# What is the difference between flow work and kinetic energy?

1. May 6, 2012

### Red_CCF

Hi

I was wondering, for the first law most books have basically Q - W = deltaUsystem + deltaH + deltaKE + deltaPE. Assuming steady state and negligible PE change, we just have deltaH and deltaKE. I was wondering, since H includes flow work (in the form of PV), why is KE still part of the equation? What is the difference between flow work and kinetic energy?

I'm also wondering, with regards to stagnation temperature (http://en.wikipedia.org/wiki/Stagnation_temperature),
If the end state is suppose to be stagnant, why does the derivation result in the balance h0 = h + V^2/2, which implies there's flow in and out at stagnation state? Does this mean that one can have enthalpy/flow work without kinetic energy?

Thanks

2. May 7, 2012

### Andrew Mason

Can you give us a cite to such a text book? The first law is:

$Q = \Delta U + W$ where Q is the heat flow into the system, ΔU is the change in internal energy of the system, and W is the work done BY the system.

AM

3. May 7, 2012

### Red_CCF

Hi

I'm referencing Fundamentals of Thermodynamics, by Moran and Shapiro. I think the formulation you gave is the same as mine, except W is moved over to the other side and enthalpy, KE, and PE are added

4. May 7, 2012

### Andrew Mason

KE and PE are included in the internal energy. H also includes internal energy. So your expression does not make sense. Can you post a copy of the page in Moran and Shapiro that sets this out? Something is not adding up here.

AM

5. May 7, 2012

### Red_CCF

Hi

I actually had a typo, the Usystem should have been an Esystem; in Moran and Shapiro Esystem includes KE and PE of the system but this doesn't really address your question. I have posted the equation, which is written in terms of the rate of change of the energy in the system. The KE and PE presented in the text represents the KE and PE of the flow in and out (which is the basis of my question), while PE and KE of the system are included in the dE/dt term.

Thanks

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6. May 7, 2012

### Andrew Mason

This is not a statement of the first law of thermodynamics. The first law of thermodynamics deals with thermodynamic changes between equilibrium states.

Your text is dealing with the energy balance in a dynamic fluid flow ie. a system that is not in equilibrium. Those little dots above the m, Q and W refer to the first time derivative of these quantities.

This question has to do with fluid dynamics. The Bernouilli equation deals with conservation of energy in dynamic fluid flow. The first law deals with conservation of energy in between states of thermodynamic equilibrium. This equation appears to combine Bernouilli's equation and the first law of thermodynamics for a dynamic flow in which heat flow and thermodynamic work occurs.

AM

7. May 8, 2012

### Red_CCF

Hi

I think the differences is a semantic issue. In the thermo course I took the prof actually used what the author had but took out the dots/time derivative so it became a pure energy balance rather than a rate balance that accounts for flow and called it the first law, which he used synonymously with energy balance. Now I know that he may have came up with his own version of first law and not the standard definition, but I think both would work fine in an application.

So now that's cleared up, I'm wondering if you can answer some of my question in the OP. Specifically, I'm confused on what the difference between flow work and kinetic energy of the flow is and what stagnation state really is?

Thank you

8. May 8, 2012

### Andrew Mason

It is not a semantic issue. What you have stated is NOT to be confused with the first law of thermodynamics. Your equation applies to non-equilibrium states ie. fluid flow. The first law of thermodynamics deals with systems in different states of thermodynamic equilibrium.

You should post this as a fluid dynamics problem re: stagnation flow.

Thermodynamic work is δW = PdV. Kinetic energy of the flow, dKE = (dm)ρv^2/2. If the system is a flowing fluid and there is heatflow into or out of the fluid, to account for all the energy in the system you have to account for both.

AM

9. May 8, 2012

### Red_CCF

Hi

I was wondering if you can give a physical interpretation of what flow work and kinetic energy of flow are. Is flow work related to the force the fluid has to exert to push some fluid volume into the system and kinetic energy the inertia of this fluid volume into the system?

With regards to stagnation state, my question is kind of a thermodynamic one, in that the word stagnation implies that there is no fluid flow, so how come enthalpy, which includes a flow work, is the result of the energy balance (http://en.wikipedia.org/wiki/Stagnation_temperature) from move to stagnation state?

Thanks very much

10. May 8, 2012

### Andrew Mason

Flow work is done by exerting a force over a distance. The force comes from a pressure difference exerted on a volume of fluid. Bernoulli's equation predicts how pressure changes with changes in potential and kinetic energy of the fluid. If I want to push fluid through a pipe up a hill, I have to do work on it to get it to move and then do work against gravity to move it up the hill. This work is done by applying an unbalanced pressure to the fluid. Pressure x volume = Force/area x volume = Force x distance = work

AM

11. May 8, 2012

### Red_CCF

Hi

Thanks very much for the response.

Assuming no PE change, basically flow work is the energy gained by the fluid due to pressure forces, and kinetic energy is the inertial energy of the fluid, and it is added because this goes into the system? Is the reason net pressure (which is usually zero) isn't used because the other side of the volume is inside the boundary of the CV?

