Claude Liquefaction Process Claude Liquefaction Process: Equations [PDF]

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9/27/2012



Claude Liquefaction Process



Claude Liquefaction Process



• Support the throttle valve by an expander: – Gas expander: saturated or slightly superheated vapor – cooled and throttled to produce liquefaction (as in the Linde process) – unliquefied portion mixes with the expander exhaust and returns for recycle. Fig 9.7



Equations



Mass relations: m1= m2 + m15; m2= m3 = m4= m5; m5 = m6 + m11; m6 = m7 = m8; m8 = m9 + m10; m13= m10 + m12 = m14 = m15= m2



Natural gas, assumed here to be pure methane, is liquefied in a Claude process. Compression is to 60 bar and precooling is to 300 K. The expander and throttle exhaust to a pressure of 1 bar. Recycle methane at this pressure leaves the exchanger system at 295 K. Assume no heat leaks into the system from the surroundings, an expander efficiency of 75%, and an expander exhaust of saturated vapor. For a draw-off to the expander of 25% of the methane entering the exchanger system, what fraction of the methane is liquefied, and what is the temperature of the high-pressure steam entering the throttle valve? For superheated methane: H 4  1140.0



kJ kg



(at 300 K and 60 bar )



kJ kg



(at 295 K and 1 bar )



H15  1188.9



For saturated liquid: H 9  285.4 Claude process



Linde Process: x = 0



kJ kg



For saturated vapor: H12  796.9



(T sat  111.5 K and 1 bar )



kJ kJ , S12  9.521 (T sat  111.5 K and 1 bar ) kg kg  K



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9/27/2012



An energy balance on the right of the dashed vertical line: m 9 H 9  m 15H15  m 4 H 4  Wout



Claude Liquefaction Process



The expander operates adiabatically: Wout  m 12 ( H12  H 5 ) z  m 9 / m 4  12 / m 4 xm A mass balance: m 15  m 4  m 9



60 bar



60 bar H4= 1140



z 1 bar



60 bar



60 bar



xH12  H 5   H 4  H15 H 9  H15



  H5  The equation defining expander efficiency: H  H12  H 5   H S   H12



1 bar 60 bar



1 bar



Guess T5 → H5, S5 → isentropic expansion → H’12 → H12 → check if satisfied?



H15=1188.9



1 bar



T5  253.6 K , H 5  1009.8



1 bar



1 bar H12 = 796.9 S12 = 9.521



z



kJ (at 60 bar ) kg



xH12  H 5   H 4  H15 0.25(796.9  1009.8)  1140.0  1188.9   0.113 H 9  H15 285.4  1188.9



H9 =289.4



11.3 % of the methane entering the exchanger system is liquefied!



An energy balance on the exchanger I: m 4 ( H 5  H 4 )  m 15 ( H15  H14 )  0 A mass balance: m 15  m 4  m 9 H14 



9 / m 4 zm



H5  H 4 1009.8  1140.0 kJ  H15   1188.9  1042.1 1 z 1  0.113 kg T14  227.2 K



(at 60 bar )



An energy balance on the exchanger II: m 7 ( H 7  H 5 )  m 14 ( H14  H12 )  0 A mass balance: m 7  m 4  m 12



 14  m 4 m 9 m



1 z H14  H12   719.8 kJ H7  H5  1 x kg



x  T7 



T7  197.6 K



For the Linde system, x = 0: z 



(at 60 bar )



xH12  H 5   H 4  H15 H 9  H15



z  0.0541



5.41 % of the methane entering the throttle valve emerges as liquid! kJ H 7  H 4  (1  z )H15  H10   769.2 kg T7  206.6 K (at 60 bar )



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