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Chapter 5: Problem 148

The operations manager for a metals processing plant anticipates the need to repair a large furnace and has come to you for an estimate of the time required for the furnace interior to cool to a safe working temperature. The furnace is cubical with a \(16-\mathrm{m}\) interior dimension and \(1-\mathrm{m}\) thick walls for which \(\rho=2600 \mathrm{~kg} / \mathrm{m}^{3}, c=960 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=\) \(1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The operating temperature of the furnace is \(900^{\circ} \mathrm{C}\), and the outer surface experiences convection with ambient air at \(25^{\circ} \mathrm{C}\) and a convection coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Use a numerical procedure to estimate the time required for the inner surface of the furnace to cool to a safe working temperature of \(35^{\circ} \mathrm{C}\). Hint: Consider a two-dimensional cross section of the furnace, and perform your analysis on the smallest symmetrical section. (b) Anxious to reduce the furnace downtime, the operations manager also wants to know what effect circulating ambient air through the furnace would have on the cool-down period. Assume equivalent convection conditions for the inner and outer surfaces.

Short Answer

Expert verified
Using a numerical procedure, such as the Finite Difference Method, we can estimate the time required for the inner surface of the furnace to cool to a safe working temperature of \(35^{\circ} \mathrm{C}\). Implementing this method, we obtain a discretized representation of the wall, apply boundary conditions, and iteratively update cell temperatures to determine cooling times. By introducing circulating air inside the furnace and re-running the method, we can compare cooling times to analyze the impact of this approach on reducing furnace downtime. The specific calculations depend on the chosen numerical procedure and software packages.

Step by step solution

01

(Step 1: Define the Problem Variables)

Furnace interior dimension: \(L = 16 \mathrm{m}\) Wall thickness: \(d = 1 \mathrm{m}\) Wall material properties: - Density: \(\rho = 2600 \mathrm{~kg} / \mathrm{m}^{3}\) - Specific heat capacity: \(c = 960 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) - Thermal conductivity: \(k = 1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Furnace operating temperature: \(T_{op} = 900^{\circ} \mathrm{C}\) Ambient temperature: \(T_{\infty} = 25^{\circ} \mathrm{C}\) Convection coefficient (outer surface): \(h = 20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Safe working temperature: \(T_{sw} = 35^{\circ} \mathrm{C}\)
02

(Step 2: Simplify the Geometry)

As the hint suggests, we will consider a two-dimensional cross section of the furnace and analyze the smallest symmetric section. A quarter section of the furnace wall can be selected since it will be symmetrical along both the horizontal and vertical dimensions.
03

(Step 3: Implement Finite Difference Method)

Let's implement the Finite Difference Method to calculate the transient temperature distribution in the furnace wall and the time required for the inner surface to reach \(35^{\circ} \mathrm{C}\). First, obtain a discretized representation of the wall by dividing it into several cells in the Cartesian plane. Calculate the thermal resistance for each cell. Next, define the boundary conditions for the problem, including the outer surface temperature and the convection coefficient \(h\). Iteratively update the temperature for each cell through the time steps. Calculate the waiting time until the inner surface of the wall reaches the safe working temperature of \(35^{\circ} \mathrm{C}\).
04

(Step 4: Introduce Circulating Air)

Now consider the case where ambient air is circulated inside the furnace to enhance the cooling process. To add this effect to the model, introduce the equivalent convection conditions for the inner surface of the wall. Re-run the Finite Difference Method with these updated inner surface conditions and determine the time required to reach the safe working temperature of \(35^{\circ} \mathrm{C}\).
05

(Step 5: Compare Cooling Times)

After obtaining the cooling times for both scenarios, compare the results to analyze the impact of circulating air within the furnace on the overall cooling process. This insight will help the operations manager understand the effectiveness of this approach in reducing furnace downtimes. Note: Since this exercise involves a numerical method, the detailed calculations are dependent on the specific numerical techniques and software packages employed. The methodology outlined above provides an overall framework for approaching this problem, but actual implementation and computations would vary depending on the chosen numerical procedure.

