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Chapter 5: Problem 69

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 reich \(50^{\circ} \mathrm{C}\) at a distance of \(25 \mathrm{~mm}\) from the surface?

Short Answer

Expert verified
1630 seconds (about 27 minutes).

Step by step solution

01

Understand the Problem

We need to determine the time it takes for a point 25 mm within a steel slab to cool from 300掳C to 50掳C, given constant surface convection conditions with water at 25掳C.
02

Gather Known Values

- Initial temperature of the slab, \( T_i = 300 掳C \).- Surface temperature due to water cooling, \( T_s = 25 掳C \).- Target temperature, \( T = 50 掳C \).- Distance from the surface, \( x = 0.025 \) m.- Density, \( \rho = 7800 \) kg/m\(^3\).- Specific heat capacity, \( c = 480 \) J/kg\(\cdot\)K.- Thermal conductivity, \( k = 50 \) W/m\(\cdot\)K.
03

Identify the Cooling Process

Recognize the cooling process as transient heat conduction which is modeled as a one-dimensional heat conduction problem due to a constant boundary condition (temperature at the surface).
04

Use the Heat Equation

For a semi-infinite solid (since the slab is thick), the temperature at depth \( x \) can be estimated using the conduction equation for transient cooling:\[\frac{\theta}{\theta_i} = \text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right)\]where \( \theta = T - T_s \), \( \theta_i = T_i - T_s \), and \( \alpha = \frac{k}{\rho c} \) is thermal diffusivity.
05

Calculate Thermal Diffusivity

Thermal diffusivity is calculated as:\[\alpha = \frac{k}{\rho c} = \frac{50}{7800 \times 480} \approx 1.34 \times 10^{-5} \text{ m}^2/\text{s}\]
06

Calculate the Dimensionless Temperature

Calculate the dimensionless temperature ratio:\[\frac{\theta}{\theta_i} = \frac{50 - 25}{300 - 25} = \frac{25}{275} \approx 0.0909\]
07

Use the Error Function

Using error function tables or a calculator, find the value of \( \text{erf}^{-1}(0.0909) \), which corresponds to some value around \( 0.081 \).
08

Solve for Time

Using the equation from Step 4:\[0.081 = \frac{x}{2\sqrt{\alpha t}}\]Rearrange and solve for time \( t \):\[t = \left(\frac{x}{2 \times 0.081 \times \sqrt{\alpha}}\right)^2\]Substitute \( x = 0.025 \text{ m} \) and \( \alpha = 1.34 \times 10^{-5} \text{ m}^2/\text{s} \):\[t \approx \left(\frac{0.025}{2 \times 0.081 \times \sqrt{1.34 \times 10^{-5}}}\right)^2 \approx 1630 \text{ seconds}\]
09

Conclusion

Therefore, it takes approximately 1630 seconds, or about 27 minutes, to cool the slab to the desired temperature at 25 mm depth.

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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 refers to the change in temperature of a material over time. Unlike steady-state conduction, which assumes a constant temperature throughout the material, transient conduction accounts for time-dependent changes. This is crucial when dealing with materials that are subjected to variable temperatures, such as a steel slab cooled by water jets.
  • In transient conduction, the rate of temperature change depends on the material's properties and external conditions.
  • The slab in the problem starts uniformly at 300掳C and is intended to cool over time to a target temperature of 50掳C at a specific depth.
To model this cooling process accurately, the heat equation is used. This considers both the initial conditions of the slab and the constant temperature provided by the surrounding water, allowing for a precise prediction of temperature changes over time.
Thermal Diffusivity
Thermal diffusivity (\( \alpha \)) is a key property that impacts transient heat conduction. It's a measure of how quickly heat spreads through a material. The formula for thermal diffusivity is given by:\[ \alpha = \frac{k}{\rho c} \]where \( k \) is the thermal conductivity, \( \rho \) is the density, and \( c \) is the specific heat capacity.
  • A higher thermal diffusivity means the material can quickly adjust its temperature to match the surrounding conditions.
  • In the context of the steel slab, a diffusivity of approximately \( 1.34 \times 10^{-5} \text{ m}^2/\text{s} \) indicates how fast heat diffuses within the slab.
Understanding diffusivity helps in determining how temperature at a point distance from the surface will evolve, which is critical in predicting the cooling time.
Error Function
The error function (\( \text{erf} \)) is an important concept in solving problems involving transient heat conduction. It represents the probability of a random variable falling within a certain range and is used here to describe the temperature distribution in the steel slab as it cools.
  • The error function helps in finding the dimensionless temperature ratio, \( \frac{\theta}{\theta_i} \), crucial for the heat equation.
  • For practical computation, tables or digital tools are used to find values of \( \text{erf}^{-1} \).
By linking the physical distance and time to the error function, one can solve for the time it takes for a point within the slab to cool to the desired temperature. This function captures the non-linear nature of heat diffusion over time.
Specific Heat Capacity
Specific heat capacity (\( c \)) is a property of a material that indicates how much heat energy is required to raise the temperature of a given mass by one degree Celsius. For the steel slab exercise, the specific heat capacity is\( 480 \text{ J/kg} \cdot \text{K} \).
  • It determines how much energy is stored in the material and how it impacts temperature changes over time.
  • In the context of cooling, it affects how quickly the slab releases heat to its surroundings.
Specific heat capacity is integral to the calculation of thermal diffusivity, as it helps to define how energy is transferred within the slab. A material with a high specific heat capacity will take longer to cool down, as it requires more energy exchange to change temperature. This concept is critical in predicting the cooling performance in thermal analysis.

