 22: a. Carnot engine operating between hot gases and the ambient; b. Carnot engine operating between water and the ambient.

100 ×(1 – 298÷1000) ≈ 70 kJ, as illustrated in Figure 22a . Therefore, the quality of energy at 1000 K is 70%. The entropy change of the hot gases is –0.1 kJ K –1

(= –100 ÷ 1000), while the entropy gain for the ambient is 0.1 kJ K (i.e., 30 kJ÷300 K). Alternatively, we can cool the hot gases using water to transfer the 100 kJ of energy. Assume that during this process the water temperature rises by 2 K from 399 K to 401 K (with the average water temperature being 400 K, as shown in Figure 22b ). If a Carnot engine is placed between the water at 400 K and the ambient at 300 K, then for the same 100 kJ of heat removed from radiator water, we can extract only 100 ×(1 – 298÷400) ≈ 25 kJ, and 75 kJ is

rejected to the ambient. In this case, the energy quality is only 25% of the extracted heat. Figure 23 illustrates the processes depicted in Figure 22a and b using a T–S diagram. The cy- cle A–B–C–D–A in Figure 23 represents the Carnot engine (CE) of Figure 22a , while area ABJIA and EGKIE represent heat transfer from engine and to hot water, respectively, for Figure 22b , while the area C–D–I–H represents the rejected heat of CE for the first case, and the area D-C-H–K–J–I–D that for the latter case. Since more heat is rejected for the second case, the work potential or the quality of the thermal energy is degraded to a smaller value at the lower temperature. This is due to the irreversible heat transfer or the temperature gradients between hot gases and radiator water (as shown in Figure 22b ). In general property gradients cause entropy generation.

Now, one might ask about the Maxwell-Boltzmann distribution of molecular veloci- ties. Consider a monatomic gas within a container with rigid adiabatic walls. A “pseudo” tem- perature distribution exists for the monatomic gas. The question is whether with collision and transfer of energy, there can be degradation of energy or generation of entropy. First, tem- perature is a continuum property and the temperature cannot be associated with a group of molecules. Secondly, after frequent collisions, at that location where frequent transfers occur, the intensive state is not altered over a time period much larger than collision time. Thus, no gradient exists and there is no entropy generation.

a. Adiabatic Reversible Processes Recall that for any process within a closed or fixed mass system, dS = δQ/T b + δσ.

For any reversible process δσ =0 so that dS = δQ/T. For an adiabatic reversible process, δQ = δσ = 0, so that

dS = 0. Consequently, the entropy remains unchanged for an adiabatic reversible process. These

processes are also known as isentropic processes.