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Thermodynamics filled in class notes_Part_14

Thermodynamics filled in class notes_Part_14 - 35 2.3...

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Unformatted text preview: 35 2.3. RANKINE 5 Turbine Boiler 7 6 4 Feedwater Heater Condenser 2 Pump P2 3 Pump P1 1 Figure 2.7: Schematic for Rankine cycle with regeneration and open feedwater heating. First consider the low pressure pump. h2 = = = h1 + v1 (P2 − P1 ), kJ m3 191.8 + 0.00101 kg kg kJ 192.194 kg (2.157) ((400 kP a) − (10 kP a)) , (2.158) (2.159) The pump work is wP 1 = v1 (P1 − P2 ), m3 = 0.00101 kg kJ = −0.3939 . kg (2.160) ((10 kP a) − (400 kP a)) , (2.161) (2.162) Note, a sign convention consistent with work done by the fluid is used here. At this point the text abandons this sign convention instead. CC BY-NC-ND. 18 November 2011, J. M. Powers. 36 CHAPTER 2. CYCLE ANALYSIS Now consider the turbine dmcv dt = m5 − m6 − m7 , ˙ ˙ ˙ (2.163) = m6 + m7 , ˙ ˙ m7 ˙ m6 ˙ + , m5 ˙ m5 ˙ (2.164) ˙ ˙ Qcv −Wcv + m5 h5 − m6 h6 − m7 h7 , ˙ ˙ ˙ (2.166) =0 m5 ˙ 1= dEcv dt = (2.165) =0 =0 ˙ Wcv = m5 h 5 − m6 h 6 − m7 h 7 . ˙ ˙ ˙ (2.167) On a per mass basis, we get, wt ˙ m6 ˙ m7 ˙ Wcv = h5 − h6 − h7 , m5 ˙ m5 ˙ m5 ˙ m6 ˙ m6 ˙ h7 , h6 − 1 − h5 − m5 ˙ m5 ˙ m6 ˙ m6 ˙ h5 − h6 + h6 − h7 , h6 − 1 − m5 ˙ m5 ˙ m6 ˙ m6 ˙ − 1− h7 , h5 − h6 + h6 1 − m5 ˙ m5 ˙ m6 ˙ h5 − h6 + 1 − (h6 − h7 ). m5 ˙ = = = = = (2.168) (2.169) (2.170) (2.171) (2.172) Now consider the feedwater heater. The first law for this device gives dEcv dt = =0 =0 h3 = = = 604.73 kJ kg m6 ˙ m5 ˙ ˙ ˙ ˙ ˙ ˙ Qcv − Wcv +m2 h2 + m6 h6 − m3 h3 , = = (2.173) =0 m6 ˙ m2 ˙ h2 + h6 , m3 ˙ m3 ˙ m7 ˙ m6 ˙ h2 + h6 , m5 ˙ m5 ˙ m6 ˙ m6 ˙ h2 + h6 , 1− m5 ˙ m5 ˙ m6 ˙ kJ 1− 192.194 m5 ˙ kg 0.165451. CC BY-NC-ND. 18 November 2011, J. M. Powers. (2.174) (2.175) (2.176) + m6 ˙ m5 ˙ 2685.6 kJ kg , (2.177) (2.178) ...
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