/*! 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 130 A brass rod \(100 \mathrm{~mm}\)... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 130

A brass rod \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter extends horizontally from a casting at \(200^{\circ} \mathrm{C}\). The rod is in an air environment with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=30\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the temperature of the rod 25,50 , and \(100 \mathrm{~mm}\) from the casting?

Short Answer

Expert verified
Using the given information and following the steps, you can calculate the temperature of the rod at 25 mm, 50 mm, and 100 mm from the casting. First, find the cross-sectional area of the rod using the supplied diameter. Then, calculate the temperature difference at each distance from the casting. Next, determine the heat transfer at each distance using the convective heat transfer formula. Finally, calculate the temperature of the rod at each specified distance using the relation \(T_{\mathrm{Rod}}(x) = T_c - Q / hA\). Upon completing these calculations, you will be able to find the temperature at each respective distance.

Step by step solution

01

Calculate the cross-sectional area of the rod

First, we need to calculate the cross-sectional area of the rod (A). The rod has a circular cross-section, so we will use the formula for the area of a circle, \(A = \pi r^2\). The radius (r) of the rod can be found by dividing the diameter (D) by 2, i.e. \(r = \frac{D}{2}\). So, the cross-sectional area (A) can be calculated as follows: \[A = \pi \left(\dfrac{D}{2}\right)^2\]
02

Calculate the temperature difference at each distance

Next, we need to calculate the temperature difference at each distance (25 mm, 50 mm, and 100 mm) from the casting. To do this, we will use the formula: 鈭員 (x) = \(T_c - T_{\infty}\), where 'x' is the distance from the casting. Note that as the rod itself is extending between \(0 \mathrm{~mm}\) to \(100 \mathrm{~mm}\), keep in mind the temperature difference is not in the brass rod but between the rod and the air environment.
03

Calculate the heat transfer at each distance

We will now calculate the heat transfer (Q) at each distance (x) from the casting. For this, we will use the convective heat transfer formula: \(Q = hA\Delta T\), where 'h' is the convective heat transfer coefficient, 'A' is the cross-sectional area of the rod, and 鈭員 is the temperature difference calculated in Step 2.
04

Calculate the temperature at each distance

Now that we have calculated the heat transfer at each distance, we can calculate the temperature at each distance (x) from the casting. To do this, we will use the following relation: \(T_{\mathrm{Rod}}(x) = T_c - Q / hA\), where \(Q\) is the heat transfer and \(hA\) is the product of the convective heat transfer coefficient and the cross-sectional area of the rod. After performing the above calculations for each of the given distances (25 mm, 50 mm, and 100 mm), you will obtain the temperature of the rod at each specified distance from the casting.

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Key Concepts

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

Conductive Heat Transfer
When we talk about conductive heat transfer, we're referring to the process where heat moves through materials that are in direct contact. Imagine a brass rod that's heated at one end; the heat doesn't just stay put鈥攊t travels along the rod to the cooler parts. This movement is due to the energy transfer from the warmer molecules to the adjacent, cooler ones.

In the context of our brass rod from the textbook exercise, as the heat flows from the hot casting into the rod, it moves toward the cooler end by conduction. The thermal conductivity of the brass is a measure of how efficiently it can conduct heat. High thermal conductivity means heat will spread more quickly along the rod.
Convective Heat Transfer
Switching our focus to convective heat transfer, this occurs when a fluid, such as air or water, is involved in the heat transfer process. For our brass rod in air, the rod's surface heats the adjacent air, which then moves away and is replaced by cooler air, creating a current; this is how heat gets transferred from the rod to the environment.

The rate at which this happens is influenced by the convective heat transfer coefficient (h), which depends on the type of fluid and its properties, along with the flow conditions. Our exercise mentioned an h value of 30 W/m虏路K, suggesting the rate at which heat is carried away by the air from the rod's surface.
Temperature Gradient
The term temperature gradient relates to the change in temperature with distance within a substance. It's like having a hill鈥攖he steeper the hill, the more rapidly the altitude changes as you move along it. Similarly, a steep temperature gradient means a significant temperature change over a short distance within the material.

For our brass rod, the temperature gradient is steepest near the casting and decreases as you move further away. This gradient is a driving force for heat flow; the bigger the difference between the hot casting and the cooler air at any point on the rod, the more heat tends to flow at that point.
Thermal Conductivity
Lastly, we'll delve into thermal conductivity, a material property that's highly relevant in both conductive and convective heat transfer contexts. It's a measure of a material's ability to conduct heat; think of it as how good a material is at letting heat flow through it. Materials like brass, copper, and silver have high thermal conductivity, meaning they're excellent at transferring heat.

In our exercise, the thermal conductivity of brass affects how quickly heat spreads from the hot casting along the length of the rod. A higher conductivity would mean a more uniform temperature along the rod's length, and a lower one would result in a more pronounced temperature drop as you move away from the casting.

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Most popular questions from this chapter

A one-dimensional plane wall of thickness \(L\) is constructed of a solid material with a linear, nonuniform porosity distribution described by \(\varepsilon(x)=\varepsilon_{\max }(x / L)\). Plot the steady-state temperature distribution, \(T(x)\), for \(k_{s}=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad k_{f}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=1 \mathrm{~m}, \quad \varepsilon_{\max }=\) \(0.25, T(x=0)=30^{\circ} \mathrm{C}\) and \(q_{x}^{\prime \prime}=100 \mathrm{~W} / \mathrm{m}^{2}\) using the expression for the minimum effective thermal conductivity of a porous medium, the expression for the maximum effective thermal conductivity of a porous medium, Maxwell's expression, and for the case where \(k_{\mathrm{efl}}(x)=k_{s}\).

A particular thermal system involves three objects of fixed shape with conduction resistances of \(R_{1}=1 \mathrm{~K} / \mathrm{W}\), \(R_{2}=2 \mathrm{~K} / \mathrm{W}\) and \(R_{3}=4 \mathrm{~K} / \mathrm{W}\), respectively. An objective is to minimize the total thermal resistance \(R_{\text {tot }}\) associated with a combination of \(R_{1}, R_{2}\), and \(R_{3}\). The chief engineer is willing to invest limited funds to specify an alternative material for just one of the three objects; the alternative material will have a thermal conductivity that is twice its nominal value. Which object (1, 2, or 3 ) should be fabricated of the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\) ? Hint: Consider two cases, one for which the three thermal resistances are arranged in series, and the second for which the three resistances are arranged in parallel.

The cross section of a long cylindrical fuel element in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod, which is of diameter \(D=25 \mathrm{~mm}\) and is wrapped in a thin aluminum cladding. (a) It is proposed that, under steady-state conditions, the system operates with a generation rate of \(\dot{q}=\) \(7 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\) and cooling system characteristics of \(T_{\infty}=95^{\circ} \mathrm{C}\) and \(h=7000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Is this proposal satisfactory? (b) Explore the effect of variations in \(\dot{q}\) and \(h\) by plotting temperature distributions \(T(r)\) for a range of parameter values. Suggest an envelope of acceptable operating conditions.

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Work Problem \(3.15\) assuming surfaces parallel to the \(x\)-direction are adiabatic.
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