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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

A wall \(0.12 \mathrm{~m}\) thick having a thermal diffusivity of \(1.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) is initially at a uniform temperature of \(85^{\circ} \mathrm{C}\). Suddenly one face is lowered to a temperature of \(20^{\circ} \mathrm{C}\), while the other face is perfectly insulated. (a) Using the explicit finite-difference technique with space and time increments of \(30 \mathrm{~mm}\) and \(300 \mathrm{~s}\), respectively, determine the temperature distribution at \(t=45 \mathrm{~min}\). (b) With \(\Delta x=30 \mathrm{~mm}\) and \(\Delta t=300 \mathrm{~s}\), compute \(T(x, t)\) for \(0 \leq t \leq t_{\mathrm{ss}}\), where \(t_{\mathrm{ss}}\) is the time required for the temperature at each nodal point to reach a value that is within \(1^{\circ} \mathrm{C}\) of the steady-state temperature. Repeat the foregoing calculations for \(\Delta t=75 \mathrm{~s}\). For each value of \(\Delta t\), plot temperature histories for each face and the midplane.

The plane wall of Problem \(2.60(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\alpha=1.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) has a thickness of \(L=40 \mathrm{~mm}\) and an initial uniform temperature of \(T_{o}=25^{\circ} \mathrm{C}\). Suddenly, the boundary at \(x=L\) experiences heating by a fluid for which \(T_{s}=50^{\circ} \mathrm{C}\) and \(h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while heat is uniformly generated within the wall at \(\dot{q}=1 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The boundary at \(x=0\) remains at \(T_{a}\). (a) With \(\Delta x=4 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~s}\), plot temperature distributions in the wall for (i) the initial condition, (ii) the steady-state condition, and (iii) two intermediate times. (b) On \(q_{x}^{\prime \prime}-t\) coordinates, plot the heat flux at \(x=0\) and \(x=L\). At what elapsed time is there zero heat flux at \(x=L\) ?

Consider the bonding operation described in Problem \(3.115\), which was analyzed under steady-state conditions. In this case, however, the laser will be used to heat the film for a prescribed period of time, creating the transient heating situation shown in the sketch. The strip is initially at \(25^{\circ} \mathrm{C}\) and the laser provides a uniform flux of \(85,000 \mathrm{~W} / \mathrm{m}^{2}\) over a time interval of \(\Delta t_{\mathrm{on}}=10 \mathrm{~s}\). The system dimensions and thermophysical properties remain the same, but the convection coefficient to the ambient air at \(25^{\circ} \mathrm{C}\) is now \(100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) and \(w_{1}=44 \mathrm{~mm}\). Using an implicit finite-difference method with \(\Delta x=4 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~s}\), obtain temperature histories for \(0 \leq t \leq 30 \mathrm{~s}\) at the center and film edge, \(T(0, t)\) and \(T\left(w_{1} / 2, t\right)\), respectively, to determine if the adhesive is satisfactorily cured above \(90^{\circ} \mathrm{C}\) for \(10 \mathrm{~s}\) and if its degradation temperature of \(200^{\circ} \mathrm{C}\) is exceeded.

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}

A constant-property, one-dimensional plane wall of width \(2 L\), at an initial uniform temperature \(T_{i}\), is heated convectively (both surfaces) with an ambient fluid at \(T_{\infty}=T_{\infty, 1}, h=h_{1}\). At a later instant in time, \(t=t_{1}\), heating is curtailed, and convective cooling is initiated. Cooling conditions are characterized by \(T_{\infty}=T_{\infty, 2}=T_{i}, h=h_{2}\) (a) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the heating phase (Phase 1). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{1}\), and \(F o\), where \(B i_{1}\) is expressed in terms of \(h_{1}\). (b) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the cooling phase (Phase 2). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{2}\), \(\mathrm{Fo}_{1}\), and \(\mathrm{Fo}\) where \(\mathrm{Fo}_{1}\) is the dimensionless time associated with \(t_{1}\), and \(B i_{2}\) is expressed in terms of \(h_{2}\). To be consistent with part (a), express the dimensionless temperature in terms of \(T_{\infty}=T_{\infty, 1^{*}}\) (c) Consider a case for which \(B i_{1}=10, B i_{2}=1\), and \(F o_{1}=0.1\). Using a finite-difference method with \(\Delta x^{*}=0.1\) and \(\Delta F o=0.001\), determine the transient thermal response of the surface \(\left(x^{*}=1\right)\), midplane \(\left(x^{*}=0\right)\), and quarter-plane \(\left(x^{*}=0.5\right)\) of the slab. Plot these three dimensionless temperatures as a function of dimensionless time over the range \(0 \leq F o \leq 0.5\). (d) Determine the minimum dimensionless temperature at the midplane of the wall, and the dimensionless time at which this minimum temperature is achieved.
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