Move work

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In thermodynamics , displacement work describes the product of pressure and volume of an amount of substance. It has the dimension of energy and, as the product of two state variables, is itself a state variable, although as work it bears the name of a process variable. It is the difference between enthalpy and internal energy of a thermodynamic system . ${\displaystyle \textstyle p}$ ${\displaystyle \textstyle V}$

In technical thermodynamics, it is useful when describing open systems through which a material flow flows, e.g. B. Compressor . There, the difference between the shifting work in front of and behind the system corresponds to the work that has to be expended to transport the amount of substance through the system.

application

Consider a fluid volume that is in the feed line (with the cross-sectional area ) of an open, flow-through system and takes up the length of the feed line there . The pressure prevails at this point in the supply line . If the flow process is quasi-stationary , then the forces at the interface between the volume under consideration and the subsequent volume are balanced and the subsequent volume exerts the force on the volume under consideration . As soon as the volume under consideration has been shifted by its own length , the subsequent volume does the work ${\displaystyle \textstyle V_{\mathrm {vor} }}$ ${\displaystyle \textstyle A_{\mathrm {ein} }}$ ${\displaystyle \textstyle s_{\mathrm {vor} }=V_{\mathrm {vor} }/A_{\mathrm {ein} }}$ ${\displaystyle \textstyle p_{\mathrm {vor} }}$ ${\displaystyle \textstyle F_{\mathrm {ein} }=p_{\mathrm {vor} }\,A_{\mathrm {ein} }}$ ${\displaystyle \textstyle s_{\mathrm {vor} }}$

${\displaystyle W_{\mathrm {ein} }=F_{\mathrm {ein} }s_{\mathrm {vor} }=(p_{\mathrm {vor} }A_{\mathrm {ein} })s_{\mathrm {vor} }=p_{\mathrm {vor} }(A_{\mathrm {ein} }s_{\mathrm {vor} })=p_{\mathrm {vor} }V_{\mathrm {vor} }}$

spent on the system.

The system turns the work accordingly on its exit surface

${\displaystyle W_{\mathrm {aus} }=p_{\mathrm {nach} }V_{\mathrm {nach} }}$

to move the amount of substance out of the system.

The difference between the two jobs is therefore the work that is necessary to transport the amount of substance through the system. ${\displaystyle W_{\mathrm {ein} }-W_{\mathrm {aus} }=p_{\mathrm {vor} }V_{\mathrm {vor} }-p_{\mathrm {nach} }V_{\mathrm {nach} }}$

Relationship with process variables

The change in the shift work depends on process variables such as the volume change work or the technical work (e.g. shaft work). This becomes clear with the total differential of : ${\displaystyle W_{\mathrm {v} }}$ ${\displaystyle W_{\mathrm {t} }}$ ${\displaystyle W=p\,V}$

${\displaystyle {\begin{aligned}\mathrm {d} (p\,V)=&~V\mathrm {d} p+p\,\mathrm {d} V\\\Rightarrow \int _{\mathrm {vor} }^{\mathrm {nach} }\mathrm {d} (p\,V)=&\int _{\mathrm {vor} }^{\mathrm {nach} }V\mathrm {d} p+\int _{\mathrm {vor} }^{\mathrm {nach} }p\,\mathrm {d} V\end{aligned}}}$

The difference in the displacement work corresponds to the technical work minus the volume change work : ${\displaystyle p_{\mathrm {nach} }\,V_{\mathrm {nach} }-p_{\mathrm {vor} }\,V_{\mathrm {vor} }}$ ${\displaystyle W_{\mathrm {v} }=-\int p\cdot \mathrm {d} V}$

${\displaystyle \Leftrightarrow p_{\mathrm {nach} }\,V_{\mathrm {nach} }-p_{\mathrm {vor} }\,V_{\mathrm {vor} }=W_{\mathrm {t,nach-vor} }-W_{\mathrm {v,nach-vor} }}$

Using the compressor as an example, the volume change work for compressing the gas flow is supplied to the system on the one hand, and the difference in displacement work must be overcome on the other hand : ${\displaystyle W_{\mathrm {nach} }-W_{\mathrm {vor} }}$

${\displaystyle \Leftrightarrow W_{\mathrm {t} ,\mathrm {nach} -\mathrm {vor} }=W_{\mathrm {v} ,\mathrm {nach} -\mathrm {vor} }+W_{\mathrm {nach} }-W_{\mathrm {vor} },}$

where the technical work z. B. is provided via an electric motor .

