A long cylindrical stainless steel rod \(\left(c_{p}=\right.\) \(450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, \varepsilon=0.30\) ) with mechanically polished surface is being conveyed through a water bath to be quenched. The \(25-\mathrm{mm}\)-diameter stainless steel rod has a temperature of \(700^{\circ} \mathrm{C}\) as it enters the water bath. \(\mathrm{A}\) length of \(3 \mathrm{~m}\) of the rod is submerged in water as it is conveyed through the water bath during the quenching process. As the stainless steel rod enters the water bath, boiling would occur at \(1 \mathrm{~atm}\). In order to prevent thermal burn on people handling the rod, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). Determine the speed of the rod being conveyed through the water bath so that it leaves the water bath without the risk of thermal burn hazard.

Before we calculate the speed, let's first calculate the mass of the stainless steel rod. Since we know its density (\(\rho = 7900\,\mathrm{kg/m^3}\)), diameter (\(D = 25\,\mathrm{mm}\)), and length (\(L = 3\,\mathrm{m}\)), we can determine the mass using the formula: Mass = Density × Volume $$m = \rho \times V$$ Where \(V\) is the volume, which can be calculated as: $$V = \pi \times r^2 \times L$$ Now, let's find the mass (m).

Now we will calculate the heat transfer rate (Q) necessary to bring the temperature of the stainless steel rod down below \(45^{\circ} \mathrm{C}\) using the formula from the analysis: $$Q = mc_p(T_{initial} - T_{final}) $$ We plug in the values given and the mass that we calculated in step 1.

Given the heat transfer rate (Q) and the length of the submerged rod, we can now calculate the speed (v) at which the rod must be conveyed through the water bath so that it leaves the water bath without the risk of thermal burn hazard. To do this, we'll use the formula: $$v = \frac{Q}{hA(T_{initial} - T_{final})L} $$ Where: - h is the heat transfer coefficient [W/m²K] - A is the surface area of the rod [m²] - L is the length of the rod submerged in the water bath [m] We already have Q, now let's calculate the surface area A, knowing the diameter D and length L: $$ A = πDL $$ Since the given ε (emissivity) is 0.30, we assume that steel rod is not a black body. For black bodies, the heat transfer coefficient (h) between the steel rod and water bath can be calculated by the equation: $$ h = 6.124 \times εv^{0.805}$$ Now that we have the required information, we can determine the speed of the stainless steel rod by plugging in the values in the formula for v. After finding the speed of the stainless steel rod, we can conclude that conveying it at this speed through the water bath will allow it to leave the bath without posing a risk of thermal burn hazard.