Question

A thermocouple with a spherical junction diameter of \(0.5 \mathrm{~mm}\) is used for measuring the temperature of hot airflow in a circular duct. The convection heat transfer coefficient of the airflow can be related with the diameter \((D)\) of the spherical junction and the average airflow velocity \((V)\) as \(h=2.2(V / D)^{0.5}\), where \(D, h\), and \(V\) are in $\mathrm{m}, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\(, and \)\mathrm{m} / \mathrm{s}$, respectively. The properties of the thermocouple junction are $k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\rho=8500 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)c_{p}=320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$. Determine the minimum airflow velocity that the thermocouple can use, if the maximum response time of the thermocouple to register 99 percent of the initial temperature difference is \(5 \mathrm{~s}\).

Short answer

Answer: The minimum airflow velocity is approximately 4.295 m/s.

Step-by-step solution

Write down the response time formula

According to the first-order thermocouple system, the response time formula can be expressed as: $$ t=\frac{\tau}{Bi}= \frac{\rho V_{s} c_{p}}{3h} $$ where \(t\) is the response time, \(\tau\) is the time constant, \(V_s\) is the volume of the sphere, \(h\) is the convection heat transfer coefficient, and \(Bi\) is the Biot number.

Calculate the Biot number

The formula for the Biot number is given by: $$ Bi = \frac{hL_c}{k} $$ In this case, the characteristic length (\(L_c\)) for a sphere is \(L_c = \frac{D}{3}\) with \(D\) being the diameter of the sphere. Substitute \(L_c = \frac{D}{3}\) and \(h=2.2(V / D)^{0.5}\) into the Biot number formula: $$ Bi = \frac{2.2(V / D)^{0.5} D}{3k} $$

Substitute the given values into the response time formula

We are given the maximum response time as \(t=5\,\mathrm{s}\), so we can substitute into the response time formula: $$ 5 = \frac{\rho V_{s} c_{p}}{3 \cdot 2.2(V / D)^{0.5} D} $$ Now, {we need to find the volume of the sphere} using \(V_{s} = \frac{4}{3}\pi r^{3}\), where \(r = \frac{D}{2}\) and \(D=0.5\,\mathrm{mm}\). The diameter of the sphere given in the problem is in millimeters, so we need to convert it to meters for consistency with the other units. $$ D = 0.5\,\mathrm{mm} \times \frac{1\,\mathrm{m}}{1000\,\mathrm{mm}} = 5 \times 10^{-4}\, \mathrm{m} $$ Thus, compute the volume of the sphere: $$ V_{s} = \frac{4}{3}\pi\left(\frac{5 \times 10^{-4}}{2}\right)^{3} = 6.54\times 10^{-11}\, \mathrm{m^3} $$ Plug \(V_{s}\), \(\rho\), \(c_p\), and \(D\) into the response time formula: $$ 5 = \frac{8500\cdot 6.54\times 10^{-11}\cdot 320}{3 \cdot 2.2(V / (5 \times 10^{-4}))^{0.5}(5\times 10^{-4})} $$

Solve for the airflow velocity

Now, we can solve for the minimum airflow velocity \(V\): $$ V = \frac{8500\cdot 6.54\times 10^{-11}\cdot 320}{3 \cdot 2.2(5) \cdot (5\times 10^{-4})^{0.5}} $$ $$ V \approx 4.295\,\mathrm{m/s} $$ Hence, the minimum airflow velocity that the thermocouple can use to register 99% of the initial temperature difference in 5 seconds is approximately \(4.295\,\mathrm{m/s}\).