Why does the total temperature not change in adiabatic flow?

[obad]

Hi guys,

I have a question to you that has been bothering me for a while now.

It is about adiabatic flow in a duct or more specifically in the air intake of a jet engine.

In lectures and textbooks it is always stated that the total temperature in an adiabatic
flow does not change. Hence the total temperature at the inlet and outlet of a jet engine
air intake is the same. However, the total pressure reduces due to friction. This kind of
flow is also known as Fanno flow.

So the common explanation is, that friction does not alter the total temperature. But
when having a look at the energy equation for viscous flows I cannot agree with this
statement.

I wrote down the energy equation in integral form for the total internal energy e.
If we say the flow is stationary, equal pressure at inlet and outlet and adiabatic, then
the instationary term on the left side and the temperature source term and temperature
conduction term as well as the pressure term on the right side can be
cancelled out.

So what remains is the left side telling me the net amount of energy that leaves my volume
and on the right side the energy term due to shear stress.
Heating of the flow due to shear stress can usually be neglected f or low Mach number flows
where the velocity gradients close to the wall are small.

But if I take it accurately then the viscous heating should change the energy flow through
my control volume, right? At least for supersonic air intakes this should make a difference,
however in textbooks the total temperature stays constant.

I appreciate your help!

 

[jack action]

An adiabatic process is one that occurs without transfer of heat or matter between a system and its surroundings. Wikipedia

But if I take it accurately then the viscous heating should change the energy flow through
my control volume, right?

If there is no heat from outside (by definition) and there is no mechanical work (it's only a duct), where do you think the energy required to produce that viscous heating comes from? The only possible answer is the flow itself, either by a lost of pressure (you assume it didn't), a decrease in temperature or a flow deceleration (both part of the total temperature). The only thing the viscous heating can do is «re-heat» the flow.

 

[obad]

Thank you Jack for your answer!

Your answer is logical.
Viscous heating is simply the conversion of kinetic energy to heat.

To recap, if we get back to the simplifications that I made in my first post, we would also have to cancel the integral that sums up the energy that enters and leaves the control volume, right? By doing that the only terms that are left are the pressure and viscous heating term. With the assumption of constant pressure at the in- and outlet the viscous heating takes its energy from decreasing the velocity that is included in the pressure integral. Is that right?

Ok so I get it that the energy does not change inside an adiabatic control volume, but there is till an open question:
Another effect of viscosity is that it leads to a total pressure loss. What is the mechanism that leads to the loss of total pressure?

Cheers!

 

[jack action]

The loss of pressure is necessary to create the force that will counteract the force of the shear stress. In the next figure, P1 and P2 must be different to equalize the shear force created:

You should study the Fanno flow. I think you will appreciate this detailed derivation of Fanno flow relationships.

 

[Chestermiller]

The viscous heating exactly cancels the cooling resulting from gas expansion. This is the same kind of thing that happens when you have a pressure drop in flow through a throttling valve or porous plug (Joule-Thompson).

Chet

 

[obad]

Hello everyone,

I would like to refresh this thread with some new thoughts.

I'm still not completely satisfied with the way viscosity effects the flow through a duct.
I understand that viscosity does not represent an energy source or sink, but only converts
kinetic energy to thermal energy. So logical thinking would suggest that the same amount of
(total) energy that enters a duct also exits it.

However, I studied the derivation of Fanno flow, which is adiabatic flow through a
duct of constant cross sectional area. And I stumbled over one equation that is simply not
compatible with the rest of the Fanno flow discussion. It is the way that the energy
equation for one-dimensional flow is derived.

I uploaded a page from Anderson's textbook "Modern Compressible Flow", on this page Anderson
describes the derivation of the 1D energy equation. Here he simply neglects the effect
of viscosity without even mentioning it...
Equation (3.9) of this page is subsequently used for the derivation of Fanno flow where
viscosity clearly is the point of interest.

Furthermore I found some Gas Dynamics lecture notes where the author points out this issue.
See http://www3.nd.edu/~powers/ame.30332/notes.pdf page 65-66 and specially the footnote at
the end of page 66.

The author of these lecture notes points out that neglecting the viscous action is common
practice, but not correct.

When looking again at Anderson's derivation and including
the viscous term, Equation (3.9) would have an additional term. This means for adiabatic
flow the total energy entering the duct wouldn't equal the energy that exits it...

According to this thinking viscosity has an effect on the total energy, which means that is acts as a source/sink.

What do you think about that?

 

[jack action]

OK, I'll take a try at this.

If I apply the change to the energy equation as proposed in the footnote of your link (and if I did not make any mistake), the only effect would be to multiply the wall shear by 2. Equation 4.36 would be:
[tex]\frac{de}{dx} − \frac{P}{\rho^2} \frac{d\rho}{dx} = (q_w + 2\tau_w u)L[/tex]
But how do we evaluate the wall shear?

In section 4.5, p. 111, the author uses the Darcy–Weisbach equation (equation 4.463), like everyone else. The Darcy–Weisbach equation was found by dimensionless analysis. This means that only a proportional relationship has been established between [itex]\tau_w[/itex] and [itex]\rho u^2[/itex] and not an equality. So the Darcy friction factor is not a measurable value, unless we all agree on the same definition.

I wonder if the author's assumption is used everywhere to determine the the Darcy friction factor, a factor of 2 won't be introduced as well to compensate?

