Editing Enthalpy (section)

===Compressors===
{{main|Compressor}}
[[File:Schematic of compressor.png|thumb|right|Schematic diagram of a compressor in the steady state. Fluid enters the system (dotted rectangle) at point&nbsp;1 and leaves it at point&nbsp;2. The mass flow is {{mvar|ṁ}}. A power {{mvar|P}} is applied and a heat flow {{mvar|Q̇}} is released to the surroundings at ambient temperature {{math|''T''{{sub|a}}}}&nbsp;.]]

A power {{mvar|P}} is applied e.g. as electrical power. If the compression is [[adiabatic]], the gas temperature goes up. In the reversible case it would be at constant entropy, which corresponds with a vertical line in the {{math|''T'' − ''s''}} diagram. For example, compressing nitrogen from 1&nbsp;bar (point '''a''') to 2 bar (point '''b''') would result in a temperature increase from 300&nbsp;K to 380&nbsp;K. In order to let the compressed gas exit at ambient temperature {{math|''T''{{sub|a}}}}, heat exchange, e.g. by cooling water, is necessary. In the ideal case the compression is isothermal. The average heat flow to the surroundings is {{mvar|Q̇}}. Since the system is in the steady state the first law gives

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

The minimal power needed for the compression is realized if the compression is reversible. In that case the [[second law of thermodynamics]] for open systems gives

<math display="block"> 0 = -\frac{\, \dot Q \,}{T_\mathsf{a}} + \dot m \, s_1 - \dot m \, s_2 ~.</math>

Eliminating {{mvar|Q̇}} gives for the minimal power

<math display="block"> \frac{\, P_\mathsf{min} \,}{ \dot m } = h_2 - h_1 - T_\mathsf{a}\left( s_2 - s_1 \right) ~.</math>

For example, compressing 1&nbsp;kg of nitrogen from 1&nbsp;bar to 200&nbsp;bar costs at least :{{nobr| {{math| {{big|(}} ''h''{{sub|'''c'''}} − ''h''{{sub|'''a'''}} {{big|)}} − ''T''{{sub|a}}( ''s''{{sub|'''c'''}} − ''s''{{sub|'''a'''}} )}} .}}
With the data, obtained with the {{nobr| {{math|''T'' − ''s''}} }} diagram, we find a value of {{nobr| {{math|(430 − 461) − 300 × (5.16 − 6.85) {{=}} 476 }}{{sfrac| kJ |kg}} .}}

The relation for the power can be further simplified by writing it as

<math display="block"> \frac{\, P_\mathsf{min} \,}{ \dot m } = \int_1^2 \left( \mathrm{d}h - T_\mathsf{a}\,\mathrm{d}s \right) ~.</math>

With
: {{nobr| {{math| d''h'' {{=}} ''T''&thinsp;d''s'' + ''v''&thinsp;d''p'' }} ,}}
this results in the final relation

<math display="block">\frac{\, P_\mathsf{min} }{ \dot m } = \int_1^2 v\,\mathrm{d}p ~.</math>