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I was reading Fundamentals of Engineering Thermodynamics by Moran and others and in the discussion of open and closed systems I found this -

In subsequent sections of this book, we perform thermodynamic analyses of devices such as turbines and pumps through which mass flows. These analyses can be conducted in principle by studying a particular quantity of matter, a closed system, as it passes through the device. In most cases it is simpler to think instead in terms of a given region of space through which mass flows. With this approach, a region within a prescribed boundary is studied. The region is called a control volume

What I interpret is that we can analyze flow devices like turbines and pumps by choosing a fixed quantity of matter, a closed system, as well. Only that an analysis performed using a closed system would be challenging as compared to if we use an open system, a control volume.

I was wondering how the analysis of flow devices can be performed by choosing a closed system. How the results that are obtained by choosing a control volume around the flow device can also be obtained by choosing a closed system?

Similar but not the same question

Harshit Rajput
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2 Answers2

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You can use integral balance equations for a material volume (Lagrangian approach, closed system), for a fixed control volume (Eulerian approach) or for a arbitrary control volume. You only need to transform the term with time derivative of the mass, momentum or total energy using Reynolds' transport theorem.

Take a look at https://basics.altervista.org/test/Physics/CM/BalanceEquations/integralBalanceEqns.html and the links therein (here a full map of the balance equations in continuum mechanics, https://basics.altervista.org/test/Physics/CM/BalanceEquations/main.html)

basics
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Moran et al have the derivation of the open system 1st law equations, based on a closed system. See section 4.4. I would have drawn Fig.4.5 a little differently, with "before" and "after" diagrams. The before diagram would show only a tiny (differential) amount of material (to the left) about to enter the control volume at time t, and no material having yet exited the control volume. The differential amount about to enter plus the material within the control volume would represent my closed system. During the time interval between t and t+dt, this differential amount of material would enter the control volume, and a tiny corresponding differential amount of material would just exit the control volume (to the right). The picture at t+dt would be the "after" picture of the closed system.

So work $p_{in} v_{in} A_{in}dt=\frac{p_{in}}{\rho_{in}}\dot{m}_{in}dt$ would be the work done on the closed system to push the material into the control volume (by the upstream fluid) during the time interval dt and the work $p_{out}v_{out}A_{out}dt=\frac{p_{out}}{\rho_{out}}\dot{m}_{out}dt$ would be the work done by the closed system to push the corresponding material out of the control volume (done on the downstream fluid) during the time interval dt.

This is only part of the work. The rest of the work is the shaft work during the time interval dt. The work to push material into and out of the control volume is treated separately from the shaft work. It is lumped together with the kinetic energy and potential energy in the differential amounts in the portion of the closed system entering and leaving the control volume during dt.

Hope this helps.

Chet Miller
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