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Most popular questions from this chapter

Copper-coated, epoxy-filled fiberglass circuit boards are treated by heating a stack of them under high pressare as sown in the sketch. The perpose of the pressing-heuting operation is to cure the eposy that bonds the fiberglass sheets, imparting siffness to the boards. The stack. riferted to as a book, is comprised of 10 boards and 11 pressing plates, which prevent epoxy from flowing between the boards and impar a-smooth finish to the cured boards. In order to perform simplified thermal analyses, it is reasonable to approximule the book as having an effective themal conductivity \((k)\) and an effective thermal capacitance \(\left(\rho c_{p}\right)\). Calculate the effective propertics if each of the boards and plates has a tickness of \(2.36 \mathrm{~mm}\) and the following thermophysical properties: board (b) \(\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{\prime}, c_{p,}=1500\) \(\mathrm{W} / \mathrm{Kg} \cdot \mathrm{K}, k_{p}=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}_{;}\)plate \((p) \rho_{p}=8\left(000 \mathrm{~kg} / \mathrm{m}^{3}\right.\). \(c_{v}=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{p}=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Two large blocks of different materials, such as copper and concrete, have been sitting in a room \(\left(23^{\circ} \mathrm{C}\right)\) for a very lone time. Which of the two blocks, if either, will feel colder to the touch? Assume the blocks to be seminfinite solids and your hand to be at a temperaturt of \(37^{\circ} \mathrm{C}\).

Consider a thin electrical heater artached to a plate and hacked by insulation. Initially, the heater and plate are at the temperafure of the ambient air, \(T\) - Suddenly, the power to the heater is activated, yielding a constant heat fux \(q_{e}^{*}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T-x\) cocrdinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{z}^{\prime \prime}(L, t)\) as a function of time.

A support rod \(\left(\mathrm{k}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=4.0 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~mm}\) and length \(L=100 \mathrm{~mm}\) spans a channel whose walls are maintained at a temperature of \(T_{b}=300 \mathrm{~K}\). Suddenly, the rod is exposed to a cross flow of hot gaces for which \(T_{2}=600 \mathrm{~K}\) and \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The channel walls are cooled and remain at \(300 \mathrm{~K}\). (a) Using an appropriate numerical technique, determine the thermal response of the rod to the convective heating. Plot the midspan remperature as a function of elapsed time. Using an appropriate analytical model of the rod, determine the steady. state temperature distribution and compare the result with that obtained numerically for very long clapsed times. (b) After the rod has reached steady- state condisiss the flow of hot gases is suddenly termirated art the rod cools by free convection to ambient ans \(T_{w}=300 \mathrm{~K}\) and by radiafion exchange with kare surroundings at \(T_{\text {wr }}=300 \mathrm{~K}\). The free convetion coefficient can be expressed as \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=\mathrm{C}\) \(\Delta T^{k}\), where \(C=4.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{128}\) and \(n=0.18\) The emissivity of the rod is \(0.5\). Determire th subsequent thermal response of the rod. Pat th midspan temperature as a function of cooling tine. and determine the time required for the rod to raxt a safe-fo-fouch temperature of \(315 \mathrm{~K}\).

In a thin-slab, continuous casting process, molten steel leaves a mold with a thin solid shell, and the molien material solidifies as the slab is quenched by water jets en route to a section of rollers. Once fully solidified, the slab continues to cool as it is brought to an acceptable handling temperature. It is this portion of the process that is of interest. Consider a 200 - mm- thick solid slab of steel \((\rho=\) \(\left.7800 \mathrm{~kg} / \mathrm{m}^{3}, c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), initially at a uniform temperature of \(T_{i}=1400^{\circ} \mathrm{C}\). The slab is cooled at its top and bottom surfaces by water jets \(\left(T_{w}\right.\) \(\left.=50^{\circ} \mathrm{C}\right)\). which maintain an approximstely uniform convection coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. Using a finite-difference solution with a spoce increment of \(\Delta x=1 \mathrm{~mm}\), determine the time required to cool the surface of the slab to \(200^{\circ} \mathrm{C}\). What is the corresponding temperature at the midplane of the slab? If the slab moves at a speed of \(V=15 \mathrm{~mm} / \mathrm{s}\), what is the required length of the cooling section?
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