Topic: Heat Transfer
The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1-mm-diameter sphere. The properties of the junction material are $k=40~W/mK$ $\rho=8500~kg/m^{3}$, and specific $heat=320~J/kgK$ and the convective heat transfer coefficient between the junction and the gas is $h=200~W/m^{2}K$. Determine how long it will take for the thermocouple to read 99 percent of the initial temperature difference.
- A. 10.5 s
- B. 5.3 s
- C. 21 s
- D. 15 s
Answer: A. 10.5 s
Explanation: Using Lumped Capacitance method, $\frac{T - T_\infty}{T_i - T_\infty} = e^{-t/\tau}$, where time constant $\tau = \frac{\rho V c}{h A_s}$.
For a sphere, $\frac{V}{A_s} = \frac{\pi D^3 / 6}{\pi D^2} = \frac{D}{6}$.
$\tau = \frac{\rho c D}{6 h} = \frac{8500 \times 320 \times (1 \times 10^{-3})}{6 \times 200} = \frac{2720}{1200} \approx 2.267~s$.
To read 99% of the initial difference means 1% of the difference remains: $\frac{T - T_\infty}{T_i - T_\infty} = 0.01$.
$0.01 = e^{-t/2.267} \Rightarrow -4.605 = -t/2.267 \Rightarrow t = 4.605 \times 2.267 \approx 10.44~s \approx 10.5~s$.