91Ó°ÊÓ

A spherical vessel used as a reactor for producing pharmaceuticals has a 5 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(D_{i}=1.0 \mathrm{~m}\). During production, the vessel is filled with reactants for which \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while exothermic reactions release energy at a volumetric rate of \(\dot{q}=10^{4} \mathrm{~W} / \mathrm{m}^{3}\). As first approximations, the reactants may be assumed to be well stirred and the thermal capacitance of the vessel may be neglected. (a) The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(h=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. If the initial temperature of the reactants is \(25^{\circ} \mathrm{C}\), what is the temperature of the reactants after \(5 \mathrm{~h}\) of process time? What is the corresponding temperature at the outer surface of the vessel? (b) Explore the effect of varying the convection coefficient on transient thermal conditions within the reactor. A spherical vessel used as a reactor for producing pharmaceuticals has a 5 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(D_{i}=1.0 \mathrm{~m}\). During production, the vessel is filled with reactants for which \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while exothermic reactions release energy at a volumetric rate of \(\dot{q}=10^{4} \mathrm{~W} / \mathrm{m}^{3}\). As first approximations, the reactants may be assumed to be well stirred and the thermal capacitance of the vessel may be neglected. (a) The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(h=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. If the initial temperature of the reactants is \(25^{\circ} \mathrm{C}\), what is the temperature of the reactants after \(5 \mathrm{~h}\) of process time? What is the corresponding temperature at the outer surface of the vessel? (b) Explore the effect of varying the convection coefficient on transient thermal conditions within the reactor.

Step by step solution

01

Calculate the energy generated by the exothermic reaction

Given the volumetric energy generation rate by the reaction, \(\dot{q}=10^{4} \mathrm{~W} / \mathrm{m}^{3}\), we need to find the total energy generated after 5 hours. First of all, calculate the volume of the reactor: \[V=\frac{4}{3} \pi \left(\frac{D_i}{2}\right)^3=\frac{4}{3} \pi \left(\frac{1.0}{2}\right)^3 \approx 0.524\,\text{m}^3\] Now, we can compute the total energy generated during the 5-hour period: \[Q=\dot{q} \cdot V \cdot t=10^4 \cdot 0.524 \cdot (5 \cdot 3600)=1.57 \times 10^8\,\text{J}\]

02

Calculate the temperature rise due to exothermic reaction

The temperature rise inside the vessel is determined by the energy generated and the heat capacity of the reactants. The mass of the reactants is \(m=\rho V = 1100 \cdot 0.524=575.4\,\text{kg}\). The heat capacity for the reactants is given as \(c=2400\,\text{J/(kg·K)}\) and using the formula, \(Q=mc\Delta T\), we can find the temperature rise caused by the reaction: \[\Delta T=\frac{Q}{mc}=\frac{1.57 \times 10^8}{575.4 \cdot 2400}=114.47\,\text{K}\]

03

Calculate the final temperature of the reactants

With the initial temperature being \(25^\circ \text{C}\), the final temperature inside the vessel after 5 hours becomes: \[T_f = T_\text{initial}+\Delta T = 25 + 114.47 = 139.47^{\circ} \text{C}\]

04

Determine the heat loss and outer surface temperature

The heat loss through the reactor wall should be considered. The heat transfer through the wall is governed by the equation: \[Q_\text{loss}=\frac{4 \pi \left(\frac{D_i}{2}\right)^2 \cdot k \cdot (T_f-T_\infty)}{x}+\frac{4 \pi \left(\frac{D_i}{2}\right)^2 \cdot h \cdot (T_f-T_\infty)}{D_i+x}\] Given the problem statement, it is plausible to assume steady-state conditions for a quick estimation and neglect the variations in heat transfer. This leads us to simplify the equation: \[\frac{Q_\text{loss}}{(T_f-T_\infty)}=4 \pi \left(\frac{D_i}{2}\right)^2 \cdot \left(\frac{k}{x} +\frac{h}{D_i+x}\right)\] Substituting the given values: \(k=17\,\text{W/(m·K)},\, h=6\,\text{W/(m}^{2}\cdot \text{K)}\), and \(x=0.005\,\text{m}\), we can calculate the heat loss: \[Q_\text{loss}=(139.47-25) \cdot 4 \pi \left(\frac{1}{2}\right)^2 \cdot \left(\frac{17}{0.005} + \frac{6}{1+0.005}\right)=3.646 \times 10^6\,\text{W}\] Neglecting the temperature rise due to this heat loss, we can find the outer surface temperature by substituting the heat loss back into the equation: \[T_\text{outer}=T_\infty + \frac{Q_\text{loss}}{4\pi\left(\frac{D_i}{2}\right)^2 \cdot h} =25 + \frac{3.646 \times 10^6}{4\pi\left(\frac{1}{2}\right)^2 \cdot 6}\approx 49.28^{\circ} \text{C}\]

