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I am trying to develop an equation to determine the cooling rate of a steel tube in air. I'm using Fourier's Law, Stefan-Boltzman Law, Newton's Law as well as the specific heat capacity equation. The tube transfers heat by radiation and convection to the air and it transfers heat through conduction to the steel cooling bed it is laying on. The equation I have determined from these is $$T_{tube, final} = T_{tube, intial}- \frac{(\dot Q_{radiation} + \dot Q_{convection} + \dot Q_{conduction})\times\Delta t}{\rho VC_{p}}$$ where, $$\dot Q_{radiation}=\epsilon\sigma A(T_{tube, initial}^4-T_{air}^4),$$ $$\dot Q_{convection}=hA(T_{tube, initial}-T_{air}),$$ $$\dot Q_{conduction}=kA\frac{T_{tube, intial}-T_{cooling bed}}{dx}.$$ The thermodynamic properties of the steel are: $$C_{p}=416\frac{J}{kgK}$$ $$\rho=7667 \frac{kg}{m^3}$$ $$k=24.2 \frac{W}{mK}$$ $$\epsilon=0.86$$ My assumptions making this equation are:

  • Air temperature and cooling bed temperature is at a constant 300 K.
  • Temperature dependent properties do not vary much and should be considered constant.
  • The heat transfer coefficient is 5 tW/(m^2K).
  • The temperature throughout the tube is consistent

I know the tube starts at about 1300 K and drops to about 1150 K in about 40 seconds, when I use this equation though I get really low final temperatures, sometimes even negative ones. Can you help point me in the right direction? I am not looking to develop a 100% precise equation. Just one that will give me a close estimate with only a few percent error.

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  • $\begingroup$ Show your working - must be an error or something somewhere... $\endgroup$ – Solar Mike Jan 30 '19 at 8:26
  • $\begingroup$ My guess is that you're using a $\Delta t$ that's way too big. If you want to estimate the temperature after 40 seconds, my advice is to use $\Delta t = .1 s$ or smaller. calculate the temperature after .1 sec, then update $T_{tube, initial}$ with the value that you get for $T_{tube,final}$. Do that 400 times (have a computer do it 400 times). If you're still getting negative temperatures, use a smaller time step. $\endgroup$ – EMiller Feb 3 '19 at 0:59

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