Question

A ball bearing manufacturing plant is using air to cool chromium steel balls \((k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The convection heat transfer coefficient for the cooling is determined experimentally as a function of air velocity to be \(h=18.05 \mathrm{~V}^{0.56}\), where \(h\) and \(V\) are in \(\mathrm{W} / \mathrm{m}^{2}, \mathrm{~K}\) and $\mathrm{m} / \mathrm{s}$, respectively. At a given moment during the cooling process with the air temperature at \(5^{\circ} \mathrm{C}\), a chromium steel ball has a surface temperature of \(450^{\circ} \mathrm{C}\). Using appropriate software, determine the effect of the air velocity \((V)\) on the temperature gradient in the chromium steel ball at the surface. By varying the air velocity from \(0.2\) to \(2.4 \mathrm{~m} / \mathrm{s}\) with increments of $0.2 \mathrm{~m} / \mathrm{s}$, plot the temperature gradient in the chromium steel ball at the surface as a function of air velocity.

Short answer

Answer: The temperature gradient in a chromium steel ball at the surface is affected by air velocity. As air velocity increases, the convection heat transfer coefficient also increases, leading to a higher rate of heat transfer by convection. This results in an increase in the temperature gradient within the steel ball at the surface. To determine the relationship between air velocity and temperature gradient, we can follow the steps discussed in the solution and plot the temperature gradient as a function of air velocity.

Step-by-step solution

Find the Convection Heat Transfer Coefficient using the given function

For air velocity \(V\), the convection heat transfer coefficient \(h\) is given by the equation: \(h = 18.05V^{0.56}\).

Apply Fourier's Law of Heat Conduction

Fourier's law states that the rate of heat conduction (\(\dot{Q}\)) is proportional to the temperature gradient (\(\frac{dT}{dx}\)). The equation is: \(\dot{Q} = -k\frac{dT}{dx}\), where \(k\) is the thermal conductivity.

Apply Newton's Law of Cooling

Newton's law states that the rate of heat transfer by convection is proportional to the temperature difference between the surface temperature (\(T_s\)) and the air temperature (\(T_\infty\)). The equation is: \(\dot{Q} = h(T_s - T_\infty)\).

Equate the rates of heat conduction and heat transfer by convection

Since the rate of heat conduction is equal to the rate of heat transfer by convection, we have: \(-k\frac{dT}{dx} = h(T_s - T_\infty)\)

Calculate the temperature gradient

First, isolate the temperature gradient term in the equation: \(\frac{dT}{dx} = -\frac{h}{k}(T_s - T_\infty)\) Now, we can calculate the temperature gradient for different air velocities (\(V\)) from \(0.2 \mathrm{~m/s}\) to \(2.4 \mathrm{~m/s}\) with increments of \(0.2 \mathrm{~m/s}\), using the given values of \(T_s = 450^{\circ} \mathrm{C}\), \(T_\infty = 5^{\circ} \mathrm{C}\), and \(k = 40 \mathrm{~W/m\cdot K}\).

Plot the temperature gradient as a function of air velocity

Using appropriate software (such as Excel, Python, or MATLAB), plot the temperature gradient in the chromium steel ball at the surface as a function of air velocity. Each point on the graph will represent the corresponding temperature gradient for the given air velocity. In conclusion, by following these steps, we can determine the effect of air velocity on the temperature gradient in the chromium steel ball at the surface, and visualize the relationship through a plot.