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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?

Short Answer

Expert verified
The time it takes for the 100-mm-thick steel plate to reach a minimum temperature of 550掳C can be found using the Lumped Capacitance Method (LCM). First, calculate the Biot number (Bi) to check if the LCM can be applied. If the Bi < 0.1, LCM is valid. Then, calculate the volume (V) and surface area (A) of the plate. Afterwards, use the LCM formula with the given values of density (蟻), specific heat capacity (cp), convective heat transfer coefficient (h), and temperatures (\(T_{\infty}\) and \(T_i\)) to find the time (t):\(t = \frac{\rho c_p \frac{V}{A}}{h}(1 - \frac{T(t) - T_{\infty}}{T_i - T_{\infty}})\). Solve the equation to find the time t required for the plate to reach a minimum temperature of 550掳C.

Step by step solution

01

1. Understand the given information and the type of problem

Here we have a problem of transient heat transfer in a 1D slab. The properties of the plate and the heating environment are given. The plate and the furnace temperatures are also provided.
02

2. Determine the Biot number to check the type of heat transfer condition

Calculate the Biot number (Bi) using the formula: \(Bi = \frac{hL_c}{k}\), where \(h = 500 \frac{W}{m^2K}\) - convective heat transfer coefficient, \(k = 45 \frac{W}{mK}\) - thermal conductivity, and, \(L_c = \frac{thickness}{2}= \frac{0.1}{2}m\) - geometric length (half-thickness). Bi < 0.1 is required for lumped capacitance method (LCM) to be valid.
03

3. Execute the Lumped Capacitance Method (LCM)

Now, we will implement the LCM method if the condition mentioned above is satisfied. For LCM, use the formula: \(t = \frac{\rho c_p V}{hA}(1 - \frac{T(t) - T_{\infty}}{T_i - T_{\infty}})\), where \(\rho = 7800 \frac{kg}{m^3}\) - density, \(c_p = 500 \frac{J}{kg K}\) - specific heat capacity, V - volume of the plate, A - surface area of the plate, \(T(t)\) - temperature of the plate at time t, \(T_{\infty} = 700^{\circ}C\) - furnace temperature, and, \(T_i = 300^{\circ}C\) - initial temperature. From the problem, the minimum temperature to be reached is \(T(t) = 550^{\circ}C\).
04

4. Calculate the volume and surface area of the plate

Calculate the plate's volume and surface area using the thickness and dimensions (assuming the plate is a rectangular shape with length L and width W): Volume: V = L * W * thickness = L * W * 0.1 m Surface area: A = 2 * (L * W + L * thickness + W * thickness) Since our goal is to determine the time t, we can simplify this further by calculating the ratio: \(\frac{V}{A} = \frac{L*W*0.1}{2*(L*W+L*0.1+W*0.1)} = \frac{0.1}{2*(0.1 + \frac{W}{L} * 0.1)}\)
05

5. Calculate the time for the plate to reach a minimum temperature of 550掳C

Now use the LCM formula to calculate the time required to reach the minimum temperature of 550掳C: \(t = \frac{7800 \cdot 500 \cdot \frac{0.1}{2*(0.1 + \frac{W}{L} * 0.1)}}{500 \cdot 2 \cdot \frac{0.1}{2}}(1 - \frac{550 - 700}{300 - 700})\) Solve the equation to find the time t. The solution for t will be the time it takes for the plate to reach a minimum temperature of 550掳C.

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Key Concepts

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

Biot Number
When studying the behavior of heat transfer in objects, the Biot number (Bi) is an important dimensionless parameter. It relates the rate of heat conduction within an object to the rate of convective heat transfer across the object's boundary to its surroundings.

The Biot number is defined mathematically as:
\[\[\begin{align*}Bi & = \frac{h L_c}{k},\end{align*}\]\]where
  • \( h \) is the convective heat transfer coefficient,
  • \( L_c \) is the characteristic length of the object, and
  • \( k \) is the thermal conductivity of the material.
The characteristic length is typically considered as the volume of the object divided by its surface area. A low Biot number (Bi < 0.1) suggests that the temperature within the object can reasonably be approximated to vary uniformly in space, making lumped system analysis suitable for analysis.
Lumped Capacitance Method
The lumped capacitance method (LCM) is a simplification of the heat transfer analysis used when there's a low Biot number. LCM assumes that the temperature distribution inside the body is uniform at any given time due to the rapid internal conduction compared to the rate of heat transfer at the boundary.

