Question

A spherical thermocouple junction of diameter $0.706mm$ is to be used for the measurement of temperature of a gas stream. The convective heat transfer coefficient on bead surface is $400W/m^{2}K.$ Thermophysical properties of thermocouple material are $k=20W/mK,c=400J/kgK$ and $\rho =8500kg/m^3.$ If the thermocouple initially at ${30}^{\circ }C$ is placed in a hot stream of ${300}^{\circ }C,$ the time taken by the bead to reach ${298}^{\circ }C,$ is

• A.
  $2.35s$
• B.
  $4.9s$
• C.
  $14.7s$
• D.
  $29.4s$

Answer 1

The correct option is B

$4.9s$
Given data:

$h=400W/m^{2}K:k=20W/mK:c=400J/kgK\rho =8500kg/m^3:T\left(t\right)={298}^{\circ }C$
Now, $L=\frac{V}{A}=\frac{1/6\pi D^3}{\pi D^2}=\frac{D}{6}=\frac{0.706\times {10}^{-3}}{6}=1.76\times {10}^{-4}m$

Biot number,
$Bi=\frac{hL}{k}=\frac{400\times 1.76\times {10}^{-4}}{20}=0.0023SinceBi<0.1$

Lumped system analysis can be used
$\frac{T-T_{\infty }}{T_i-T_{\infty }}=e^{-\left(\frac{hAt}{\rho cV}\right)}$
$\frac{298-300}{30-300}=e^{-\left(\frac{400\times 6\times t}{8500\times 0.706\times {10}^{-3}\times 400}\right)}$

$7.407\times {10}^{-3}=e^{-t}$
Taking $lo{g}_e$ both sides, we get
$lo{g}_{e}7.407\times 1=-t$
or $t=4.90seconds$