Monday, July 03, 2006

An Encounter with V.D

They're dealing with foully high Reynolds Numbers in my lab: to the tune of 400,000. You read right. 400,000. And they're not using water; they're using air. Needless to say, a plethora of compressors are being used to produce such high flow rates. Mach Numbers, incidentally are still incompressible.

Just to put things in perspective, I'll tell you that I'm working on Gas Turbine cooling. And also that gas turbine blades are cooled by air bled from the compressor. The blades are essentially porous. "Cool" air passes through them and stops them from melting. This gives designers latitude to allow higher temperatures in the turbine; letting the same do its bit in dealing with the imminent energy crisis. The flow within the blade can be likened to that through a channel ( thanks to similitude, a subject beaten to death in undergraduate fluid mechanics courses). The higher the heat transfer, the better. It is found that a ribbed channel would sustain higher turbulence, thereby increasing the heat transfer.

So, we're studying a ribbed channel for enhancement in heat transfer. The experiment essentially uses copper plates (infinite thermal conductivity for all practical purposes) to provide "regionally averaged" heat transfer data. Lots of copper plates (seperated by insulating gaskets) make the channel. These plates have thermocouples stuck on them everywhere.

It really perplexed me to learn that they were conducting an "adiabatic" test for this experiment. And adiabatic test is the following: air from the ambient is sucked in through the compressor; made to flow through the channel. No heat of any sort is supplied to the channel in this test. Temperatures are read by the thermocouple. The readings from this test are then subtracted from the actual heat transfer case.

I looked online for precedents. I looked at old work. I asked friends. I finally realized that this test was to account for the viscous dissipation. So, I set out to quantify the same using the data that I had. I looked at temperature differences in the channel; I found that the last plate was heated up by roughly 0.9 C in a smooth channel due to viscous dissipation and by 1.5C for a ribbed channel. This figures, as ribs would certainly create more entropy.

And I presented this to my guide. He nodded, and said that aerodynamic heating was not expected to be significant in my case. When you know everything in the subject, nothing probably excites you anymore.

Incidentally, one aspires to reach such a situation. And reaching such a situation entails a lot of excitement on the way.

9 Comments:

At 1:10 AM, Anonymous Arunn said...

I guess one can expect that the VD would be very low for air, whose dynamic viscosity is very low (when compared to say, water) even before doing anything. Usually, unless you tweak the "mu" not much significance would be gained by the VD term in the Energy transport equation, as VD is essentially a product of "mu" and all the velocity derivatives. It takes a great degree of spatial gradient in the velocity to climb over the vry less "mu" of air, to give any significance to VD.

Let me know if you can come up with a case where this happens actually for air.

Alternately, there have been papers on porous medium natural convection inside enclosures with VD artificially tweaked up through (fictitious values of) Eckert number. Some of these artificial tweaking would actually violate first law! Check on an interesting paper by V. Coasta (sorry, spelling may be wrong) in IJHMT 2005 on this topic.

And for your last remark: Yes, it is true that happiness, excitement are path functions. That too, when realized through a reversible process with zero VD, the realization is maximum. The states that you cross to realize this happiness would be simply levels of ignorance, both ways!

 
At 1:30 AM, Anonymous Arunn said...

I forgot to add:

One way to tweak (decrease or increase significantly) "mu" is to model it as mu(T, P) and choose a fluid whose mu is a strong function of T and/or P.

If you notice papers that use temperature dependent properties while solving the transport equations, tehy would take VD into account in their energy equation.

Of course, this is not that straight forward in PM flows, but that's another story to betold on another occassion...;)

And finally: if you still remember the paper that we didn't send( due to...OK OK I will not go into this..;), our VD, even for water was very low. Of course, the Re itself is low and we haven't modeled the flow variation. But even if we do, unless we tweak the mu of water as mu(T), VD may not affect the HT. If you remember, that is what we thought of doing later on (i.e. to increase the irreversibility of the system through VD, to expand the paper...;)

 
At 9:27 AM, Blogger Akhilesh said...

