Question

class: experimental methots

Problem 3 (25 pts.) A J-type thermocouple is calibrated against an RTD standard to within 0.02°C between 0° and 200°C. The emf is measured with an A/D converter having a resolution of 0.075 mV. The reference junction temperature is provided by an ice bath. The calibration procedure yields the following linear equation: T(°C) = 0.11 + 19.2V ,where the units of voltage Vis mV. This first order curve fit has an error of 0.5°C. Part a) 15 pts. Find the uncertainty in the temperature measurement within the range of temperature values. Part b) 10 pts. Suppose the ice bath has melted and the reference junction is now at room temperature of 25°C. What is the temperature sensed by the thermocouple if the emf at the measuring junction is 4 mV?

Answer 1

1. Uncertainity is range of values in which the true value lies. No measurement device is perfect. The accuracy for every instrument is limited. For example, while measuring length of a line, some may read it as 11.22mm while other 11.23 and third one 11.21. Now, we can surely say that the value is 11.2X . This X is nothing but the uncertainity. So, the measured value should be expressed as 11.22 ±0.01. Thus, rather measuring a fixed reading, we get a range of values in which the true value lies.

The RTD reading is ±0.02°C and curve fit has error of 0.5°C.

Hence, max possible error = +0.02 + 0.5 = 0.52

Thus, when we measure using this thermocouple as 50°C, the actual value will be 50±0.52 °C.

Thus, the uncertainty is ±0.52 °C. .........Answer 1


2. T = 0.11 + 19.2V
Given that V = 4mV = 0.004 V

T = 0.11 + 19.2 * 0.004  = 0.1868 °C

The thermocouple will sense a temperature of 0.1868°C ..........Answer 2


Note: Please upvote if this helps you. In case of doubt, comment below. Good luck!