Abstract
In the previous chapters, the law of conservation of energy, which enables thermodynamic systems to be energetically evaluated, has been discussed in detail. A distinction has been made between closed and open systems.
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Notes
- 1.
Such as \(\left[ u\right] ={1}\,{\frac{\mathrm{kJ}}{\mathrm{kg}}}\) and \(\left[ h\right] ={1}\,{\frac{\mathrm{kJ}}{\mathrm{kg}}}\).
- 2.
If two independent state values unambiguously define the thermodynamic state, all other state values must also unambiguously belong to this state and be determinable.
- 3.
The indices at the brackets indicate that this variable is kept constant.
- 4.
At constant specific volume v.
- 5.
At constant pressure p.
- 6.
The thermodynamic proof is given in Sect. 12.3.2.
- 7.
Note that the system is adiabatic, i.e. \(Q_{12}=0\), and that no work is exchanged across the system boundary, i.e. \(W_{12}=0\). The work to open the valve is neglected.
- 8.
Yet this would violate Eq. 12.17.
- 9.
If T remains constant, the specific internal energy does not change. This is exactly what was observed in the experiment.
- 10.
Frankly, it is even difficult to find a physical meaning for the inner energy. Even though you have probably accepted its existence by now.
- 11.
Also known as integrating factor.
- 12.
Similar to the first law of thermodynamics: The state value internal energy is influenced by the process values work and heat.
- 13.
For the Gibbs free energy there is a physical motivation for chemical reactive systems, e.g. fuel cells or Lithium Ion batteries. This is treated in Part III of this book, see Sect. 24.3.
- 14.
No path information is required to solve the integral.
- 15.
A constant specific heat capacity is assumed here. How to proceed if the specific heat capacity is not constant is shown in Sect. 12.4.4.
- 16.
Assuming that \(c_{v}=\text {const.}\) and \(c_{p}=\text {const.}\)
- 17.
This is indicated by the index v.
- 18.
This is indicated by the index p.
- 19.
Very slow is a synonym for no turbulence inside, see also Theorem 7.16.
- 20.
The acceleration due to gravity g is not relevant because the piston is operated horizontally.
- 21.
The change in kinetic energy can be neglected because the change of state is quasi-static. Furthermore, there is no change in potential energy because the cylinder is horizontal.
- 22.
According to \(V=\frac{mRT}{p}\).
- 23.
Note that \(H=mh\).
- 24.
Fluid movement would have to be triggered by a moving piston, for example.
- 25.
Note that \(U=mu\).
- 26.
This part is intended for advanced readers who are already familiar with the T, s-diagram, see Sect. 13.4.
- 27.
Assuming, that \(c_{p}=\text {const.}\)
- 28.
Including arithmetic and logarithmic averaged values.
- 29.
A temperature difference is needed for the calculation, so it makes no difference whether one takes \(\varDelta \vartheta \) or \(\varDelta T\), see Eq. 12.183.
- 30.
The specific enthalpy remains constant.
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Schmidt, A. (2022). Caloric Equations of State. In: Technical Thermodynamics for Engineers. Springer, Cham. https://doi.org/10.1007/978-3-030-97150-2_12
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DOI: https://doi.org/10.1007/978-3-030-97150-2_12
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