I am conducting few trials for checking the placement of a thermocouple on a valve for measuring the temperature of the valve and the material is an X45 alloy. You can see the slot taken on the valve and the thermocouple is inserted in the slot (picture is attached).Thermocouple placed on the neck of the valve


  1. Turned on the muffle furnace with a set point of 300 deg C

  2. After the furnace temperature is stable at 300 deg C, kept the valve inside the furnace.

  3. Kept the valve inside the furnace and let it attain an equilibrium.

  4. Took the valve outside after set amount of time and measured the temperature using a microcontroller (arduino + MAX6675). IT took about 10 seconds to measure the temperature and I could not run the wires inside the furnace as I did not have access to high temperature withstanding cables for the thermocouple. Also, as the valve is taken out for measuring the temperature, there will be some heat loss due to atmospheric exposure.

  5. After 2 hours of this trial, I found the temperature of the material to be around 250 deg C.

    Due to the huge difference between setpoint and the material temperature, I wanted to raise a question.

NOTE: The thermocouple and the furnace are calibrated.

My question is, if a material is placed in a furnace with a setpoint temperature, would the temperature of the material be same as the setpoint after a certain amount of time? My guess is there will be some temperature difference but since I am not a mechanical engineer, I want to clarify this in this wonderful forum.

Thank you!


  • $\begingroup$ In your third step you said you "let it attain an equilibrium." Do you mean you let the furnace reach equilibrium or the material? If it was the material, how do you know it's at equilibrium without measuring its temperature? $\endgroup$ Aug 9 '17 at 15:07
  • $\begingroup$ Alright I should not have used the word equilibrium. I just placed the material inside the furnace for quite a while. Thank you for the reply! $\endgroup$
    – Vish_evo
    Aug 9 '17 at 15:44
  • $\begingroup$ how do you know the furnace calibration is any good? $\endgroup$
    – agentp
    Aug 13 '17 at 21:58

The standard technique is to weld a thermocouple to the work. Then the time to temperature depends on the furnace, size , heat source, fan , etc. A serious furnace will have a survey ( like Mil H 6872 ) of temperature variations available from the manufacturer. If this is a typical small lab muffle , just weld the thermocouple to the valve you already sacrificed. For such a small part you should be looking at much less than one hour.

  • $\begingroup$ Thank you. But does that mean the valve will reach the setpoint temperature? The furnace is a small lab furnace and valve itself is about 96mm in length and 5 mm in thickness. $\endgroup$
    – Vish_evo
    Aug 10 '17 at 0:20
  • $\begingroup$ It will reach the temperature of the furnace ;The controller may not be accurate. I looked up engine valve alloys in an old text and found no designations like "X 45" , but several mentions of a 45 degree seat on X =exhaust valves. $\endgroup$ Aug 10 '17 at 1:46

You can use Newton's Law of Cooling to model both the heating and cooling of your valve. You'd use the data you already collected to establish the model, then use the model to see how long it takes the valve to approach equilibrium in the furnace. You can use the same model to estimate the temperature behavior between the time you removed the part from the furnace and when the Arduino was able to get a fix on the thermocouple.

Here's the detailed process for modeling the heating/cooling of the part. I'll be working with the following sample temperature ($T$) vs time ($t$) data. The data represents temperature data taken every minute, starting 5 minutes after the part is removed from the furnace.

t (min)   T (C)
5         234
6         222
7         211
8         201
9         191
10        182

Take the natural logarithm of the temperature data, it should be roughly linear with respect to time. Now do a linear regression through $t$ vs. $ln(T)$ and you should get a slope ($\alpha$) and y-intercept of -0.05 and 5.704 respectively. I used Excel to do the regression; LibreOffice Calc, Matlab, Octave, et al. should also do the job.

The temperature of the part just as it came out of the furnace ($T_0$) will be approximately equal to $e$ raised to the power of the y-intercept: $e^{5.704}$, or 300 C in this case.

You can use the slope and initial temperature from the linear fit combined with Newton's Law to model the cooling process:

$$ T(t)=T_0\cdot e^{\alpha t} $$

You can use the same equation to model the heating process in the furnace by flipping the sign of $\alpha$, as long as the heat transfer modes within the furnace are similar to those during cooling, e.g. natural convection and radiation.


Another answer : parts about the size of your valve, smallish ( 3 cubic feet) electric furnace with strong circulation .Parts heated to 2150 F required less than 25 minutes but this included a couple minutes soaking time ( at temperature , ie. essentially no difference in temperature between parts and furnace).


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