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

The initial values of the heat flux and the temperature gradient in the steel ball at the surface are \(7735.6 \mathrm{W}/\mathrm{m}^2\) and \(-515.7067 \mathrm{K}/\mathrm{m}\), respectively.

Step-by-step solution

Calculate the convection heat transfer coefficient

Using the formula \(h = 17.9 V^{0.54}\), we can find the convection heat transfer coefficient for the given air velocity \(V = 1.5 \mathrm{~m}/\mathrm{s}\): \(h = 17.9 (1.5)^{0.54} = 26.68 \mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K}\)

Apply Fourier's law to find the heat flux

Fourier's law of heat conduction states that the heat flux \(q\) is proportional to the negative of the temperature gradient at the surface, with the constant of proportionality being the thermal conductivity \(k\): \(q = -k \frac{\mathrm{d}T}{\mathrm{d}r}\) Since we want to find the heat flux at the surface of the steel ball, we need to find the temperature gradient at the surface. To do this, we'll first need to use Newton's law of cooling.

Use Newton's law of cooling to find the temperature gradient at the surface

Newton's law of cooling states that the heat transfer rate (also known as the heat flux) is proportional to the difference in temperature between the surface of the object and the surrounding fluid (in this case, air): \(q = h(T_s - T_\infty)\) where \(T_s\) is the surface temperature of the object, and \(T_\infty\) is the surrounding fluid temperature. Since we know the initial temperature of the steel ball and the surrounding air temperature, we can plug in the values: \(q = 26.68 (300 - 10) \mathrm{W}/\mathrm{m}^2 = 7735.6 \mathrm{~W}/\mathrm{m}^2\) Now, we can use this value in Fourier's law to find the temperature gradient at the surface.

Calculate the temperature gradient at the surface

Using the heat flux \(q\) from Step 3 and the thermal conductivity \(k = 15 \mathrm{~W}/\mathrm{m} \cdot \mathrm{K}\), we can find the temperature gradient at the surface of the steel ball: \(\frac{\mathrm{d}T}{\mathrm{d}r} = -\frac{q}{k} = -\frac{7735.6}{15} = -515.7067 \mathrm{K}/\mathrm{m}\) The negative sign indicates the temperature decreases as the radius increases away from the surface, which is expected since the steel ball is cooling. The initial values of the heat flux and temperature gradient in the steel ball at the surface are \(7735.6 \mathrm{~W}/\mathrm{m}^2\) and \(-515.7067 \mathrm{K}/\mathrm{m}\), respectively.