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2022 | OriginalPaper | Chapter

21. Steady State Flow Processes

Author : Achim Schmidt

Published in: Technical Thermodynamics for Engineers

Publisher: Springer International Publishing

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Abstract

Steady state flow processes have already been discussed in the introduction of the first law of thermodynamics for open systems, see Sect. 11.3. Figure 21.1 shows an example for a simple open system with a single inlet and a single outlet.

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Footnotes
1
The total pressure is composed of the static pressure p to which a person following the flow at its velocity c would be subjected, the dynamic pressure \(\frac{\rho }{2}c^{2}\) and the hydraulic pressure \(g\rho z\).
 
2
The flow cross-section does not have to be constant.
 
3
In general, the potential energies can be neglected as long as the vertical distance between inlet and outlet is not very large.
 
4
The velocity c is always positive, since the direction of the velocity vector has already been taken into account in the sketches by the flow in or out.
 
5
This is due to \(\vartheta _{1}>\vartheta '(p_{1})\).
 
6
No matter what fluid is used, cf. Sect. 13.​6.
 
7
Note the units. The specific enthalpies should be converted to \({\frac{{\text {J}}}{{\text {kg}}}}\).
 
8
A nozzle is a work-insulated, i.e. passive, system.
 
9
Imagine a pressure pulse moving from right to left with a velocity \(a={100}\,{\frac{{\text {km}}}{{\text {h}}}}\) and is subjected to a headwind of \(c={20}\,{\frac{{\text {km}}}{{\text {h}}}}\). Thus, the fluid hits the pulse front with a velocity of \({120}\,{\frac{{\text {km}}}{{\text {h}}}}\). However, the velocity of the fluid behind the impulse shall be slightly higher, e.g. \(c+\text {d}c={22}\,{\frac{{\text {km}}}{{\text {h}}}}\), so that the velocity increases by \(\text {d}c={2}\,{\frac{{\text {km}}}{{\text {h}}}}\). If the air is to hit the pulse front with exactly the velocity \(a={100}\,{\frac{{\text {km}}}{{\text {h}}}}\), the coordinate system must move to the left with a velocity of \(a-c={80}\,{\frac{{\text {km}}}{{\text {h}}}}\). In this case, the velocity of the air leaving the coordinate system on the right side is \(a+dc={102}\,{\frac{{\text {km}}}{{\text {h}}}}\). Note that any coordinate system can be used for balancing. The solution must never depend on the choice of coordinate system.
 
10
Gino Girolamo Fanno ( https://static-content.springer.com/image/chp%3A10.1007%2F978-3-030-97150-2_21/MediaObjects/482881_2_En_21_Figa_HTML.gif 18 November 1882 in Conegliano, https://static-content.springer.com/image/chp%3A10.1007%2F978-3-030-97150-2_21/MediaObjects/482881_2_En_21_Figb_HTML.gif 23 March 1962 in Pegli).
 
11
The potential energies are neglected. Also, no technical work occurs across the system boundary.
 
12
For instance by its internal state \(p_{1}\), \(T_{1}\) and its velocity \(c_{1}\).
 
13
\(T_{\text {ref}}\), \(p_{\text {ref}}\), \(h_{\text {ref}}\) and \(s_{\text {ref}}\) represent a arbitrary reference state.
 
14
Compare with Problem 12.​5.
 
15
Sure, it can remain constant, but then there would be no driver for any changes of the flow.
 
16
\(T_{\text {ref}}\), \(p_{\text {ref}}\), \(h_{\text {ref}}\) and \(s_{\text {ref}}\) represent an arbitrary reference state.
 
17
Thus, at high subsonic velocities.
 
18
For a given \(p_{0}\) it is not possible to reach a pressure smaller than \(p_{\text {crit}}\) in a converging nozzle, see Fig. 21.25.
 
19
The flow is adiabatic and reversible, i.e. it is isentropic.
 
20
Simplified, it can be said that at constant mass flux, the mass flux density must become greatest at the narrowest cross-section.
 
21
This is equivalent with a constant total pressure. Note that no dissipation occurs.
 
Metadata
Title
Steady State Flow ProcessesProcesssteady state
Author
Achim Schmidt
Copyright Year
2022
Book
Technical Thermodynamics for Engineers

Print ISBN: 978-3-030-97149-6

Electronic ISBN: 978-3-030-97150-2

Copyright Year: 2022

DOI
https://doi.org/10.1007/978-3-030-97150-2_21

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