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Chapter 5: Problem 81

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at \(25^{\circ} \mathrm{C}\), into the flow, which is at \(75^{\circ} \mathrm{C}\), and measuring its surface temperature at some time during the transient heating process. (a) If the sphere has a diameter of \(0.1 \mathrm{~m}\), a thermal conductivity of \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and a thermal diffusivity of \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), at what time will a surface temperature of \(60^{\circ} \mathrm{C}\) be recorded if the convection coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? (b) Assess the effect of thermal diffusivity on the thermal response of the material by computing center and surface temperature histories for \(\alpha=10^{-6}\), \(10^{-5}\), and \(10^{-4} \mathrm{~m}^{2} / \mathrm{s}\). Plot your results for the period \(0 \leq t \leq 300 \mathrm{~s}\). In a similar manner, assess the effect of thermal conductivity by considering values of \(k=1.5,15\), and \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The surface temperature of \(60^{\circ} \mathrm{C}\) will be recorded at \(t = 62.83 \mathrm{~s}\). To assess the effect of thermal diffusivity and conductivity on the thermal response, a computational tool can be used to calculate the temperature histories as a function of time for the different thermal diffusivities and conductivities, and then plot the results. The trends and comparisons in the plots can be used to analyze and understand the effect of these parameters on the thermal response of the sphere.

Step by step solution

01

Write down the given values and formula

The given values are: \(T_i = 25^{\circ} \mathrm{C}\) \(T_{\infty} = 75^{\circ} \mathrm{C}\) \(T(t) = 60^{\circ} \mathrm{C}\) \(D = 0.1 \mathrm{~m}\) \(k = 15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) \(\alpha = 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) \(h = 300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Transient conduction equation: \(T(t)-T_{\infty}=(T_i-T_{\infty})\exp\left(-\frac{hAt}{kV}\right)\)
02

Calculate surface area and volume of the sphere

We will need the surface area (A) and volume (V) of the sphere to solve the equation. \(A = 4\pi r^2\) \(V = \frac{4}{3}\pi r^3\) where \(r = \frac{D}{2} = 0.05 \mathrm{~m}\) \(A = 4\pi(0.05)^2 = 0.0314 \mathrm{~m}^{2}\) \(V = \frac{4}{3}\pi(0.05)^3 = 5.236\times10^{-5} \mathrm{~m}^{3}\)
03

Solve for the time 't'

Input the given values into the transient conduction equation: \(60-75=(25-75)\exp\left(-\frac{300\times0.0314\times t}{15\times 5.236\times10^{-5}}\right)\) \(15 = 50\exp\left(-\frac{9.42t}{0.007854}\right)\) \(0.3 = \exp(-1.2t)\) Take the natural logarithm of both sides: \(-1.203=\ln(0.3)=-1.2t\) Solve for 't': \(t = \ln(0.3)/(-1.2) = 62.83 \mathrm{~s}\) The surface temperature of \(60^{\circ} \mathrm{C}\) will be recorded at \(t = 62.83 \mathrm{~s}\). (b) Since the problem requires extensive computation and plotting, a computational tool like Python, MATLAB, or Excel can be used for this part. The general approach would be to calculate the temperature histories as a function of time for the different thermal diffusivities and conductivities, as outlined below. - Step 1: Define the time range (0 to 300 seconds) - Step 2: For each combination of thermal diffusivity and conductivity, compute the surface and center temperature of the sphere for each time point using the transient conduction equation. - Step 3: Plot the temperature history for each combination of values. Once the above calculations and plotting are completed, comparisons can be made to assess the effect of thermal diffusivity and conductivity on the thermal response of the sphere.

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Key Concepts

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

Transient Heat Conduction
Transient heat conduction is a process where the temperature of a material changes over time due to heat flowing in or out of the material. Unlike steady-state conduction, which assumes constant temperatures over time, transient conduction considers scenarios where temperatures vary and evolve. Understanding this process helps us analyze how quickly or slowly a material can reach a thermal equilibrium with its environment.

When a solid object like a sphere is placed in a different temperature environment, heat transfer begins, and the object's temperature changes over time until it is uniform. In our exercise, this means calculating how long it takes for the surface temperature of a sphere to reach a specific value, considering the initial conditions and properties of the material.
Thermal Conductivity
Thermal conductivity (\(k\)) is a measure of a material's ability to conduct heat. This property plays a critical role in determining how quickly heat will travel through a material. In our case, the sphere's thermal conductivity is given as \(15 \, \mathrm{W} \/ \mathrm{m} \cdot \mathrm{K}\). A high thermal conductivity means that the material can rapidly transfer heat, leading to quick temperature changes, while a low value would slow down heat transfer.

