A thermocouple, with a spherical junction diameter of \(0.5 \mathrm{~mm}\), is used for measuring the temperature of hot air flow in a circular duct. The convection heat transfer coefficient of the air flow can be related with the diameter \((D)\) of the duct and the average air flow 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 air flow velocity that the thermocouple can be used, if the maximum response time of the thermocouple to register 99 percent of the initial temperature difference is \(5 \mathrm{~s}\).

First, let's find the heat transfer coefficient based on the given relationship: \[h = 2.2\left(\frac{V}{D}\right)^{0.5}\] Here, \(h\) is the heat transfer coefficient of the air flow, \(V\) is the air flow velocity, and \(D\) is the diameter of the duct.

Next, we will write the expression for response time \(t\): \[t =\frac{\rho c_{p}r^2}{6h}\] Here, \(t\) is the response time, \(\rho\) is the density of thermocouple, \(c_{p}\) is the specific heat capacity, \(r\) is the radius of the thermocouple, and \(h\) is the heat transfer coefficient.

We want the response time to be equal to 5 seconds (99% of the initial temperature difference): \[5 s =\frac{8500 \cdot 320 \cdot (0.5 \cdot 10^{-3} / 2)^2}{6h}\] Now, we need to express \(h\) in terms of \(V\) and \(D\). From step 1, we have: \[h = 2.2\left(\frac{V}{D}\right)^{0.5}\] So we can rewrite the previous equation as: \[5 s =\frac{8500 \cdot 320 \cdot (0.5 \cdot 10^{-3} / 2)^2}{6 \cdot 2.2\left(\frac{V}{D}\right)^{0.5}}\] Now we need to solve for the minimum air flow velocity \(V\): \[5 =\frac{8500 \cdot 320 \cdot (0.5 \cdot 10^{-3} / 2)^2}{6 \cdot 2.2 \cdot (V / D)^{0.5}}\] \[\Rightarrow V^{0.5} = \frac{6 \cdot 2.2}{5} \cdot \frac{8500 \cdot 320 \cdot (0.5 \cdot 10^{-3} / 2)^2}{D}\] Finally, for the minimum air flow velocity, \(V = \left(\frac{6 \cdot 2.2}{5} \cdot \frac{8500 \cdot 320 \cdot (0.5 \cdot 10^{-3}/ 2)^2}{D}\right)^2\) Note that to find the exact \(V\) value, you will need to know the diameter of the duct (\(D\)).