/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 132 One end of a stainless steel (AI... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(t_{1}\).

Short Answer

Expert verified
To find the time required for the midlength of the stainless steel rod to reach \(100^{\circ} \mathrm{C}\), we use the explicit finite-difference technique with a space increment of \(\Delta x = 0.016 ~\mathrm{m}\). After iterating, we determine the time required for the midlength of the rod to reach \(100^{\circ} \mathrm{C}\). Next, we compute the temperature distribution 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 \(50^{\circ} \mathrm{C}\), using \(\Delta x = 0.016 ~\mathrm{m}\) and \(\Delta t = 10 ~\mathrm{s}\). Finally, we plot the temperature distribution for \(t = 0 ~\mathrm{s}\), 200 s, 400 s, and \(t_{1}\).

Step by step solution

01

Diameter of the rod (D) \(10 ~\mathrm{mm}\) to meters

\(D = 10 \times 10^{-3} ~\mathrm{m}\)
02

Length of the rod (L) \(0.16 ~\mathrm{m}\)

\(L = 0.16 ~\mathrm{m}\)
03

Initial temperature of the rod (T_i) \(200^{\circ} \mathrm{C}\)

\(T_i = 200 ~\mathrm{K}\)
04

Ambient air temperature (T_a) \(25^{\circ} \mathrm{C}\)

\(T_a = 25 ~\mathrm{K}\)
05

Convection heat transfer coefficient (h) \(30 ~\mathrm{W}/\mathrm{m}^{2} \cdot \mathrm{K}\)

\(h = 30 ~\mathrm{W}/\mathrm{m}^{2} \cdot \mathrm{K}\) Step 2: Set up the explicit finite-difference technique
06

Space increment \(\Delta x = 0.016 ~\mathrm{m}\)

\(\Delta x = 0.016 ~\mathrm{m}\) Step 3: Iterate using the explicit finite-difference method until the midlength reaches the desired temperature
07

Calculate the time for the midlength of the rod to reach \(100^{\circ} \mathrm{C}\)

Iterate using the explicit finite-difference method and record the time required for the midlength of the rod to reach 100 degrees Celsius. Step 4: Compute the temperature distribution 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 \(50^{\circ} \mathrm{C}\)
08

Calculate the time for the midlength of the rod to reach \(50^{\circ} \mathrm{C}\) (t1)

Iterate using the explicit finite-difference method and record the time required for the midlength of the rod to reach 50 degrees Celsius (t1).
09

Compute the temperature distribution

With the given \(\Delta x = 0.016 ~\mathrm{m}\) and \(\Delta t = 10 ~\mathrm{s}\), iterate and compute the temperature distribution T(x,t) as the midlength of the rod reaches different temperatures. Step 5: Plot the temperature distribution for \(t=0 ~\mathrm{s}\), 200 s, 400 s, and \(t_{1}\)
10

Temperature distribution plot

Plot the temperature distribution by using the computed temperature distribution values at the given time points.

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Key Concepts

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

Convection Heat Transfer
Understanding convection heat transfer is essential for solving problems related to temperature changes in solids exposed to fluid environments. Convection is the mode of heat transfer that occurs between a solid surface and a moving fluid when they are at different temperatures. In our exercise, when the insulating sleeve is removed, the rod is exposed to air at a lower temperature, causing the rod to transfer heat to the air. The rod's surface experiences convective cooling due to the motion of the ambient air.

The rate at which convection occurs is governed by the convective heat transfer coefficient, denoted by 'h'. This coefficient is a measure of the convection heat transfer ability of a fluid in contact with a solid and is influenced by various factors including fluid velocity, viscosity, and surface roughness. In our case, the value of 'h' is given as 30 W/m^2∙K. A higher 'h' value would indicate a more efficient heat transfer, thus cooling the rod more quickly.

To calculate the heat lost from the rod through convection, we use Newton's law of cooling which states that the rate of heat loss is proportional to the temperature difference between the solid's surface and the fluid. Here, the rate of heat loss per unit surface area, 'Q', is given by the equation: \[ Q = h \times A \times (T_{\text{surface}} - T_{\text{fluid}}) \]

where 'A' represents the surface area of the rod, and the temperatures are those of the surface and the air, respectively. Understanding this concept is crucial for proceeding with explicit finite-difference calculations to predict temperature changes over time.
Temperature Distribution
The temperature distribution within a solid refers to how temperature varies across the material. Initially, the rod in the exercise has a uniform temperature. However, once exposed to cooler air, a temperature gradient develops along the rod's length due to heat being convected away from its surface. The exercise aims to determine how this temperature distribution changes over time, especially in the middle section of the rod.

In the step-by-step solution, an explicit finite-difference method is employed to estimate this change, allowing us to compute the temperature at discrete points along the rod's length and at successive time intervals. The finite-difference method divides the rod into small elements and uses their heat balance to find the temperatures. Typical outputs of these calculations would be a set of temperature profiles at each time step, showing how the rod cools.

Plotting the temperature distribution at various times gives a visual representation of the cooling process. For an accurate estimate, the space and time increments (\(\Delta x\) and \(\Delta t\)) should be chosen carefully. The exercise specifies computing the temperature distribution for various timesteps until the midlength temperature reaches a specified value, illustrating how the distribution evolves as the rod cools down.
Transient Heat Conduction
Transient heat conduction, also known as unsteady-state heat conduction, occurs when temperature within an object changes as a function of time. This is opposed to steady-state conduction, where temperature throughout the material remains constant over time. In the given exercise, we are dealing with a transient heat conduction problem; the temperature throughout the rod changes over time due to convection effects when the insulating sleeve is removed.

Solving transient heat conduction problems requires understanding how temperature propagates through a material due to thermal conductivity and how this is affected by external cooling or heating. The explicit finite-difference method is a technique used to solve such problems numerically. This method utilizes a time-marching procedure where the next time step's temperature distribution is calculated based on the previous step, using the thermal properties of the material and the boundary conditions. For our stainless steel rod, this involves setting an initial temperature, then updating the temperature of each segment based on the heat lost due to convection on the surface and the heat conducted from adjacent segments.

In summary, to solve for the time required for the midlength of the rod to reach a certain temperature, we must account for transient conduction within the rod as well as convective cooling at the surface. Through iterative calculations, we can model the change in temperature over time and accurately predict when certain temperature conditions are reached.

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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 \(800 \mathrm{~K}\) ?

A tile-iron consists of a massive plate maintained at \(150^{\circ} \mathrm{C}\) by an embedded electrical heater. The iron is placed in contact with a tile to soften the adhesive, allowing the tile to be easily lifted from the subflooring. The adhesive will soften sufficiently if heated above \(50^{\circ} \mathrm{C}\) for at least \(2 \mathrm{~min}\), but its temperature should not exceed \(120^{\circ} \mathrm{C}\) to avoid deterioration of the adhesive. Assume the tile and subfloor to have an initial temperature of \(25^{\circ} \mathrm{C}\) and to have equivalent thermophysical properties of \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=1.5 \times 10^{6}\) \(\mathrm{J} / \mathrm{m}^{3} \cdot \mathrm{K}\) Tile, 4-mm thickness Subflooring (a) How long will it take a worker using the tile-iron to lift a tile? Will the adhesive temperature exceed \(120^{\circ} \mathrm{C} ?\) (b) If the tile-iron has a square surface area \(254 \mathrm{~mm}\) to the side, how much energy has been removed from it during the time it has taken to lift the tile?

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

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.

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