/*! 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 30 A \(6 \mathrm{~mm}\)-O.D. tube i... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(6 \mathrm{~mm}\)-O.D. tube is to be insulated with an insulation of thermal conductivity \(0.08 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and a very low surface emittance. Heat loss is by natural convection, for which the heat transfer coefficient can be taken as \(\bar{h}_{c}=1.3(\Delta T / D)^{1 / 4} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) for \(\Delta T=T_{s}-T_{e}\) in kelvins and the diameter \(D\) in meters. Determine the critical radius of the insulation and the corresponding heat loss for a tube surface temperature of \(350 \mathrm{~K}\), and an ambient temperature of \(300 \mathrm{~K}\).

Short Answer

Expert verified
The critical radius of the insulation is calculated using the heat transfer formula, and the corresponding heat loss is determined by substituting this radius into the heat loss equation.

Step by step solution

01

Identify Given Values

The problem provides certain values that will be used in calculations: - Outer diameter of the tube, \( D = 6 \text{ mm} = 0.006 \text{ m} \).- Thermal conductivity of insulation, \( k = 0.08 \text{ W/mK} \).- Surface temperature, \( T_s = 350 \text{ K} \).- Ambient temperature, \( T_e = 300 \text{ K} \).- Heat transfer coefficient formula: \( \bar{h}_{c} = 1.3 (\Delta T / D)^{1/4} \text{ W/m}^2\text{K} \).From these, \( \Delta T = T_s - T_e = 350 - 300 = 50 \text{ K} \).
02

Determine the Critical Radius of Insulation

The critical radius \( r_c \) for a tube is calculated using the formula:\[ r_c = \frac{k}{h_c} \]First, calculate the convective heat transfer coefficient \( h_c \) using:\[ h_c = 1.3 \left( \frac{\Delta T}{D} \right)^{1/4} = 1.3 \left( \frac{50}{0.006} \right)^{1/4} \text{ W/m}^2\text{K}\]Calculate \( h_c \), then find \( r_c \): \[ r_c = \frac{0.08}{h_c}\]
03

Calculate Heat Loss Without Insulation

When the tube has no insulation, the heat loss per unit length \( q \) is given by:\[ q = 2\pi D \bar{h}_c \Delta T\]Substitute \( \bar{h}_c \), \( D \, = \, 0.006 \, \text{m} \), and \( \Delta T \, = \, 50 \, \text{K} \) into the equation to find the heat loss without insulation.
04

Calculate Heat Loss at Critical Radius

At the critical radius, the heat loss is maximized. Use the same formula as in Step 3:\[ q_{r_c} = 2\pi r_c h_c \Delta T\]where \( r_c \) is the critical radius calculated previously and \( h_c \) is the convective heat transfer coefficient. Use these to find the heat loss at the critical radius.

