What we really wanted was a simplification for q cv

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require significant development. What we really wanted was a simplification for ˙ Q cv which we could use in the energy equation, considered next: dE cv dt bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =0 = ˙ Q cv bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright = ˙ m R 2 1 T ds ˙ W cv + ˙ m parenleftbigg h 1 + 1 2 v 2 1 + gz 1 parenrightbigg ˙ m parenleftbigg h 2 + 1 2 v 2 2 + gz 2 parenrightbigg , (9.33) ˙ W cv ˙ m = integraldisplay 2 1 Tds + ( h 1 h 2 ) + 1 2 ( v 2 1 v 2 2 ) + g ( z 1 z 2 ) . (9.34) Now one form of the Gibbs equation, Eq. (8.64), has Tds = dh vdP , so integraldisplay 2 1 Tds = integraldisplay 2 1 dh integraldisplay 2 1 vdP, (9.35) integraldisplay 2 1 Tds = h 2 h 1 integraldisplay 2 1 vdP, (9.36) integraldisplay 2 1 Tds + ( h 1 h 2 ) = integraldisplay 2 1 vdP. (9.37) Now substitute Eq. (9.37) into Eq. (9.34) to get w cv = integraldisplay 2 1 Tds + ( h 1 h 2 ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = R 2 1 vdP + 1 2 ( v 2 1 v 2 2 ) + g ( z 1 z 2 ) , (9.38) w cv = integraldisplay 2 1 vdP + 1 2 ( v 2 1 v 2 2 ) + g ( z 1 z 2 ) . (9.39) CC BY-NC-ND. 2011, J. M. Powers.
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272 CHAPTER 9. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Now if w cv = 0, we get 0 = integraldisplay 2 1 vdP + 1 2 ( v 2 2 v 2 1 ) + g ( z 2 z 1 ) . (9.40) 9.2.1 Incompressible limit In the important limit for liquids in which v is approximately constant, we recall that ρ = 1 /v and write 0 = v integraldisplay 2 1 dP + 1 2 ( v 2 2 v 2 1 ) + g ( z 2 z 1 ) , (9.41) 0 = v ( P 2 P 1 ) + 1 2 ( v 2 2 v 2 1 ) + g ( z 2 z 1 ) , (9.42) 0 = P 2 P 1 ρ + 1 2 ( v 2 2 v 2 1 ) + g ( z 2 z 1 ) . (9.43) We can rewrite this as P ρ + 1 2 v 2 + gz = constant, (9.44) Bernoulli’s equation for an incompressible liquid. In a different limit, that in which changes in kinetic and potential energy can be neglected, Eq. (9.39) reduces to w cv = integraldisplay 2 1 vdP. (9.45) This integral is not the area under the curve in P v space. It is the area under the curve in v P space instead. Contrast this with the system result where we get 1 w 2 = integraltext 2 1 Pdv . In the important operation of pumping liquids, v is nearly constant, and we can say w pump = v ( P 2 P 1 ) , (9.46) ˙ W pump = ˙ mv ( P 1 P 2 ) . (9.47) 9.2.2 Calorically perfect ideal gas limit Let us consider the Bernoulli equation for a CPIG undergoing a reversible adiabatic process. For such a process, we have Pv k = C . Thus v = ( C/P ) 1 /k . Let us consider Eq. (9.40) for CC BY-NC-ND. 2011, J. M. Powers.
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9.2. BERNOULLI’S PRINCIPLE 273 this case: 0 = integraldisplay 2 1 parenleftbigg C P parenrightbigg 1 k dP + 1 2 ( v 2 2 v 2 1 ) + g ( z 2 z 1 ) , (9.48) 0 = k k 1 bracketleftBigg P parenleftbigg C P parenrightbigg 1 k bracketrightBigg P 2 P 1 + 1 2 ( v 2 2 v 2 1 ) + g ( z 2 z 1 ) , (9.49) 0 = k k 1 ( P 2 v 2 P 1 v 1 ) + 1 2 ( v 2 2 v 2 1 ) + g ( z 2 z 1 ) . (9.50) Thus for a CPIG obeying Bernoulli’s principle, we can say, taking v = 1 , k k 1 P ρ + 1 2 v 2 + gz = constant, (9.51) Bernoulli’s equation for a CPIG.
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