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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

Carbon steel (AISI 1010) shafts of 0.1-m diameter are heat treated in a gas- fired furnace whose gases are at \(1200 \mathrm{~K}\) and provide a convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the shafts enter the furnace at \(300 \mathrm{~K}\), how long must they remain in the furnace to achieve a centerline temperature of \(800 \mathrm{~K}\) ?

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a \(100-\mathrm{mm}\)-thick steel plate \(\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=550 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and is to be heated to a minimum temperature of \(550^{\circ} \mathrm{C}\). Heating is effected in a gas-fired furnace, where products of combustion at \(T_{\infty}=800^{\circ} \mathrm{C}\) maintain a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on both surfaces of the plate. How long should the plate be left in the furnace?

An electronic device, such as a power transistor mounted on a finned heat sink, can be modeled as a spatially isothermal object with internal heat generation and an external convection resistance. (a) Consider such a system of mass \(M\), specific heat \(c\), and surface area \(A_{s}\), which is initially in equilibrium with the environment at \(T_{\infty}\). Suddenly, the electronic device is energized such that a constant heat generation \(\dot{E}_{g}(\mathrm{~W})\) occurs. Show that the temperature response of the device is $$ \frac{\theta}{\theta_{i}}=\exp \left(-\frac{t}{R C}\right) $$ where \(\theta \equiv T-T(\infty)\) and \(T(\infty)\) is the steady-state temperature corresponding to \(t \rightarrow \infty ; \theta_{i}=T_{i}-T(\infty)\); \(T_{i}=\) initial temperature of device; \(R=\) thermal resistance \(1 / \bar{h} A_{s} ;\) and \(C=\) thermal capacitance \(M c\). (b) An electronic device, which generates \(60 \mathrm{~W}\) of heat, is mounted on an aluminum heat sink weighing \(0.31 \mathrm{~kg}\) and reaches a temperature of \(100^{\circ} \mathrm{C}\) in ambient air at \(20^{\circ} \mathrm{C}\) under steady-state conditions. If the device is initially at \(20^{\circ} \mathrm{C}\), what temperature will it reach \(5 \mathrm{~min}\) after the power is switched on?

Stainless steel (AISI 304) ball bearings, which have uniformly been heated to \(850^{\circ} \mathrm{C}\), are hardened by quenching them in an oil bath that is maintained at \(40^{\circ} \mathrm{C}\). The ball diameter is \(20 \mathrm{~mm}\), and the convection coefficient associated with the oil bath is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If quenching is to occur until the surface temperature of the balls reaches \(100^{\circ} \mathrm{C}\), how long must the balls be kept in the oil? What is the center temperature at the conclusion of the cooling period? (b) If 10,000 balls are to be quenched per hour, what is the rate at which energy must be removed by the oil bath cooling system in order to maintain its temperature at \(40^{\circ} \mathrm{C}\) ?

A thick steel slab \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, c=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is initially at \(300^{\circ} \mathrm{C}\) and is cooled by water jets impinging on one of its surfaces. The temperature of the water is \(25^{\circ} \mathrm{C}\), and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach \(50^{\circ} \mathrm{C}\) at a distance of \(25 \mathrm{~mm}\) from the surface?

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