/*! 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 90 Exposure to high concentration o... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 90

Exposure to high concentration of gaseous ammonia can cause lung damage. To prevent gaseous ammonia from leaking out, ammonia is transported in its liquid state through a pipe \(\left(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}\right.\), \(D_{o}=4 \mathrm{~cm}\), and \(L=10 \mathrm{~m}\) ). Since liquid ammonia has a normal boiling point of \(-33.3^{\circ} \mathrm{C}\), the pipe needs to be properly insulated to prevent the surrounding heat from causing the ammonia to boil. The pipe is situated in a laboratory, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficients of the liquid hydrogen and the ambient air are \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine the insulation thickness for the pipe using a material with \(k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to keep the liquid ammonia flowing at an average temperature of \(-35^{\circ} \mathrm{C}\), while maintaining the insulated pipe outer surface temperature at \(10^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: To determine the required insulation thickness, we need to analyze the heat transfer for the ammonia flowing through the insulated pipe and use the given parameters and criteria. After calculating the heat transfer coefficient, the mechanical energy balance, and the heat flow, we can use the following equation to find the insulation thickness: $$ L_\text{ins} = \frac{k_{ins} A_\text{ins} \Delta T}{q} $$ By substituting the known values and solving for \(L_\text{ins}\), we can obtain the required insulation thickness to maintain the desired temperature conditions.

Step by step solution

01

Calculate the heat transfer coefficient for the pipe

Given the parameters for the pipe, we can calculate the overall heat transfer coefficient (U) for the pipe following the expression: $$ \frac{1}{U A}= \frac{\ln{(D_{o}/D_{i})}}{2 \pi k L}+\frac{1}{h_{i} A_{i}}+\frac{1}{h_{o} A_{o}} $$ Where: \(D_{o}\) = outer diameter (4 cm) \(D_{i}\) = inner diameter (2.5 cm) \(k\) = pipe's thermal conductivity (25 W/(m路K)) \(L\) = pipe's length (10 m) \(h_{i}\) = heat transfer coefficient for ammonia (100 W/(m虏路K)) \(h_{o}\) = heat transfer coefficient for ambient air (20 W/(m虏路K)) In order to calculate \(U\), we also need the inner area \(A_{i}\) and outer area \(A_{o}\), which are given by: $$ A_{i}= \pi D_{i} L \\ A_{o}= \pi D_{o} L $$
02

Calculate the mechanical energy balance

Considering the mechanical energy balance equation: $$ \underbrace{q}_{\text { heat flow }}=U A \Delta T $$ We want to keep the liquid ammonia at an average temperature of -35掳C, while maintaining the insulated pipe outer surface temperature at 10掳C. Therefore, $$ \Delta T = 10^{\circ} \mathrm{C} - (-35^{\circ} \mathrm{C}) = 45 K $$
03

Determine the required insulation thickness

Now, we can use the heat flow equation to determine the required insulation thickness (thickness of the insulation layer). We know the insulation material's thermal conductivity (\(k_{ins}\)) is 0.75 W/(m路K). To find the thickness, we can use the following relationship between heat flow through the insulation and the temperature difference across it: $$ q = k_{ins} A_\text{ins} \frac{\Delta T}{L_\text{ins}} $$ Where: \(A_\text{ins}\) = effective heat transfer area of the insulation \(L_\text{ins}\) = insulation thickness Note that the heat flow \(q\) is the same through the pipe and the insulation layers. We can relate the insulation thickness to the pipe dimensions, using the area of the insulation (assuming cylindrical): $$ A_\text{ins} = 2 \pi (r_\text{ins} - r_{o}) L $$ Where: \(r_\text{ins}\) = insulation layer's radius \(r_{o}\) = pipe outer radius Since we already have the heat flow \(q\), we can use this along with the given parameters to find the insulation thickness (\(L_\text{ins}\)): $$ L_\text{ins} = \frac{k_{ins} A_\text{ins} \Delta T}{q} $$ Calculating all the needed parameters, we can find the required insulation thickness to maintain the desired temperature conditions.

