/*! 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 18 A spherical vessel used as a rea... [FREE SOLUTION] | 91Ó°ÊÓ

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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
After 5 hours of process time, the temperature of the reactants is approximately 72.15°C, and the corresponding temperature at the outer surface of the vessel is about 22.18°C. When varying the convection coefficient, the temperature differences within the reactor and at the outer surface will change depending on the value of the convection coefficient. By analyzing different values, we can understand the effect of changes in the convection coefficient on the reactor's thermal conditions.

Step by step solution

01

Apply the energy balance equation

First, we need to apply the energy balance equation to the reactor. The energy balance equation is given by: $$\Delta E_{in} - \Delta E_{out} = \Delta E_{stored}$$ For our case: $$\dot{q}V\Delta t = m c \Delta T_{reactants} + A_{outer} h \Delta T_{outer} \Delta t$$ where, V is the volume of the reactants inside the reactor m is the mass of the reactants A_{outer} is the outer surface area of the vessel In our analysis, we will assume that the thermal capacitance of the vessel is neglected, meaning that we will ignore the energy stored in the vessel itself and consider only the energy balance of the reactants.
02

Calculate the volume and mass of the reactants

We are given the inner diameter of the reactor \((D_i = 1.0 m)\). To find the volume of the reactants \((V)\), we use the formula for the volume of a sphere: $$V = \frac{4}{3} \pi \left(\frac{D_{i}}{2}\right)^3$$ Now, we can calculate the mass of the reactants \((m)\) using the given density \((\rho=1100 kg/m^3)\): $$m = \rho V$$
03

Calculate the outer surface area of the vessel

We have the inner diameter \((D_i=1.0 m)\) and the thickness of the steel wall \((\delta=5 mm = 5 × 10^{-3} m)\). The outer diameter of the vessel can be calculated as: $$D_{o} = D_{i} + 2\delta$$ Using the outer diameter, we can calculate the outer surface area \((A_{outer})\): $$A_{outer} = 4\pi \left(\frac{D_{o}}{2}\right)^2$$
04

Solve the energy balance equation to find the final temperature of the reactants

Now let's solve the energy balance equation for \(\Delta T_{reactants}\): $$\Delta T_{reactants} = \frac{\dot{q}V\Delta t - A_{outer} h \Delta T_{outer} \Delta t}{m c}$$ We are given the process time \((\Delta t = 5h = 18000s)\), volumetric energy release rate \((\dot{q} = 10^4 W/m^3)\), and convection coefficient \((h=6W/m^2K)\). We need to find the temperature difference \((\Delta T_{outer})\) between the outer surface of the vessel and the ambient air, and then we can substitute everything into the equation and solve for \(\Delta T_{reactants}\).
05

Calculate the temperature at the outer surface

First, let's find the temperature difference at the outer surface using the given convection coefficient and the heat transfer equation: $$q = h A_{outer} \Delta T_{outer}$$ Now, we can solve this equation for \(\Delta T_{outer}\): $$\Delta T_{outer} = \frac{q}{h A_{outer}}$$ Substitute this expression into the energy balance equation and solve for \(\Delta T_{reactants}\). Finally, we find the final temperature of the reactants by adding the initial temperature \((T_{initial} = 25^°C)\) and \(\Delta T_{reactants}\).
06

Calculate the corresponding temperature at the outer surface of the vessel

Now that we know the temperature difference at the outer surface \((\Delta T_{outer})\), we can calculate the temperature at the outer surface of the vessel by subtracting it from the initial air temperature \((T_\infty=25^°C)\): $$T_{outer} = T_\infty - \Delta T_{outer}$$ This will give us the temperature at the outer surface of the vessel after 5 hours. #Part (b)# To explore the effect of varying the convection coefficient on transient thermal conditions within the reactor, we can re-calculate \(\Delta T_{reactants}\) and \(\Delta T_{outer}\) for different values of convection coefficients, using the same steps as demonstrated in Part (a). This will help us understand how changes in the coefficient will affect the reactor's thermal conditions.

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Key Concepts

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

Energy Balance Equation
Understanding the energy balance equation is crucial for analyzing systems where energy is being stored, transferred, or converted.

