/*! 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 60 An uninsulated, thin-walled pipe... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 60

An uninsulated, thin-walled pipe of \(100-\mathrm{mm}\) diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of \(-15^{\circ} \mathrm{C}\) and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is \(3^{\circ} \mathrm{C}\) and a convection coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained at the inner surface of the ice, which is at \(0^{\circ} \mathrm{C}\), what is the thickness of the ice layer?

Short Answer

Expert verified
The thickness of the ice layer formed on the inner surface of the pipe is approximately 1.12 mm.

Step by step solution

01

Identify Known Variables

We are given the following information: - Pipe diameter: \(D = 100~\mbox{mm}\) - Pipe wall temperature: \(T_{w} = -15^{\circ}\mbox{C}\) - Mean water temperature: \(T_{m} = 3^{\circ}\mbox{C}\) - Convection coefficient: \(h = 2000~\mbox{W/m}^2\cdot\mbox{K}\) - Temperature at inner surface of ice: \(T_{i} = 0^{\circ}\mbox{C}\)
02

Write down the heat transfer equation

We will use the equation of heat transfer by conduction through the pipe wall to the ice layer, and by convection inside the pipe due to the water. For steady-state heat transfer, the heat transfer by conduction through the ice layer should be equal to the heat transfer by convection to the water. The equation for the heat transfer by conduction through a cylindrical surface is given by: \[q_{cond} = kA\frac{\Delta T}{\Delta x}\] Where: - \(q_{cond}\) is the heat transfer by conduction (W) - \(k\) is the thermal conductivity of the material (W/m路K) - \(A\) is the surface area for heat transfer (m虏) - \(\Delta T\) is the temperature difference across the material (K) - \(\Delta x\) is the thickness of the material (m) The equation for the heat transfer by convection for a cylindrical surface is given by: \[q_{conv} = hA\Delta T\] Where: - \(q_{conv}\) is the heat transfer by convection (W) - \(h\) is the convection coefficient (W/m虏路K) - \(A\) is the surface area for heat transfer (m虏) - \(\Delta T\) is the temperature difference between the inner surface and the surrounding fluid (K)
03

Calculate the surface area

To proceed, we first need to calculate the surface area of the inner surface of the ice layer where the convection is occurring. The surface area for a cylinder is given by: \[A = 2\pi rL\] Where: - \(r\) is the radius of the cylinder (m) - \(L\) is the length of the cylinder (m) We are interested in the surface area per unit length, so we can express it as: \[A = 2\pi r\] With r being the radius to the inner surface of the ice. We have the diameter of the pipe, so we can calculate its radius as follows: \[r_{pipe} = \frac{D}{2} = \frac{100~\mbox{mm}}{2} = 50~\mbox{mm} = 0.05~\mbox{m}\] Let the thickness of the ice layer be \(x\), then the radius to the inner surface of the ice will be \(r_{ice} = r_{pipe} - x\).
04

Equate the conduction and convection equations

At steady-state, the heat transfer by conduction through the ice layer should be equal to the heat transfer by convection to the water: \[q_{cond} = q_{conv}\] Substitute the equation for conduction and convection: \[kA\frac{\Delta T}{\Delta x} = hA\Delta T\] The surface area and temperature difference will be the same for both sides of the equation, so we can divide both sides of the equation by \(A\Delta T\) to simplify: \[\frac{k}{\Delta x} = h\] We need to solve the equation for the ice thickness, \(\Delta x\).
05

Solve for the thickness of the ice layer

Rearrange the equation to solve for \(\Delta x\): \[\Delta x = \frac{k}{h}\] We know the convection coefficient \(h\), but we need to find the thermal conductivity of the ice (k). According to the literature, the thermal conductivity of ice is around \(k = 2.24~\mbox{W/m} \cdot \mbox{K}\). Now substitute the values of \(k\) and \(h\) in the equation: \[\Delta x = \frac{2.24~\mbox{W/m} \cdot \mbox{K}}{2000~\mbox{W/m}^2 \cdot \mbox{K}} = 0.00112~\mbox{m}\] So, the thickness of the ice layer is: \[\Delta x = 1.12~\mbox{mm}\] As a result, the thickness of the ice layer formed on the inner surface of the pipe is approximately 1.12 mm.

