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Chapter 5: Problem 91

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

Short Answer

Expert verified
(a) After 3 minutes have elapsed: - The temperature at the surface (x=0): \(T(x=0) \approx 168.93^{\circ} \mathrm{C}\) - The temperature at a depth of 45mm (x=0.045m): \(T(x=0.045) \approx 307.59^{\circ} \mathrm{C}\) For part (b), compute and plot the temperature histories for parametric variations of thermal diffusivity (\(\alpha\)) and thermal conductivity (\(k\)). The plot will show how different values of \(\alpha\) and \(k\) affect the cooling rate and final temperature of the slab.

Step by step solution

01

Define the given variables and constants

The problem statement provides the following information: - Thermal diffusivity, \(\alpha = 5.6 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) - Thermal conductivity, \(k = 20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Initial temperature, \(T_i = 325^{\circ} \mathrm{C}\) - Coolant temperature, \(T_{\infty} = 15^{\circ} \mathrm{C}\) - Convection heat transfer coefficient, \(h = 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Time elapsed, \(t = 3 \mathrm{~min} = 180 \mathrm{~s}\) - Depth, \(x = 45 \mathrm{~mm} = 0.045 \mathrm{~m}\)
02

Determine the dimensionless parameters

Define the dimensionless parameters useful for this problem: Biot number (Bi) and Fourier number (Fo). 1. Biot Number: Bi = \(\frac{hL_c}{k}\) Here, \(L_c\) is the characteristic length. For a slab, we can approximate \(L_c\) with the given depth \(x\). Therefore, \(L_c = 0.045 \mathrm{~m}\). The Biot number represents the ratio of conduction resistance to convection resistance. 2. Fourier Number: Fo = \(\frac{\alpha t}{L_c^2}\) The Fourier number represents the ratio of conduction heat transfer rate within the slab to the total heat transfer rate. Now, we calculate the values of Biot number and Fourier number using the given values: - Bi = \(\frac{100 \times 0.045}{20} = 0.225\) - Fo = \(\frac{5.6 \times 10^{-6} \times 180}{0.045^2} = 0.504\)
03

Temperature at the surface after 3 minutes

To find the temperature at the surface after 3 minutes, we can use the first term of the semi-infinite medium one-term approximation: \(T(x=0) = T_{\infty} + (T_i - T_{\infty}) \cdot e^{-h \, t / \rho c_p}\) Here, \(\rho\) is the density and \(c_p\) is the specific heat capacity of the material. We can find the value of \(\rho c_p\) by using the relation: \(\rho c_p = \frac{k}{\alpha}\) Now, substitute the known values and calculate the temperature at the surface (x=0): \(T(x=0) = 15 + (325-15) \cdot e^{-100 \times (180) / (20 / (5.6\times 10^{-6}))} \approx 15 + 310 \times e^{-0.1008} \approx 168.93^{\circ} \mathrm{C}\)
04

Temperature at 45mm depth after 3 minutes

We can use the one-term approximation for the temperature at a depth of 45mm: \(T(x=0.045) = T_{\infty} + (T_i - T_{\infty}) \cdot e^{-(\frac{h x}{2k})^2 \cdot \mathrm{erf}^{-1}(2Fo)}\) Now, substitute the known values and calculate the temperature at the depth of 45mm: \(T(x=0.045) = 15 + (325-15) \cdot e^{-(\frac{100 \times 0.045}{2 \times 20})^2 \cdot \mathrm{erf}^{-1}(2 \times 0.504)} \approx 15 + 310 \times e^{-(0.1125)^2 \cdot (-0.103)} \approx 307.59^{\circ} \mathrm{C}\)
05

Answer: (a)

(a) After 3 minutes have elapsed: - The temperature at the surface (x=0): \(T(x=0) \approx 168.93^{\circ} \mathrm{C}\) - The temperature at a depth of 45mm (x=0.045m): \(T(x=0.045) \approx 307.59^{\circ} \mathrm{C}\)
06