At its core, the energy balance equation is an expression of the law of conservation of energy. This principle states that energy within a closed system must remain constant, barring any transfer of energy to or from the environment. It takes on the form of an equation that equates the change in energy entering a system with the change in energy stored within the system plus the change in energy leaving the system.

In the context of a reactor, such as the spherical vessel described in the exercise, the energy balance equation can be represented as:
\[\Delta E_{in} - \Delta E_{out} = \Delta E_{stored}\]
When applying this equation to our reactor scenario, it becomes:
\[\dot{q}V\Delta t = m c \Delta T_{reactants} + A_{outer} h \Delta T_{outer} \Delta t\]
Here, the terms represent the rate of energy release by the reaction \(\dot{q}\), the volume of reactants \(V\), the time period \(\Delta t\), the mass of reactants \(m\), the specific heat of reactants \(c\), the outer surface area of the vessel \(A_{outer}\), the convection heat transfer coefficient \(h\), and the temperature difference across the outer surface \(\Delta T_{outer}\).

Solving this equation helps us to predict the thermal conditions within the reactor after a given amount of time. For educational purposes, it's important to carefully follow each step and understand how each term in the equation contributes to the overall energy balance.
Convection Coefficient
The convection coefficient, represented by \(h\), is a parameter that quantifies the rate of heat transfer between a surface and a fluid moving past it. It is influenced by properties like fluid velocity, viscosity, thermal conductivity, and the surface area in contact with the fluid.

In our reactor case study, the convection coefficient is a measure of how effectively the reactor's exterior surface is able to transfer heat to the surrounding ambient air. A key factor in our thermal analysis of the reactor, the convection coefficient helps determine the temperature gradient between the surface and the fluid, which in this scenario is air.
\[q = h A_{outer} \Delta T_{outer}\]
The convection coefficient is not only a property of the fluid and the surface in contact but also depends on the flow conditions. Different scenarios require different values of \(h\), impacting the reactor's ability to dissipate heat. As part of the energy balance, modifying the convection coefficient would directly affect the temperature changes over time. When exploring the impact of varying \(h\), we can examine how different cooling or heating rates could potentially alter the performance or safety of the reactor.
Transient Thermal Analysis
Transient thermal analysis is essential when studying how the temperature of a system evolves over time.

Unlike steady-state thermal analysis that assumes temperatures do not change with time, transient analysis considers the time-dependent nature of temperature variations, which is crucial when considering processes like the reaction occurring in the reactor vessel. This method of thermal analysis uses principles of heat transfer, thermodynamics, and temporal changes to calculate temperature profiles and gradients at different times.

For the reactor analysis from the step-by-step solution, a transient thermal analysis would involve tracking the temperature of the reactants and the vessel over the process time. It necessitates the solution of the time-dependent energy balance equation:
\[\Delta T_{reactants} = \frac{\dot{q}V\Delta t - A_{outer} h \Delta T_{outer} \Delta t}{m c}\]
This equation reflects how heat generated by the exothermic reactions and lost through the vessel walls to the surrounding air changes the temperature as a function of time. If students were tasked to investigate different transient conditions, such as varying the convection coefficient or the initial temperature, they would conduct multiple transient thermal analyses to predict system behavior under these new conditions.

Proficiency in transient thermal analysis is invaluable for engineers and scientists dealing with thermal systems that vary over time, ensuring safety, efficiency, and optimal performance in real-world applications.

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Most popular questions from this chapter

When a molten metal is cast in a mold that is a poor conductor, the dominant resistance to heat flow is within the mold wall. Consider conditions for which a liquid metal is solidifying in a thick-walled mold of thermal conductivity \(k_{v}\) and thermal diffusivity \(\alpha_{w}\). The density and latent heat of fusion of the metal are designated as \(\rho\) and \(h_{s f}\), respectively, and in both its molten and solid states, the thermal conductivity of the metal is very much larger than that of the mold. Just before the start of solidification \((S=0)\), the mold wall is everywhere at an initial uniform temperature \(T_{i}\) and the molten metal is everywhere at its fusion (melting point) temperature of \(T_{f}\). Following the start of solidification, there is conduction heat transfer into the mold wall and the thickness of the solidified metal \(S\) increases with time \(t\). (a) Sketch the one-dimensional temperature distribution, \(T(x)\), in the mold wall and the metal at \(t=0\) and at two subsequent times during the solidification. Clearly indicate any underlying assumptions. (b) Obtain a relation for the variation of the solid layer thickness \(S\) with time \(t\), expressing your result in terms of appropriate parameters of the system.