05

Exploring the effect of convection coefficient variations

Increasing the convection coefficient corresponds to improving heat dissipation from the vessel's outer surface. This would result in a lower outer surface temperature and also a lower final temperature for the reactants inside the vessel, as more heat would be lost to the surrounding environment. Conversely, decreasing the convection coefficient would lead to higher temperatures for both the reactants and the outer surface, as less heat would be dissipated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical Vessel

A spherical vessel is often used in industrial processes due to several advantages. It provides a uniform distribution of stress due to its symmetrical shape.
The surface area to volume ratio is minimized in a sphere, allowing for efficient use of materials. In our scenario, we have a spherical vessel with an inner diameter of 1.0 meters. This measurement helps determine the volume, which is calculated using the formula:

• \[ V = \frac{4}{3} \pi \left(\frac{D_i}{2}\right)^3 \]

This formula gives the volume as approximately 0.524 cubic meters. Understanding the properties of the vessel helps determine the heat transfer processes, as the entire surface interacts with the surrounding environment, affecting how energy is retained or lost.

Convection Coefficient

The convection coefficient, denoted as \( h \), plays a crucial role in determining how efficiently heat is transferred between the vessel and the surrounding air. It is expressed in units of \( \, \text{W/m}^{2}\cdot \text{K} \).
In this exercise, the initial value used is 6 \( \, \text{W/m}^{2}\cdot \text{K} \), indicating the rate at which heat is transferred from the vessel’s outer surface into the ambient air.

• Higher convection coefficients imply better heat dissipation, resulting in lower temperatures inside and outside the vessel.
• Lower coefficients indicate poorer heat transfer, causing temperatures to rise both inside and outside.

Considering changes in this coefficient can help understand and optimize the thermal performance of the reactor.

Stainless Steel Wall

The stainless steel wall of the reactor vessel serves as a critical barrier, not only to contain the reactants but also to facilitate heat transfer. Stainless steel is chosen for its durability and thermal conductivity, which in this case is given as 17 \( \, \text{W/m}\cdot \text{K} \).
This high value of thermal conductivity means that the stainless steel wall effectively conducts heat from the inner to the outer surface of the vessel. The wall thickness is an important factor, mentioned as 5 mm in the problem, which affects how quickly heat moves through it.
To calculate heat loss through the vessel wall, we use the equation:

• \[ Q_{\text{loss}} = 4 \pi \left(\frac{D_i}{2}\right)^2 \left(\frac{k}{x} + \frac{h}{D_i+x}\right)(T_f-T_\infty) \]

This equation helps find the rate of heat transfer and assess thermal conditions within the stainless steel wall.

Exothermic Reactions

Exothermic reactions release energy during a chemical process, increasing the temperature of the system. In our reactor, the exothermic reaction produces heat at a volumetric rate of \( 10^4 \mathrm{~W/m}^3 \).
This energy release is a key part of how the temperature within the reactor rises.

• The heat generated is calculated using the formula: \[ Q = \dot{q} \times V \times t \] This formula considers the volumetric energy generation rate, the volume of the spherical vessel, and the duration of the process (5 hours).
• The temperature rise inside the vessel is linked to this energy, thereby informing how effectively the heat must be managed by the system.

Understanding exothermic reactions is essential in designing reactors to ensure safety and efficiency. They dictate how cooling mechanisms, such as improved convection, must be employed to keep temperatures in check.