Question

Consider a long solid cylinder of radius \(r_{o}=4 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the cylinder uniformly at a rate of $\dot{e}_{\mathrm{gen}}=35 \mathrm{~W} / \mathrm{cm}^{3}$. The side surface of the cylinder is maintained at a constant temperature of \(T_{s}=80^{\circ} \mathrm{C}\). The variation of temperature in the cylinder is given by $$ T(r)=\frac{e_{\mathrm{gen}} r_{o}^{2}}{k}\left[1-\left(\frac{r}{r_{o}}\right)^{2}\right]+T_{s} $$ Based on this relation, determine \((a)\) if the heat conduction is steady or transient, \((b)\) if it is one-, two-, or threedimensional, and \((c)\) the value of heat flux on the side surface of the cylinder at \(r=r_{a r}\)

Short answer

The heat conduction is steady. (b) Is the heat conduction one-dimensional, two-dimensional, or three-dimensional? The heat conduction is one-dimensional. (c) What is the heat flux on the side surface of the cylinder at \(r=r_o\)? The heat flux on the side surface of the cylinder at \(r=r_o\) is 280 W/cm².

Step-by-step solution

Determine if the heat conduction is steady or transient

The given temperature equation for the cylinder is: $$T(r)=\frac{e_{\mathrm{gen}} r_{o}^{2}}{k}\left[1-\left(\frac{r}{r_{o}}\right)^{2}\right]+T_{s}$$ Since the temperature equation has no dependency on time, the heat conduction is steady.

Determine if the heat conduction is one-, two-, or three-dimensional

The temperature equation only has one variable \(r\), which is the radial distance. Hence, the heat conduction is one-dimensional along the radial direction.

Calculate the heat flux on the side surface of the cylinder at \(r=r_o\)

The radial heat flux can be calculated using Fourier's law:$$q_r = -k\frac{dT}{dr}$$Now, differentiate the given temperature equation with respect to \(r\):$$\frac{dT}{dr} = -\frac{2e_{\mathrm{gen}} r}{k} \cdot \left(\frac{r}{r_{o}}\right)$$Next, substitute the given values and the radial position \(r=r_o\) into the above equation:$$\frac{dT}{dr}\Big|_{r=r_o} = -\frac{2e_{\mathrm{gen}} r_o}{k}$$Finally, substitute the obtained value into Fourier's law:$$q_r = -k\left(-\frac{2e_{\mathrm{gen}} r_o}{k}\right) = 2e_{\mathrm{gen}} r_o$$Substitute the given values \(e_{\mathrm{gen}}=35 \mathrm{~W} / \mathrm{cm}^{3}\) and \(r_o=4 \mathrm{~cm}\):$$q_r = 2(35\,\mathrm{W/cm^3})(4\,\mathrm{cm}) = 280\,\mathrm{W/cm^2}$$The heat flux on the side surface of the cylinder at \(r=r_o\) is \(280\,\mathrm{W/cm^2}\).