heat transfer in a steel rod ??

One assumes a little knowledge on the part of the reader. You act as if I just invented condensation. A little while ago you were sure 'inducing heat flux' was the mechanism what happened to that?

the steam" Please be more clear.

Well, the rest of the sentence was the part that made it clear. The original poster had made it clear he understood the extents of the envelope of live steam and was keeping his distance. The original question to him was if he understood his hand would be scalded why did he have trouble understanding that that part of the rod in contact with the steam was being heated. After your failure to comprehend that question, my reply to you was: Your hand would get hot from the steam if you were dumb enough to place your hand (in the steam).

You didn't ask you - instead you pontificated some nonsense about "inducing heat flux".

If there were no pressure differential there would be no movement of steam it would just stay in a little ball around the red hot end.

which only means it takes very little pressure drop to cause a rapid flow.

That would depend on how much surface area contacts the steam.

because "some of it condensing" represents a huge transfer of energy.

The steam would be rapidly expanding outward from the source it will also respond to any pressure differential that exist that includes atmospheric pressure differentials and pressure differential caused by condensation.

And only a fraction of the total heat will be transferred to the other end.

-jim

This effect will happen (perhaps even more intense) if the rod is held horizontal and the red hot any place into a stream of running tap water. For a really dramatic demonstration of the effects of steam - heat the end of a length of water pipe to cherry red and dunk that into water.

-jim

As I mentioned before, there's practically no expansion or contraction because the pressure drop is about .5 PSI MAX. That amounts to a max volume change of of 3%. There's no "rapidly" expanding steam involved.

Let's add some numbers to your theory, shall we? The heat of vaporization of water is 2260J/g, and the heat capacity of steel is

0=2E448 J/g-=B0C. Let's make the following assumptions: 1=2E Let's assume a full 10cc's of water (about 250 drops) condense on the rod. 2=2E Let's assume that ALL of that energy is concentrated onto the surface of the rod 6" away from your hand at time=3D0 seconds. 3=2E No heat escapes the rod: i.e.adiabatic

The temperature at your first finger, 6" away from where the energy went in, will notice an increase in temp of 1 degree at time =3D 87.5 seconds. This is because steel is a relatively poor conductor (k=3D50W/mK).

At 2 minutes, your first finger would be only 3 degrees hotter. Now in the original post, Henry stated "...I would plunge it into the water because someone asked me to do something else..." So I seriously doubt he would be holding onto the bar that long.

Dave

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Are you assuming convection heating of the rod by the steam via natural convection?

How long was the rod, or do we need to think about characteristic length b/c the OP stated it was at an angle?

The only eqn. I know (or recall well enough to look up) is the Churchill-Chu eqn, and it's pretty ugly, using solving for the Nusselt ave. number, using Rayleigh and Prandtl numbers, correcting for the angle by taking g in the Rayleigh # as g * cos (theta), and then using Nu = (hL/k). Also taking those numbers to some pretty odd fractional powers. That's all just to solve for a heat transfer coefficient, h, in watts per (meter^2 * Kelvin).

Do you know an easier way? I thought there was, but upon pondering it some more, I'm not sure if there is.

regards, k wallace