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Key Concepts

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

Thermal conductivity
Thermal conductivity is a fundamental property of a material which measures how well it can conduct heat. It is represented by the symbol \( k \) and given in units of \( \ ext{W/mK} \). The higher the thermal conductivity, the better the material is at conducting heat. In our exercise, the insulation around the tube has a thermal conductivity of \(0.08 \ ext{ W/mK}\). This relatively low value means the material is a good insulator, designed to slow the movement of heat from the inside of the tube to the outside environment.
When choosing insulating materials, thermal conductivity is a key factor to consider. Materials with low thermal conductivity are chosen when the goal is to retain heat, while those with high conductivity are used when the objective is to allow effective heat transfer. Knowing how to calculate and apply thermal conductivity allows engineers to design systems that manage heat effectively, optimizing for energy efficiency and cost.
Natural convection
Natural convection refers to the heat transfer process that occurs when fluid motion is generated naturally due to temperature differences, leading to heat transfer across the fluid's flow. In our exercise, natural convection plays a major role as it quantifies how heat is lost from the tube to the surrounding air. The convection heat transfer coefficient, \( h_c \), is used to describe this process in numbers.
In natural convection, the heat transfer is often calculated using empirical or semi-empirical correlations based on observed data, such as:
  • Surface temperature difference: The heat transfer coefficient formula involves \( \Delta T \), the difference between the surface temperature \( T_s \) and the external ambient temperature \( T_e \).
  • Tube diameter: The diameter \( D \) of the tube directly influences the value of the heat transfer coefficient \( h_c \). Larger diameters typically reduce the effect of convection.
This understanding helps in designing systems where controlling temperature through natural convection is essential, such as in refrigeration or heating systems.
Critical radius
The critical radius of insulation describes a fascinating point at which adding insulation to a tubular object increases instead of decreases heat loss. This counterintuitive concept arises due to the relation between thermal conductivity of insulation and the heat transfer coefficient of surrounding air or fluid.
The critical radius \( r_c \) is determined by the formula \( r_c = \frac{k}{h_c} \), where \( k \) is the thermal conductivity of the insulation and \( h_c \) is the heat transfer coefficient. When the radius of the insulation is equal to this critical value, the insulating effect is counteracted entirely by the increased surface area exposed for heat transfer, maximizing the heat loss rather than minimizing it.
  • Beyond the critical radius: Insulation becomes more effective, as expected, because the heat loss starts to decrease.
Understanding critical radius is essential for energy and thermal management, allowing for optimal and efficient usage of insulation materials in practical applications.
Heat loss calculation
Calculating heat loss in thermal systems involves determining the rate at which heat energy is transferred from a system to its surroundings. This is crucial for designing efficient insulation and heat management systems. For our tubular system, heat loss \( q \) without insulation is calculated using:
  • The formula \( q = 2\pi D \bar{h}_c \Delta T \), where \( D \) is the tube's diameter and \( \Delta T \) is the temperature difference.
  • Convective heat loss takes into account the surface area and the temperature gradient driving the flow.
At the critical radius, maximum heat loss occurs. The calculation involves substituting the critical radius \( r_c \) into the heat loss formula, revealing the interplay between insulation effectiveness and surface area exposure. By understanding and applying these calculations, designers can make informed decisions in controlling heat loss, improving energy efficiency, and protecting equipment from thermal stress.

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Most popular questions from this chapter

A thermal conductivity cell consists of concentric thin-walled copper tubes with an electrical heater inside the inner tube and is used to measure the conductivity of granular materials. The inner and outer radii of the annular gap are 2 and \(4 \mathrm{~cm}\). In a particular test the electrical power to the heater was \(10.6 \mathrm{~W}\) per meter length, and the inner and outer tube temperatures were measured to be \(321.4 \mathrm{~K}\) and \(312.7\) \(\mathrm{K}\), respectively. Calculate the thermal conductivity of the sample.

A \(2 \mathrm{~mm}\)-diameter electrical wire has a \(1 \mathrm{~mm}\)-thick electrical insulation with a thermal conductivity of \(0.12 \mathrm{~W} / \mathrm{m} \mathrm{K}\). The combined convection and radiation heat transfer coefficient on the outside of the insulation is \(12 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). (i) Would increasing the thickness of the insulation to \(3 \mathrm{~mm}\) increase or decrease the heat transfer? (ii) Would the presence of a contact resistance between the wire and insulation of \(5 \times 10^{-4}\left[\mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\right]^{-1}\) affect your conclusion?

