Question

A thermocouple junction, which may be approximated as a sphere, is to be used for temperature measurement in a gas stream. The convection coefficient between the junction surface and the gas is known to be h = 400 W/m2 ?K, and the junction thermophysical properties are k = 20 W/m?K, Cp = 400 J/kg?K, and ? = 8500 kg/m3 . Determine the junction diameter needed for the thermocouple to have a time constant of 1 s. If the junction is at 25 oC and is placed in a gas stream that is at 200 oC, how long will it take for the junction to reach 199 oC?

Answer 1

Answer:

t=5.2seconds

Explanation:

First, we state our assumptions:$1. Temperature \ of \ the \ junction \ is \ uniform \ at \ any \ instant\\2. Radiation \ is \ negligible\\3. Constant \ properties$

Given $T_i=25\textdegree C, k=20W/m.K\ , C_p=400J/kg.K \ \rho=8500kg/m^3$

and gas stream$T_-=25\textdegree C , h=400W/m.K$

We use the time constant to find the diameter of the junction:

$\tau _t=\frac{1}{hA}\rhoVC_p=\frac{1}{h\pi D^2}\times \frac{\rho \pi \D^3}{6}C_p$$=>D=0.706mm$

Now, check the validity of the lumped system analysis. With $L_c=r_o/3$

$Bi=\frac{hL_c}{k}=2.35\times 10^-^4\leq 0.1$    #Lumped Analysis is OK

Bi<<0.1, therefore the lumped approximation is excellent. The time required for the junction to reach $T=199\textdegree C$:

$\frac{T(t)-T_\infty}{T_i-T_\infty}=e^-^b^t\\b=\frac{hA}{\rho VC}\\\\t=\frac{1}{b}In \frac{T_i-T_\infty}{T(t)-T_\infty}\\t=5.2s$

It takes 5.2seconds for the junction to reach 199 oC