4.11 - 4 4.11 (a) This is a Joule-Thomson expansion H (70...

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4.11 (a) This is a Joule-Thomson expansion ⇒= () == ° ≈= = ± ? ± . ± . HTH T H T 70 bar, 10133 400 1 400 32782 bar, C bar, C kJ kg ° and T = , ° 447 C ± . S = 6619 kJ kg K (b) If turbine is adiabatic and reversible , then ² S gen = 0 ch ±± . SS out in kJ kg K 6619 and P = 1013 . bar. This suggests that a two-phase mixture is leaving the turbine Let fraction vapor kJ kg K kJ kg K V L x S S = = = ± . ± . 73594 13026 Then xx 7 3594 1 13026 .. . +− ( ) = kJ kg K or x = 08778 . . Therefore the enthalpy of fluid leaving turbine is ± H HH + ×= 08788 26755 1 08778 417 46 2399 6 VL sat’d, 1 bar sat’d, 1 bar kJ kg . ² ± ² ± MH M H in in out out Energy balance 0 =+ + ² Q 0 ² WP s dV dt 0 but ²² MM in out =− =−= ² ² . W M s in kJ kg 2399 6 8786 (c) Saturated vapor at 1 bar ± .; ± . ² . % . . . ² ² . SH W M S M s −= = () = × = 7 3594 26755 26755 602 7 602 7 100 686% 7 3594 0740 kJ kg K kJ kg kJ kg Efficiency kJ Kh in Actual gen in
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(d) 0 0 0 12 2 1 11 2 2 =+⇒ = =− + + + + ±± ± ± ± ²² ±± ± ²² ± ± MM M M MH H W Q P dV dt MS S Q T S s ch
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4.11 - 4 4.11 (a) This is a Joule-Thomson expansion H (70...

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