Since it was an experiment, it is impossible to "tweak" the mu. I could try putting water in the channel instead of air, but I am afraid I will have to pay for the damages from my modest RA stipend.

VD does indeed go as the product
2 \mu s_{ij} s_{ij} (following the convention that repeated indices are summed over). It does become significant at higher velocities, though, for the following reason: The higher the Reynolds number; the smaller the smallest turbulent length scale, the larger the velocity gradient and the greater the VD.

An easy "order of maginitude" estimate of VD can be arrived at by doing an enthalpy balance.

I'll use Latex notation from now on. Easy to understand.

h = const (for steady flow adiabatic process).

h = p/{\rho} + c_v {\delta}T

Pressure drops can be estimated by good 'ol moody diagrams.

Larger pressure drop ==> larger delta T!

For Re = 400,000 this works out to roughly 4 deg C!

 
At 9:29 AM, Blogger Akhilesh said...

Of course, the numbers pertain to the experimental setup I am currently working on. I might need to do some further non - dimentionalization.

 
At 4:51 PM, Anonymous Arunn said...

Since it was an experiment, it is impossible to "tweak" the mu.

You can always pre-heat the flow to reduce the mu or pre-cool it to increase the mu relatively in an experiment, which is still isothermal. It is not impossible. Just perhaps not worth it.

I could try putting water in the channel instead of air, but I am afraid I will have to pay for the damages from my modest RA stipend.

nowhere in my comments I have actually "suggested" you to use water. I would very much like you to successfully complete your Ph. D.

Now, agreeing with what you write about why at higher velocities VD should increase, kidnly explain these things
h = p/{\rho} + c_v {\delta}T

what is this equation? is it a definition for h or something else. I don't understand the notation. h = u + pv is all I know.

Larger pressure drop ==> larger delta T!

if you are reasoning this from the constancy of h for flow across a "adiabatic" channel, then does the converse also is true? (why isn't so?)

larger delta T ===> larger pressure drop?

So, in an adiabatic channel for flow to happen does this mean all I need to do is to impose a delta T across the channel? For instance, 4 deg C delta T in your expt. setup will result in Re 400,000?

please start with the steady flow energy equation (SFEE) and explain this to me. I will wait for your reply before proceeding with further discussions.

 
At 10:14 PM, Blogger Akhilesh said...

Looks like I can't get away with too much sophistry while "arguing" with a thermodynamics expert.

1. u = c_v * T
2. pv = p * specific volume = p/rho

And therefore the equation.

Of course, I could very well have used Cp*delta T instead of Cv*deltat + pv, so perhaps using the sfee isn't such a good idea after all - since the flow is still incompressible. (Mach < 0.2)

Since the flow is incompressible, we can write the Bernoulli equation (with "head loss"),
p1+v1^2/2 = p2 + v2^2/2 + hl, and hl = cp*deltat , and thus correlate deltat with delta p. I guess my motives on using the sfee were the same.
For a higher friction factor, the turbulence produced is higher, the dissipation is higher.

 
At 10:33 PM, Blogger Akhilesh said...

The converse is not true at all.
I'm just saying the irreversible energy conversion from work to heat is roughly 4C for my flow.
Forcing a 4C gradient will not establish such a flow!
It's like saying, if I supply 50W of heat to my laptop heat-sink, will it power itself on?
Consider a pipe flow.
If you were holding the most upsteam section at T1 C and most downstream section at T1+4C, and you insulated the surface of the pipe, you would get a linear temperature profile, with heat being conducted from downstream to upstream. This will have a minor effect on the flow-field, since it will create a natural convection ONLY when the piple is vertical. Otherwise, a temperature gradient is more or less powerless in generating a flow field.

 
At 6:00 AM, Anonymous Arunn said...