 

[Nidum]

Came across this by chance while I was searching for something else :

http://uk.mathworks.com/help/aerotbx/examples/analyzing-flow-with-friction-through-an-insulated-constant-area-duct.html

 

[obad]

Hi Jack,

I had a look at Equation (4.31) since it is used for the derivation of Equation (4.46). This equation is used in the derivation of the equations that describe Fanno Flow.
I inserted wall shear into the energy equation as the author suggests it in the footnote by adding [itex] \tau_w L \Delta_x u \Delta_t [/itex] to Equation (4.29). With this I get a modified version of Equation (4.46) (with [itex] q_w = 0 [/itex]), that is [tex] h + \frac{u^2}{2} - (h_0 + \frac{u_0^2}{2} ) = \int_0^1 \frac{\tau_w L}{\rho A} dx [/tex]

According to this the shear stress has an effect on the energy, right?

 

[Chestermiller]

Hi Jack,

I had a look at Equation (4.31) since it is used for the derivation of Equation (4.46). This equation is used in the derivation of the equations that describe Fanno Flow.
I inserted wall shear into the energy equation as the author suggests it in the footnote by adding [itex] \tau_w L \Delta_x u \Delta_t [/itex] to Equation (4.29). With this I get a modified version of Equation (4.46) (with [itex] q_w = 0 [/itex]), that is [tex] h + \frac{u^2}{2} - (h_0 + \frac{u_0^2}{2} ) = \int_0^1 \frac{\tau_w L}{\rho A} dx [/tex]

According to this the shear stress has an effect on the energy, right?

This equation does not look correct to me. (See equation 2 in this document: www.chimeracfd.com/professional/technote/fanno_flow.pdf) According to the open system version of the first law, the right hand side of your equation should be zero for steady state operation (i.e., neglecting kinetic energy, the enthalpy change should be zero). For an incompressible liquid, the viscosity effect would be captured by the fact that vΔp is negative (the pressure decreases as a result of viscous dissipation of mechanical energy) so the temperature rises in order to maintain the enthalpy constant (neglecting kinetic energy). For a compressible ideal gas, the viscous heating is exactly compensated by expansion cooling, so here again, the enthalpy still doesn't change, but in this case, since it is an ideal gas, the temperature must remain constant (since enthalpy is a function only of temperature). What happens for an ideal gas then is that the pressure decreases and the volume increases such that Δ(pv)=0. Of course, if kinetic energy effects are present, then the enthalpy does change, and this results in the temperature changing.

For a real gas, the enthalpy is a function not only of temperature but also of pressure. So, in this case, the temperature can change while maintaining constant enthalpy between the stream flowing into the system and the stream flowing out of the system. Thus, both the pressure change (from viscous effects) and the temperature change combine to keep the enthalpy change constant.

So, to summarize, viscous effects are present in all the cases, but they are must be consistent with the fact that the enthalpy does not change (again neglecting kinetic energy change).

Incompressible liquid: Δ(pv)<0, so Δu>0, so ΔT>0, all because Δh = 0 (guaranteed no kinetic energy change)

Compressible ideal gas: Δ(pv)=0, so Δu =0, so ΔT=0, all because Δh=0 (case of no kinetic energy change)

Real compressible gas: Δh is still equal to zero, but T can change (even with no kinetic energy change)

Chet

 

[jack action]

@Chestermiller,

The equation described by @obad should correctly represent the theory as explained in the footnote on page 66 of http://www3.nd.edu/~powers/ame.30332/notes.pdf:

In neglecting work done by the wall shear force, I have taken an approach which is nearly universal, but fundamentally difficult to defend. At this stage of the development of these notes, I am not ready to enter into a grand battle with all established authors and probably confuse the student; consequently, results for flow with friction will be consistent with those of other sources. The argument typically used to justify this is that the real fluid satisfies no-slip at the boundary; thus, the wall shear actually does no work. However, one can easily argue that within the context of the one-dimensional model which has been posed that the shear force behaves as an external force which reduces the fluid’s mechanical energy. Moreover, it is possible to show that neglect of this term results in the loss of frame invariance, a serious defect indeed. To model the work of the wall shear, one would include the term [itex]((\tau_w L \Delta x)\overline{u}\Delta t)[/itex] in the energy equation.

I can really discuss this theory as I really like the argument against it («real fluid satisfies no-slip at the boundary; thus, the wall shear actually does no work») and I don't understand what he means with «Moreover, it is possible to show that neglect of this term results in the loss of frame invariance, a serious defect indeed.»

I also like the way @obad defined viscosity:

viscosity does not represent an energy source or sink, but only converts kinetic energy to thermal energy

If viscosity (i.e. friction) convert kinetic energy to heat, then where does the work comes from?

I would say work is done ON the fluid, which creates friction, which converts it into heat. Friction is a reactive force: it could never exist by itself, hence how could it create work?

 

[Chestermiller]

@Chestermiller,

The equation described by @obad should correctly represent the theory as explained in the footnote on page 66 of http://www3.nd.edu/~powers/ame.30332/notes.pdf:I can really discuss this theory as I really like the argument against it («real fluid satisfies no-slip at the boundary; thus, the wall shear actually does no work») and I don't understand what he means with «Moreover, it is possible to show that neglect of this term results in the loss of frame invariance, a serious defect indeed.»

What can I say Jack. I totally agree with you.

Chet