Editing Enthalpy (section)

==Diagrams==
[[File:Ts diagram of N2 02.jpg|500px|thumb|right|{{nobr| {{math|''T'' − ''s''}} }} diagram of nitrogen.<ref>Figure composed with data obtained with [[REFPROP]], NIST Standard Reference Database 23.</ref> The red curve at the left is the melting curve. The red dome represents the two-phase region with the low-entropy side the saturated liquid and the high-entropy side the saturated gas. The black curves give the {{nobr| {{math| ''T'' − ''s'' }} }} relation along isobars. The pressures are indicated in bar. The blue curves are isenthalps (curves of constant enthalpy). The values are indicated in blue in {{sfrac| kJ |kg}}. The specific points&nbsp;'''a''', '''b''', etc., are treated in the main text.]]

The enthalpy values of important substances can be obtained using commercial software. Practically all relevant material properties can be obtained either in tabular or in graphical form. There are many types of diagrams, such as {{nobr|{{math|''h'' − ''T''}} }} diagrams, which give the specific enthalpy as function of temperature for various pressures, and {{nobr| {{math|''h'' − ''p''}} }} diagrams, which give {{mvar|h}} as function of {{mvar|p}} for various {{mvar|T}}. One of the most common diagrams is the temperature–specific entropy diagram ({{nobr| {{math|''T'' − ''s''}} }} diagram). It gives the melting curve and saturated liquid and vapor values together with isobars and isenthalps. These diagrams are powerful tools in the hands of the thermal engineer.

{{clear}}

===Some basic applications===
The points&nbsp;{{math|'''a'''}} through {{math|'''h'''}} in the figure play a role in the discussion in this section.
:{| class="wikitable"  style="text-align:center"
|- 
|Point
! {{mvar|T}} !! {{mvar|p}} !! {{mvar|s}} !! {{mvar|h}} 
|- style="background:#EEEEEE;"
| Unit || [[kelvin (unit)|K]] || [[bar (unit)|bar]] || {{sfrac| kJ | kg K }} || {{sfrac| kJ | kg }} 
|-
| {{math|'''a'''}} || 300 || 1 || 6.85 || 461
|-
| {{math|'''b'''}} || 380 || 2 || 6.85 || 530
|-
| {{math|'''c'''}} || 300 || 200 || 5.16 || 430
|-
| {{math|'''d'''}} || 270 || 1 || 6.79 || 430
|-
| {{math|'''e'''}} || 108 || 13 || 3.55 || 100
|-
| {{math|'''f'''}} || 77.2 || 1 || 3.75 || 100
|-
| {{math|'''g'''}} || 77.2 || 1 || 2.83 || 28
|-
| {{math|'''h'''}} || 77.2 || 1 || 5.41 || 230
|}
Points&nbsp;{{math|'''e'''}} and {{math|'''g'''}} are saturated liquids, and point&nbsp;{{math|'''h'''}} is a saturated gas.

===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%.

===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>