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Chapter 5: Problem 51

The 150 -mm-thick wall of a gas-fired furnace is constructed of fireclay brick \((k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2600\) \(\left.\mathrm{kg} / \mathrm{m}^{3}, c_{p}=1000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) and is well insulated at its outer surface. The wall is at a uniform initial temperature of \(20^{\circ} \mathrm{C}\), when the burners are fired and the inner surface is exposed to products of combustion for which \(T_{\infty}=950^{\circ} \mathrm{C}\) and \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) How long does it take for the outer surface of the wall to reach a temperature of \(750^{\circ} \mathrm{C}\) ? (b) Plot the temperature distribution in the wall at the foregoing time, as well as at several intermediate times.

Short Answer

Expert verified
It takes approximately 5671 seconds, or 1.57 hours, for the outer surface of the furnace wall to reach 750掳C. The temperature distribution across the wall at different times can be plotted by using the Heisler chart, varying the dimensionless position (\(\xi\)) and the Fourier number (\(\eta\)), and calculating the temperature, \(T(X, \theta)\), at those positions.

Step by step solution

01

Analyze given information and find the Biot number

We need to analyze the given information first: - Wall thickness: \(L = 0.15m\) - Thermal conductivity of the fireclay brick, k: \(1.5 W/m \cdot K\) - Density of the fireclay brick, 蟻: \(2600 kg/m^3\) - Specific heat of the fireclay brick, \(c_p\): \(1000 J/kg \cdot K\) - Initial temperature of the wall, \(T_i\): \(20^{\circ}C\) - Combustion gas temperature, \(T_{\infty}\): \(950^{\circ}C\) - Convective heat transfer coefficient, h: \(100 W/m^2 \cdot K\) First, we need to find the Biot number, Bi: \[Bi = \frac{hL}{k}\]
02

Calculate Biot number

Using the given values for h, L, and k, we can calculate the Biot number: \[Bi = \frac{100 \times 0.15}{1.5} = 10\] Since the Biot number is greater than 0.1, the lumped capacitance method is not valid for this exercise. Therefore, we need to use the Heisler chart to determine the time it takes for the outer surface to reach 750掳C.
03

Use Heisler chart

To use the Heisler chart, we need to define some parameters. First, we can express the temperature at a given location, X, and time, 胃, on the chart as follows: \[\frac{T(X, \theta) - T_{\infty}}{T_i - T_{\infty}} = \phi(\xi, \eta)\] Where \(\phi(\xi, \eta)\) is the dimensionless temperature function on the Heisler chart, \(\xi = X/L\) is the dimensionless position, and \(\eta\) is the Fourier number, defined as: \[\eta = \frac{\alpha \theta}{L^2}\] And \(\alpha\) is the thermal diffusivity of the fireclay brick, defined as: \[\alpha = \frac{k}{\rho c_p}\]
04

Calculate thermal diffusivity and Fourier number

Calculate the thermal diffusivity with given k, 蟻, and \(c_p\): \[\alpha = \frac{1.5}{2600 \times 1000} = 5.77 \times 10^{-7}m^2/s\] Now we can find the Fourier number when the outer surface of the wall reaches 750掳C. From the given information: \[\frac{750 - 950}{20 - 950} = \phi(1, \eta)\] Using the Heisler chart, we get the value of \(\phi(1, \eta) \approx 0.798\). Now we need to find the Fourier number, \(\eta\).
05

Find time required to reach 750掳C

From the Heisler chart, we obtain the value of \(\eta\) corresponding to \(\phi(1, \eta) \approx 0.798\), which is \( \eta \approx 0.875\). Now we can find the time required to reach 750掳C using the Fourier number: \[\theta = \frac{L^2 \eta}{\alpha} = \frac{0.15^2 \times 0.875}{5.77 \times 10^{-7}} = 5670.8s\] Hence, it takes approximately 5671 seconds, or 1.57 hours, for the outer surface of the wall to reach 750掳C.
06

Plot temperature distribution

To plot the temperature distribution across the wall at different times, we can use the \(\phi(\xi, \eta)\) function from the Heisler chart and vary the values of \(\xi\) and \(\eta\). By choosing several time points (and corresponding \(\eta\) values) and several positions within the wall (and corresponding \(\xi\) values), we can calculate the temperature at those positions using: \[T(X, \theta) = T_{\infty} + \phi(\xi, \eta)(T_i - T_{\infty})\] By plotting T(X, 胃) at different positions within the wall (varying \(\xi\)) and at different times (varying \(\eta\)), we obtain the temperature distribution across the wall.

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Key Concepts

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

Biot Number
The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations to relate the conductive heat resistance within an object to the convective heat transfer across its boundary. The formula to calculate the Biot number is given by \[Bi = \frac{hL}{k}\]
where:
  • \(h\) is the convective heat transfer coefficient, measured in watts per square meter kelvin (W/m^2鈰匥),
  • \(L\) is the characteristic length, typically the thickness of the object, measured in meters (m), and
  • \(k\) is the thermal conductivity of the material, measured in watts per meter kelvin (W/m鈰匥).

In our exercise, the Biot number was determined to be greater than 0.1, which indicates that the lumped capacitance method is not suitable. Instead, we need more complex models like transient heat conduction analysis or the use of Heisler charts to determine temperature profiles and heat transfer rates over time in the medium.
Fourier Number
The Fourier number (Fo) is another dimensionless number, which is used to characterize heat conduction. It represents the ratio of heat conduction rate to heat storage rate within a material and is defined as:
\[Fo = \frac{\alpha \theta}{L^2}\]
where:
  • \(\alpha\) is the thermal diffusivity of the material (measured in square meters per second, m^2/s),
  • \(\theta\) is the time (s), and
  • \(L\) is the characteristic length (m).

