Question

In metal processing plants, workers often operate near hot metal surfaces. Exposed hot surfaces are hazards that can potentially cause thermal burns on human skin. Metallic surfaces above \(70^{\circ} \mathrm{C}\) are considered extremely hot. Damage to skin can occur instantaneously upon contact with metallic surfaces at that temperature. In a plant that processes metal plates, a plate is conveyed through a series of fans to cool its surface in an ambient temperature of \(30^{\circ} \mathrm{C}\). The plate is \(25 \mathrm{~mm}\) thick and has a thermal conductivity of $13.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The temperature at the bottom surface of the plate is monitored by an infrared (IR) thermometer. Obtain an expression for the variation of temperature in the metal plate. The IR thermometer measures the bottom surface of the plate to be \(60^{\circ} \mathrm{C}\). Determine the minimum value of the convection heat transfer coefficient needed to keep the top surface below \(47^{\circ} \mathrm{C}\).

Short answer

Answer: The minimum value of the convection heat transfer coefficient needed to keep the top surface below \(47^{\circ}\mathrm{C}\) is 409.41 W/m²K.

Step-by-step solution

Assume a one-dimensional heat conduction problem

For simplicity, we can assume that heat transfer takes place only in the vertical direction (through the thickness direction), and the temperature variation is only in that direction. This allows us to use the one-dimensional heat conduction equation to find the temperature distribution in the metal plate.

Apply Fourier's Law of heat conduction

Fourier's Law of heat conduction states that the heat flux (q) is proportional to the temperature gradient (dT/dx) and the proportionality constant is the thermal conductivity (k): q = -k * (dT/dx)

General heat conduction equation in one dimension

For a one-dimensional steady-state conduction without heat generation, the general heat conduction equation simplifies to: d^2T/dx^2 = 0 This implies that the temperature distribution in the plate is linear. T(x) = C1*x + C2, where C1 and C2 are constants to be determined from boundary conditions.

Apply boundary conditions

The bottom surface temperature is given as \(60^{\circ} \mathrm{C}\). At the bottom surface (x=0), the temperature distribution equation is: T(0) = C2 = 60 The plate's thermal conductivity (k) is given as \(13.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Additionally, the plate's thickness is given as 25 mm, which is equivalent to 0.025 m.

Calculate convection heat transfer coefficient

The convective heat transfer on the top surface (x=0.025) can be expressed as: q = h * (T_top - T_ambient) where h is the convection heat transfer coefficient we need to find, T_top is the temperature of the top surface that we want to be below \(47^{\circ} \mathrm{C}\), and T_ambient is the ambient temperature given as \(30^{\circ} \mathrm{C}\). We also have: q = -k * (dT/dx) at x = 0.025 To maintain T_top less than or equal to \(47^{\circ} \mathrm{C}\): h * (47 - 30) = -13.5 * (dT/dx)

Find the minimum value of the convection heat transfer coefficient

Now we need to find the minimum value of the convection heat transfer coefficient needed to accomplish this temperature goal. The temperature gradient, (dT/dx), can be found from the temperature distribution equation: dT/dx = C1 We also have T(x) = C1*x + C2 => C1 = (T(x) - C2) / x At x = 0.025 and T(0.025) = 47, we have C1 = (47 - 60) / 0.025 = -520 Now we can find the minimum value of the convection heat transfer coefficient: h * (47 - 30) = -13.5 * (-520) h = (13.5 * 520) / 17 h = 409.41 W/m²K The minimum value of the convection heat transfer coefficient needed to keep the top surface below \(47^{\circ} \mathrm{C}\) is 409.41 W/m²K.