Question

During air cooling of steel balls, the convection heat transfer coefficient is determined experimentally as a function of air velocity to be \(h=17.9 V^{0.54}\) for \(0.5

Short answer

Question: Calculate the initial values of heat flux and temperature gradient in the steel ball at the surface given the following parameters: air velocity, diameter, initial temperature, and thermal conductivity of the steel ball; air temperature and air velocity during cooling. Answer: The initial values of heat flux and temperature gradient in the steel ball at the surface are 7167.76 W/m² and -477.85 K/m, respectively.

Step-by-step solution

To determine the convection heat transfer coefficient (h) based on the given formula and air velocity, simply substitute the given air velocity (V) into the formula: \(h = 17.9 V^{0.54}\) From the exercise, we have \(V=1.5\mathrm{~m} / \mathrm{s}\), so: \(h = 17.9 (1.5)^{0.54} = 24.72 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) #Step 2: Calculate the heat flux using Newton's Law of Cooling#

Newton's Law of Cooling states that the heat flux (q) is proportional to the difference in temperature between the surface and the air. Mathematically, this can be represented as: \(q = h (T_s - T_\infty)\) where \(T_s\) is the surface temperature of the steel ball, \(T_\infty\) is the air temperature, and h is the convection heat transfer coefficient. From the exercise, we have: \(T_s = 300 ^\circ \mathrm C\) \(T_\infty = 10 ^\circ \mathrm C\) Substitute these values and the value of h into Newton's Law of Cooling: \(q = 24.72 (300-10) = 7167.76 \mathrm{~W} / \mathrm{m}^2\) #Step 3: Calculate the temperature gradient using Fourier's Law#

According to Fourier's Law, the temperature gradient is given by: \(\frac{dT}{dx} = -\frac{q}{k}\) where \(\frac{dT}{dx}\) is the temperature gradient, q is the heat flux, and k is the thermal conductivity of the steel ball. From the exercise, we have: \(k = 15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Substitute the values of q and k into Fourier's Law: \(\frac{dT}{dx} = -\frac{7167.76}{15} = -477.85 \mathrm K / \mathrm m\) #Step 4: Report the initial values of heat flux and temperature gradient#

Based on our calculations, the initial values of the heat flux and temperature gradient in the steel ball at the surface are as follows: Heat flux (q): \(7167.76 \mathrm{~W} / \mathrm{m}^2\) Temperature gradient (\(\frac{dT}{dx}\)): \(-477.85 \mathrm{~K} / \mathrm{m}\)

Key concepts

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, typically denoted as "h," is a measure of how effectively heat is transferred between a solid surface and a fluid in motion, such as air or water. It is influenced by various parameters including the velocity of the fluid. In the given exercise, the convection heat transfer coefficient is calculated using:

• Formula: \( h = 17.9 V^{0.54} \)
• Where \( V \) is the velocity of air, measured in meters per second (m/s).

This formula indicates that as the air velocity increases, the convection heat transfer coefficient also increases, implying a more efficient heat transfer process. By substituting the given velocity value into the formula, we learn how the airflow around the steel ball affects the cooling process. Using a convection heat transfer coefficient effectively models how temperatures are equalized through movement and contact in practical applications like cooling metallic spheres.

Newton's Law of Cooling

Newton's Law of Cooling is a principle that helps us understand how heat transfer occurs through convection. It asserts that the rate of heat loss or gain of an object is directly proportional to the temperature difference between the object and the environment. Mathematically, it is expressed as:\[ q = h (T_s - T_\infty) \] where:

• \( q \) is the heat flux, or rate of heat transfer per unit area.
• \( T_s \) is the temperature of the surface (here, the steel ball).
• \( T_\infty \) is the temperature of the surrounding fluid (air).

Newton's Law of Cooling details the importance of the temperature difference in determining how much heat will naturally flow between a system and its environment. In our exercise, substituting the known values into the formula helps predict the initial rate at which the steel ball loses heat, providing insight into the cooling efficiency under specified conditions.

Fourier's Law

Fourier's Law is key to understanding the conduction aspect of heat transfer. It is represented by the formula:\[ \frac{dT}{dx} = -\frac{q}{k} \] where:

• \( \frac{dT}{dx} \) is the temperature gradient across the material.
• \( q \) is the heat flux.
• \( k \) is the thermal conductivity of the material, showing its ability to conduct heat.

In the context of the exercise, applying Fourier's Law allows us to calculate the temperature gradient at the surface of the steel ball, which informs us how temperature changes across its material. This is crucial for applications requiring precise temperature control since the gradient directly influences how heat spreads through the material.

Temperature Gradient

The temperature gradient is a critical concept in heat transfer analysis, representing how temperature changes from one point to another along a material. In mathematical terms, it is given by \( \frac{dT}{dx} \), which is derived using Fourier's Law. The magnitude of the temperature gradient indicates how quickly temperature changes within the material. A steep gradient implies rapid temperature change over a short distance, reflecting fast heat conduction. In the case of a steel ball cooling in air, knowing the temperature gradient allows us to understand how the internal heat dissipates to the surface. This concept is vital for designing systems that depend on precise thermal management, ensuring components remain within safe temperature operating ranges.

Thermal Conductivity

Thermal conductivity is a material property that tells us how well a material can conduct heat. It is denoted by "k" and has units of \( \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \). Higher thermal conductivity values indicate that a material can more efficiently transfer heat. In the context of the provided exercise, the thermal conductivity of the steel ball is given as 15 \( \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \).Understanding thermal conductivity helps in predicting how quickly a material can reach thermal equilibrium with its surroundings during heating or cooling. It's essential for selecting materials in applications ranging from heat sinks in electronics to cookware and building insulation. In this given scenario, the steel ball's thermal conductivity affects how fast it can cool down as air flows over its surface, directly impacting the temperature gradient and cooling rate.