# I Can't understand how flow work is possible in a control volume

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1. Dec 21, 2017

### kibestar

I am struggling to fully grasp the concept of flow work for a non-deformabable control volume.

Nearly every source puts it in this way: flow work is the work required to push fluid into and out of the control volume and as such is defined as Pv on a unit mass basis. But how can this work be accomplished if the control volume's boundary is fixed in space and has no width? As I see it, fluid that is outside the control volume cannot exert a force due to pressure on fluid inside the control volume through a finite distance, since any differential displacement would put the upstream parcel of fluid inside the control volume. If flow work then refers to work done on fluid that hasn't yet crossed the boundary, why wouldn't this energy transfer be accounted for by the internal energy?

How to reconcile this? Where am I wrong?

2. Dec 21, 2017

### Staff: Mentor

This is a difficult concept to explain. Suppose that, at time t, we were to supplement the material in the control volume by the tiny amount of mass that is about to enter between times t and $t+\Delta t$. The mass of this material would be $\rho v A \Delta t=\dot{m}\Delta t$ and the work to push it in would be $pvA\Delta t$, where p is the pressure at the entrance. Combining these two results to eliminate vA, we would have $$dW_{in}=\left[\dot{m}\frac{p}{\rho}\right]_{in}\Delta t$$. By time $t + \Delta t$, a corresponding supplemental amount of mass will have come out of the control volume, and the work to push this mass out would be
$$dW_{out}=\left[\dot{m}\frac{p}{\rho}\right]_{out}\Delta t$$So the net amount of work done by the material in the supplemented control volume on its surroundings during the time interval would be $$dW_{net}=\left(\left[\dot{m}\frac{p}{\rho}\right]_{in}-\left[\dot{m}\frac{p}{\rho}\right]_{out}\right)\Delta t$$If we now let the supplemental amount of material approach zero by taking the limit as the time interval becomes very small, we obtain the net rate at which the material in the control volume is doing work on its surroundings at the inlet and outlet:
$$\frac{dW}{dt}=\left[\dot{m}\frac{p}{\rho}\right]_{in}-\left[\dot{m}\frac{p}{\rho}\right]_{out}$$The represents the work to push material out of the control volume against the "piston" formed by the material ahead of it minus the work to push material into the control volume by the "piston" formed by the material behind it.

3. Dec 21, 2017

### kibestar

I can definitely understand it mathematically, but physically it just seems like some "trick". As soon as those delta's are made to approach zero I just can't spiral out of "if it is supposed to be a form of boundary work, how does a non-moving boundary provide it??". At the same time, I realize that if enthalpies weren't taken at the inlets and outlets, the first law wouldn't hold. I guess I'm looking for a physically conceptual explanation that doesn't revolve around "if it weren't like this, the outcome would be absurd".

Am I the only one who finds this extremely unintuitive and (albeit wrongly) senseless?

4. Dec 21, 2017

### Staff: Mentor

No. Everyone struggles with this.

Just because we have an imaginary boundary across the entrance or the exit of a control volume (that we draw with a dashed line in a figure) does not mean that no work can be occurring across the boundary. The material on both sides of the boundary communicate with one another in terms of pressure, and the actual material is moving, so that there are displacements taking place. This is much different than the case of a rigid boundary, where no work can occur. An imaginary boundary does nothing physical.

What I've done in post #2 is to modify the system slightly so that we are back to dealing with a closed system, consisting of the material within the actual control volume at time t, plus the amount of material about to enter between times t and t +delta t. This is the same as if we had two pistons, one adjacent to the inlet and one adjacent to the outlet exchanging work with the material within the closed system. As the time interval becomes vanishingly small, we approach the limit of the actual control volume, and obtain a "rate of doing work" at the two boundaries.

5. Dec 21, 2017

### kibestar

The thing is, to me, when the interval becomes vanishingly small, the situation stops making sense physically. Following that train of thought, I can only see it as being logically consistent if time evolved in discrete steps. This is how my mind reads this:

The only moment at which fluid out of the control volume is exerting a force on fluid in the control volume is when those two parcels are separated by the imaginary boundary. The slightest displacement would cause the contact plane between both parcels to enter the controle volume. From this point on, whatever work is being done by the upstream fluid on the downstream fluid is being offset by the work done by the downstream fluid on the upstream fluid, which are equal in magnitude but opposite in sign. On a continuous operation, I see this flow work as being the infinite sum, at all points in time, of the work done by the upstream fluid on downstream fluid. Except there isn't an infinitesimal displacement during which fluid from out of the control volume transfers energy to fluid in the control volume. The "displacement" is exactly equal to 0, which is the width of the boundary of the control volume.

Am I missing some fundamental aspect?

6. Dec 21, 2017

### Staff: Mentor

Well, I tried my best to explain it. So maybe I should give someone else at shot. I'm out of ideas.

7. Dec 21, 2017

### jbriggs444

Instead of trying to reason about infinitesimals as if they were exactly zero, try taking a limit. You have a very small incremental volume moving into the control region over some very small increment of time. It is clearly subject to a net force and moves a net distance. You compute how much work is done moving this small volume and how much time it takes to move. That gives you a number with units of power.

If you repeat the process for a sequence of finite time increments, you can plot this and get a bar graph of power versus time.

Now take the limit as the time increment gets smaller and smaller. The graph smooths out and, in the limit, becomes a continuous graph of power over time.

[All of which is practically the definition of the process of differentiation]

8. Dec 21, 2017

### vela

Staff Emeritus
Do you have the same difficulty with the concept of instantaneous velocity? If not, what is it you think is different between the two cases?

9. Dec 21, 2017

### kibestar

Now this is uncanny; I just thought about exactly that! Both situations are really analogous. I think my discomfort with this boils down to time being continuous, which, for velocity, never presented any issue for me; I suspect it's simply because we are just so used to thinking of velocities in our daily lives, whereas not so much about control volumes and the first law.

I guess I could just keep going back to Zeno's paradoxes with this reasoning; things just are what they are. And you're right, if I don't have difficulty with one, I shouldn't have with the other. Time being continuous (is it really though?) is really weird.

Thanks for all the responses.

10. Dec 21, 2017

### Staff: Mentor

The non-moving boundary doesn’t provide the work, the moving fluid does. I often find it easier to think about power than work. Just as $W=f \cdot d$ we have $P = f \cdot v$. Since $f$ and $v$ are non zero (and not orthogonal) then $P$ is also non zero.