Question

The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1.2-mm-diameter sphere. The properties of the junction are k = 35 W/m⋅K, rho = 8500 kg/m3, and cp = 320 J/kg⋅K, and the heat transfer coefficient between the junction and the gas is h = 110 W/m2⋅K. Determine how long it will take for the thermocouple to read 98.5 percent of the initial temperature difference

Answer 1

Answer:

First compute the characteristic length and the Biot number to see if the lumped analysis is applicable

Lc = V/A = (pie*D3/6) / (pie * D2)= 1.2/6 = 0.0012/6= 0.0002m

Bi = hLc/K = (110W/m2.oC)(0.0002)(moC/35W)= 110*0.0002/35 = 0.0006 less than 0.1

Since the Biot number is less than 0.1, we can use the lumped parameter analysis.

In such an analysis, the time to reach a certain temperature is given by

t = -1/bIn(T-Tinfinite/T - Tinfinite)

From the data in the problem we can compute the parameter, b, and then compute the time for the ratio (T – Tinfinite/(Ti – Tinfinite) to reach the desired value.

b = hA/pCpV = h/pCpLc = 110/8500*0.0002 *320*s

b = 110/544s = 0.2022/s

The problem statement is interpreted to read that the measured temperature difference T – Tinfinite has eliminated 98.5% of the transient error in the initial temperature reading Ti – Tinfinite so the value of value of (T – Tinfinite)/(Ti – Tinfinite) to be used in this equation is 0.015

t = -1/bIn(T-Tinfinite/T - Tinfinite)

t = -s/0.1654 (In0.015)

t = (-s*-4.1997)/0.2022

t = 20.77s

It will take the thermocouple 20.77s to reach 98.5% of the initial temperature