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Key Concepts

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

Insulation Thickness Calculation
Insulation thickness plays a crucial role in preventing heat flow into sensitive or volatile materials. In this scenario, the pipe carrying liquid ammonia must be insulated effectively to avoid boiling due to ambient temperature. The ultimate goal is to calculate the thickness of insulation required to achieve this.When calculating insulation thickness, we need to consider:
  • The thermal conductivity of the insulation material.
  • The temperature difference between the pipe surface and the ambient air.
  • Any existing convective heat transfers.
These factors are incorporated into a formula, which relies on understanding the interplay between the pipe's dimensions and the surrounding conditions. The heat through the insulation layer and pipe can be calculated using the heat flow formula. This involves knowing the inner and outer radii of the pipe, as well as the insulated pipe's effective heat transfer area. By determining parameters like the effective conductivity (\(k_{ins}\)) and insulation thickness (\(L_{ins}\)), engineers can establish the necessary thickness of insulation.
Thermal Conductivity
Thermal conductivity (\(k\)) is a material's intrinsic ability to conduct heat. It's a key factor when dealing with insulation, as it determines how heat moves through a material. For the given exercise:
  • The pipe itself has a high thermal conductivity (\(25 \, \text{W} / \text{m} \cdot \text{K}\)), meaning it easily allows heat to pass through.
  • The insulation used has a much lower thermal conductivity, (\(0.75 \, \text{W} / \text{m} \cdot \text{K}\)), reflecting its efficiency in reducing heat flow.
Understanding the comparison between these values is essential. A higher thermal conductivity implies more heat transfer, requiring either more insulation or materials with lower thermal conductivities to mitigate heat transfer.Engineers use this property to choose materials that properly balance cost, insulation effectiveness, and material thickness for optimal thermal management.
Convective Heat Transfer Coefficient
Convective heat transfer coefficient (\(h\)) represents how effectively heat is transferred between a solid surface and a fluid interacting with it. It varies with fluid velocity, the nature of the fluid, and the surface.In our exercise, we have:
  • Internal convective heat transfer coefficient for ammonia (\(100 \text{ W}/\text{m}^{2}\cdot \text{K}\)).
  • External convective heat transfer coefficient for air (\(20 \text{ W}/\text{m}^{2}\cdot \text{K}\)).
These coefficients guide how much heat can enter or exit the pipe. Higher coefficients mean more heat exchange between the mediums, affecting the insulation calculations as these additional fluxes must be balanced to maintain the desired conditions within the insulated pipe.By optimizing the coefficients through design or selecting specific materials and coverings, efficient thermal management is achieved. Understanding these coefficients allows for more accurate determination of the necessary insulation and system design to control heat transfer effectively.

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Most popular questions from this chapter

Two 3-m-long and \(0.4-\mathrm{cm}\)-thick cast iron \((k=\) \(52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) steam pipes of outer diameter \(10 \mathrm{~cm}\) are connected to each other through two 1 -cm-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

A hot plane surface at \(100^{\circ} \mathrm{C}\) is exposed to air at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of \(0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is (a) \(0.1 \mathrm{~cm}\) (b) \(0.5 \mathrm{~cm}\) (c) \(1.0 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(5 \mathrm{~cm}\)

Consider two identical people each generating \(60 \mathrm{~W}\) of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1 -mm-thick leather \((k=\) \(0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) that covers half of the body while the second one is wearing clothes made of 1 -mm-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that covers the body completely. The ambient air is at \(30^{\circ} \mathrm{C}\), the heat transfer coefficient at the outer surface is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the inner surface temperature of the clothes can be taken to be \(32^{\circ} \mathrm{C}\). Treating the body of each person as a 25 -cm-diameter, \(1.7-\mathrm{m}\)-long cylinder, determine the fractions of heat lost from each person by perspiration.

Obtain a relation for the fin efficiency for a fin of constant cross-sectional area \(A_{c}\), perimeter \(p\), length \(L\), and thermal conductivity \(k\) exposed to convection to a medium at \(T_{\infty}\) with a heat transfer coefficient \(h\). Assume the fins are sufficiently long so that the temperature of the fin at the tip is nearly \(T_{\infty}\). Take the temperature of the fin at the base to be \(T_{b}\) and neglect heat transfer from the fin tips. Simplify the relation for \((a)\) a circular fin of diameter \(D\) and \((b)\) rectangular fins of thickness \(t\).

Hot water at an average temperature of \(70^{\circ} \mathrm{C}\) is flowing through a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat transfer coefficient at the inner surface of the pipe is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by \(3^{\circ} \mathrm{C}\) as it passes through the basement.
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