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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 measure of a material's ability to conduct heat. It is a crucial factor in heat transfer processes, especially in situations involving conduction, such as heat passing through a solid material.
It is denoted by the symbol \( k \) and typically expressed in units of \( ext{W/m} \cdot \text{K} \). This unit tells us how much heat (in watts) is conducted through a material with a thickness of one meter when there is a temperature difference of one Kelvin across it.
Some key factors affecting thermal conductivity include:
  • Material Composition: Metals generally have high thermal conductivity, whereas materials like wood or polystyrene have low thermal conductivity.
  • Temperature: In most solids, thermal conductivity changes with temperature, often decreasing as temperature increases.
This concept is critical in determining how efficient a material is at conducting heat. In our exercise, the thermal conductivity of ice is needed to calculate the thickness of the ice formed around the water pipe. Understanding this helps in solving scenarios where heat needs to be efficiently managed across materials.
Convection Coefficient
The convection coefficient, often symbolized as \( h \), represents the thermal exchange between a solid surface and a fluid when they are in contact. It quantifies how effectively heat is transferred between the surface and the fluid due to convection.
Measured in \( ext{W/m}^2 \cdot \text{K} \), the convection coefficient's value depends on several factors:
  • Fluid Velocity: Higher velocity increases the convection coefficient because it enhances the heat-carrying capacity of the fluid.
  • Surface Roughness: Rough surfaces create turbulence, which can increase the convection coefficient.
  • Fluid Properties: Viscosity, density, and specific heat of the fluid play roles in defining the value of \( h \).
In the given problem, a convection coefficient of \( 2000 \, \text{W/m}^2 \cdot \text{K} \) is used for the water inside the pipe and the adjoining ice layer. It reflects the efficiency of heat transfer taking place between the cold ice surface and the flowing water, allowing for the calculation of ice thickness in a consistent manner.
Steady-State Heat Transfer
Steady-state heat transfer means that the amount of heat entering a system is equal to the amount of heat leaving it, so the temperature at any point in the system does not change with time. This concept simplifies the mathematical model of heat transfer considerably because it assumes constant temperatures, so we can focus on spatial temperature differences.
This assumption allows for using the simplified form of heat equations, where temporal derivatives drop out of consideration. For cylindrical heat conduction, such as in the small pipe exercise:
  • The heat conducted through the pipe walls is equal to the heat convected into the water.
  • By equating conduction and convection equations, we can find steady solutions like the ice thickness in the problem.
Applying the steady-state assumption lets us draw meaningful conclusions quickly. In this scenario, it helps us ensure that the temperature differential across the ice layer is constant over time, allowing the use of simple ratio calculations to determine ice thickness from given parameters like thermal conductivity and convection coefficient.

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Most popular questions from this chapter

A technique for measuring convection heat transfer coefficients involves bonding one surface of a thin metallic foil to an insulating material and exposing the other surface to the fluid flow conditions of interest. By passing an electric current through the foil, heat is dissipated uniformly within the foil and the corresponding flux, \(P_{\text {elec }}^{\prime \prime}\), may be inferred from related voltage and current measurements. If the insulation thickness \(L\) and thermal conductivity \(k\) are known and the fluid, foil, and insulation temperatures \(\left(T_{\infty}, T_{s}, T_{b}\right)\) are measured, the convection coefficient may be determined. Consider conditions for which \(T_{\infty}=T_{b}=25^{\circ} \mathrm{C}, P_{\text {elec }}^{\prime \prime}=2000\) \(\mathrm{W} / \mathrm{m}^{2}, L=10 \mathrm{~mm}\), and \(k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) With water flow over the surface, the foil temperature measurement yields \(T_{s}=27^{\circ} \mathrm{C}\). Determine the convection coefficient. What error would be incurred by assuming all of the dissipated power to be transferred to the water by convection? (b) If, instead, air flows over the surface and the temperature measurement yields \(T_{s}=125^{\circ} \mathrm{C}\), what is the convection coefficient? The foil has an emissivity of \(0.15\) and is exposed to large surroundings at \(25^{\circ} \mathrm{C}\). What error would be incurred by assuming all of the dissipated power to be transferred to the air by convection? (c) Typically, heat flux gages are operated at a fixed temperature \(\left(T_{s}\right)\), in which case the power dissipation provides a direct measure of the convection coefficient. For \(T_{s}=27^{\circ} \mathrm{C}\), plot \(P_{\text {elec }}^{\prime \prime}\) as a function of \(h_{o}\) for \(10 \leq h_{o} \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What effect does \(h_{o}\) have on the error associated with neglecting conduction through the insulation?

A bonding operation utilizes a laser to provide a constant heat flux, \(q_{o}^{\prime \prime}\), across the top surface of a thin adhesivebacked, plastic film to be affixed to a metal strip as shown in the sketch. The metal strip has a thickness \(d=1.25 \mathrm{~mm}\), and its width is large relative to that of the film. The thermophysical properties of the strip are \(\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=435 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal resistance of the plastic film of width \(w_{1}=40 \mathrm{~mm}\) is negligible. The upper and lower surfaces of the strip (including the plastic film) experience convection with air at \(25^{\circ} \mathrm{C}\) and a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The strip and film are very long in the direction normal to the page. Assume the edges of the metal strip are at the air temperature \(\left(T_{\infty}\right)\). (a) Derive an expression for the temperature distribution in the portion of the steel strip with the plastic film \(\left(-w_{1} / 2 \leq x \leq+w_{1} / 2\right)\). (b) If the heat flux provided by the laser is 10,000 \(\mathrm{W} / \mathrm{m}^{2}\), determine the temperature of the plastic film at the center \((x=0)\) and its edges \(\left(x=\pm w_{1} / 2\right)\). (c) Plot the temperature distribution for the entire strip and point out its special features.

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

A thin flat plate of length \(L\), thickness \(t\), and width \(W \geqslant L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{o}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q_{o}^{\prime \prime}\) (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.
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