Thursday, July 27, 2006

Reversing the Heat Equation

Today's class on Partial differential equations gave me a whole new perspective on the ubiquitous heat equation (whose forms in various coordinate systems, I "solved" numerically for my Masters thesis). I'm going to share that perspective here.

Let's start with something simple and 1d. A rod (with temperature independent conductivity), perhaps. You're given a rod with a non-steady state temperature profile. One end of the rod has been maintained at a constant high temperature for 20 minutes - and you are handed over the rod after the 20 minutes. The entire rod is convecting (and has always been) with the ambient (with a constant heat transfer coefficient). Eventually, it would have reached a steady state - but that would have taken more than 20 mintues. So, essentially, you're given a rod whose temperatur distribution still being governed by the heat equation rather than the Laplace equation. In other words, the profile is still evolving.

Is it possible to predict the initial temperature profile, given an instantaneous temperature profile as detailed above?

The answer, well, is yes and no.

Yes, because, if a solution does exist, it is unique - fact that can be proven by some math [1]. So, essentially, for a given final temperature profile T_f(x), the ambient temperature (T_inf), the boundary conditions (T_o and T_(x=L)) . Starting with the T_f(x) and going back in time (by solving the reverse heat equation), one can, in principle, recover the initial temperature profile.

No, because, well, it diverges. A primitive variable seperable closed-form expression for the solution makes things quite clear.

where, the a's are the Fourier coefficients of T_f(x). In the case of the forward heat equation, the exponent would have been negative, rendering the higher harmonics more or less neglegible But, in the reverse case, this term would tend to blow up at higher values of n. If T_f(x) were purely sinusoidal (only one a_n), then the equation would be quite friendly. The equation poses no problem for a small number of harmonics.

But suppose there were a "little" noise. Noise usually has many frequencies, some quite high. The corresponding terms would blow up. Evaluating the above series is not going to be easy.

This gives us another elegant insight: the "sharper" a profile (a profile with a discontinuity, for instance would yield infinite terms in the fourier space), the more difficult it is to reconstruct the original temperature profile. This figures: the larger the local temperature gradient, the faster it tends to smooth itself out! Minor spatial fluctuations in temperature are removed faster by the diffusion equation.

Let us now consider a 5 cm x 5cm x 5cm copper block. (Copper because it's always at more or less the same temperature at ambient heat transfer coefficients). Suppose we heat it up to a temperature and cool it in air, and measure its temperature by a thermocouple with an accuracy of 0.1C. Suppose we measure time with a digital watch with an accuracy of 1 sec. Suppose, also, that we know the heat transfer coefficients accurately. Given the time since the start of convective cooling (measured by the stop-watch, actually) and the final temperature , can the initial temperature be determined?

Yes, you would say. Exponenital cooling curve. Easy to fit it.

Suppose we look at a situation where the instruments would say that a steady state has been reached. There's no way to get back the original temperature --- there's many possible solutions!!!!!!! The only way out is it use more accurate devices.

And now, add to this the complexity of temperature profiles. Suppose we were using steel piece (lower conductivity) , and somehow "magically" obtained its temperautre profile at a certain time throughout its volume. Suppose the profile was somewhat jagged (but steadying out gradually) ......... Well, with a lot of computers one can arrivie at an estimate ... with an error bar.

And this wisdom can be applied to the Navier stokes too. The only process in the entire N/S equations that "loses" information is, actually, viscous dissipation. Turbulence generates entropy faster by wrinkling the instantaneous temperature and velocity profiles to bring about high amounts of heat transfer and viscous dissipation respectively.

Maybe I'm getting way too carried away here.....

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.