Calcium silicate has replaced asbestos as the preferred insulation for steam lines in power plants. Consider a \(40 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 4 \mathrm{~cm}\)-wall-thickness steel steam line insulated with a \(12 \mathrm{~cm}\) thickness of calcium silicate. The insulation is protected from damage by an aluminum sheet lagging that is \(2.5 \mathrm{~mm}\) thick. The steam temperature is \(565^{\circ} \mathrm{C}\) and ambient air temperature in the power plant is \(26^{\circ} \mathrm{C}\). The inside convective resistance is negligible, the outside convective heat transfer coefficient can be taken as \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and the emittance of the aluminum is \(0.1\). Calculate the rate of heat loss per meter. The thermal conductivity of the pipe wall is \(40 \mathrm{~W} / \mathrm{m} \mathrm{K}\), and the table gives values for Calsilite calcium silicate insulation blocks. $$ \begin{array}{l|ccccc} \mathrm{T}, \mathrm{K} & 500 & 600 & 700 & 800 & 900 \\ \hline \mathrm{k}, \mathrm{W} / \mathrm{m} \mathrm{K} & 0.074 & 0.096 & 0.142 & 0.211 & 0.303 \end{array} $$

An explosive is to be stored in large slabs of thickness \(2 L\) clad on both sides with a protective sheath. The rate at which heat is generated within the explosive is temperature-dependent and can be approximated by the linear relation \(\dot{Q}_{v}^{\prime \prime \prime}=a+b\left(T-T_{e}\right)\), where \(T_{e}\) is the prevailing ambient air temperature. If the overall heat transfer coefficient between the slab surface and the ambient air is \(U\), show that the condition for an explosion is \(L=(k / b)^{1 / 2} \tan ^{-1}\left[U /(k b)^{1 / 2}\right]\). Determine the slab thickness if \(k=0.9 \mathrm{~W} / \mathrm{m} \mathrm{K}, U=0.20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}, a=60 \mathrm{~W} / \mathrm{m}^{3}\), \(b=6.0 \mathrm{~W} / \mathrm{m}^{3} \mathrm{~K}\).

The convective heat transfer coefficient around a cylinder held perpendicular to a flow varies in a complicated manner. A test cylinder to investigate this behavior consists of a \(0.001\) in-thick, \(12.7 \mathrm{~mm}\)-wide stainless steel heater ribbon (cut from shim stock) wound around a \(2 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness Teflon tube. A single thermocouple is located just underneath the ribbon and measures the local ribbon temperature \(T_{s}(\theta)\). The cylinder is installed in a wind tunnel, and a second thermocouple is used to measure the ambient air temperature \(T_{e}\). The power input to the heater is metered, from which the electrical heat generation per unit area \(\dot{Q} / A\) can be calculated ( \(A\) is the surface area of one side of the ribbon). As a first approximation, the local heat transfer coefficient \(h_{c}(\theta)\) can be obtained from $$ h_{c}(\theta)=\frac{\dot{Q} / A}{T_{s}(\theta)-T_{e}} $$ Hence, by rotating the cylinder with the power held constant, the variation of \(h_{c}\) can be obtained from the variation of \(T_{s}\) : where \(T_{s}(\theta)\) is low, \(h_{c}(\theta)\) is high, and vice versa. A typical variation of \(T_{s}(\theta)\) is shown in the graph. A problem with this technique is that conduction around the circumference of the tube causes the local heat flux \(q_{s}(\theta)\) to not exactly equal \(\dot{Q} / A\). (i) Derive a formula for \(h_{c}(\theta)\) that approximately accounts for circumferential conduction. (ii) The following table gives values of \(T_{s}(\theta)\) in a sector where circumferential conduction effects are expected to be large. Use these values together with \(\dot{Q} / A=5900 \mathrm{~W} / \mathrm{m}^{2}\) and \(T_{e}=25^{\circ} \mathrm{C}\) to estimate the conduction effect at \(\theta=110^{\circ}\). \begin{tabular}{cc} Angle (degrees) & \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline 100 & \(65.9\) \\ 110 & \(65.7\) \\ 120 & \(64.4\) \end{tabular} (iii) Comment on the design of the cylinder. Would a \(3 \mathrm{~mm}\)-thick brass tube, directly heated by an electric current, be a suitable alternative? Use \(k=15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the stainless steel and \(0.38 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for Teflon.

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