For a expanding piston cylinder at constant pressure I believe that the energy balance is Q-PΔV= ΔU and I believe this turns into Q = muf-mui+PVf-PVi which turns into Q = mΔh. In this case, enthalpy is used for a non-flow situation, so is there a generalized explanation on what the term PV represents outside of being flow work? I'm just trying to understand why a PV can be present for a system with no flow (the stagnation state).

So regarding the first law again, it seems that my understanding beforehand was completely wrong.

In Moran and Shapiro they say that first law is used to describe a thermodynamic cycle, but within the control volume that contains the cycle, there are fluid flow, so how come this system (a control volume that encompasses the whole cycle) can be analyzed by first law? Is the way I define the system determine whether first law can be used to analyze it or not?

In the case of a closed piston cylinder with a gas, if I only take a control volume of half of the cylinder, can I still analyze the energy balance with first law? How would I identify whether first law or the general energy balance equation applies for a particular situation, you mentioned that first law only applies for a system with two end states in thermodynamic equilibrium, I can only think of closed systems that satisfy this is this correct?

Last edited: May 9, 2012
12. May 10, 2012

### Andrew Mason

PV is not to be confused with PΔV. PV describes a particular state of pressure and volume - it represents a form of energy or ability to do work. PΔV relates to a transition between two states in which work PΔV is done at constant pressure.

The first law applies to determine the change in internal energy, heat flow and work done in a transition between two states of equilibrium of the system.

If the piston moves slowly, you can analyze everything using the first law. In that case, the mechanical (kinetic) energy of the gas and piston will be negligible. If the mechanical energy of the system and control volume is not negligible, you have to account for it using the kind of equation in your text.

AM

13. May 10, 2012

### Studiot

Whilst Physicists and Chemists use the first law packaged as Andrew has stated, Engineers (Chemical and Mechanical) use the law as presented originally by Red.
Rayner Joel (Basic Engineering Thermodyamics) for instance introduces it as The Energy Equation and he distinguishes flow and non flow versions. This is common in steam and process flow engineering.

Users should be aware of this difference of notation and practice.

14. May 11, 2012

### Red_CCF

Thanks for the information. I did in fact that engineering thermodynamics and not realize that there's this fundamental difference. I think in engineering the emphasis was applying the calculations correctly so I think my prof didn't really bother with differentiating nomenclature and lumped first law with energy balance.

I did some more reading and although not explicitly stated, I get the impression that first law is defined with a closed system in mind, as http://en.wikipedia.org/wiki/First_law_of_thermodynamics implies in the first paragraph. Is this a valid way of determining whether first law applies?

Is it correct that a steady state open system cannot be analyzed with first law? What about an open system that is initially in steady state, and then some change occurs and it reaches steady state again at another T and P?

Thanks very much for your help!

15. May 11, 2012

### Red_CCF

In my previous post I actually separated PΔV into PVf-PVi, both V are total volume. Would this be valid now for combination with the Uf - Ui into enthalpy (Q = muf+PVf-mui-PVi)?

Sorry but I'm not getting this, why does the fact the piston moves quasistatically affect whether first law can be applied or not as even if piston expands slowly the total mass of gas inside the CV at the end state will go down regardless of how fast it moves?

Thanks again

16. May 11, 2012

### Andrew Mason

Only if P was constant. H = U + PV so ΔH = ΔU + Δ(PV) = ΔU + VΔP + PΔV. If P is constant then ΔH = ΔU + PΔV. Otherwise, ΔH includes that VΔP term.

Because the first law applies to transitions between states of equilibrium. So if it is moving slowly (ie. so the rate of energy change due to the incoming and outgoing mass is negligible) the first law will account for all the changes in energy.

The equation that is set out in your text says this:

change in total energy = heat flow into the control volume - work done by the control volume + (enthalpy of the incoming mass + kinetic energy energy of incoming mass + potential energy of incoming mass) - (enthalpy of the exiting mass + kinetic energy energy of exiting mass + potential energy of exiting mass).

If you only have the control volume (no incoming or outgoing mass flow), you are left with just:

ΔE = heat flow into the control volume - work done by the control volume = Q - W, which is the first law.

AM

17. May 11, 2012

### Red_CCF

I'm actually curious why the PV isn't always included. For a rigid tank of gas that is heated, the initial and final pressure are different while V is constant, so how come this effect isn't included in energy balance?

Hi

So basically first law applies to closed systems? In the case of quasistatic piston motion, first law can be used to balance between one state and another equilibrium state after the piston moved a small amount, and I can do this repeatedly to get the total energy change from one state to a state where the piston moved a significant amount and the effect of negligible mass leaving for multiple repetitions doesn't add up?

I'm also wondering if first law applies for the case where, there is flow at steady state, and suddenly the incoming fluid temperature goes up and eventually the system reaches another steady state. Can first law analyze the change in energy content of the CV between the two steady state conditions?

Thanks

Last edited: May 11, 2012