Compute and plot the temperature histories

For part (b), we need to compute and plot temperature histories for the given parametric variations of thermal diffusivity and thermal conductivity. 1. Prepare an empty plot with Time (\(t\)) in the x-axis and Temperature (\(T\)) in the y-axis. 2. Use the given values of \(\alpha\) and \(x\) to compute temperature history for each combination: (\(\alpha=5.6 \times 10^{-7}, k=2\)), (\(\alpha=5.6 \times 10^{-6}, k=20\)), (\(\alpha=5.6 \times 10^{-5}, k=200\)). 3. For each parametric combination, calculate the Fourier number and Biot number as in Steps 2 and 3. 4. For each parametric combination, plot the temperature history (0 \(\leq t \leq\) 300 s) at \(x=0\) and \(x=0.045\) using the one-term approximation equations from Steps 3 and 4. 5. Analyze the temperature profiles for different combinations of \(\alpha\) and \(k\), and their effects on the cooling rate and surface temperature. The plot would show how different thermal conductivity and thermal diffusivity values affect the cooling rate and final temperature of the slab.

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Key Concepts

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

Thermal Diffusivity
Thermal diffusivity is a measure of how quickly heat moves through a material. It's an important property in heat transfer problems because it blends thermal conductivity, density, and specific heat capacity into one parameter. This makes it easier to understand how materials will behave when there's a temperature change.

Thermal diffusivity is denoted by the symbol \( \alpha \), and it's calculated by dividing the material's thermal conductivity \( k \) by the product of its density \( \rho \) and specific heat capacity \( c_p \). The formula for thermal diffusivity is:
  • \( \alpha = \frac{k}{\rho c_p} \)
Units for thermal diffusivity are square meters per second (\( \mathrm{m}^2/\mathrm{s} \)).

Materials with high thermal diffusivity, like metals, respond quickly to temperature changes because they conduct heat rapidly. Conversely, materials with low thermal diffusivity, like wood or plastic, change temperature more slowly. In our exercise, the thermal diffusivity of the slab was given as \( 5.6 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s} \). This value indicates the slab's moderate ability to conduct heat relative to its capacity to store it.
Convection Heat Transfer
Convection heat transfer occurs when a fluid, such as air or water, flows over a surface and transfers heat between the surface and the fluid. It's an essential concept in many engineering applications, as it impacts the rate at which heat is removed or added to a system.

The effectiveness of convection heat transfer is quantified using the convection heat transfer coefficient, denoted as \( h \). The higher the value of \( h \), the more efficient the heat transfer process.
  • Typical units for \( h \) are \( \mathrm{W/m^2 \, K} \).
In the given exercise, the convection heat transfer coefficient \( h \) is \( 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \). This indicates a moderate rate of heat transfer from the slab surface to the coolant.

Convection can be either natural, where the fluid movement is due to density differences caused by temperature gradients, or forced, where external devices like fans or pumps direct the fluid flow.
Biot Number
The Biot number (\( \mathrm{Bi} \)) is a dimensionless parameter that provides insight into the resistance to heat flow within a body compared to the resistance to heat flow from the body's surface to its surroundings. It's crucial in determining whether surface temperature assumptions in analytical solutions are valid.

It is calculated using the formula:
  • \( \mathrm{Bi} = \frac{hL_c}{k} \)
where \( h \) is the convection heat transfer coefficient, \( L_c \) is the characteristic length (often the thickness of the slab or the radius of a sphere), and \( k \) is the thermal conductivity of the material.

A lower Biot number (\( \mathrm{Bi} < 0.1 \)) indicates that the internal resistance to heat conduction is insignificant compared to the external convection resistance. This suggests that the entire body can be assumed to change temperature uniformly, which is a simplification often used in theoretical analyses.

In our exercise, the Biot number was computed as \( 0.225 \), suggesting some consideration of internal resistance in the analysis.
Fourier Number
The Fourier number \( (\mathrm{Fo}) \) is another vital dimensionless parameter in heat transfer analysis. It helps in assessing how the heat conducted through the material compares to the time elapsed. In essence, it gives a sense of the progress of conduction through a material over time.
  • It is calculated by the formula \( \mathrm{Fo} = \frac{\alpha t}{L_c^2} \),
where \( \alpha \) is the thermal diffusivity, \( t \) is the time elapsed, and \( L_c \) is the characteristic length of the material.

The Fourier number is crucial in transient heat conduction problems. A larger Fourier number means that the heat has penetrated deeper into the material relative to time, indicating a higher degree of thermal diffusion.

For this exercise, the calculated Fourier number was \( 0.504 \), indicating a moderate level of heat penetration and diffusion within the 3-minute period analyzed. This figure helps to determine the temperature distribution within the slab and evolve time for better understanding of the transient heat conduction.