example

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An effect known from practice, which goes back to the displacement work, occurs when a gas cylinder is emptied or filled . First of all, let the gas bottle with the volume be closed with a valve. The gas inside, with the indiv. Gas constant and the spec. isochoric heat capacity , is under pressure and has the ambient temperature . In this case, the energy and mass for the gas are also known. With the thermal equation of state applies to the gas mass ${\displaystyle V}$ ${\displaystyle R}$ ${\displaystyle c_{v}}$ ${\displaystyle p}$ ${\displaystyle T_{1}}$

${\displaystyle m_{1}={\frac {p\,V}{R\,T_{1}}}}$

and for the internal energy applies:

${\displaystyle U_{1}=m_{1}\,c_{v}\,T_{1}={\frac {c_{v}}{R}}\,p\,V}$

In this case, despite the expression " ", no shift work occurs, as this is only defined at the system boundary and therefore only occurs in open systems. The product of volume and pressure is expressed as internal energy (for an ideal gas according to the thermal equation of state). ${\displaystyle pV}$

If you now open the valve and the pressure inside is greater than the ambient pressure, the gas will escape. The following applies to the mass balance of the open system

${\displaystyle {\frac {\mathrm {d} m}{\mathrm {d} t}}=-{\dot {m}}}$

whereby the mass flow flows over the system boundary. At the same time, the energy inside the gas bottle will also decrease. The specific internal energy is first dissipated with the mass flow: ${\displaystyle {\dot {m}}}$ ${\displaystyle c_{v}\,T}$

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} U}{\mathrm {d} t}}=&-{\dot {m}}\,c_{v}\,T\\[0,1cm]{\frac {\mathrm {d} m}{\mathrm {d} t}}\,c_{v}\,T+m\,c_{v}\,{\frac {\mathrm {d} T}{\mathrm {d} t}}=&~{\frac {\mathrm {d} m}{\mathrm {d} t}}\,c_{v}\,T\\[0,1cm]{\frac {\mathrm {d} T}{\mathrm {d} t}}=&~0\\\end{aligned}}}$

You can see: With this change in energy, the temperature remains constant. However, this does not correspond to experience, because the gas actually has to perform additional displacement work, which is expressed in a change in temperature. Taking into account the specific shift work, the following still applies: ${\displaystyle p\,v}$

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} U}{\mathrm {d} t}}=&-{\dot {m}}\,c_{v}\,T-{\dot {m}}\,p\,v\\[0,1cm]{\frac {\mathrm {d} m}{\mathrm {d} t}}\,c_{v}\,T+m\,c_{v}\,{\frac {\mathrm {d} T}{\mathrm {d} t}}=&~{\frac {\mathrm {d} m}{\mathrm {d} t}}\,c_{v}\,T+{\frac {\mathrm {d} m}{\mathrm {d} t}}\,R\,T\\[0,1cm]{\frac {c_{v}}{R}}\,{\frac {1}{T}}\,{\frac {\mathrm {d} T}{\mathrm {d} t}}=&~{\frac {1}{m}}\,{\frac {\mathrm {d} m}{\mathrm {d} t}}\\[0,1cm]{\frac {c_{v}}{R}}\,\int _{1}^{2}{\frac {\mathrm {d} T}{T}}=&~\int _{1}^{2}{\frac {\mathrm {d} m}{m}}\\\end{aligned}}}$

With integration over the change in mass within the gas bottles, in the interval [1,2], a relationship for the gas temperature is obtained:

${\displaystyle T_{2}=T_{1}\left({\frac {m_{2}}{m_{1}}}\right)^{\frac {R}{c_{v}}}}$

This shows that the gas inside the bottle cools down when it is emptied.

See also

• Gay Lussac attempt
• Joule-Thomson effect

literature

• E. Hahne: Technical Thermodynamics: Introduction and Application. Oldenbourg Verlag, Munich 2010, ISBN 978-3-486-59231-3 .
• W. Schneider, St. Haas, K. Ponweiser: Repetitorium Thermodynamik. Oldenbourg Verlag, Munich 2012, ISBN 978-3-486-70779-3 .

Individual evidence

1. ↑ R. Pischinger, M. Klell, T. Sams: Thermodynamics of the internal combustion engine. Springer-Verlag, 2009, ISBN 978-3-211-99277-7 , chap. 1.2, p. 3.
2. ^ E. Doering, H. Schedwill, M. Dehle: Fundamentals of technical thermodynamics. Vieweg + Teubner Verlag, Wiesbaden 2008, ISBN 978-3-8351-0149-4 , limited preview in the Google book search
3. ^ A. Dittmann, S. Fischer, J. Huhn, J. Klinger: Repetitorium der Technischen Thermodynamik Teubner Verlag, Wiesbaden 1995, ISBN 3-519-06354-9 .