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Heat Transfer in a bar of steel(9)

Qshake (Structural) 
20 Apr 11 22:33 
Hello MEs. I'm a long time member who usually stays in my own backyard  over in the structures area. But I have a situation where a relatively quick calculation (I hope) is appreciated from those of you with course work in heat transfer. I have a 10' mild steel rod that at one free end is heated by a typical oxyacetylene torch for say a period of 5 minutes. Let's say that the temperature for the 5 minute period is about 700 degrees F. Say the bar has an area of 1 square inch. How far along the bar's length will the heat travel? Many thanks. Regards, Qshake
EngTips Forums:Real Solutions for Real Problems Really Quick. 

Well, the silly answer is "all the way." But, that's also more or less the correct answer. Are you trying to find out where you can touch it, or what a quasisteady state temperature might be? For where, probably no farther that about 3 ft from the hot end duing the 5 minutes, since it's unlikely to propagate that far that quickly. Over time, the hot end will still be the hottest, but the heat will try to equilibrate down the rod, while radiating and convecting at the same time. The far end will probably climb 50F, maybe? TTFN
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ione (Mechanical) 
21 Apr 11 3:15 

Temperature profile will be a function of distance x (assume one dimension for a rod), and time, t. If the sides of the rod are insulated, its an easy partial dif eqn solved via Fourier series. (I remember doing those in class  by hand.) The zinger is the convective heat transfer coef. Problem then becomes a little more complicated. If the rod is in still air, you can assume one. Sorry i can't help, except if the rod is insulated, but maybe one the wizzes here can. "Very interesting indeed, but tell me, Mr. Faraday, of what use is this electricity?" "Sir, you will soon be able to tax it." 

It's relatively simple if you can quantify a heat flux as an input rather than a temperature.
A crude finite element calculation could be done in a spreadsheet.
Heat into each element = heat stored + heat lost by radiation + heat lost by convection + heat left to be input to the next element.
The heat stored term can then be used to calculate a temperature rise.
If you have an initial temperature at the first element you can iteratively obtain a solution for heat input to the first element. 

It sounds like you're thinking more along the lines of a transient type of problem than a steadystate type of problem: is that accurate? Whenever time becomes a variable, things obviously get more complicated.
The steady state solution is pretty straightforward: You have a fixed temperature boundary and convective and radiation losses along the rod. By discretizing the problem in, say, Octave or Matlab, you should be able to calculate the temperature distribution with Newtonian cooling, radiation loss to ambient, and Fourier's law for each discrete section. Excel would even work fine, although I think that Octave/Matlab is more suited for providing an accurate answer (I suppose that you could write an Excel macro and do just as well).
Of course, things become more complicated when the problem becomes transient. You need to factor in heat energy transferred vs time to your discrete parts. If your initial state is a 700F temperature at one end and a room temperature state at the other, you'll have to determine the amount of heat energy transferred for a given time interval and step through time (i.e. "X" W/m2 for 1 sec = "Y" Joules; "Y" Joules is proportional to "Z" delta Celcius using the specific heat of the material).
It sounds from your description of the problem that a transient analysis may be most appropriate. Are you interested in what happens when the torch is applied, or after it's taken away? 

(3) zekeman (Mechanical) 
21 Apr 11 12:08 
The solution taken from Carslaw and Yaeger, " Conduction of Heat in Solids" is
TmT0=
0.5(TmT)*exp(x*sqrt(nu/kappa))*erfc[.5x/sqrt(kappa*t)+sqrt(kappa*t)]+ 0.5(TmT)*exp(x*sqrt(nu/kappa))*erfc[.5x/sqrt(kappa*t)sqrt(kappa*t)]
Tm = temperature at one end T0 = initial temperature of rod and environment nu=hp/cpa h= surface conductance c= specific heat of rod p =perimeter of rod a= cross section of rod k= thermal conductivity rho density kappa=k/(rho*c) t=time T= temperature at x and t erfc = error function complement (tabulated )
I will evaluate this later


zekeman (Mechanical) 
21 Apr 11 12:12 
Just noticed, the solution I gave is for an infinite rod. I will post the finite rod solution subsequently. 

