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Unformatted text preview: 1 Meeting 20, Sections 710 713 What we covered Friday Entropy change of a Thermal Energy Reservoir Derived the Tds equations Solved the Tds equations applying different assumptions: Incompressible and compressible Constant specific heats and variable specific heats Isentropic and nonisentropic. 2 Topics for today Actual work vs. Ideal work. Isentropic efficiencies. Entropy balance. 3 4 Work in an internally reversible flow system Earlier we had This was true for a quasiequilibrium process. = pdv w 5 Work for an internally reversible flow system. Quasiequilibrium processone for which departures from equilibrium are infinitesimally small. All states through which a system passes in a quasiequilibrium process may be considered to be themselves equilibrium states. 6 Work for an internally reversible flow system A reversible process must proceed through a series of equilibrium states. Otherwise, there would be a tendency for the system to change spontaneously, which is irreversible. Therefore, a quasiequilibrium process is an (internally) reversible process 7 Work for an internally reversible flow system Consider an internally reversible steady flow system: Second law: = 2 1 rev int Tds q 8 Work for an internally reversible flow system The first law (not limited to internally reversible processes at this point) says ( 29 ( 29 ) z z ( g 2 2 h h w q 1 2 2 1 2 2 1 2 + + = V V 9 Work for an internally reversible flow system If we do limit it to internally reversible processes, which we will emphasize with subscripts, Then the heat transfer term can be replaced ( 29 ( 29 ) z z ( g 2 2 h h w q 1 2 2 1 2 2 1 2 rev int rev int + + = V V 10 Work for an internally reversible flow process Combining the laws yields Rearranging ( 29 ( 29 ) z z ( g 2 2 h h w Tds 1 2 2 1 2 2 1 2 rev int 2 1 + + = V V ( 29 ( 29 ( 29 ) z z ( g 2 2 h h Tds w 2 1 2 2 2 1 1 2 2 1 rev int + +  = V V Remember this term 11 Work for an internally reversible flow process Now, use a Tds relation: Tds = dh vdp  = 2 1 2 1 2 1 vdp dh Tds  = 2 1 1 2 2 1 vdp ) h h ( Tds 12 Work for an internally reversible flow process ( 29 ( 29 ) z z ( g 2 2 vdp w 2 1 2 2 2 1 2 1 rev int + + = V V An expression for internally reversible work in a steady flow process. 13 In the absence of KE and PE effects (a stationary system), On a pv diagram, Work for an internally reversible flow process  = 2 1 rev int vdp w p 2 1 v 14 Work For open systems, For closed systems  = 2 1 rev int vdp w = 2 1 rev int pdv w 15 Compressor work By using the relationship pvk = constant (polytropic and isentropic) and solving for , the expression or as the book uses can be integrated to get the expressions on p. 373  = 2 1 rev int vdp w + = 2 1 rev int vdp w k p c v 1 = 16...
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 Summer '07
 RAMUSSEN

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