Editing Enthalpy (section)

===Throttling===
{{main|Joule–Thomson effect}}
[[File:Schematic of throttling.png|thumb|right|Schematic diagram of a throttling in the steady state. Fluid enters the system (dotted rectangle) at point 1 and leaves it at point 2. The mass flow is {{mvar|ṁ}}&nbsp;.]]

One of the simple applications of the concept of enthalpy is the so-called throttling process, also known as [[Joule–Thomson effect|Joule–Thomson expansion]]. It concerns a steady adiabatic flow of a fluid through a flow resistance (valve, porous plug, or any other type of flow resistance) as shown in the figure. This process is very important, since it is at the heart of domestic [[refrigerator]]s, where it is responsible for the temperature drop between ambient temperature and the interior of the refrigerator. It is also the final stage in many types of [[liquefier]]s.

For a steady state flow regime, the enthalpy of the system (dotted rectangle) has to be constant. Hence

<math display="block"> 0 = \dot m h_1 - \dot m h_2 ~.</math>

Since the mass flow is constant, the specific enthalpies at the two sides of the flow resistance are the same:

<math display="block"> h_1 = h_2 \; ,</math>

that is, the enthalpy per unit mass does not change during the throttling. The consequences of this relation can be demonstrated using the {{nobr| {{mvar|T − s}} }} diagram above.

====Example 1====
Point&nbsp;'''c''' is at 200&nbsp;bar and room temperature (300&nbsp;K). A Joule–Thomson expansion from 200&nbsp;bar to 1&nbsp;bar follows a curve of constant enthalpy of roughly 425&nbsp;{{sfrac| kJ |kg}} (not shown in the diagram) lying between the 400 and 450&nbsp;{{sfrac| kJ |kg}} isenthalps and ends in point&nbsp;'''d''', which is at a temperature of about 270&nbsp;K&nbsp;. Hence the expansion from 200&nbsp;bar to 1&nbsp;bar cools nitrogen from 300&nbsp;K to 270&nbsp;K&nbsp;. In the valve, there is a lot of friction, and a lot of entropy is produced, but still the final temperature is below the starting value.

====Example 2====
Point&nbsp;'''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&nbsp;bar ends in the two-phase region (point&nbsp;'''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%.