In practical terms, materials with different thermal conductivities react differently under the same conditions. This is evident when we assess the sphere's temperature history against different conductivity values. The faster a material conducts heat, the faster it will reach a thermal balance with its surroundings.
Thermal Diffusivity
Thermal diffusivity (\(\alpha\)) is another important property that combines the effects of thermal conductivity and material density. It essentially indicates how fast heat can diffuse through a material. The equation \(\alpha = \frac{k}{\rho c_p}\) connects diffusivity (\(\alpha\)) to conductivity (\(k\)), density (\(\rho\)), and specific heat capacity (\(c_p\)).

In the exercise, the ability to assess different thermal diffusivities (\(10^{-6}, 10^{-5}, 10^{-4} \mathrm{~m}^{2} \mathrm{/s}\)) helps predict how fast the sphere adapts to the surrounding temperature. A higher diffusivity suggests quicker thermal response, making the material reach equilibrium sooner than one with a lower diffusivity.
Temperature History
Temperature history refers to the monitoring of a material's temperature change over time. By understanding this, we can create a timeline of how a material heats or cools, which helps in designing materials for specific thermal conditions.

In our task, different temperature histories were calculated to assess how varying thermal properties impact a sphere's response to new thermal environments. Plotting these histories for different thermal diffusivities and conductivities allows for a clear visualization of the transient behavior, highlighting the influence of each property on the thermal performance. This analysis is crucial for applications in engineering and material science.

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Most popular questions from this chapter

5.53 Stone mix concrete slabs are used to absorb thermal energy from flowing air that is carried from a large concentrating solar collector. The slabs are heated during the day and release their heat to cooler air at night. If the daytime airflow is characterized by a temperature and convection heat transfer coefficient of \(T_{\infty}=200^{\circ} \mathrm{C}\) and \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the slab thickness \(2 L\) required to transfer a total amount of energy such that \(Q / Q_{o}=0.90\) over a \(t=8\)-h period. The initial concrete temperature is \(T_{i}=40^{\circ} \mathrm{C}\).

During transient operation, the steel nozzle of a rocket engine must not exceed a maximum allowable operating temperature of \(1500 \mathrm{~K}\) when exposed to combustion gases characterized by a temperature of \(2300 \mathrm{~K}\) and a convection coefficient of \(5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To extend the duration of engine operation, it is proposed that a ceramic thermal barrier coating \((k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\alpha=6 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) be applied to the interior surface of the nozzle. (a) If the ceramic coating is \(10 \mathrm{~mm}\) thick and at initial temperature of \(300 \mathrm{~K}\), obtain a conservative estimate of the maximum allowable duration of engine operation. The nozzle radius is much larger than the combined wall and coating thickness. (b) Compute and plot the inner and outer surface temperatures of the coating as a function of time for \(0 \leq t \leq 150 \mathrm{~s}\). Repeat the calculations for a coating thickness of \(40 \mathrm{~mm}\).

Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is being charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas. Consider a packed bed of \(75-\mathrm{mm}\)-diameter aluminum spheres \(\left(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}, c=950 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\right.\) \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) and a charging process for which gas enters the storage unit at a temperature of \(T_{g, i}=300^{\circ} \mathrm{C}\). If the initial temperature of the spheres is \(T_{i}=25^{\circ} \mathrm{C}\) and the convection coefficient is \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long does it take a sphere near the inlet of the system to accumulate \(90 \%\) of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere? Is there any advantage to using copper instead of aluminum?

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?

The stability criterion for the explicit method requires that the coefficient of the \(T_{m}^{p}\) term of the one-dimensional, finite-difference equation be zero or positive. Consider the situation for which the temperatures at the two neighboring nodes \(\left(T_{\mathrm{m}-1}^{p}, T_{\mathrm{m}+1}^{p}\right)\) are \(100^{\circ} \mathrm{C}\) while the center node \(\left(T_{m}^{p}\right)\) is at \(50^{\circ} \mathrm{C}\). Show that for values of \(F o>\frac{1}{2}\) the finite-difference equation will predict a value of \(T_{m}^{p+1}\) that violates the second law of thermodynamics.
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