What is the temperature at thermal junction T if the EMF output is 1.8mV?

[FurryMcFlurry]

Hi guys! First time posting here so please be nice. :)

Homework Statement

A thermocouple T is connected in the following circuit with two different types of metal wires to measure a temperature ranging from 0°C to 500°C:

[img=http://s12.postimg.org/9y4nuly15/mechatronics_thermocouple.png]

This thermocouple's EMF output E_AB is shown in the following graph:

[img=http://s12.postimg.org/hfdv9tnk9/mechatronics_thermocouple_graph.png]

Suppose that the measurement from the digital voltage meter (DVM) is 1.8mV, estimate the measured temperature at thermal junction T (Answer must match to 1 decimal place).

Homework Equations

E_AB = αT + βT²

The Attempt at a Solution

What I've done is that I've taken both the points on the graph (T = 100°C and T = 200°C) and solved for values of α and β by using simultaneous equations. From here, I was able to get α = 4.4333 x 10^-5 and β = -1.3333..3x10^-8

4. Where I'm stuck
My current stab at the solution is basically the substitution of E_AB = 1.8mV (the measurement from the DVM) into the equation and substituting the values of α and β into the equation, and then solving for the roots of T. My values as of such have been completely unrealistic (x10^-4). I can't shake off the feeling that that step is wrong... I've been looking through my lecture notes (which are, by the way, are very much incomplete and only mentions that one equation and doesn't have any examples), and I haven't been able to find anything that could help me. Any help would be very much appreciated.

 

[rude man]

You need to sum voltage drops around the entire circuit and set = 0. This includes the DVM.

Write this equation.

One question: Does E_AB mean E_B - E_A or E_A - E_B?

 

[FurryMcFlurry]

Thanks for the reply!

This wasn't mentioned in the question (and was never explained in the lectures for that matter) so I'm assuming that E_AB means E_B-E_A.

Summing the loops, I've gotten the voltage loop (clockwise) as:
V_T=50°C + 1.8mV + V_T=150°C - V_T = 0

After substituting T=50°C and T=150°C into the equation, I got E_T=50°C = 2.2mV and E_T=150°C = 6.3mV.

Substituting it all into the loop, I get V_T = 1.8mV + 2.2mV + 6.3mV = 10.3mV.

Would this be correct?

 

[rude man]

Thanks for the reply!

This wasn't mentioned in the question (and was never explained in the lectures for that matter) so I'm assuming that E_AB means E_B-E_A.

Summing the loops, I've gotten the voltage loop (clockwise) as:
V_T=50°C + 1.8mV + V_T=150°C - V_T = 0

There should be two V₁₅₀ terms in your loop, not just one. A quick look at the schematic diagram tells you that.

Why not start your voltage loop at the low side of the dvm:

(E_B - E_A)₁₅₀ + (E_A - E_B)_T + (E_B - E_A)₁₅₀ + (E_A - E_B)50 = 1.8 mV.

Notice that you need to distinguish between V and -V for each drop across dissimilar metals. I think you missed that fact. Your V₅₀ and V₁₅₀ have the same polarity which is incorrect, again as can be seen on the schematic diagram.



/QUOTE]

 

[rezeew]

Hello! Welcome to the forum.

First of all, great job on solving for the values of α and β! That is the correct first step in solving this problem.

Now, for the next step, you are correct in substituting the value of EAB into the equation and solving for the roots of T. However, there are a few things you need to keep in mind:

1. Make sure that all units are consistent. The units for EAB are millivolts (mV), so the units for α and β should also be in mV.
2. Remember that the equation is EAB = αT + βT^2, so when you substitute EAB = 1.8mV, you should have 1.8mV = αT + βT^2.
3. Since this is a quadratic equation, there will be two roots for T. One of the roots will be the correct temperature at thermal junction T, while the other root will be an extraneous solution that should be discarded.
4. Make sure to convert the final temperature from Kelvin to Celsius, since the equation is in terms of temperature difference (ΔT = T2 - T1).

Hope this helps! Let me know if you have any other questions. Good luck!