Question

Problem 3 (30 pts.) A J-type thermocouple is calibrated against an RTD standard within 20.02°C between 0º and 200°C. The emf is measured with a potentiometer having 0.005 mV resolution and an uncertainty of 0.045 mV over its operating range. The reference junction temperature is provided by an ice bath. The calibration procedure yields the following equation: T(°C) = 0.11 + 19.16V - 0.06V2, where the units of voltage V is mV. This second order curve fit has an error of 0.45°C. Part a) 12 pts. Find the maximum uncertainty in the temperature measurement within the range of temperature values. Part b) 8 pts. What is the uncertainty in temperature at 50°C? Part c) 10 pts. Suppose the ice bath has melted and the reference junction is now at 25°C. What is the temperature sensed by the thermocouple if the emf at the measuring junction is 5 mV? Neglect errors in the devices.

Answer 1

Given T(°C) = 0.11 + 19.16V - 0.06V2 a) The standard error of the fit for the second order polynomial is 0.45 Error contributions are RTD - +0.02°C Potentiometer = v (0.0052+0.0452) = 0.0453mV A J-type thermocouple has a sensitivity of 55 V/ °C which is equivalent to a temperature uncertainty of (0.0453 mV) / (1/0.055mV/OC) = +0.820C and the value of se taken to be 0.34, yielding UT = V( 0.022 + 0.822 + 0.452) = 0.8750C c) By the law of intermediate temperatures the output emf for a thermocouple circuit having two junctions, one at 0OC and the other at T1, would be the sum of the emfs for a thermocouple circuit between 00C and 250C and between 250C and T1.

emfo-25 + emf25-T1 = emfo-T1 This relationship allows the voltage reading from the nonstandard reference temperature to be converted to a 00C reference temperature by adding emfo-25 =1.277 to the existing reading. This results in an equivalent output, referenced to 0oC as 1.277+ 5 = 6.277mV From the table temperature corresponce to 6.277 is 1180C