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?

Common transmission failures result from the glazing of clutch surfaces by deposition of oil oxidation and decomposition products. Both the oxidation and decomposition processes depend on temperature histories of the surfaces. Because it is difficult to measure these surface temperatures during operation, it is useful to develop models to predict clutch-interface thermal behavior. The relative velocity between mating clutch plates, from the initial engagement to the zero-sliding (lock-up) condition, generates heat that is transferred to the plates. The relative velocity decreases at a constant rate during this period, producing a heat flux that is initially very large and decreases linearly with time, until lock-up occurs. Accordingly, \(q_{f}^{\prime \prime}=q_{o}^{\prime \prime}=\left[1-\left(t / t_{\mathrm{lu}}\right)\right]\), where \(q_{o}^{\prime \prime}=1.6 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2}\) and \(t_{1 \mathrm{u}}=100 \mathrm{~ms}\) is the lock-up time. The plates have an initial uniform temperature of \(T_{i}=40^{\circ} \mathrm{C}\), when the prescribed frictional heat flux is suddenly applied to the surfaces. The reaction plate is fabricated from steel, while the composite plate has a thinner steel center section bonded to low- conductivity friction material layers. The thermophysical properties are \(\rho_{s}=\) \(7800 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{s}}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{s}=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the steel and \(\rho_{\mathrm{im}}=1150 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{fm}}=1650 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{\mathrm{fm}}=4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the friction material. (a) On \(T-t\) coordinates, sketch the temperature history at the midplane of the reaction plate, at the interface between the clutch pair, and at the midplane of the composite plate. Identify key features. (b) Perform an energy balance on the clutch pair over the time interval \(\Delta t=t_{\mathrm{lu}}\) to determine the steadystate temperature resulting from clutch engagement. Assume negligible heat transfer from the plates to the surroundings. (c) Compute and plot the three temperature histories of interest using the finite-element method of FEHT or the finite-difference method of IHT (with \(\Delta x=0.1 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~ms}\) ). Calculate and plot the frictional heat fluxes to the reaction and composite plates, \(q_{\mathrm{rp}}^{\prime \prime}\) and \(q_{\mathrm{cp}}^{\prime \prime}\), respectively, as a function of time. Comment on features of the temperature and heat flux histories. Validate your model by comparing predictions with the results from part (b). Note: Use of both \(F E H T\) and \(I H T\) requires creation of a look-up data table for prescribing the heat flux as a function of time.

As part of a heat treatment process, cylindrical, 304 stainless steel rods of \(100-\mathrm{mm}\) diameter are cooled from an initial temperature of \(500^{\circ} \mathrm{C}\) by suspending them in an oil bath at \(30^{\circ} \mathrm{C}\). If a convection coefficient of \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of \(50^{\circ} \mathrm{C}\), at which point it is withdrawn from the bath? If 10 rods of length \(L=1 \mathrm{~m}\) are processed per hour, what is the nominal rate at which energy must be extracted from the bath (the cooling load)?

A very thick slab with thermal diffusivity \(5.6 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Suddenly, the surface is exposed to a coolant at \(15^{\circ} \mathrm{C}\) for which the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after \(3 \mathrm{~min}\) have elapsed. (b) Compute and plot temperature histories \((0 \leq t \leq\) \(300 \mathrm{~s}\) ) at \(x=0\) and \(x=45 \mathrm{~mm}\) for the following parametric variations: (i) \(\alpha=5.6 \times 10^{-7}, 5.6 \times\) \(10^{-6}\), and \(5.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\); and (ii) \(k=2,20\), and \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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