/*! 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 stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of \(36 \mathrm{~mm}\) and a wall thickness of \(2 \mathrm{~mm}\). The pharmaceutical and ambient air are at temperatures of \(6^{\circ} \mathrm{C}\) and \(23^{\circ} \mathrm{C}\), respectively, while the corresponding inner and outer convection coefficients are \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat gain per unit tube length? (b) What is the heat gain per unit length if a \(10-\mathrm{mm}\) thick layer of calcium silicate insulation \(\left(k_{\text {ins }}=\right.\) \(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tube?

Consider cylindrical and spherical shells with inner and outer surfaces at \(r_{1}\) and \(r_{2}\) maintained at uniform temperatures \(T_{s, 1}\) and \(T_{s, 2}\), respectively. If there is uniform heat generation within the shells, obtain expressions for the steady-state, one-dimensional radial distributions of the temperature, heat flux, and heat rate. Contrast your results with those summarized in Appendix C.

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a composite wall consisting of an exterior sheet of \(6.35-\mathrm{mm}\)-thick plain carbon steel, an intermediate layer of \(100-\mathrm{mm}\)-thick cork insulation, and an inner sheet of \(6.35\)-mm-thick aluminum alloy (2024). Adhesive interfaces between the insulation and the metallic strips are each characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are \(22^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\), respectively?

The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?
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