Thermodynamics filled in class notes_Part_12

Thermodynamics filled in class notes_Part_12 - 31 2.3....

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Unformatted text preview: 31 2.3. RANKINE The necessary heat transfer rate in the boiler is then ˙ Qin = m(h3 − h2 ), ˙ kg = 104.697 s = 269706 kW , = (2.123) 2758 kJ kg − 181.94 kJ kg , (2.124) (2.125) 269.7 M W. (2.126) In the condenser, one finds ˙ Qout = m(h1 − h4 ), ˙ kg = 104.697 s = −169705 kW, = (2.127) 173.88 kJ kg − 1794.8 kJ kg , (2.128) (2.129) −169.7 M W. (2.130) Note also for the cycle that one should find ˙ ˙ ˙ Wcycle = Qin + Qout = (269.7 M W ) − (169.7 M W ) = 100 M W. (2.131) For the condenser mass flow rate now perform a mass balance: dEcv dt = ˙ ˙ Qcv − Wcv +mc (hin − hout ) + m(h4 − h1 ), ˙ ˙ =0 =0 =0 0= mc ˙ = mc (hin − hout ) + m(h4 − h1 ), ˙ ˙ m(h4 − h1 ) ˙ − , hin − hout = = = − kg s 1794.8 62.99 104.697 = (2.132) kJ kg kg , s kg 2027.79 s kJ kg (2.133) (2.134) − 173.88 − 146.68 kJ kg , (2.135) (2.136) 2027.79 7.3 × 106 kJ kg 3600 s hr , (2.137) kg . hr (2.138) The enthalpy for the cooling water was found by assuming values at the saturated state at the respective temperatures of 15 ◦ C and 35 ◦ C . Example 2.4 Compute the exergy at various points in the flow of a Rankine cycle as considered in the previous example problem. For that example, we had at state 1, the pump inlet that h1 = 173.88 kJ , kg s1 = 0.5926 kJ . kg K (2.139) CC BY-NC-ND. 18 November 2011, J. M. Powers. 32 CHAPTER 2. CYCLE ANALYSIS After the pump, at state 2, we have kJ , kg h2 = 181.94 s2 = 0.5926 kJ . kg K (2.140) After the boiler, at state 3, we have h3 = 2758 kJ , kg s3 = 5.7432 kJ . kg K (2.141) After the turbine, at state 4, we have h4 = 1794.8 kJ , kg s4 = 5.7432 kJ . kg K (2.142) Now for this example, kinetic and potential energy contributions to the exergy are negligible, so we can say in general that ψ = (h − To s) − (ho − To so ). (2.143) Now for this problem, we have To = 298.15 K , ho = 104.89 kJ/kg, so = 0.3674 kJ/kg/K . We have estimated ho and so as the enthalpy and entropy of a saturated liquid at 25 ◦ C = 298.15 K . So the exergies are as follows. At the pump entrance, we get ψ1 = (h1 − To s1 ) − (ho − To so ) , kJ kJ − (298.15 K ) 0.5926 = 173.88 kg kg K kJ kJ − (298.15 K ) 0.3674 − 104.89 kg kg K = 1.84662 (2.144) , kJ . kg (2.145) (2.146) After the pump, just before the boiler, we have ψ2 = (h2 − To s2 ) − (ho − To so ) , kJ kJ = 181.94 − (298.15 K ) 0.5926 kg kg K kJ kJ − (298.15 K ) 0.3674 − 104.89 kg kg K = 9.90662 (2.147) , kJ . kg (2.148) (2.149) So the exergy has gone up. Note here that ψ2 − ψ1 = h2 − h1 because the process is isentropic. Here ψ2 − ψ1 = h2 = h1 = 181.94 − 173.88 = 8.06 kJ/kg . After the boiler, just before the turbine, we have ψ3 = (h3 − To s3 ) − (ho − To so ) , kJ kJ = 2758 − (298.15 K ) 5.7432 kg kg K kJ kJ − (298.15 K ) 0.3674 − 104.89 kg kg K = 1050.32 kJ . kg CC BY-NC-ND. 18 November 2011, J. M. Powers. (2.150) , (2.151) (2.152) ...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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