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Isotherms reversible cycle with

As shown in the diagram, we can embed the path (heavy solid line) in a grid of isotherms and adiabats (dotted lines), dividing the path area into a grid of mini-Carnot cycles of infinitesimal width. The ith mini-Carnot cycle (light solid line) is shown delimited by isotherms Tif (forward direction) and Tir (reverse direction), with associated differential heat exchanges dqif, dqir satisfying... [Pg.135]

Hint Assume that they do intersect, and complete the cycle with a line representing a reversible, isothermal process. Show that peformance of this cycle violates the second law.)... [Pg.416]

But, generally, such a cycle with adiabatic and isothermal irreversible processes may be realized with real gas (or even liquid). Those with real gas approximate the reversible Carnot cycle with ideal gas by a double limiting process as follows (i.e., we form the ideal cyclic process from set A (and also B and C), see motivation of postulate U2 in Sect. 1.2) running this cycle slower and slower... [Pg.281]

A heat engine, as mentioned in Sec. 4.2, is a closed system that converts heat to work and operates in a cycle. A Carnot engine is a particular kind of heat engine, one that performs Carnot cycles with a working substance. A Camot cycle has four reversible steps, alternating isothermal and adiabatic see the examples in Figs. 4.3 and 4.4 in which the working substances are an ideal gas and H2O, respectively. [Pg.105]

During each stage of the experimental process with nonzero heat, we allow the Carnot engine to undergo many infinitesimal Carnot cycles with infinitesimal quantities of heat and work. In one of the isothermal steps of each Carnot cycle, the Carnot engine is in thermal contact with the heat reservoir, as depicted in Fig. 4.8(a). In this step the Carnot engine has the same temperature as the heat reservoir, and reversibly exchanges heat dg with it. [Pg.115]

The Carnot engine operates on a two-stroke cycle that is called the Carnot cycle. We begin the cycle with the piston at top dead center and with the hot reservoir in contact with the cylinder. We break the expansion stroke into two steps. The first step is an isothermal reversible expansion of the system at the temperature of the hot reservoir. The final volume of the first step is chosen so that the second step, which is an adiabatic reversible expansion, ends with the system at the temperature of the cold reservoir and with the piston at bottom dead center. The compression stroke is also broken into two steps. The third step of the cyclic process is a reversible isothermal compression with... [Pg.107]

Figure 3.4 A Reversible Cycle of Isotherms and Adiabats. (a) The original cycle, (b) The cycle with an added process. Figure 3.4 A Reversible Cycle of Isotherms and Adiabats. (a) The original cycle, (b) The cycle with an added process.
Theorem.—A process yields the maximum amount of available energy when it is conducted reversibly.—Proof. If the change is isothermal, this is a consequence of Moutier s theorem, for the system could be brought back to the initial state by a reversible process, and, by the second law, no work must be obtained in the whole cycle. If it is non-isothermal, we may suppose it to be constructed of a very large number of very small isothermal and adiabatic processes, which may be combined with another corresponding set of perfectlyJ reversible isothermal and adiabatic processes, so that a complete cycle is formed out of a very large number of infinitesimal Carnot s cycles (Fig 11). [Pg.67]

Then if any two phases are separately in equilibrium with a third phase, they are also in equilibrium when placed in contact, so that if any one phase (e.y., the vapour) is taken as a test-phase, and the other phases are separately in equilibrium with this, the whole system will be in equilibrium. Under the conditions imposed, it is sufficient that the vapour pressure, or osmotic pressure, of each component has the same value at all the interfaces, for we may consider each component separately by intruding across the interface a diaphragm permeable to that compo- -nent alone. Then if the vapour, or osmotic pressures, are not equal at the third interface to their values at the first and second interfaces, i.e., at the interfaces on the test-phase, we could carry out a reversible isothermal cycle in which any quantity of a specified component is taken from the test-phase to the phase of higher pressure, then across the interface to the phase of lower pressure, and then back to the test-phase. In this cycle, work would be obtained, which however is impossible. Hence the two phases which are separately in equilibrium with the test-phase are also in equilibrium with each other. This may be called the Law of the Mutual Compatibility of Phases (cf. 106). [Pg.388]

If we remove the mol of gas from the low pressure space, compress it isothermally and reversibly until its pressure rises to 2h> and then introduce it into the high pressure space, the cycle will be completed. The work done with the gas is ... [Pg.464]

Figure 2.10 (a) A schematic Carnot cycle in which isotherms at empirical temperatures 6 and 62 alternate with adiabatics in a reversible closed path. The enclosed area gives the net work produced in the cycle, (b) The area enclosed by a reversible cyclic process can be approximated by the zig-zag closed path of the isothermal and adiabatic lines of many small Carnot cycles. [Pg.59]

We wish to show that no points to the leftbb of 2 on the isotherm 62 are accessible from point 1 via any adiabatic path, reversible or irreversible. Suppose we assume that some adiabatic path does exist between 1 and 2. We represent this path as a dotted curve in Figure 2.11a. We then consider the cycle I —>2 —> 1 — 1. The net heat associated with this cycle would be that arising from the last step 1 — 1, since the other two steps are defined to be adiabatic. We have defined the direction 1 — 1 to correspond to an absorption of heat, which we will call qy. From the first law, the net work vv done in the cycle, is given by w = —q, since AU for the cycle is zero. Thus, for this process, iv is negative (and therefore performed by the system), since qy is positive, having been absorbed from the reservoir. The net effect of this cycle, then, is to completely convert heat absorbed at a high temperature reservoir into work. This is a phenomenon forbidden by the Kelvin-Planck statement of the Second Law. Hence, points to the left of 2 cannot be reached from point 1 by way of any adiabatic path. [Pg.70]

