Temperature entropy diagram

The Carnot refrigeratiou cycle is reversible and consists of adiabatic (iseutropic due to reversible character) compression (1-2), isothermal rejection of heat (2-3), adiabatic expansion (3-4) and isothermal addition of heat (4-1). The temperature-entropy diagram is shown in Fig. 11-70. The Carnot cycle is an unattainable ideal which serves as a standard of comparison and it provides a convenient guide to the temperatures that should be maintained to achieve maximum effectiveness. 

The Joule-Brayton (JB) constant pressure closed cycle is the basis of the cyclic gas turbine power plant, with steady flow of air (or gas) through a compressor, heater, turbine, cooler within a closed circuit (Fig. 1.4). The turbine drives the compressor and a generator delivering the electrical power, heat is supplied at a constant pressure and is also rejected at constant pressure. The temperature-entropy diagram for this cycle is also... 

Temperature - entropy diagram Fig. 1.4. Joule-Brayton cycle (after Ref. [1]). 

The second law of thermodynamics may be used to show that a cyclic heat power plant (or cyclic heat engine) achieves maximum efficiency by operating on a reversible cycle called the Carnot cycle for a given (maximum) temperature of supply (T ax) and given (minimum) temperature of heat rejection (T jn). Such a Carnot power plant receives all its heat (Qq) at the maximum temperature (i.e. Tq = and rejects all its heat (Q ) at the minimum temperature (i.e. 7 = 7, in) the other processes are reversible and adiabatic and therefore isentropic (see the temperature-entropy diagram of Fig. 1.8). Its thermal efficiency is... 

Fig. 1.8. Temperature-entropy diagram for a Carnot cycle (after Ref. (11).

Fig. 1.10. Temperature-entropy diagram. showing reheat, intercooling and recuperation.

Fig. 4.3. Temperature-entropy diagram for two step cooling, (a) Cooling air taken at appropriate pre.ssures and (b) LP cooling air throttled from compressor exit.

Figure 1 IB. Temperature-entropy diagram for water and steam 814...

A three-stage compressor is required to compress air from 140 kN/m2 and 283 K to 4000 kN/m2. Calculate fee ideal intermediate pressures, the work required per kilogram of gas, and fee isothermal efficiency of fee process. Assume the compression to be adiabatic and the interstage cooling to cool the air to the initial temperature. Show qualitatively, by means of temperature-entropy diagrams, fee effect of unequal work distribution and imperfect intercooling, on the performance of the compressor. 

Calculate the ideal intermediate pressures and the work required per kilogram of gas. Assume compression to be isentropic and the gas to behave as an ideal gas. Indicate on a temperature-entropy diagram the effect of imperfect imercooling on the work done at each stage. 

Temperature-entropy diagram, water and steam 814 Temperature gradient, flow over plane surface 688... 

Figure 6.9. Gibbs temperature-entropy diagram for a Camot cycle.

For an isolated (adiabatic) system, AS > 0 for any natural (spontaneous) process from State a to State b, as was proved in Section 6.8. An alternative and probably simpler proof of this proposition can be obtained if we use a temperature-entropy diagram (Fig. 6.13) instead of Figure 6.8. In Figure 6.13, a reversible adiabatic process is represented as a vertical line because AS = 0 for this process. In terms of Figure 6.13, we can state our proposition as follows For an isolated system, a spontaneous process from a to b must lie to the right of the reversible one, because AS = Sb Sa> 0. 

Figure 6.13. Reversible and spontaneous changes of state on a temperature-entropy diagram.

Figure 6.6-7. Pump process in the temperature-entropy diagram...

The temperature-entropy diagram for the Camot cycle, corresponding to the pressure-volume diagram is shown in Fig. 2. 

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