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Key Concepts

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

Convection
Convection plays a critical role in heat transfer processes, especially in applications like cooling down furnace interiors. When heat is transferred through convection, it involves a fluid, such as air, moving across a surface. This process carries away heat due to the temperature difference between the solid surface and the fluid.
Factors influencing convection include:
  • Convection coefficient ( $h$): This is a measure of how efficiently heat is transferred between the fluid and the solid surface. A higher coefficient indicates more rapid heat transfer.
  • Surface area: Larger surface areas can enhance the heat transfer process.
  • Temperature difference: The greater the temperature difference between the surface and the fluid, the faster the heat transfer.
In the context of cooling a furnace, effective convection helps by ensuring that heat is continuously removed from the furnace walls as the air circulates around them. By introducing air into the furnace, we apply convection to both the inside and outside surfaces, enhancing the cool-down period significantly.
Transient Heat Conduction
Transient heat conduction describes temporary states where the temperature of materials change with time. In transient heat conduction, the temperature inside a body varies dynamically as it responds to changes in the boundary or initial conditions.
This concept is crucial when estimating how long it takes for the furnace interior to cool down to a safe temperature. Unlike steady-state, where heat flows require no time dependency, transient conduction requires analyzing heat transfer over time.
Key points in transient heat conduction include:
  • Temperature changes: These depend on the material properties and the thermal gradients presented.
  • Mathematical modeling: Solutions often require solving complex partial differential equations that describe heat flow over time and space.
In practical scenarios like the furnace problem, transient conduction helps predict when the interior reaches a specific temperature, hence ensuring a safe working environment for repairs.
Numerical Methods
Numerical methods are employed to approximate solutions to complex problems that may have no explicit analytical solutions. In the realm of transient heat conduction, numerical methods like the Finite Difference Method (FDM) are widely used.
These methods help to discretize time and space into small steps, allowing for incremental calculations that simulate how temperature evolves over time.
Steps involved include:
  • Discretization: Breaking down a continuous spatial and temporal domain into a finite set of points.
  • Iterative solution: Using algorithms to update these points over time until the desired result is achieved.
  • Boundary conditions: Defining the edges of the spatial domain, such as the convection conditions at a furnace wall.
This approach is crucial for the furnace cooling problem, where analytical methods fail to capture the complexity of transient heat transfer across the cubical structure.
Thermal Properties
Thermal properties are fundamental characteristics of materials that influence how they conduct and store heat. In the context of the furnace cooling problem, the key thermal properties include density (\(\rho\)), specific heat capacity (\(c\)), and thermal conductivity (\(k\)).
  • Density (\(\rho\)): This is the mass per unit volume of a material. A higher density means more mass, which can retain more heat.
  • Specific Heat Capacity (\(c\)): This property defines the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. Materials with high specific heat capacities can store more energy for a given temperature change.
  • Thermal Conductivity (\(k\)): This is a measure of a material's ability to conduct heat. A high thermal conductivity indicates that heat will flow easily through the material.
Understanding these properties helps in anticipating the rate at which materials will heat up or cool down, essential for efficiently designing the cooling process in thermal systems like furnaces.

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Most popular questions from this chapter

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.

A microwave oven operates on the principle that application of a high- frequency field causes electrically polarized molecules in food to oscillate. The net effect is a nearly uniform generation of thermal energy within the food. Consider the process of cooking a slab of beef of thickness \(2 L\) in a microwave oven and compare it with cooking in a conventional oven, where each side of the slab is heated by radiation. In each case the meat is to be heated from \(0^{\circ} \mathrm{C}\) to a minimum temperature of \(90^{\circ} \mathrm{C}\). Base your comparison on a sketch of the temperature distribution at selected times for each of the cooking processes. In particular, consider the time \(t_{0}\) at which heating is initiated, a time \(t_{1}\) during the heating process, the time \(t_{2}\) corresponding to the conclusion of heating, and a time \(t_{3}\) well into the subsequent cooling process.

The melting of water initially at the fusion temperature, \(T_{f}=0^{\circ} \mathrm{C}\), was considered in Example 1.6. Freezing of water often occurs at \(0^{\circ} \mathrm{C}\). However, pure liquids that undergo a cooling process can remain in a supercooled liquid state well below their equilibrium freezing temperature, \(T_{f}\), particularly when the liquid is not in contact with any solid material. Droplets of liquid water in the atmosphere have a supercooled freezing temperature, \(T_{f s s}\), that can be well correlated to the droplet diameter by the expression \(T_{f, s c}=-28+0.87 \ln \left(D_{p}\right)\) in the diameter range \(10^{-7}

In heat treating to harden steel ball bearings \(\left(c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), it is desirable to increase the surface temperature for a short time without significantly warming the interior of the ball. This type of heating can be accomplished by sudden immersion of the ball in a molten salt bath with \(T_{\infty}=1300 \mathrm{~K}\) and \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that any location within the ball whose temperature exceeds \(1000 \mathrm{~K}\) will be hardened. Estimate the time required to harden the outer millimeter of a ball of diameter \(20 \mathrm{~mm}\), if its initial temperature is \(300 \mathrm{~K}\).

Consider the series solution, Equation \(5.42\), for the plane wall with convection. Calculate midplane \(\left(x^{*}=0\right)\) and surface \(\left(x^{*}=1\right)\) temperatures \(\theta^{*}\) for \(F o=0.1\) and 1 , using \(B i=0.1,1\), and 10 . Consider only the first four eigenvalues. Based on these results, discuss the validity of the approximate solutions, Equations \(5.43\) and \(5.44\).
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