/*! 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 154 A 3-m-diameter spherical tank co... [FREE SOLUTION] | 91Ó°ÊÓ

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A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.

Short Answer

Expert verified
Answer: The rate of heat transfer from the tank to the ground is approximately 665.7 W.

Step by step solution

01

Identify the dimensions of the tank and ground

The diameter of the spherical tank is given as 3 meters. The radius of the tank can be calculated as half of the diameter, which is 1.5 meters. The distance between the top surface of the tank and the ground surface is given as 4 meters.
02

Calculate the conductive thermal resistance

The heat transfer through the ground can be modeled as a conductive thermal resistance. The conductive thermal resistance R_cond can be calculated as: \(R_\mathrm{cond} = \dfrac{r_\mathrm{outer}}{4 \pi \cdot k \cdot r_\mathrm{inner}}\) where r_inner is the inner radius of the tank, r_outer is the outer radius of the ground covering the tank, and k is the thermal conductivity of the ground. In this case, r_inner = 1.5 m, r_outer = 1.5 m + 4 m = 5.5 m, and k = 1.4 W/(m·K). Substituting these values, we get: \(R_\mathrm{cond} = \dfrac{5.5}{4 \pi \cdot 1.4 \cdot 1.5} = 0.1879 \, \mathrm{K} / \mathrm{W}\)
03

Determine temperature difference

The temperature difference between the surface of the tank (T_tank) and the ground (T_ground) is given as: \(\Delta T = T_\mathrm{tank} - T_\mathrm{ground}\) Substituting the given values, we get: \(\Delta T = 140 - 15 = 125^{\circ} \mathrm{C}\)
04

Calculate the rate of heat transfer

The rate of heat transfer (Q) can be determined using the overall thermal resistance (R_cond) and the temperature difference. The formula is as follows: \(Q = \dfrac{\Delta T}{R_\mathrm{cond}}\) Substituting the values, we get: \(Q = \dfrac{125}{0.1879} = 665.7 \, \mathrm{W}\) Therefore, the rate of heat transfer from the tank to the ground is approximately 665.7 W.

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Key Concepts

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

Conductive Thermal Resistance
Conductive thermal resistance is a key concept when analyzing heat transfer, especially through solid materials. It describes how resistant a material is to the flow of heat. Simply put, it's how hard it is for heat to pass through. The lower the thermal resistance, the better the material conducts heat.
To calculate the conductive thermal resistance, we use the formula:
  • \(R_\mathrm{cond} = \dfrac{r_\mathrm{outer}}{4 \pi \cdot k \cdot r_\mathrm{inner}}\)
Here, \(r_\mathrm{outer}\) and \(r_\mathrm{inner}\) are the outer and inner radii, respectively, and \(k\) represents the thermal conductivity. The units of this measure are typically \(\mathrm{K/W}\).
This resistance plays a crucial role in determining how efficiently heat can be transferred from the warmer body to the cooler surroundings. Understanding this concept is essential for designing systems that need to manage heat flow efficiently.
Spherical Tank
When discussing heat transfer, the geometry of the object, such as a spherical tank, is significant. A spherical shape is efficient for containing materials like gases and liquids. It evenly distributes stress and minimizes the surface area for a given volume, which can affect heat transfer dynamics.
The spherical tank in our problem has a defined diameter of 3 meters, which is straightforward to calculate the radius from. The radius is key in determining the overall thermal resistance, as seen in the previous section. For this particular example, the spherical tank is buried underground to safeguard its contents and manage the thermal interactions with the environment. This configuration affects how heat is conducted through the earth, influencing the rate of heat transfer.
Temperature Difference
Understanding temperature difference, often denoted as \(\Delta T\), is essential in calculating heat transfer. It reflects how much warmer one body is compared to another. This difference is a driving force behind the heat flow. Heat naturally moves from a high temperature to a low temperature region, aiming to reach thermal equilibrium.
For our spherical tank example, the temperature difference is quite significant. The tank's surface is much hotter at \(140^\circ \mathrm{C}\), whereas the ground surface is cooler at \(15^\circ \mathrm{C}\). Thus, the calculated \(\Delta T\) is \(125^\circ \mathrm{C}\).
This large temperature gradient facilitates a higher rate of heat transfer, and hence, it is important in determining the amount of heat exchanged with the surroundings efficiently.
Thermal Conductivity
Thermal conductivity is a material's inherent ability to conduct heat. A higher thermal conductivity means the material can transfer heat more efficiently. It is a constant that varies among materials and influences how thermal energy moves through substances.
Defined by the symbol \(k\), its units are \(\mathrm{W/m \cdot K}\). In the context of our problem, the soil surrounding the spherical tank has a thermal conductivity \(k = 1.4\, \mathrm{W/m \cdot K}\), meaning it's not as conductive as metals, but adequate enough for typical ground conditions.
Thermal conductivity, combined with the specific geometric orientation of materials, governs how quickly heat can pass through a layer, such as the earth around our buried tank. By understanding and calculating thermal conductivity, we can predict heat transfer rates more accurately and design better for thermal management.

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Most popular questions from this chapter

The \(700 \mathrm{~m}^{2}\) ceiling of a building has a thermal resistance of \(0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is \(-10^{\circ} \mathrm{C}\) and the interior is at \(20^{\circ} \mathrm{C}\) is (a) \(23.1 \mathrm{~kW} \quad\) (b) \(40.4 \mathrm{~kW}\) (c) \(55.6 \mathrm{~kW}\) (d) \(68.1 \mathrm{~kW}\) (e) \(88.6 \mathrm{~kW}\)

Consider a short cylinder whose top and bottom surfaces are insulated. The cylinder is initially at a uniform temperature \(T_{i}\) and is subjected to convection from its side surface to a medium at temperature \(T_{\infty}\), with a heat transfer coefficient of \(h\). Is the heat transfer in this short cylinder one- or twodimensional? Explain.

Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.04 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-cm-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

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