Question

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is \(10^{\circ} \mathrm{C}\), while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.

Short answer

The calculated convection heat transfer coefficient is -133.33 W/m²·K, which is not a reasonable value for forced convection of air as the expected range is between 25 to 250 W/m²·K. The given values in the problem statement might not be consistent for a realistic scenario.

Step-by-step solution

Identify the formula to calculate the heat transfer through the plate

We will first calculate the heat transfer through the plate using thermal conductivity. To do this, we will use Fourier's law of heat conduction: \(q = -kA\frac{dT}{dx}\) Where \(q\) is the heat transfer rate [W], \(k\) is the thermal conductivity [W/m·K], A is the cross-sectional area [m²], \(dT\) is the temperature difference [K], and \(dx\) is the thickness of the plate [m].

Calculate the heat transfer rate q through the plate

We are given the thermal conductivity (\(k = 80 \mathrm{~W / m \cdot K}\)) and the temperature difference across the plate (\(dT = T_{lower} - T_{upper} = 60 - 50 = 10^{\circ}\mathrm{C}\)). Also, we know that the thickness of the plate (\(dx = 15 \mathrm{~cm} = 0.15 \mathrm{~m}\)). The cross-sectional area (A) doesn't matter since it will cancel out later. Therefore, we can now apply Fourier's law to find the heat transfer rate (q): \(q = -k\frac{dT}{dx}\) \(q = -80 \frac{10}{0.15}\) Calculating the value of q: \(q = -5333.33 \mathrm{~W}\) Note that the negative sign indicates that the heat is flowing from the lower to the upper surface.

Identify the formula to calculate the convection heat transfer coefficient (h)

Now, let's consider the convection heat transfer at the upper surface. We'll use the convection heat transfer equation: \(q = hA (T_{s} - T_{∞})\) Where \(q\) is the heat transfer rate [W], h is the convection heat transfer coefficient [W/m²·K], A is the area [m²], \(T_s\) is the surface temperature [°C], and \(T_{∞}\) is the air temperature [°C].

Calculate the convection heat transfer coefficient (h)

Since the heat transfer rate (q) is the same throughout the plate in steady-state conditions, we can use the value obtained in Step 2 to determine the value of h. Rearranging the convection heat transfer equation to find h: \(h = \frac{q}{A (T_{s} - T_{∞})}\) Plugging in values for \(q\), \(T_s\), and \(T_{∞}\): \(h = \frac{-5333.33}{A (50 - 10)}\) \(h = \frac{-5333.33}{A (40)}\) Since the area (A) cancels out, we are left with only the value of h: \(h = -133.33 \mathrm{~W / m^2 \cdot K}\)

Determine whether the value of h is reasonable for forced convection of air

The convection heat transfer coefficient (h) for forced convection of air typically ranges between 25 to 250 W/m²·K. However, in our case, we obtained a negative value for h (-133.33 W/m²·K), which is not possible for convection. This means that the given values in the problem statement might not be consistent for a realistic scenario. In conclusion, the calculated convection heat transfer coefficient (-133.33 W/m²·K) is not a reasonable value for forced convection of air, indicating that the given problem statement's values may not be consistent.

Key concepts

Fourier's Law of Heat Conduction

Fourier's Law of Heat Conduction describes how heat energy moves through a material. It is an essential principle in the study of thermodynamics and plays a critical role in understanding how heat transfer works. According to Fourier's Law, the rate at which heat flows through a material is proportional to the negative gradient of the temperature and the cross-sectional area through which heat is flowing, and is inversely proportional to the thickness of the material.

The equation derived from Fourier's Law is expressed as:
\[q = -kA\frac{dT}{dx}\]
Here, \(q\) represents the heat transfer rate in watts (W), \(k\) is the thermal conductivity of the material in W/m·K, \(A\) is the cross-sectional area through which heat is conducted, \(dT\) is the temperature difference across the material, and \(dx\) is the thickness of the material. The negative sign indicates the direction of heat flow from hot to cold regions.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It is denoted by the symbol \(k\) and has units of watts per meter-kelvin (W/m·K). Materials with high thermal conductivity, like metals, are good conductors of heat and can transfer heat quickly. In contrast, materials with low thermal conductivity, such as wood or foam insulation, are considered thermal insulators because they transfer heat more slowly.

In the context of the given problem, the thermal conductivity of the solid plate is \(80 \mathrm{~W/m \cdot K}\), which is a value you would expect for a good thermal conductor. This property enables us to use Fourier's Law to calculate the rate at which heat is transferred through the plate.

Forced Convection

Forced convection occurs when a fluid flow is generated by external means, such as a pump, fan, or wind, which enhances the transfer of heat compared to natural convection. In forced convection, the heat transfer coefficient (\(h\)) is influenced by factors like the fluid's velocity, its properties, and the geometry of the surface. The value of \(h\) can greatly vary depending on the conditions but typically falls within a certain range for a given setup.

For air, the value of \(h\) in a forced convection scenario generally ranges from 25 to 250 W/m²·K. When comparing the heat transfer coefficient derived from calculated heat transfer rates to accepted values, engineers can determine the efficiency of cooling or heating processes and whether the situation is physically plausible, as attempted in the exercise.

Steady-State Heat Transfer

Steady-state heat transfer refers to the condition where the temperature distribution in a material does not change with time. Once a steady state is achieved, the amount of heat entering a section of the material is equal to the amount of heat leaving that section.

In the exercise, the assumption of steady-state heat transfer allows us to assert that the rate of heat conducted through the plate is equal to the rate of heat transferred from the upper surface to the air, facilitating the calculation of the convection heat transfer coefficient (\(h\)). This assumption is critical in simplifying complex transient thermal analyses into more manageable ones with steady-state conditions.