The Fourier number basically indicates how 'far' the heat has diffused into the material at a certain moment in time. For the furnace wall in the given problem, the Fourier number is crucial for predicting at what time the outer surface temperature reaches 750掳C. By using the Fourier number and the Heisler chart, we can determine the necessary time for the wall to reach a specified temperature.
Heisler Chart
Heisler charts are a set of graphs that provide a solution to the transient heat conduction problem in cylindrical and spherical objects, as well as a slab (a thin, flat object), for various geometries of the object. These charts help us determine the temperature distribution within an object over time without solving the complex partial differential equations of heat conduction.
Using the Heisler charts, we identify the \(\phi(\xi, \eta)\) function, where \(\xi\) is the dimensionless position and \(\eta\) is the Fourier number. By finding the appropriate \(\phi\) value for the given Fourier number, we can then backtrack to determine the time at which a certain temperature is reached. In our problem, the Heisler chart was used to establish the time required for the outer surface of a gas-fired furnace wall to reach 750掳C.
Thermal Conductivity
Thermal conductivity (\(k\)), a fundamental property of materials, indicates a material's ability to conduct heat. It quantifies the rate at which heat energy is transferred through a material due to a temperature gradient. High thermal conductivity means the material is a good conductor of heat, while low thermal conductivity indicates the material is more of an insulator.
In the context of our problem, the fireclay brick has a thermal conductivity of 1.5 W/m鈰匥. This is a key parameter in determining both the Biot number and the thermal diffusivity, which in turn are essential for analyzing the heat transfer within the furnace wall.
Thermal Diffusivity
Thermal diffusivity (\(\alpha\)) measures the rate at which heat spreads through a material and is calculated by dividing the thermal conductivity (\(k\)) by the product of density (\(\rho\)) and specific heat capacity (\(c_p\)). The formula is given as:
\[\alpha = \frac{k}{\rho c_p}\]
where:
  • \(k\) is the thermal conductivity,
  • \(\rho\) is the density, and
  • \(c_p\) is the specific heat capacity of the material.

This property essentially describes how quickly a material responds to changes in temperature. For the fireclay brick of our exercise, once we calculate the thermal diffusivity, we can proceed to evaluate the Fourier number and then use the Heisler chart to determine how heat is transferred over time through the wall.

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Most popular questions from this chapter

Circuit boards are treated by heating a stack of them under high pressure, as illustrated in Problem 5.45. The platens at the top and bottom of the stack are maintained at a uniform temperature by a circulating fluid. The purpose of the pressing-heating operation is to cure the epoxy, which bonds the fiberglass sheets, and impart stiffness to the boards. The cure condition is achieved when the epoxy has been maintained at or above \(170^{\circ} \mathrm{C}\) for at least \(5 \mathrm{~min}\). The effective thermophysical properties of the stack or book (boards and metal pressing plates) are \(k=0.613 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=2.73 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} \cdot \mathrm{K}\) (a) If the book is initially at \(15^{\circ} \mathrm{C}\) and, following application of pressure, the platens are suddenly brought to a uniform temperature of \(190^{\circ} \mathrm{C}\), calculate the elapsed time \(t_{e}\) required for the midplane of the book to reach the cure temperature of \(170^{\circ} \mathrm{C}\). (b) If, at this instant of time, \(t=t_{e}\), the platen temperature were reduced suddenly to \(15^{\circ} \mathrm{C}\), how much energy would have to be removed from the book by the coolant circulating in the platen, in order to return the stack to its initial uniform temperature?

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 \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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 \(300 \mathrm{~K}\).

Derive the explicit finite-difference equation for an interior node for three- dimensional transient conduction. Also determine the stability criterion. Assume constant properties and equal grid spacing in all three directions.

A plane wall \(\left(\rho=4000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of thickness \(L=20 \mathrm{~mm}\) initially has a linear, steady-state temperature distribution with boundaries maintained at \(T_{1}=0^{\circ} \mathrm{C}\) and \(T_{2}=100^{\circ} \mathrm{C}\). Suddenly, an electric current is passed through the wall, causing uniform energy generation at a rate \(\dot{q}=2 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The boundary conditions \(T_{1}\) and \(T_{2}\) remain fixed. (a) On \(T-x\) coordinates, sketch temperature distributions for the following cases: (i) initial condition \((t \leq 0)\); (ii) steady-state conditions \((t \rightarrow \infty\) ), assuming that the maximum temperature in the wall exceeds \(T_{2}\); and (iii) for two intermediate times. Label all important features of the distributions. (b) For the system of three nodal points shown schematically (1, \(m, 2)\), define an appropriate control volume for node \(m\) and, identifying all relevant processes, derive the corresponding finitedifference equation using either the explicit or implicit method. (c) With a time increment of \(\Delta t=5 \mathrm{~s}\), use the finitedifference method to obtain values of \(T_{m}\) for the first \(45 \mathrm{~s}\) of elapsed time. Determine the corresponding heat fluxes at the boundaries, that is, \(q_{x}^{\prime \prime}\) \((0,45 \mathrm{~s})\) and \(q_{x}^{\prime \prime}(20 \mathrm{~mm}, 45 \mathrm{~s})\). (d) To determine the effect of mesh size, repeat your analysis using grids of 5 and 11 nodal points ( \(\Delta x=5.0\) and \(2.0 \mathrm{~mm}\), respectively).
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