/*! 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 71 The wall of a spherical tank of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 71

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

Short Answer

Expert verified
The required thickness of urethane foam insulation to reduce the exterior temperature to \(40^\circ \mathrm{C}\) is \(3.6 \mathrm{~cm}\). The percentage reduction in heat rate achieved by using the insulation is approximately \(40.6\% \).

Step by step solution

01

Identify the thermal resistance formula

We can use the thermal resistance formula to find the relationship between the insulation thickness and temperature. The formula for thermal resistance, R, of a spherical shell is given by: \(R = \frac{r_2 - r_1}{4 \pi \cdot k \cdot r_1 \cdot r_2}\) where \(r_1\) is the inner radius of the sphere, \(r_2\) is the outer radius of the sphere, and \(k\) is the thermal conductivity of the insulation material. Urethane foam has a thermal conductivity of \(k = 0.029 \mathrm{~W/m \cdot K}\).
02

Calculate the heat transfer rate for the initial condition

We will now find the heat transfer rate for the initial condition, which is when the exterior temperature is \(40^\circ \mathrm{C}\). The heat transfer rate, \(q\), can be calculated using the formula: \(q = h \cdot A \cdot \Delta T\) where \(h\) is the convection coefficient, \(A\) is the surface area, and \(\Delta T\) is the difference in temperature between the wall surface and the ambient air. We are given the convection coefficient as \(h = 20 \mathrm{~W/m^2 \cdot K}\), and the temperature difference can be calculated as: \(\Delta T = 200 - 40 = 160 \) The surface area of a sphere can be calculated as: \(A = 4 \pi \cdot r_1^2\) Considering that diameter of the tank is \(1-m\), its radius would be: \(r_1 = \frac{1}{2} \mathrm{m}\) Plugging these values into the formula, we have: \(q = 20\cdot 4 \pi \cdot \left(\frac{1}{2}\right)^2 \cdot 160\) Calculating q, \(q = 25133 \mathrm{~W}\)
03

Determine the required insulation thickness

Now, we will use the thermal resistance formula to find the required insulation thickness. The equation we derived in step 1 can be rearranged to solve for the thickness \(t = r_2 - r_1\): \(t = \frac{R \cdot 4 \pi \cdot k \cdot r_1 \cdot r_2}{r_2 - r_1}\) We also know that: \(R = \frac{1}{h} \cdot \frac{T_\mathrm{exterior} - T_\mathrm{ambient}}{T_\mathrm{wall} - T_\mathrm{exterior}}\) Plugging in the values, we get: \(R = \frac{1}{20} \cdot \frac{40 - 25}{200 - 40}\) Calculating R, \(R = 0.0025 \mathrm{~m^2 \cdot K/W}\) Now, we plug in the values of \(R\), \(k\), \(r_1\), and \(r_2\) into the formula for \(t\), and solve for \(t\): \(t = \frac{0.0025 \cdot 4 \pi \cdot 0.029 \cdot \frac{1}{2} \cdot r_2}{r_2 - \frac{1}{2}}\) Now we have a non-linear equation in terms of \(r_2\). We can solve this equation using numerical methods, such as Newton's method or by using a solver tool: \(r_2 = 0.536 \mathrm{~m}\) The thickness of the insulation required is: \(t = r_2 - r_1 = 0.536 - 0.5 = 0.036\) So, a thickness of \(3.6 \mathrm{~cm}\) of urethane foam insulation is required to reduce the exterior temperature to \(40^\circ\mathrm{C}\).
04