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Most popular questions from this chapter

For each of the following cases, determine an appropriate characteristic length \(L_{c}\) and the corresponding Biot number \(B i\) that is associated with the transient thermal response of the solid object. State whether the lumped capacitance approximation is valid. If temperature information is not provided, evaluate properties at \(T=300 \mathrm{~K}\). (a) A toroidal shape of diameter \(D=50 \mathrm{~mm}\) and cross-sectional area \(A_{c}=5 \mathrm{~mm}^{2}\) is of thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface of the torus is exposed to a coolant corresponding to a convection coefficient of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) A long, hot AISI 304 stainless steel bar of rectangular cross section has dimensions \(w=3 \mathrm{~mm}\), \(W=5 \mathrm{~mm}\), and \(L=100 \mathrm{~mm}\). The bar is subjected to a coolant that provides a heat transfer coefficient of \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at all exposed surfaces. (c) A long extruded aluminum (Alloy 2024) tube of inner and outer dimensions \(w=20 \mathrm{~mm}\) and \(W=24 \mathrm{~mm}\), respectively, is suddenly submerged in water, resulting in a convection coefficient of \(h=37 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the four exterior tube surfaces. The tube is plugged at both ends, trapping stagnant air inside the tube. (d) An \(L=300-m m\)-long solid stainless steel rod of diameter \(D=13 \mathrm{~mm}\) and mass \(M=0.328 \mathrm{~kg}\) is exposed to a convection coefficient of \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (e) A solid sphere of diameter \(D=12 \mathrm{~mm}\) and thermal conductivity \(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suspended in a large vacuum oven with internal wall temperatures of \(T_{\text {sur }}=20^{\circ} \mathrm{C}\). The initial sphere temperature is \(T_{i}=100^{\circ} \mathrm{C}\), and its emissivity is \(\varepsilon=0.73\). (f) A long cylindrical rod of diameter \(D=20 \mathrm{~mm}\), density \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\), specific heat \(c_{p}=1750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suddenly exposed to convective conditions with \(T_{\infty}=20^{\circ} \mathrm{C}\). The rod is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and reaches a spatially averaged temperature of \(T=100^{\circ} \mathrm{C}\) at \(t=225 \mathrm{~s}\). (g) Repeat part (f) but now consider a rod diameter of \(D=200 \mathrm{~mm}\).

A chip that is of length \(L=5 \mathrm{~mm}\) on a side and thickness \(t=1 \mathrm{~mm}\) is encased in a ceramic substrate, and its exposed surface is convectively cooled by a dielectric liquid for which \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=20^{\circ} \mathrm{C}\). In the off-mode the chip is in thermal equilibrium with the coolant \(\left(T_{i}=T_{\infty}\right)\). When the chip is energized, however, its temperature increases until a new steady state is established. For purposes of analysis, the energized chip is characterized by uniform volumetric heating with \(\dot{q}=9 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming an infinite contact resistance between the chip and substrate and negligible conduction resistance within the chip, determine the steady-state chip temperature \(T_{f}\). Following activation of the chip, how long does it take to come within \(1^{\circ} \mathrm{C}\) of this temperature? The chip density and specific heat are \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

Steel is sequentially heated and cooled (annealed) to relieve stresses and to make it less brittle. Consider a 100 -mm-thick plate \(\left(k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c_{p}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) that is initially at a uniform temperature of \(300^{\circ} \mathrm{C}\) and is heated (on both sides) in a gas-fired furnace for which \(T_{\infty}=700^{\circ} \mathrm{C}\) and \(h=\) \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). How long will it take for a minimum temperature of \(550^{\circ} \mathrm{C}\) to be reached in the plate?

A solid steel sphere (AISI 1010 ), \(300 \mathrm{~mm}\) in diameter, is coated with a dielectric material layer of thickness \(2 \mathrm{~mm}\) and thermal conductivity \(0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The coated sphere is initially at a uniform temperature of \(500^{\circ} \mathrm{C}\) and is suddenly quenched in a large oil bath for which \(T_{\infty}=100^{\circ} \mathrm{C}\) and \(h=3300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the time required for the coated sphere temperature to reach \(140^{\circ} \mathrm{C}\). Hint: Neglect the effect of energy storage in the dielectric material, since its thermal capacitance \((\rho c V)\) is small compared to that of the steel sphere

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.
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