Question

A metal plate $\left(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, and \)c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\( ) with a thickness of \)1 \mathrm{~cm}$ is being cooled by air at \(5^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the plate is \(300^{\circ} \mathrm{C}\), determine the plate temperature gradient at the surface after 2 minutes of cooling. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

Short answer

Answer: To find the temperature gradient at the surface of the metal plate after 2 minutes of cooling, follow these steps: 1. Calculate the Biot number: Bi = 0.00167 (lumped system analysis can be applied since Bi < 0.1). 2. Calculate the dimensionless time: τ = 0.0102 3. Apply the lumped system analysis to find the surface temperature: \(θ = e^{-τ}\), Ts = T∞ + \(θ\)(Ti - T∞) 4. Calculate the heat flux at the surface: \(\mathrm{q}' = h(T_s - T_{\infty})\) 5. Calculate the temperature gradient at the surface: \( \frac{dT}{dy} \Big|_{y=0} = \frac{\mathrm{q}'}{k}\) By solving these equations, the temperature gradient at the surface of the metal plate after 2 minutes of cooling can be determined.

Step-by-step solution

01. Calculate the Biot number

The first step is to determine if the lumped system analysis method is appropriate for this problem. Check the Biot number (Bi) using the following formula: Bi = \(\frac{hL_{c}}{k}\), where \(h\) is the convection heat transfer coefficient (30 W/m²⋅K), \(L_{c}\) is the characteristic length (the thickness of the plate, which is 1 cm), and \(k\) is the thermal conductivity of the metal plate (180 W/m⋅K). Since the plate is 1 cm thick and all the values are in the metric unit, we need to convert the thickness into meters: \(L_{c}\) = \(\frac{1}{100}\) m = 0.01 m Now, calculate the Biot number: Bi = \(\frac{30 * 0.01}{180}\)

02. Verify the application of the lumped system analysis

If the Biot number is less than 0.1, then the lumped system analysis can be applied. In this case: Bi = \(\frac{30 * 0.01}{180}\) = 0.00167 Since Bi < 0.1, we can use lumped system analysis to solve this problem.

03. Calculate the dimensionless time

Now, we need to calculate the dimensionless time (τ) using the following formula: τ = \(\frac{ht}{ρc_{p}L_{c}}\) where \(t\) is the time after which we need to know the temperature gradient (2 minutes or 120 seconds), \(ρ\) is the density of the metal plate (2800 kg/m³), and \(c_{p}\) is the specific heat capacity of the metal plate (880 J/kg⋅K). Calculate the dimensionless time: τ = \(\frac{30 * 120}{2800 * 880 * 0.01}\)

04. Apply the lumped system analysis to find the surface temperature

Using the lumped system analysis, we can find the surface temperature as follows: \(θ = e^{-τ}\) where \(θ\) is the ratio of the difference between the surface temperature (Ts) and the surrounding temperature (T∞) to the difference between the initial temperature (Ti) and the surrounding temperature (T∞). In this case: T∞ = 5°C, Ti = 300°C Calculate the value of \(θ\): \(θ = e^{-τ}\) Now, we can find the surface temperature of the metal plate (Ts): Ts = T∞ + \(θ\)(Ti - T∞)

05. Calculate the temperature gradient at the surface

Finally, we can determine the temperature gradient at the surface of the metal plate using the formula: \( \frac{dT}{dy} \Big|_{y=0} = \frac{\mathrm{q}'}{k}\) where \(\mathrm{q}'\) is the heat flux at the surface, which we can find using Newton's Law of Cooling: \(\mathrm{q}' = h(T_s - T_{\infty})\) Calculate the heat flux: \(\mathrm{q}' = 30(T_s - 5)\) Now, calculate the temperature gradient: \( \frac{dT}{dy} \Big|_{y=0} = \frac{\mathrm{q}'}{180}\)