/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A spherical vessel used as a rea... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 19

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.

Short Answer

Expert verified
The temperature of the reactants after 5 hours of process time is approximately \(139.47^{\circ} \text{C}\), and the corresponding temperature at the outer surface of the vessel is approximately \(49.28^{\circ} \text{C}\). Varying the convection coefficient affects the transient thermal conditions within the reactor: increasing the convection coefficient leads to better heat dissipation, resulting in lower temperatures for both the reactants and outer surface, while decreasing the convection coefficient leads to higher temperatures by reducing heat dissipation.

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 \(c=2400\,\text{J/(kg路K)}\) 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: \(k=17\,\text{W/(m路K)},\, h=6\,\text{W/(m}^{2}\cdot \text{K)}\), 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鈥檚 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.

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Most popular questions from this chapter

Steel-reinforced concrete pillars are used in the construction of large buildings. Structural failure can occur at high temperatures due to a fire because of softening of the metal core. Consider a 200 -mm-thick composite pillar consisting of a central steel core \((50 \mathrm{~mm}\) thick) sandwiched between two 75 -mm-thick concrete walls. The pillar is at a uniform initial temperature of \(T_{i}=\) \(27^{\circ} \mathrm{C}\) and is suddenly exposed to combustion products at \(T_{\infty}=900^{\circ} \mathrm{C}, h=40 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) on both exposed surfaces. The surroundings temperature is also \(900^{\circ} \mathrm{C}\). (a) Using an implicit finite difference method with \(\Delta x=10 \mathrm{~mm}\) and \(\Delta t=100 \mathrm{~s}\), determine the temperature of the exposed concrete surface and the center of the steel plate at \(t=10,000 \mathrm{~s}\). Steel properties are: \(k_{s}=55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho_{s}=7850 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{s}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Concrete properties are: \(k_{c}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho_{c}=2300 \mathrm{~kg} / \mathrm{m}^{3}, c_{c}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(\varepsilon=0.90\). Plot the maximum and minimum concrete temperatures along with the maximum and minimum steel temperatures over the duration \(0 \leq t \leq 10,000 \mathrm{~s} .\) (b) Repeat part (a) but account for a thermal contact resistance of \(R_{t, c}^{\prime}=0.20 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) at the concretesteel interface. (c) At \(t=10,000 \mathrm{~s}\), the fire is extinguished, and the surroundings and ambient temperatures return to \(T_{\infty}=T_{\text {sur }}=27^{\circ} \mathrm{C}\). Using the same convection heat transfer coefficient and emissivity as in parts (a) and (b), determine the maximum steel temperature and the critical time at which the maximum steel temperature occurs for cases with and without the contact resistance. Plot the concrete surface temperature, the concrete temperature adjacent to the steel, and the steel temperatures over the duration \(10,000 \leq t \leq 20,000 \mathrm{~s} .\)

In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

Consider a thin electrical heater attached to a plate and backed by insulation. Initially, the heater and plate are at the temperature of the ambient air, \(T_{\infty}\). Suddenly, the power to the heater is activated, yielding a constant heat flux \(q_{o}^{\prime \prime}\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T \leftarrow x\) coordinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{x}^{\prime \prime}(L, t)\) as a function of time.

In a thin-slab, continuous casting process, molten steel leaves a mold with a thin solid shell, and the molten material solidifies as the slab is quenched by water jets en route to a section of rollers. Once fully solidified, the slab continues to cool as it is brought to an acceptable handling temperature. It is this portion of the process that is of interest. Consider a 200-mm-thick solid slab of steel \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, \quad c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \quad k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), initially at a uniform temperature of \(T_{i}=1400^{\circ} \mathrm{C}\). The slab is cooled at its top and bottom surfaces by water jets \(\left(T_{\infty}=50^{\circ} \mathrm{C}\right)\), which maintain an approximately uniform convection coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. Using a finite-difference solution with a space increment of \(\Delta x=1 \mathrm{~mm}\), determine the time required to cool the surface of the slab to \(200^{\circ} \mathrm{C}\). What is the corresponding temperature at the midplane of the slab? If the slab moves at a speed of \(V=15 \mathrm{~mm} / \mathrm{s}\), what is the required length of the cooling section?

A support rod \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=4.0 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~mm}\) and length \(L=100 \mathrm{~mm}\) spans a channel whose walls are maintained at a temperature of \(T_{b}=300 \mathrm{~K}\). Suddenly, the rod is exposed to a cross flow of hot gases for which \(T_{\infty}=600 \mathrm{~K}\) and \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The channel walls are cooled and remain at \(300 \mathrm{~K}\). (a) Using an appropriate numerical technique, determine the thermal response of the rod to the convective heating. Plot the midspan temperature as a function of elapsed time. Using an appropriate analytical model of the rod, determine the steadystate temperature distribution, and compare the result with that obtained numerically for very long elapsed times. (b) After the rod has reached steady-state conditions, the flow of hot gases is suddenly terminated, and the rod cools by free convection to ambient air at \(T_{\infty}=300 \mathrm{~K}\) and by radiation exchange with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\). The free convection coefficient can be expressed as \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=C\) \(\Delta T^{n}\), where \(C=4.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{1.188}\) and \(n=0.188\). The emissivity of the rod is \(0.5\). Determine the subsequent thermal response of the rod. Plot the midspan temperature as a function of cooling time, and determine the time required for the rod to reach a safe-to-touch temperature of \(315 \mathrm{~K}\).
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