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Analysis of Organic Rankine Cycles for a Boiler Station

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Analysis of Organic Rankine Cycles for a Boiler Station ( analysis-organic-rankine-cycles-boiler-station )

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Table 2.1 Description of technical terms relating to thermodynamic processes Adiabatic Isobaric Isentropic Polytropic No heat transfer Constant pressure Constant entropy pV n = Constant Isothermal Isochoric Isenthalpic Reversible process Constant temperature Constant volume Constant specific enthalpy Zero entropy generation work, and deposits the remaining heat to the cold reservoir. The Carnot cycle consists of four processes, occuring in the following order [33]: (1) →− (2) Isothermal expansion. Heat is transfered from the hot source into the system at constant pressure whereby the volume of the system increases and work is performed by the system on its boundary. (2) →− (3) Reversible adiabatic expansion. The system is continuing to expand but now during a state of reduced pressure, but constant temperature. The system boundary is increasing in volume and work is being performed by the system on its boundary. (3) →− (4) Reversible Isothermal compression. The surroundings do work on the system boundary, increasing the pressure. The system is rejecting heat into the cold source, at constant pressure while it is being compressed. (4) →− (1) Reversible adiabatic compression. The surroundings continue to do work on the system, compressing it further resulting in a pressure and temperature increase of the system. A pressure-volume diagram can be used to illustrate the changes in volume and pressure of a system, shown in figure 2.1. For cyclic processes, the work output can be estimated as the area enclosed by the curve when travelling through the sequence of processes. The temperature-entropy diagram is another common diagram to visualize the cycle, shown in figure 2.2. The work output of during one cycle can thus be found by integrating the pressure over the volume. By knowing that the internal energy of the system is the same after one cycle, i.e. that 􏳼 dU = 0, the work is 􏳣􏳣􏳣􏳣 W= P·dV= (dQ−dU)= (T·dS−dU)= T·dS=(Th−Tc)∆S Reasoning using the second law of thermodynamics the work output may also be found. From the hot source an amount of energy amount equivalent to TH ∆S is extracted and into the cold source TH∆S is deposited. From this the maximum amount of work output is found as the difference 19

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