Question

A metal plate \(\left(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\right.\), and \(\left.c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) 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: The temperature gradient at the surface of the metal plate after 2 minutes of cooling is approximately -4.06 K/m.

Step-by-step solution

Check Applicability of the Lumped System Analysis

The lumped system analysis is applicable only if the Biot number (Bi) is less than 0.1. The Biot number is calculated as follows: Bi = hL_c / k where h = convection heat transfer coefficient (30 W/m²K) L_c = characteristic length (thickness / 2, in meters) k = thermal conductivity of the metal plate (180 W/mK) Let's first calculate the characteristic length of the plate, and then the Biot number.

Calculate the Characteristic Length

The thickness of the plate is 1 cm, so the characteristic length is half of the thickness: L_c = 0.01 m / 2 = 0.005 m

Calculate the Biot Number

Now let's calculate the Biot number using the given values of h, L_c, and k, Bi = (30 W/m²K) * (0.005 m) / (180 W/mK) = 0.00833 Since Biot number (Bi) is less than 0.1, the lumped system analysis is applicable in this case.

Determine the temperature drop

To determine the temperature gradient of the plate at the surface after 2 minutes of cooling, we'll first find the temperature at that time. We can use the following equation for the lumped system analysis: T(t) - T_∞ = (T_i - T_∞) * exp(-t / τ) where T(t) is the temperature of the plate at time t T_∞ is the temperature of the environment (5°C) T_i is the initial temperature of the plate (300°C) t is the time (2 minutes = 120 seconds) τ is the time constant, defined as τ = ρ * c_p * L_c / h We should first find the time constant (τ) and then calculate the temperature at time t.

Calculate the Time Constant

Using the given values of ρ, c_p, L_c, and h, we can calculate τ as follows: τ = (2800 kg/m³) * (880 J/kgK) * (0.005 m) / (30 W/m²K) = 518.67 seconds

Calculate the Temperature at Time t

Now, we have all the values to find the temperature of the plate at time t = 120 seconds. Let's use the lumped system analysis equation: T(120) - 5°C = (300°C - 5°C) * exp(-120s / 518.67s) T(120) - 5°C ≈ 243.76°C T(120) ≈ 248.76 °C

Calculate the Temperature Gradient

Finally, we can calculate the temperature gradient at the surface of the plate. The temperature gradient (dT/dx) at the surface can be found using the Fourier's law equation: q = -k * dT/dx We can rearrange it to find dT/dx: dT/dx = -q / k We also have the equation for the heat transfer rate by convection: q = h * A * (T - T_∞) Combining these two equations, we get: dT/dx = -h * (T - T_∞) / k By substituting the given values, we get: dT/dx = -(30 W/m²K) * (248.76°C - 5°C) / (180 W/mK) ≈ -4.06 K/m The temperature gradient of the plate at the surface after 2 minutes of cooling is approximately -4.06 K/m.

Key concepts

Biot Number

In the context of heat transfer, the Biot number (Bi) plays a critical role in the lumped system analysis. It is a dimensionless quantity that compares the rate of heat conduction within an object to the rate of heat convection from its surface to the surrounding fluid.

Specifically, the Biot number is defined by the ratio of the convection heat transfer resistance to the conduction heat transfer resistance. It's given by the formula:
\[ \text{Bi} = \frac{hL_c}{k} \] where \( h \) is the convection heat transfer coefficient, \( L_c \) is the characteristic length, and \( k \) is the thermal conductivity of the object. The characteristic length is typically the volume to surface area ratio of the object.

For the lumped system analysis to be valid, the Biot number should be less than 0.1, indicating that the temperature varies significantly more across the fluid than within the object. This means the object can be assumed to have a uniform temperature throughout its volume, which simplifies the analysis considerably.

In the given exercise, the Biot number was calculated to be 0.00833, confirming that the lumped system analysis is applicable for determining the temperature gradient of the cooling metal plate.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, denoted by \( h \), is a measure of the convective heat transfer between a solid surface and a fluid moving past it. It quantifies how effectively heat is transferred from the object to the fluid, or vice versa.

This coefficient is influenced by properties of the fluid such as its velocity, viscosity, and thermal conductivity, as well as by the object's surface roughness and shape. The units of \( h \) are watts per square meter per Kelvin (W/m²K). In heat transfer problems, a higher value of \( h \) indicates that convection is more efficient at transferring heat.

The convection heat transfer coefficient is essential in calculating the heat transfer rate by convection using Newton's law of cooling:
\[ q = hA(T_s - T_\infty) \] where \( q \) is the heat transfer rate, \( A \) is the surface area, \( T_s \) is the surface temperature of the object, and \( T_\infty \) is the temperature of the surrounding fluid.

In our example, the convection heat transfer coefficient is 30 W/m²K. This value is used to calculate heat loss from the plate to the air, and it plays a significant role in determining the Biot number and the effectiveness of the lumped system analysis.

Temperature Gradient

The temperature gradient, often represented as (dT/dx), refers to the rate of temperature change with respect to distance within a material. The gradient signifies how swiftly temperatures decrease or increase from one point to another.

Negative gradients indicate a temperature decrease with distance, while positive gradients show temperature increases. In the field of heat transfer, understanding and calculating the temperature gradient is crucial because it directly relates to the direction and rate of heat flow through a substance.

According to Fourier's law of heat conduction, the heat transfer rate (q) through a material is proportional to the negative of the temperature gradient and the material's area and thermal conductivity:
\[ q = -kA\frac{dT}{dx} \] where \( -k \) is the negative of the thermal conductivity. The negative sign indicates that heat flows from higher to lower temperatures.

In our exercise, the temperature gradient at the surface of the cooling plate is calculated using the formula:
\[ \frac{dT}{dx} = -\frac{h(T - T_\infty)}{k} \] The final calculated temperature gradient of approximately -4.06 K/m reflects the rate of temperature decrease per meter from the plate's surface into its interior after 2 minutes of cooling.