To apply LCM, we use the following formula to calculate the time for temperature changes: \[\[\begin{align*}t & = \frac{\rho c_p V}{hA}\left(1 - \frac{T(t) - T_{\infty}}{T_i - T_{\infty}}\right),\end{align*}\]\]where
  • \( \rho \) is the density of the material,
  • \( c_p \) is the specific heat capacity,
  • V is the volume of the object,
  • A is the surface area exposed to convection,
  • \( T(t) \) is the temperature at time t,
  • \( T_{\infty} \) is the ambient temperature,
  • and \( T_i \) is the initial temperature of the object.
LCM enables quick and easy calculations in situations where precise spatial temperature variations inside the object are not required.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient (h) represents the convective heat transfer per unit area per unit temperature difference between the surface and the fluid far away from the surface. It is a measure of how well heat is transferred from the object to the surrounding fluid, such as air or water. The coefficient's unit is watts per square meter-kelvin \( W/m^2K \).

The value of \( h \) is affected by various factors such as the velocity of the fluid, the viscosity, the thermal properties of the fluid, and the surface geometry of the object. In heat transfer problems, \( h \) is either given or estimated using empirical correlations or calculations based on fluid dynamics.
Thermal Conductivity
Thermal conductivity (k) is a material property that quantifies the ability of the material to conduct heat. It is defined as the amount of heat, in watts, that can be conducted through a one-meter thickness of the material, with a one square meter cross-sectional area, per degree Kelvin of temperature difference.<\br>
In mathematical terms, it can be given as: \[\[\begin{align*}q & = -k abla T,\end{align*}\]\]where <\br>
  • q represents the heat transfer per unit area,
  • \( abla T \) represents the temperature gradient.
The thermal conductivity of a material is affected by its composition, temperature, and physical state (solid, liquid, or gas). In the given exercise, the steel plate's high thermal conductivity facilitates the assumption of uniform temperature distribution, which is essential for the lumped capacitance method.

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Most popular questions from this chapter

As permanent space stations increase in size, there is an attendant increase in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necessary to transfer the dissipated heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (LDR). The heat is first transferred to a high vacuum oil, which is then injected into outer space as a stream of small droplets. The stream is allowed to traverse a distance \(L\), over which it cools by radiating energy to outer space at absolute zero temperature. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(\varepsilon=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{i}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {sur }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejected by each droplet?

Small spherical particles of diameter \(D=50 \mu \mathrm{m}\) contain a fluorescent material that, when irradiated with white light, emits at a wavelength corresponding to the material's temperature. Hence the color of the particle varies with its temperature. Because the small particles are neutrally buoyant in liquid water, a researcher wishes to use them to measure instantaneous local water temperatures in a turbulent flow by observing their emitted color. If the particles are characterized by a density, specific heat, and thermal conductivity of \(\rho=999 \mathrm{~kg} / \mathrm{m}^{3}\), \(k=1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively, determine the time constant of the particles. Hint: Since the particles travel with the flow, heat transfer between the particle and the fluid occurs by conduction. Assume lumped capacitance behavior.

A cold air chamber is proposed for quenching steel ball bearings of diameter \(D=0.2 \mathrm{~m}\) and initial temperature \(T_{i}=400^{\circ} \mathrm{C} .\) Air in the chamber is maintained at \(-15^{\circ} \mathrm{C}\) by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that \(70 \%\) of the initial thermal energy content of the ball above \(-15^{\circ} \mathrm{C}\) be removed. Radiation effects may be neglected, and the convection heat transfer coefficient within the chamber is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the residence time of the balls within the chamber, and recommend a drive velocity of the conveyor. The following properties may be used for the steel: \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=2 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and \(c=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Consider the fuel element of Example 5.11. Initially, the element is at a uniform temperature of \(250^{\circ} \mathrm{C}\) with no heat generation. Suddenly, the element is inserted into the reactor core, causing a uniform volumetric heat generation rate of \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\). The surfaces are convectively cooled with \(T_{\infty}=250^{\circ} \mathrm{C}\) and \(h=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the explicit method with a space increment of \(2 \mathrm{~mm}\), determine the temperature distribution \(1.5 \mathrm{~s}\) after the element is inserted into the core.

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.

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