Chromatography Forum

Can anyone suggest me how to apply correction factor for calibrated thermometer?

For example, if the set temperature is -10.0,10.0,50.0,100.0 and 150.0°C and our thermometer reading is -10.2,10.1,49.9, 100.2, and 149.8°C respectivly.

for example
If our meaured temperature is 75.1, then what will be actual temperature with correction factor?

Is there any referance available for this?

Please help

Regards

Sunjay,
How much precision do you need. Is this an actual glass/liquid thermometer or is it digital. If this is a glass/liquid thermometer the precision over that range isn't bad. If it is digital, there are two ways to approch the problem. One, is to do a linear regression of the data, set versus read temperature and interpolate the result from the equation. The other way is to space your set points evenly across the range of temperatures you wish to measure and use a correction factor based on the deviation from the set temperature closest to the temperature that you are reading. I haven't resarched it, but I would imagine NIST or ASTM would have a procedure for this.

Sunjay,

What is the tolerance of your reference thermometer vs. the accuracy you need for your process? Here most of our liquid in glass thermometers have a tolerance of +/- 1 division and if they are within that range range then they are meeting their specs and we use them without a correction factor. If you need very accurate measurements for your process you may need to look into using a thermometer that is more accurate then your process requirements. We don't readily use correction factors for our thermometers.

Jason

Graph paper

Thanks for ur reply.

While searching internet i found useful referance (article name is: Liquid-in-glass thermometers in the ISO-certified laboratory by Deanne Niller Emory) for calculation for correction factor the link is given below if anyone wants to do the same.

www.iscpubs.com/articles/aln/n9812emo.pdf

It is (b-a)(f-d)/(c-a)+d
b is observed temperature for example 35.0
Set Temp Obs Reading
0.00 –0.02
10.00 10.02
20.00 20.03
30.00 a 30.04 d
40.00 c 39.94 f
50.00 49.96

(35.00–30.00) (39.94–30.04)/(40.00–30.00)+30.04

=34.99
So if the observed reading with our thermometer is 35°C, actually it will be 34.99 with this formula.

Any comments....
Thanks

Regards