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Time for a steel die to reach temperature

  1. Nov 19, 2015 #1
    Hello. I will be doing some experiments with boron steel. The idea is to heat a steel specimen (200x20x1,5 mm dimensions) to 950°C and then quickly transfer it from the oven to the die at room temperature. When the specimen reaches the steel die it will supposedly be at around 800°C. At this temperature, the upper and lower die close and come into contact with the specimen which rapidly cools and quenches in 15 seconds. Since the tool is not cooled, I am now trying to calculate an approximate time when the dies will be at certain temperature (lets say around 60- 80°C) beacuse of the heat conduction from the specimen. The upper and lower die combined are approximately 200x200x150mm in size. Since I will be doing this experiment repeatedly I need at least some kind of orientation how long the experiments will last. Thank you for any help or advice.
     
  2. jcsd
  3. Nov 19, 2015 #2
    Assuming that the die material has the same thermal capacity (specific heat) as the specimen can be roughly estimated temperature rise in one experiment:
    $$\Delta t= 1+\frac{v}{V}\cdot (t_e-t_r)=1+\frac{200\cdot 20 \cdot 1,5}{200\cdot 200\cdot 150}\cdot (800-20) $$
    ie less than one degree if I understand well.
    This estimate does not account for the heating associated with stamping. :)
     
  4. Nov 19, 2015 #3
    Well the die material has a bit higher specific heat, but I think the difference is not cruical for the approximate calculation. Correct me if I'm wrong, the result of this equation means, that becuse of cooling down the specimen from 800°C, the die roughly heats up by only 1.78°C in 1 experiment?
     
  5. Nov 19, 2015 #4
    Forgive me for haste. The formula had to be:
    $$t=\frac{t_r\cdot V + t_e\cdot v}{V+v}$$
    and as V>>v:
    $$t=t_r+ \frac{t_e\cdot v}{V}=20+0,8=20,8$$
    so in the next experiment tr=20,8...
    That means $$\Delta t = t - t_r = \frac{t_e\cdot v}{V}=0,8 $$
    :)
     
  6. Nov 19, 2015 #5
    Did not think the difference would be so small :) May I just ask where did you get this equation, since I was searching in a lot of different sources and have never seen so simple solution elsewhere? But thank you for the reply!
     
  7. Nov 19, 2015 #6
    You have, I sure. You just did not recognize it. The conservation heat energy (balance):
    $$c\cdot m\cdot t _m+ c\cdot M\cdot t_M = c\cdot (m+M)\cdot t$$
    when two bodies are made of the same material or have equal spesific heats (c). Than assuming the densities are close too we can change the masses by volumes. That is because here it is important the ratio as it is a proportion (the resulting equation for temperature of both bodies t - right side )).
     
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