Elementary Principles 261 - 7.33 (a) P = 5 bar Table B.6...

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7-16 7.33 (a) PT o == 5 bar C Table B.6 saturation 1518 . . At 75°C the discharge is all liquid (b) Inlet : T =350°C, P =40 bar Table B.7 in = 3095 kJ / kg ± H , ± V in 3 = 0.0665 m / kg Outlet : T =75°C, P =5 bar Table B.7 out = 314.3 kJ / kg ± H , ± V out -3 3 =1.03 10 m /kg × u V A u V A in in in 3 22 out out out 3 kg 1 min m / kg min 60 s 0.075) / 4 m m/s kg 1 min 0.00103 m / kg min 60 s 0.05) / 4 m = = ² . ( . ² ( . 200 0 0665 5018 200 175 π Energy balance: ²² ² ² ² ( ±± ) ² () QW H E mH H m uu sk −≈ + = + ΔΔ 21 2 2 1 2 2 2 2 200 kg 1 min (314-3095) kJ 200 kg 1 min (1.75 -50.18 ) m min 60 s kg 2 min 60 s s 13,460 kW ( 13,460 kW transferred from the turbi s −= + =− ² ² ne) 7.34 (a) Assume all heat from stream transferred to oil 1.00 10 kJ 1 min min 60 s kJ s 4 ² Q = × = 167 25 bars, sat'd 100 kg oil/min ² m (kg H 2 O (v) /s) 25 bars, sat'd 135°C 100 kg oil/min 185°C ² m (kg H 2 O (l) /s) Energy balance on H O:
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