Question

Problem 6 (15 points) In a thermocouple, the combination of two dissimilar materials (A and B) is used, the Seebeck potential is determined from a differential Seebeck coefficient α48 ar, The net voltage of the Junction is dl.,-«, dT and Bu is the differential coefficient. The voltage as a function of a temperature gradient for a K-type thermocouple with a high degree of accuracy can be approximated by a second order equation: v,,-4+a,T+a,T2-0.0 + 39.35T + 1.72*10"r' where v," is in μν. Find the Seebeck voltage when the voltage gradient is 100 C. For the case where the thermocouple is to measure the top oil temperature of a transformer and the themocouple is brought out to the transformer control cabinet which is at an ambient temperature of 40 degrees C Find the alarm voltage if you wish to alarm when the top oil temperature is 1050. Determine a simple circuit model for the above themocouple. (Assume a negligible internal resistance relative to the voltmeter

Answer 1

A known thermo couple. V_AB = 39.35T + 2.72 times 10^-2 T^2 mu V. VT here corresponds to temperature cordient when 100 degrees C is the gradient then V_AB = 39.35 times 100 + 2.72 times 10^-2 times 100 times 100 = 39.35 times 100 + 2.72 times 100 = (3935 + 272)mu V = 4207 mu V when the top 0 it temp = 105 degrees C & ambient temp = 40 degrees C Delta T = (105 - 40)degrees C Delta T = 65 Then V_AB = 39.35 times Delta T + 2.72 times 10^-2 Delta T^2 = 39.35 times 35 + 2.72 times 10^-2 times 65^2 = 2557.75 + 2.72 times 10^-2 times 652 = 2672.67 mu V \r\n So alarm voltage will be 2672.67 mu V