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Enthalpy
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====Example 2==== Point '''e''' is chosen so that it is on the saturated liquid line with {{nobr| {{math|''h'' {{=}} 100 }}{{sfrac| kJ |kg}} .}} It corresponds roughly with {{nobr| {{math|''p'' {{=}} 13}} bar }} and {{nobr| {{math|''T'' {{=}} 108}} K .}} Throttling from this point to a pressure of 1 bar ends in the two-phase region (point '''f'''). This means that a mixture of gas and liquid leaves the throttling valve. Since the enthalpy is an extensive parameter, the enthalpy in {{nobr|'''f''' {{math|( ''h''{{sub|'''f'''}} )}} }} is equal to the enthalpy in {{nobr| '''g''' {{math|( ''h''{{sub|'''g'''}} )}} }} multiplied by the liquid fraction in {{nobr| '''f''' {{math|( ''x''{{sub|'''f'''}} )}} }} plus the enthalpy in {{nobr| '''h''' {{math|( ''h''{{sub|'''h'''}} )}} }} multiplied by the gas fraction in {{nobr|'''f''' {{math| (1 − ''x''{{sub|'''f'''}} )}} .}} So <math display="block"> h_\mathbf\mathsf{f} = x_\mathbf\mathsf{f} h_\mathbf\mathsf{g} + (1 - x_\mathbf\mathsf{f})h_\mathsf\mathbf{h} ~.</math> With numbers: : {{nobr|{{math| 100 {{=}} ''x''{{sub|'''f'''}} × 28 + {{big|(}}1 − ''x''{{sub|'''f'''}}{{big|)}} × 230 }} ,}} so {{nobr|{{math|''x''{{sub|'''f'''}} {{=}} 0.64}} .}} This means that the mass fraction of the liquid in the liquid–gas mixture that leaves the throttling valve is 64%.
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