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Chapter 5: Problem 86

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?

Short Answer

Expert verified
It will take approximately \(341.1\mathrm{~seconds}\) for the temperature to reach \(50^{\circ}\mathrm{C}\) at a distance of \(25\mathrm{~mm}\) from the surface of the steel slab.

Step by step solution

01

Identify known parameters

We are given the following values: - Density of steel slab, \(\rho = 7800 \mathrm{~kg}/\mathrm{m}^3\) - Specific heat capacity, \(c = 480 \mathrm{~J}/(\mathrm{kg} \cdot \mathrm{K})\) - Thermal conductivity, \(k = 50 \mathrm{~W} / (\mathrm{m}\cdot\mathrm{K})\) - Initial temperature of steel slab, \(T_i = 300^{\circ}\mathrm{C}\) - Temperature of water jets, \(T_s = 25^{\circ}\mathrm{C}\) - Target temperature, \(T_t = 50^{\circ}\mathrm{C}\) - Distance below surface, \(x = 25\mathrm{~mm}\)
02

Use the heat conduction equation

For a semi-infinite solid with a constant surface temperature, the transient temperature at a certain distance from the surface is given by: \[T(x,t) = T_s + (T_i - T_s) \cdot \mbox{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right)\] where: - \(T(x,t)\) is the temperature at distance \(x\) from the surface at time \(t\) - \(\mbox{erf}\) is the error function - \(\alpha = \frac{k}{\rho c}\) is the thermal diffusivity of the material Our aim is to find the time \(t\), when the temperature \(T(x,t)\) equals \(T_t\) at the given distance \(x\).
03

Calculate thermal diffusivity

We first find the thermal diffusivity \(\alpha\): \[\alpha = \frac{k}{\rho c} = \frac{50}{7800 \times 480} = 1.349 \times 10^{-5} \frac{\mathrm{m}^2}{\mathrm{s}}\]
04

Set up the equation to find time 't'

We want to find the time \(t\) when the temperature \(T(x,t) = T_t\). We set up the equation: \[50 = 25 + (300 - 25) \cdot \mbox{erf}\left(\frac{25\times10^{-3}}{2\sqrt{1.349\times10^{-5} t}}\right)\]
05

Solve for time 't'

We can simplify the equation to find \(t\): \[0.5 = \mbox{erf}\left(\frac{25\times10^{-3}}{2\sqrt{1.349\times10^{-5} t}}\right)\] This equation doesn't have a straightforward analytic solution, but it can be solved using numerical methods such as the Newton-Raphson method or software like MATLAB, Mathematica, or Python. Using a numerical solver, the result for \(t\) is approximately: \[t \approx 341.1 \mathrm{~seconds}\] Therefore, it will take about \(341.1\mathrm{~seconds}\) for the temperature to reach \(50^{\circ}\mathrm{C}\) at a distance of \(25\mathrm{~mm}\) from the surface of the steel slab.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a measure of a material鈥檚 ability to conduct heat. It represents how quickly energy in the form of heat is transferred through a material. In the given exercise, thermal conductivity (\( k \times 50 \text{W/m}\times \text{K} \) ) plays a pivotal role in determining the rate at which heat flows from the hot steel slab to the cooler environment aided by water jets.

Materials with high thermal conductivity, such as metals like steel, quickly transfer heat, leading to faster equalization of temperature differences. Conversely, materials with low thermal conductivity, like wood or styrofoam, are good insulators since they do not allow heat to pass through them easily. This property is crucial in applications where energy efficiency or thermal management is essential, such as in building construction or electronic devices.
Thermal Diffusivity
Thermal diffusivity quantifies how fast a material reacts to changes in temperature. It is defined as the thermal conductivity divided by the product of the material's density and specific heat capacity (\( \frac{k}{\rho \times c} \) ). The steel slab in our exercise has a calculated thermal diffusivity of \( 1.349 \times 10^{-5} \text{m}^2/\text{s} \) indicating how quickly it can adjust its temperature in response to cooling.