Figure 3.4 Carnot cycle for the expansion and compression of an ideal gas. Isotherms alternate with adiabats in a reversible closed path. The shaded area enclosed by the curves gives the net work in the cyclic process. Figure 3.4 Carnot cycle for the expansion and compression of an ideal gas. Isotherms alternate with adiabats in a reversible closed path. The shaded area enclosed by the curves gives the net work in the cyclic process.
Figure 5. Top Adsorption isotherms of C02 for 1-en at the indicated temperatures. Bottom Adsorption-desorption cycling of C02 for 1-en showing reversible uptake from (a) simulated air (0.39 mbar C02 and 21% 02 balanced with N2) and from (b) simulated flue gas (0.15 bar C02 balanced with N2). (c) time-dependent C02 adsorption for porous materials (A = 1-en, B = mmen-Mg2(dobpdc), C = 1, D = Mg-MOF-74, E = Zeolite 13X, F = MOF-5). (d) C02 adsorption ratio of 1-en in flue gas (after 6 min exposure to 100% RH at 21 °C) to 1-en in flue gas (Adapted from [192]). Figure 5. Top Adsorption isotherms of C02 for 1-en at the indicated temperatures. Bottom Adsorption-desorption cycling of C02 for 1-en showing reversible uptake from (a) simulated air (0.39 mbar C02 and 21% 02 balanced with N2) and from (b) simulated flue gas (0.15 bar C02 balanced with N2). (c) time-dependent C02 adsorption for porous materials (A = 1-en, B = mmen-Mg2(dobpdc), C = 1, D = Mg-MOF-74, E = Zeolite 13X, F = MOF-5). (d) C02 adsorption ratio of 1-en in flue gas (after 6 min exposure to 100% RH at 21 °C) to 1-en in flue gas (Adapted from [192]).
Hysteresis was generally observed in the compression-expansion cycles of the force-area isotherms, indicating that the timescale for relaxation of the fully compressed film back to its expanded state was slower than the movement of the barrier of the Langmuir trough. Our studies, like many others, imply that monolayers are metastable and that reversible thermodynamics can only be applied to their analysis with caution. [Pg.134]

The hydrolysis of methyl acetate (A) in dilute aqueous solution to form methanol (B) and acetic acid (C) is to take place in a batch reactor operating isothermally. The reaction is reversible, pseudo-first-order with respect to acetate in the forward direction (kf = 1.82 X 10-4 s-1), and first-order with respect to each product species in the reverse direction (kr = 4.49 X10-4 L mol-1 S l). The feed contains only A in water, at a concentration of 0.050 mol L-1. Determine the size of the reactor required, if the rate of product formation is to be 100 mol h-1 on a continuing basis, the down-time per batch is 30 min, and the optimal fractional conversion (i.e., that which maximizes production) is obtained in each cycle. [Pg.446]

Most of the thermobalances have built-in, simple heating programs which allow e.g. linear heating and cooling with different rates, change to isothermal conditions, or to cycle the temperature between two preselected values (Fig. 14 E). The application of the latter is to check the reversibility of certain decompositions, or the reproducibility of DTA-peaks. [Pg.97]

Furthermore, in the four steps of the cycle (Fig. 6.8) three are adiabatic (one irreversible, two reversible). Hence, Qcycie is identical with Q of the isothermal step, that is, Q of Equation (6.104). If g > 0, then W < 0 that is, work would have been performed by the system. In other words, if Q were positive, we would have carried out a cyclical process in which heat at a constant temperature had been converted completely into work. According to the Kelvin-Planck statement of the second law, such a process cannot be carried out. Hence, Q cannot be a positive number. As Q must be either negative or zero, it follows from Equation (6.104) that... [Pg.140]

During process 4-1, heat is transferred isothermally from the working substance to the low-temperature reservoir at Tl. This process is accomplished reversibly by bringing the system in contact with the low-temperature reservoir whose temperature is equal to or infinitesimally lower than that of the working substance. The amount of heat transfer during the process is 641= f TdS = Ti Si — S4), which can be represented by the area 1-4-5-6-1 Q41 is the amount of heat removed from the Carnot cycle to a low-temperature thermal reservoir. [Pg.25]

We can achieve the desired equality along each small path segment (with reversible heat exchange dqpath) by choosing isotherm Tt of the mini-Carnot cycle i (with reversible heat exchange dqisothGrm) such that... [Pg.136]

CARNOT CYCLE. An ideal cycle or four reversible changes in the physical condition of a substance, useful in thermodynamic theory. Starting with specified values of die variable temperature, specific volume, and pressure, the substance undergoes, in succession, an isothermal (constant temperature) expansion, an adiabatic expansion (see also Adiabatic Process), and an isothermal compression to such a point that a further adiabatic compression will return the substance to its original condition. These changes are represented on the volume-pressure diagram respectively by ub. he. ctl. and da in Fig. I. Or the cycle may he reversed ad c h a. [Pg.300]

A particular quantity of an ideal gas [Cv = (5/2) R] undergoes the following mechanically reversible steps that together form a cycle. The gas, initially at 1 bar and 300 K, is compressed isothermally to 3 bar. It is then heated at constant P to a temperature of 900 K. Finally, it is cooled at constant volume to its initial state with the extraction of 1,300 J as heat. Determine Q and IV for each step of the cycle and for the complete cycle. [Pg.58]


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See also in sourсe #XX -- [ Pg.115 , Pg.115 , Pg.116 ]




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