Help: Where Does the Energy Enter?

I have always been unsure about the following issue, I was hoping someone could clear it up. Suppose you have a straight section of pipe carrying a viscous flow ( water for instance). I want to avoid assuming the flow is incompressible for the time being, you'll see why near the end of the discussion.

Starting with and energy rate balance across the inlet and outlet incorporating the Reynolds Transport Theorem:

Q˙Ws˙=ddtcv(u+v22+gz)ρdV+cs(u+v22+gz+pρ)ρvdA \displaystyle \dot{Q} - \dot{W_s} = \frac{d}{dt} \int_{cv} \left( u+ \frac{v^2}{2} + gz \right) \rho dV + \int_{cs} \left( u+ \frac{v^2}{2} + gz + \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A}

Q˙ \dot{Q} is the net efflux of heat crossing the boundary per unit time.

Ws˙ \dot{W_s} is the shaft work between the inlet/outlet.

v \mathbf{v} velocity vector of the flow ( it may vary across the pipe ).

v v scalar flow velocity

A \mathbf{A} vector describing the cross sectional area of the pipe.

u u internal energy of the flow

ρ \rho mass density of the flow

z z Centerline elevation of the flow from some reference

p p Pressure at inlet/outlet respectively

Simplifications between inlet/outlet:

Steady State =ddtcv(u+v22+gz)ρdV=0 \Rightarrow = \frac{d}{dt} \int_{cv} \left( u+ \frac{v^2}{2} + gz \right) \rho dV = 0

Ws˙=0 \dot{W_s} = 0 No shaft work present.

Δz=0 \Delta z = 0 No change in elevation.

We are left with:

Q˙=cs(u+v22+pρ)ρvdA \displaystyle \dot{Q} = \int_{cs} \left( u+ \frac{v^2}{2} + \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A}

The Second Law of Thermodynamics

Q˙>0 \displaystyle \dot{Q} > 0 Heat is generated in the process of fluid flowing between inlet and outlet.

Which implies:

cs(u+v22+pρ)ρvdA>0 \displaystyle \int_{cs} \left( u+ \frac{v^2}{2} + \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} > 0

Here is where my dilemma begins. If we assume incompressible flow and a uniform velocity distribution over A\mathbf{A} , the equation further reduces to:

cs(u+pρ)ρvdA>0 \displaystyle \int_{cs} \left( u+ \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} > 0

cs(u+pρ)ρvdA=m˙(hohi)>0 \displaystyle \int_{cs} \left( u+ \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} = \dot{m}( h_o - h_i ) > 0

Where, h=u+pρ h = u+ \frac{p}{\rho} which is commonly known as the Enthalpy.

I see two things changing as the flow moves between inlet and outlet. The Pressure p p is decreasing from inlet to outlet, and the internal energy u u is increasing ( the temp of the flow is increasing from inlet to outlet ).

So it seems like these two remaining forms of energy ( p p and u u ) are just changing hands, but the result implies that the Enthalpy is not constant. It is growing in the direction of flow.

So where is this energy to change the enthalpy being added to the system?

If we take a step back to the incompressibility assumption, no flow is truly incompressible. As such, there exists a tiny velocity gradient between the inlet and outlet. Is the added energy to change the enthalpy in actuality necessarily coming from the expansion of the flow?

I can't see how accounting for the expansion changes anything.

cs(u+pρ)ρvdA>cs(v22)ρvdA \displaystyle \int_{cs} \left( u+ \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} > -\displaystyle \int_{cs} \left( \frac{v^2}{2} \right) \rho \mathbf{v} \cdot d \mathbf{A}

It still seems like the equation above is saying an imbalance of energy exists ( albeilt smaller than its incompressible counterpart), so what am I missing here?

Thanks!

Note by Eric Roberts
1 month, 3 weeks ago

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I have a limited understanding of the subject as a result of which I ask the following:

The second law of thermodynamics dictates that the net entropy change of the universe is greater than or equal to zero. For a reversible process the entropy change is ds=dQTds = \frac{dQ}{T}. Reversible processes are an idealization. My question is that how do you conclude using the second law of thermodynamics that heat is generated in the system?

Suggestion: Google search 'steady flow energy equation'. Its a simplified version of the energy balance equation you have typed.

My understanding: QQ is the rate of heat transfer into the system and it is not the rate of heat generated in the CV. How this heat is utilised is: It partly gets used up in increasing the fluid temperature (Its enthalpy). It is partly used for useful work WW, and the changes in kinetic and potential energies are dependent on the nature of the flow which also factors into the energy balance equation.