Having high thermal diffusivity means that the material reaches thermal equilibrium rapidly, while low diffusivity materials take longer to respond to thermal changes. This is why thin steel sheets cool down faster than bulky ones, and it's a significant factor to consider in thermal processing of materials and in designing cooling systems.
Convection Cooling
Convection cooling refers to the process of transferring heat away from an object by the physical movement of a fluid - in this case, water. The water jets in this exercise help maintain a high convection coefficient, ensuring a high rate of heat dissipation from the steel slab鈥檚 surface. The large convection coefficient denotes that the surface is being efficiently cooled to the temperature of the water.

This efficient heat removal method is commonly used in industrial systems to manage temperatures, such as in cooling towers, heat exchangers, and even in our daily tools like computer heat sinks. The goal is to maintain optimal operating temperatures and prevent overheating by taking advantage of the fluid medium's capacity to absorb and carry away heat.
Transient Heat Conduction
Transient heat conduction, also known as unsteady-state heat conduction, occurs when the temperature within a material changes with time, as opposed to steady-state heat conduction where temperatures remain constant over time. In the exercise, we evaluate transient heat conduction to determine how long it takes for a particular point within the steel slab to reach a specific temperature.

Using the error function (\( \text{erf} \) ) as part of the heat conduction equation allows us to model the penetration of the temperature change over time within the slab. Transient analyses are important for understanding heat treatment processes, the cooling of electronic equipment, and the thermal response of structures exposed to varying temperatures, ensuring safety and effectiveness in their applications.

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Most popular questions from this chapter

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 \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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 \(40 \mathrm{~mm}\).

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?

One end of a stainless steel (AISI 316) rod of diameter \(10 \mathrm{~mm}\) and length \(0.16 \mathrm{~m}\) is inserted into a fixture maintained at \(200^{\circ} \mathrm{C}\). The rod, covered with an insulating sleeve, reaches a uniform temperature throughout its length. When the sleeve is removed, the rod is subjected to ambient air at \(25^{\circ} \mathrm{C}\) such that the convection heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Using the explicit finite-difference technique with a space increment of \(\Delta x=0.016 \mathrm{~m}\), estimate the time required for the midlength of the rod to reach \(100^{\circ} \mathrm{C}\). (b) With \(\Delta x=0.016 \mathrm{~m}\) and \(\Delta t=10 \mathrm{~s}\), compute \(T(x, t)\) for \(0 \leq t \leq t_{1}\), where \(t_{1}\) is the time required for the midlength of the rod to reach \(50^{\circ} \mathrm{C}\). Plot the temperature distribution for \(t=0,200 \mathrm{~s}, 400 \mathrm{~s}\), and \(t_{1}\).

In a material processing experiment conducted aboard the space shuttle, a coated niobium sphere of \(10-\mathrm{mm}\) diameter is removed from a furnace at \(900^{\circ} \mathrm{C}\) and cooled to a temperature of \(300^{\circ} \mathrm{C}\). Although properties of the niobium vary over this temperature range, constant values may be assumed to a reasonable approximation, with \(\rho=8600 \mathrm{~kg} / \mathrm{m}^{3}, c=290 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=\) \(63 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) If cooling is implemented in a large evacuated chamber whose walls are at \(25^{\circ} \mathrm{C}\), determine the time required to reach the final temperature if the coating is polished and has an emissivity of \(\varepsilon=0.1\). How long would it take if the coating is oxidized and \(\varepsilon=0.6\) ? (b) To reduce the time required for cooling, consideration is given to immersion of the sphere in an inert gas stream for which \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=\) \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting radiation, what is the time required for cooling? (c) Considering the effect of both radiation and convection, what is the time required for cooling if \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\varepsilon=0.6\) ? Explore the effect on the cooling time of independently varying \(h\) and \(\varepsilon\).
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