Calculate the percentage reduction in heat rate

To determine the percentage reduction in heat rate, we need to first calculate the heat transfer rate after applying the insulation. The heat transfer rate for the insulated condition can be calculated with the same formula as before, but with the added insulation thermal resistance: \(q_\mathrm{insulated} = \frac{T_\mathrm{wall} - T_\mathrm{exterior}}{R_\mathrm{insulation} + \frac{1}{h}}\) \(R_\mathrm{insulation}\) can be calculated: \(R_\mathrm{insulation} = R - \frac{1}{h}\cdot \frac{T_\mathrm{exterior} - T_\mathrm{ambient}}{T_\mathrm{wall} - T_\mathrm{exterior}}\) Plugging in the values, we get: \(R_\mathrm{insulation} = 0.0025 - \frac{1}{20} \cdot \frac{40 - 25}{200 - 40}\) Calculating \(R_\mathrm{insulation}\), \(R_\mathrm{insulation} = 0.00125 \mathrm{~m^2 \cdot K/W}\) Now, we can find the heat transfer rate after insulation: \(q_\mathrm{insulated} = \frac{200 - 40}{0.00125 + \frac{1}{20}}\) Calculating \(q_\mathrm{insulated}\), \(q_\mathrm{insulated} = 14933 \mathrm{~W}\) The percentage reduction in heat rate can be found using the formula: \(\text{Percentage reduction} = \frac{q - q_\mathrm{insulated}}{q} \times 100\%\) Plugging in the values, \(\text{Percentage reduction} = \frac{25133 - 14933}{25133} \times 100\%\) Calculating the percentage reduction, \(\text{Percentage reduction} \approx 40.6\% \) Thus, the percentage reduction in heat rate achieved by using the insulation is approximately \(40.6\%\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Spherical Tank
A spherical tank is a type of container with a shape that resembles a sphere. It's commonly used in the storage of liquids and gases.
The geometry of a sphere is favorable for such uses because it minimizes surface area for a given volume, reducing material costs and heat loss.
  • A sphere's surface area, which influences heat transfer, is determined using the formula: \[ A = 4 \pi r^2 \]where \( A \) is the surface area and \( r \) is the radius.
  • In this problem, the spherical tank has a diameter of 1 meter, giving it a radius \( r = 0.5 \) meters.
  • Temperature management in spherical tanks is crucial, especially when containing exothermic reactions, which release heat as a result of chemical processes.
Thermal Resistance
Thermal resistance is an essential concept in heat transfer, representing a material's resistance to heat flow. It's analogous to electrical resistance in a circuit.
  • For a spherical shell, thermal resistance is calculated using: \[ R = \frac{r_2 - r_1}{4 \pi \cdot k \cdot r_1 \cdot r_2} \]where \( r_1 \) and \( r_2 \) are the inner and outer radii, and \( k \) is the thermal conductivity.
  • Thermal conductivity \( k \) is a material property indicating how well it conducts heat. For urethane foam, \( k = 0.029 \, \text{W/m}\cdot\text{K} \) suggests it is a good insulator.
Lower thermal resistance means better heat transfer. By increasing insulation, the thermal resistance increases, reducing heat flow and improving energy efficiency.
Exothermic Reaction
Exothermic reactions are chemical reactions that release heat as products are formed. In a spherical tank containing such a reaction, the heat generated contributes to an increase in temperature within the tank.
  • These reactions are significant in many industrial processes, such as chemical reactors and energy production.
  • The heat released by the reaction in the tank raises the need for proper temperature management to prevent overheating, which can damage the tank or alter reaction conditions.
  • Understanding the heat release in exothermic reactions is crucial for effective design and insulation of storage containers to maintain desired operational temperatures.
Insulation Thickness
Determining the correct insulation thickness is key to controlling heat loss or gain in a system.
The insulation needs to be thick enough to ensure that the outside temperature of the tank is at the desired level.
  • In the given problem, the insulation thickness required to reduce the external tank surface temperature from \( 200^\circ\text{C} \) to \( 40^\circ\text{C} \) is calculated by manipulating thermal resistance equations.
  • After calculations, the required insulation thickness is found to be \( 3.6\text{ cm} \).
  • Thicker insulation improves temperature regulation but can also increase material costs and affect design constraints.
  • The effectiveness of insulation is measured by the reduction in heat transfer, with significant energy saving in industrial applications.
Choosing the optimal insulation thickness balances cost, efficiency, and performance requirements of the system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?

A wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

The outer surface of a hollow sphere of radius \(r_{2}\) is subjected to a uniform heat flux \(q_{2}^{\prime \prime}\). The inner surface at \(r_{1}\) is held at a constant temperature \(T_{s, 1}\). (a) Develop an expression for the temperature distribution \(T(r)\) in the sphere wall in terms of \(q_{2}^{\prime \prime}, T_{s, 1}, r_{1}, r_{2}\), and the thermal conductivity of the wall material \(k\). (b) If the inner and outer tube radii are \(r_{1}=50 \mathrm{~mm}\) and \(r_{2}=100 \mathrm{~mm}\), what heat flux \(q_{2}^{\prime \prime}\) is required to maintain the outer surface at \(T_{s, 2}=50^{\circ} \mathrm{C}\), while the inner surface is at \(T_{s, 1}=20^{\circ} \mathrm{C}\) ? The thermal conductivity of the wall material is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Consider a composite wall that includes an 8-mm-thick hardwood siding, 40 -mm by 130 -mm hardwood studs on \(0.65-\mathrm{m}\) centers with glass fiber insulation (paper faced, \(28 \mathrm{~kg} / \mathrm{m}^{3}\) ), and a 12 -mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is \(2.5 \mathrm{~m}\) high by \(6.5 \mathrm{~m}\) wide (having 10 studs, each \(2.5 \mathrm{~m}\) high)? Assume surfaces normal to the \(x\)-direction are isothermal.
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Atoms and Radioactivity

Read Explanation

Circular Motion and Gravitation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Medical Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.