Question

Consider a spherical reactor of \(5-\mathrm{cm}\) diameter operating at steady conditions with a temperature variation that can be expressed in the form of \(T(r)=a-b r^{2}\), where \(a=850^{\circ} \mathrm{C}\) and $b=5 \times 10^{5} \mathrm{~K} / \mathrm{m}^{2}\(. The reactor is made of material with \)c=200 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=9000 \mathrm{~kg} / \mathrm{m}^{3}$. If the heat generation of the reactor is suddenly set to $9 \mathrm{MW} / \mathrm{m}^{3}$, determine the time rate of temperature change in the reactor. Is the heat generation of the reactor suddenly increased or decreased to $9 \mathrm{MW} / \mathrm{m}^{3}$ from its steady operating condition?

Short answer

#Answer# The heat generation of the reactor was increased from its steady operating condition.

Step-by-step solution

Calculate the initial heat generation rate

Using the temperature variation equation, \(T(r) = a - b\cdot r^2\) and the material properties (\(c, k, \rho\)), we first calculate the heat generation rate at steady-state operating conditions. The heat conduction equation for a spherical reactor at steady-state is given by: $$ \frac{d}{dr}\left( r^2 \cdot k\cdot \frac{d T}{dr}\right) = \rho\cdot Q_{gen}, $$ where \(Q_{gen}\) is the heat generation rate. Find \(\frac{d T}{dr}\) from the given function \(T(r)\), and substitute it into the equation to find the heat generation rate. Differentiating \(T(r)\), $$ \frac{d T}{dr} = -2\cdot b\cdot r. $$ Now, substitute this into the heat conduction equation: $$ \frac{d}{dr}\left( r^2\cdot k\cdot (-2\cdot b\cdot r)\right) = \rho\cdot Q_{gen}. $$ Differentiating once again, $$ (-6\cdot k\cdot b)\cdot r^2 = \rho\cdot Q_{gen}. $$ Now, we have an equation for the heat generation rate, \(Q_{gen}\). Before finding the value of \(Q_{gen}\), let's derive the equation for the time rate of temperature change.

Derive the time rate of temperature change equation

We use the following unsteady heat conduction equation for a spherical reactor: $$ \frac{1}{\alpha}\cdot \frac{\partial T}{\partial t} = \frac{1}{r^2}\cdot \frac{\partial}{\partial r}\left( r^2\cdot \frac{\partial T}{\partial r}\right) + \frac{Q_{gen}}{c\cdot \rho}, $$ where \(\alpha = \frac{k}{c\cdot \rho}\) is the thermal diffusivity. We have already found the derivative of \(T(r)\), so now we will substitute the \(\frac{d T}{dr}\), \(\frac{Q_{gen}}{c\cdot \rho}\), and the value for \(\alpha\) into the equation: $$ \frac{1}{\alpha}\cdot \frac{\partial T}{\partial t} = \frac{1}{r^2}\cdot \frac{\partial}{\partial r}\left( r^2\cdot (-2\cdot b\cdot r)\right) + \frac{Q_{gen}}{c\cdot \rho}. $$ We have computed the expression inside the derivative in Step 1. Substituting it and the value for \(\alpha\) leads to: $$ \frac{c\cdot \rho}{k}\cdot \frac{\partial T}{\partial t} = \frac{(-6\cdot k\cdot b)\cdot r^2}{r^2} + Q_{gen}. $$

Determine whether heat generation has increased or decreased

From the calculation in Step 1, we know that at steady-state, the heat generation is given by: $$ (-6\cdot k\cdot b)\cdot r^2 = \rho\cdot Q_{gen, SS}. $$ Now, calculate the heat generation rate \(Q_{gen, SS}\) using the given parameters: $$ Q_{gen, SS} = \frac{(-6\cdot k\cdot b)}{\rho} = \frac{-6 \times 40 \times 5 \times 10^{5}}{9000} = -1.33 \times 10^7 \mathrm{~W}/\mathrm{m}^3 = -13.3 \ \mathrm{MW}/\mathrm{m}^3. $$ The initial heat generation rate is -13.3 MW/m³, and since the heat generation rate was suddenly set to 9 MW/m³, which is a higher value than -13.3 MW/m³, we can conclude that the heat generation of the reactor was increased from its steady operating condition.