Using a partial immersion thermometer incorrectly immersed

Well, this is a new twist but maybe not as unusual as I had thought. A customer called us today with this question: he has an NIST traceable calibrated thermometer, calibrated for partial immersion (76mm). He is going to use it only immersed to 50mm. What kind of an error will he have? The thermometer has a range of -1 to 101 °C in 0.1° divisions, and he wants to perform testing at 40, 70 & 90 °C

My response to that question was, as typical, “The magnitude of the error introduced by the decreased immersion will be modest, but good science demands that we calculate and quantify the error, and then decide if it is important (or not) to the application.”

Here’s the issue: the thermometer is designed to be immersed 76mm (3″) into the liquid being tested, and has been calibrated (certified) in that condition of immersion.

In this application, the thermometer is only immersed 50mm, so 26mm of the thermometer stem which should be exposed to the temperature to be measured is instead exposed to some unknown temperature. How does that affect the indication of the thermometer?

The calculation is almost identical to the determination of a correction for a total immersion thermometer partially immersed.

1. We need to determine 4 variables:

k is given.
k = the coefficient of expansion of the thermometric liquid and the glass, combined.

For Celsius mercury thermometers, k = 0.00016
For Fahrenheit mercury thermometers, k = 0.00009
For red liquid Celsius thermometers, k = 0.001
For red liquid Fahrenheit thermometers, k = 0.0006

n = the number of scale degrees of the thermometer column between the surface of the liquid being measured and the immersion line.
In this case, the thermometer is immersed to to 50 mm degree mark, so 26mm of stem is emergent. If we measure 26mm on the scale of this thermometer, we see that 26mm represents 5.4 degrees C. So, n = 5.4

T = the reading of the thermometer in situ
(Let’s assume that the thermometer reads exactly 40.00 °C)

t = average temperature of the emergent liquid column.
This is the ‘tricky’ variable. To obtain this value, suspend alongside the main thermometer a secondary, total immersion thermometer. Position this thermometer so that its bulb is centered halfway between the surface of the liquid and the immersion line. The temperature indicated on the second thermometer will be our best estimate of the average temperature of the emergent liquid column. For this example, we will assume a temperature of 30 °C was observed.

2. Now, find the magnitude of the correction from the following equation:

Correction = kn(T-t)

(0.00016 x 5.4) x (40.00-30) = 0.0086

Adding this value to the observed reading of the thermometer yields 40.00° + 0.0086 = 40.0086 °C which is the actual temperature of the liquid being measured.

Example 2.
Suppose the thermometer is reading 60.03° (T), n remains unchanged because the immersion is constant, and the measured temperature around the emergent stem is 37 °C

(0.00016 x 5.4) x (60.03-37) = 0.019°

Adding this value to the observed reading of the thermometer yields 60.03° + 0.019 = 60.049 °C which is the temperature of the liquid being measured.

Example 3.
Suppose the thermometer is reading 90.06° (T), n remains unchanged because the immersion is constant, and the measured temperature around the emergent stem is 54 °C

(0.00016 x 5.4) x (90.06-54) = 0.031°

Adding this value to the observed reading of the thermometer yields 90.06° + 0.031 = 90.091 °C which is the temperature of the liquid being measured.

Remember that the greater the departure of the test temperature from room temperature, the greater the correction – and the greater the uncertainty of the measurement.

The customer then asked, “The calibration report has a correction factor at (for example) 90 °C of -0.04 °C. Do I still apply that correction? Yes, absolutely. The correction from the report is for that thermometer, properly immersed. So, the actual temperature of the medium being measured in the hypothetical example above is 90.06 + 0.031 – 0.04 = 90.051 °C

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