1 inch square rod = 11/8 diameter.
700 degree (F ?) is really pretty cool  not even red hot, but you're heating it (the tip > correct) for a pretty long time. (Much longer, for example than needed to welding or flame cutting!)
I don't have such a 5 ft round rod, but heated a 13/8 hex bar 6 inches long for 5 minutes with a MAPP gas torch. The hex bar is welded to a 1/2 thick, 6 x 6 plate, held in a vice. Room temp (steel temps also) at the beginning were 72 degrees. Still air, no fan or forced draft cooling.
Mapp gas torch was measured with a IR (no touch) thermometer with a laser pointer on the metal tip of the torch at 1100 degrees.
After 5 minutes of playing the torch flame directly on the end of the bar, that face was 650 degrees F. At the 3 inch point from the flame, the steel was 320  360 degrees. At the 51/2 inch point (just before the weld to the flat plate) the steel bar was 180 degrees. The flat 6x6 plate was 120 degrees.
More experiemnts later  after I get a proper bar. 

Qshake (Structural) 
21 Apr 11 13:08 
All  First let me say thanks to all for answering this matter of mine. After reading your posts, I believe I can better state the problem I'm looking at. Based on your input I appear to be most interested in a temperature profile of the bar (finite or infinite might not matter as I didn't want to give the impression the bar was only 1' in length and after heating for 5 min at 700 deg F, I agree with IRSTUFF that the trivial solution.) for say time periods of 5 mins after the heat is applied, 10 mins after the heat is applied etc. I do realize that this is time dependent and that heat is lost through radiation and conduction and I don't have a clue as to how to determine the heat flux (H or Q) but I can read and apply some formulas (not so much in the heat transfer area though). Thanks again. Regards, Qshake
EngTips Forums:Real Solutions for Real Problems Really Quick. 

SnTMan (Mechanical) 
21 Apr 11 13:10 
I think that if I needed to know soon with a fair degree of certainty, I'd test.
Regards,
Mike 

zekeman (Mechanical) 
22 Apr 11 14:35 
"After 5 minutes of playing the torch flame directly on the end of the bar, that face was 650 degrees F. At the 3 inch point from the flame, the steel was 320  360 degrees. At the 51/2 inch point (just before the weld to the flat plate) the steel bar was 180 degrees. The flat 6x6 plate was 120 degrees. "
That's not the proper experiment. Your results are not meaningful. You can't have a plate ,since it is a heat sink. Just a bare rod , insulation clamped at one end, the end where you play the torch.


zekeman (Mechanical) 
22 Apr 11 15:35 
"It's relatively simple if you can quantify a heat flux as an input rather than a temperature."
Easier said than done. There are problems with this problem, since the heatup time to get to the 700 degrees at the end complicates the problem.
I think the best approach is to do it experimentally. with the welding equipment and rod/plate as it exists.
But to get a handle on the problem, solve the problem of the rod suddenly exposed to temperature T0 ,either analytically or piecewise.
It may be easier to take C&Y, P135, solution for the finite rod with ends at +L, suddenly exposed to Tm,with ambient = initial temperature, T0
TT0=(TmT0)*cosh(x*(nu/kappa)^.5/cosh(L*(nu/kappa)^.5+
summation, n=0 to infinity of
4*(TmT0)/Pi*(1)^n*exp{nu*t[kappa(2n+1)^2Pi^2*t/4L^2]}*cos{(2n+1)*Pix/2L)}divided by (2n+1)[1+{4nu*L^2/[(2n+1)^2*Pi^2*kappa}]
usually, it rapidly converges in a few terms so the method is not overwhelming and with a simple Excel spreadsheet, many more terms can be readily done.


dvd (Mechanical) 
22 Apr 11 18:36 
I vote for racookpe1978, it is very simply done experiment in the shop. Many options for experimental temperature measurement if you need. 

OK, OK  I got a 10 ft long, 1 inch sq bar. More info tonight.
Didn't have one before  had to use the 6 inch 13/8 hex. And the plate was welded on > so I clamped the plate in the vise as a heat sink 'cause it was easier than cutting it off. 8<) 

Quote (racookpe1978):OK, OK  I got a 10 ft long, 1 inch sq bar. More info tonight.
I love it! Actual experimental results! 

I think that racookpe1978 saw an opportunity to play with a welding torch and seized it. 