The c_V (over c_P) is fine with me - you try to distill internal energy alone, that is OK with me.

But two things: Benoulli eq. (from your third comment) is for inviscid flows only, turbulent or not. Another way of saying this is Bernoulli Eq. is equivalent to First Law of ThermoD (SFEE in CV approach) ONLY for reversible process.

(Usually we patch work this glitch by putting a head loss correcting term that accounts for the delta_P resulting because of viscous forces - Poisuille eq. - into the bernoulli eq. itself and call it again bernoulli eq. But that is digressing from what we discuss here for the moment)

VD is for viscous flows only, turbulent or not. if mu is zero, there is NO VD period. (So, NO Bernoulli Eq. interpretation is possible here, BUT SFEE is still valid)

So, going back to your original argument (your first comment) of dh = 0, hence more delta_p implies more delta_T (your first comment herein)

Please apply SFEE across the CV that you want to analye VD for. If you do that, you will realize, the delta_P that has to prevail across the CV for the flow to happen at Re = 400000 (the delta_P is to overcome the wall/viscous friction an irreversible process) is an external work done (i.e. on the system, which is your channel etc. from outside of the channel) and not realted to the flow work (a concept valid even when hte flow is inviscid and irrespective of compressible or incompressible flow) that crops up in the definition of h ( = u + pv).

In fact, (some of) the external (pump) work (that sets up the delta_P which is NOT equivalent to the delta_P in the dh equation) is what is getting dissipated as VD, if the flow is treated incompressible, resulting in a delta_T across the CV.

The difference is between the thermodynamic pressure that determines the local thermodynamic state of a system (and doesn't include viscous forces that contribute to the mechanical pressure) and the mechanical pressure that drives a flow (valid even for viscous flows). Please check Viscous Fluid Flow by White for some more details...

Coming back to our discussion...
That is: Ideally, the fluid should remain at rest (stagnant w.r.t. to an external reference frame) because of viscosity resistance. This is fine with First Law. At this stage, there is NO VD, because there is NO flow. Then when we want a flow, we impose (do work) externally a delta_P. This delta_P goes into overcoming friction caused by viscous resistance. Here, Firs Law (SFEE) is valid but not Bernoulli Eq. (uncorrected for viscous losses). But what happens when this flow prevails is that some of the delta_P (external work or energy if you multiply this delta_P with steady flow rate value) is "lost" internally, into raising the internal energy of the fluid, which reflects in the delta_T increase across the CV from entry to exit, across which is where the delta_P is imposed.

This means the flow should have an average velocity slightly lesser than what it should be, if the effects of VD were absent. So, the delta_T IS because of VD, but that energy came from an external source.

And that is the reason why delta_T delta_P cause and effect doesn't flip back. Because, the first process of imposing delta_P itself is an irreversible process - a one way process. If delta_T is increased (again, this can be done only externally), it simply goes into the internal energy of the fluid, which can still be at rest (measuring through continuum velocity values).

That is my understanding/ignorance so far...

continue with your thoughts...

I am still looking for an argument that depcits a practical case (flow in channel etc.) when even though mu is small, VD will be high (resulting in say, high delta_T), turbulent or not, incompressible or not, adiabatic or not....

 
At 12:53 PM, Blogger Akhilesh said...

1. Of course, I am aware that the Bernoulli equation is not valid for a viscous flow. In order to "force" a validity, a head loss is incorporated.
My contention is that it is this head loss that is equal to the viscous heat dissipated (if turbulent ke does not change)

2. If VD were absent, there would be no pressure drop across the pipe at all: as you say, the Bernoulli equation would be valid now.

The energy, of course comes from the pump - because the pump (or compressor) increases the pressure. It is indeed the "external source" that the pressure drop comes from. The "pressure" energy (from the compressor) is thus downgraded to heat energy.....

I'll try to send you the math I worked out soon ... maybe you could see if it makes sense.

 

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