Karan Chatrath - 1 month, 3 weeks ago

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I think that is actually where I went wrong. The statement of the Second Law would be that heat is lost to the environment in every real process ( entropy of the universe increases ), making the process irreversible. So, the net efflux of heat from the system per unit time Q˙ \dot{Q} must be negative, not positive. That changes things to the propper direction...what a goof.

cs(u+pρ)ρvA=m˙(hohi)<0 \displaystyle \int_{cs} \left( u + \frac{p}{\rho} \right) \rho \, \mathbf{v} \cdot \mathbf{A} = \dot{m}( h_o - h_i ) < 0

The Enthaply is decreasing in the direction of the flow makes sence ( I was having it the other way around which was absurd, which I knew, but I wasn't able to see the second Law blunder ).

If the flow was perfectly isolated from its surroundings, then the change in enthalpy would exactly equal 0, meaning the energy change associated with u u in the direction of flow would be exactly mirrored by the energy change associated with p p . No need to talk about compressibilty.

Nice catch! The propper way to not be distracted by the details ( as I apparently was ). Thanks for your reply!

However, now that you mention it, I do wonder if the Second Law constraint is only valid for flows initially in thermal equalibrium or at greater temperature than its surroundings? For example, if it was a flow of cold refridgerant in a room temperature surroundings the net efflux of heat across the control surface most definately could be positive between the inlet and outlet, and the enthalpy of that type of flow could increase in the direction of the flow. I'm just kind of surprised to just realize this now. I'm looking through a text on Fluid Mechanics and I'm realizing they don't seem to consider that scenario of a cold flow in a warm environment at any point in the text. Perhaps its because the authors wish to avoid the difficulties of heat transfer, or is it because the only flow property that can change in that context is the internal energy ( thermal ). The pressure gradient still must decrease in the direction of flow no matter how much heat the flow is absorbing. So perhaps it just isn't relevant to the subject. Its definately discussed in Heat Transfer and Thermodynamics but I personly never fully made the connection until now. Thanks for bringing it up.

Eric Roberts - 1 month, 3 weeks ago

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I will elaborate a bit further a bit about what I meant to say in my earlier comment. When I say universe: I mean:

Universe=System+SurroundingsUniverse = System + Surroundings

The net change of the entropy of the universe must be greater than or equal to zero as per the 2nd law of thermodynamics. This statement does not enable us to conclude the direction of heat transfer for the system (whether into or out of). It may as well be the case that the entropy of the system reduces during a process but the net change of the universe is still greater than or equal to zero. So in my view, I reiterate that a conclusion about the direction of heat transfer rate QQ cannot be made from this analysis.

I am glad that you found my comment helpful. As I said, my knowledge base is limited in this domain. So if I happen to have a conceptual fallacy in my words, please do point it out.

Karan Chatrath - 1 month, 3 weeks ago

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@Karan Chatrath Your right, the direction of heat transfer cannot be determined in the way I have done so. After, some more careful consideration I believe I see where I have gone wrong.

Q˙=cs(u+pρ)ρvA \displaystyle \dot{Q} = \int_{cs} \left( u + \frac{p}{\rho} \right) \rho \, \mathbf{v} \cdot \mathbf{A}

Distributing the properties evenly across the inlet/outlets the integral above reduces to:

Q˙=(uo+poρ)m˙(ui+piρ)m˙ \displaystyle \dot{Q} = \left( u_o + \frac{p_o}{\rho} \right) \dot{m} - \left( u_i + \frac{p_i}{\rho} \right) \dot{m}

Where the subscripts i i and o o refer to the inlet/outlet respectively. The above relationship can be arranged in the following manner.

pipoρ=(uoui)Q˙m˙ \displaystyle \frac{p_i - p_o}{\rho} = ( u_o - u_i ) - \frac{\dot{Q}}{\dot{m}}

In the direction of flow pipoρ>0 \frac{p_i - p_o}{\rho} > 0 necessarily and so [(uoui)Q˙m˙]>0 \left[ ( u_o - u_i ) - \frac{\dot{Q}}{\dot{m}} \right] > 0 and that is actually the consequense of the Second Law, not the direction of heat transfer.

Case 1: \textbf{\underline{Case 1:}}

No Heat Transfer Q˙=0 \dot{Q}= 0 :

uo>ui \displaystyle u_o > u_i

Thermal energy increases in the direction of flow.

Case 2: \textbf{\underline{Case 2:}}

Heat Transfer Out Q˙<0 \dot{Q} < 0 :

uo>uiQ˙m˙ \displaystyle u_o > u_i - \frac{\dot{Q}}{\dot{m}}

Case 3: \textbf{\underline{Case 3:}}

Heat Transfer In Q˙>0 \dot{Q} > 0 :

uo>ui+Q˙m˙ \displaystyle u_o > u_i + \frac{\dot{Q}}{\dot{m}}

The inequalites are a manifestaion of the Second Law, not the direction of heat transfer ( as you properly state ). Now I'm fairly certain what im stating is consistent. Thank you for reiterating your concerns. If you now agree/disagree feel free to let me know.

And I again thank you for your help!

Eric Roberts - 1 month, 3 weeks ago

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I moved this post

Eric Roberts - 1 month, 3 weeks ago

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