Drat! Me methods have been discovered: Never pass up the opportunity to (1) buy steel or (2) heat up steel to dangerous temperatures or (3) play with toys (er, tools).
Method. A 10 foot 1x1 bar of carbon steel was marked off at every 6 inches from one end. It was supported horizontally on a knife edge at the 36 inch point, and clamped in a vise at the far end approximately 36 inch up in a static air (well, in a closeddoor garage) at 65 degrees. (Surprisingly, the MAPP gas flame heated the garage sufficiently that I had to open the door not for ventilation or odors as I expected, but because ambient was getting too high, which was changing the reference temps of the walls and concrete floor.)
Steady State Temperatures: After 20 min, the end of the bar was at 750 degrees. At 2 inches, 600. At 6 inches, 260. At 12 inches 110. At 18 inches, 85. Room temp (70) from there to the end of the bar. After 45 minutes, temp's were essentially the same: End of bar = 750775. At 2 inches, 625. At 6 inches, 260 again. At 12 inches 115 At 18 inches, 88. No change down the rest of the bar.
Heatup temps. At 1 minute. At 2 inches, 227. At 6 inches, 82. At 12 inches, 67 At 18 inches, 67. Actually, the bar never really heated up past 12 inches, so my 6 inch approximation to start with wasn't too bad.
At 2 minute. At 2 inches, 420. At 6 inches, 90. At 12 inches, 70 At 18 inches, 67.
At 4 minute. At 2 inches, 440. At 6 inches, 115. At 12 inches, 75 At 18 inches, 67.
At 10 minute. 650 at very the tip of the bar. At 2 inches, 530. At 6 inches, 180. At 12 inches, 82 At 18 inches, 69.
Oh, by the ways ....
Measurements were with a SearsCraftsman (uncalibrated) IR thermometer, aimed with the IR laser light. Figure the numbers are pretty good at low temperatures, less so over 400 degrees (+/ 10 or so).
Tried MAPP gas, and then (later) propane. Propane only got the end of the bar up to 450 degrees, MAPP gas got it easily to 750  800 degrees.
After 2 hours cooling time, it was taking too long (and I was running out of liquid refreshment) for the bar to return to ambient, so I reversed the bar  marking the other end  and tried again.
Slight differences, but essentially the same result.
End of bar (face that the flame was hitting) reached 750  800 degrees within 10 minutes. Bar reached near steady state conditions between 18 and 25 minutes, and a very slow heating down the bar (from room temperature to room temp + 20  35 degrees) from the 12 inch point to the 18 inch point.
No real change (other than the room heating up by 10 degrees?) past the 18 inch point.
LOTS of radiant heat was felt ("measured" 1 inch under the bar) with a bare hand at the 2 inch point. Little felt at the 6 inch point.) Any FEA approximation needs to include this amount.
... 

IRstuff (Aerospace) 
23 Apr 11 14:07 

Qshake (Structural) 
23 Apr 11 23:21 
I love you guys! or gals. Clearly I'm going to expand my backyard to play over here more often. And I'm definitely going to get more toys....tools! Many thanks all! Regards, Qshake
EngTips Forums:Real Solutions for Real Problems Really Quick. 

zekeman (Mechanical) 
24 Apr 11 10:00 
racookpe197,
Excellent. Nice to see some empirical data and it looks almost like the analytic solution.
For the fun of it, I tried to compare the steady state solution, which I did for the infinite bar, ( since 10 feet acts virtually as "infinite") with the empirical
I get
TT0=(TmT0)*exp[sqrt(hp/Ka)x] h=3 BTU/hrft^2F p/a=4/1/12=48ft^1 for the square bar K=20 BTU/hrftF Tm =750 T0=70
hp/Ka=7.2 sqrt(hp/Ka)=2.68
Evaluating at x=2"=1/6' TT0=(75070)exp(2.68/6)=434 T=504
at x= 6" exp(2.68*.5)=.261 TT0=.261*680=178 T=248
at x=1ft exp(2.68*1)=0.068 TT0=.068*680=47 T=117
BTW, the time constant for this system is 1/nu=1/2.4= .4 hrs = 24 minutes (obtained from the transient solution) 

Thank you for the calc's. 



