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Chapter 2: Problem 3

A hot water pipe with outside radius \(r_{1}\) has a temperature \(T_{1}\). A thick insulation, applied to reduce the heat loss, has an outer radius \(r_{2}\) and temperature \(T_{2}\). On \(T-r\) coordinates, sketch the temperature distribution in the insulation for one-dimensional, steady-state heat transfer with constant properties. Give a brief explanation, justifying the shape of your curve.

Short Answer

Expert verified
The temperature distribution in the insulation can be represented by the linear function \(T(r) = \frac{T_2 - T_1}{r_2 - r_1} r + (T_1 - \frac{T_2 - T_1}{r_2 - r_1} r_1)\). When plotted on a \(T-r\) graph, the function forms a straight line with a negative slope, starting from \(T_1\) at \(r = r_1\) and ending at \(T_2\) at \(r = r_2\). This reflects heat conduction from the hotter pipe to the colder outer surface of the insulation, effectively reducing heat loss.

Step by step solution

01

Set up the heat conduction equation

As it's one-dimensional steady-state heat transfer with constant properties, we can model this problem through the radial heat conduction equation. The heat conduction equation for this situation is given by: \[\frac{d}{dr} \left( k \, \frac{dT}{dr} \right) = 0\] Here, \(k\) is the thermal conductivity of the insulation material, which we assume to be constant, and \(T\) is the temperature as a function of radial distance \(r\).
02

Integrate to find temperature distribution

The heat conduction equation can be rewritten as: \[k \, \frac{d^2T}{dr^2} = 0\] Since \(k\) is constant, we can integrate the equation with respect to \(r\) once to obtain: \[\frac{dT}{dr} = C_{1}\] where \(C_{1}\) is a constant of integration. Integrating once more with respect to \(r\) will give us the temperature distribution function: \[T(r) = C_{1} \, r + C_{2}\] where \(C_{2}\) is another constant of integration.
03

Apply boundary conditions

In order to find the constants \(C_{1}\) and \(C_{2}\), we need to apply the boundary conditions given in the exercise: 1. At \(r = r_1\), the temperature is \(T_1\). 2. At \(r = r_2\), the temperature is \(T_2\). First, let's apply boundary condition 1: \[T_1 = C_{1} \, r_1 + C_{2}\] Next, apply boundary condition 2: \[T_2 = C_{1} \, r_2 + C_{2}\] Now we have two equations with two unknowns \(C_{1}\) and \(C_{2}\).
04

Solve for constants and find temperature distribution

Solve the system of equations in Step 3 for \(C_{1}\) and \(C_{2}\): From the first equation, we can find relation between \(C_1\) and \(C_2\): \[C_2 = T_1 - C_1 \, r_1\] Substitute this expression for \(C_2\) into the second equation: \[T_2 = C_{1} \, r_2 + T_1 - C_1 \, r_1\] Simplify and solve for \(C_{1}\): \[C_{1} = \frac{T_2 - T_1}{r_2 - r_1}\] Substitute back the value of \(C_1\) in expression of \(C_2\): \[C_{2} = T_1 - \frac{T_2 - T_1}{r_2 - r_1} r_1\] Now, substituting the values of \(C_{1}\) and \(C_{2}\) into the temperature distribution function, we get: \[T(r) = \frac{T_2 - T_1}{r_2 - r_1} r + (T_1 - \frac{T_2 - T_1}{r_2 - r_1} r_1)\]
05

Sketch the curve and provide brief explanation

The temperature distribution function \(T(r)\) is a linear function of \(r\), which means it will be a straight line when plotted on a \(T-r\) graph. This line will start from \(T_1\) at \(r = r_1\) and end at \(T_2\) at \(r = r_2\). The curve will have a negative slope since \(T_2 < T_1\). The negative slope of the curve indicates that heat is being conducted from the hotter pipe (with temperature \(T_1\)) to the colder outer surface of the insulation (with temperature \(T_2\)), thereby reducing heat loss from the pipe, as intended.

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Key Concepts

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

Radial Heat Conduction Equation
The radial heat conduction equation plays a crucial role in understanding how heat travels through materials that are cylindrical in shape, such as pipes.

At its core, this equation is built upon the fundamental principle of heat transfer, which states that heat moves from regions of higher temperature to regions of lower temperature. In a radial configuration, heat moves outward or inward through the layers of the material.

The equation \[\frac{d}{dr} \left( k \, \frac{dT}{dr} \right) = 0\] highlights this principle in a mathematical form, signaling a steady-state condition where there's no accumulation of heat within the insulation layer, which would otherwise change the temperature over time. This is why it's vital only in steady-state scenarios.

By integrating this expression, as seen in the solutions provided, we obtain a direct relationship between the temperature gradient and the distance from the center of the cylinder. The simplicity of the equation also demonstrates the assumption of constant thermal conductivity, meaning the material's ability to conduct heat doesn't change with temperature or position.
Thermal Conductivity
Thermal conductivity, denoted by the symbol \(k\), is a property that indicates how well a material conducts heat. It is often described as the amount of heat that passes through a material with a given temperature gradient.

The higher the thermal conductivity, the more efficient the material is at transferring heat. For instance, metals typically have a high thermal conductivity, making them good conductors of heat, while materials like fiberglass or foam insulation have low thermal conductivity, making them good insulators.

In the context of the radial heat conduction equation, thermal conductivity is assumed constant. This assumption simplifies the equation and is reasonable when the material properties do not significantly change with temperature. However, in real-world applications, it's worth noting that \(k\) can vary with temperature, and accounting for this can affect the accuracy of temperature distribution calculations.
Temperature Distribution
Understanding temperature distribution is key to solving many practical problems in thermal engineering.

The temperature distribution equation \[T(r) = C_{1} \, r + C_{2}\] represents how temperature varies with radial distance \(r\) from the center of the pipe. In the given scenario, it's a reflection of how heat emanates from a hot water pipe to its cooler surroundings.

By applying the boundary conditions where the temperatures at the inner and outer radii are known, we can solve for the constants in the equation to articulate the exact temperature at any point within the insulation. The result is a linear relationship indicating a constant rate of temperature change with respect to the radial distance. This tells us that in the steady state condition, with constant thermal conductivity, the temperature drops uniformly as we move away from the hot surface.

The graphical representation of this linear temperature distribution is essential for visualizing heat flow and for designing insulation systems that are efficient in minimizing energy loss.

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Most popular questions from this chapter

Typically, air is heated in a hair dryer by blowing it across a coiled wire through which an electric current is passed. Thermal energy is generated by electric resistance heating within the wire and is transferred by convection from the surface of the wire to the air. Consider conditions for which the wire is initially at room temperature, \(T_{i}\), and resistance heating is concurrently initiated with airflow at \(t=0\). (a) For a wire radius \(r_{o}\), an air temperature \(T_{\infty}\), and a convection coefficient \(h\), write the form of the heat equation and the boundary/initial conditions that govern the transient thermal response, \(T(r, t)\), of the wire. (b) If the length and radius of the wire are \(500 \mathrm{~mm}\) and \(1 \mathrm{~mm}\), respectively, what is the volumetric rate of thermal energy generation for a power consumption of \(P_{\text {elec }}=500 \mathrm{~W}\) ? What is the convection heat flux under steady-state conditions? (c) On \(T-r\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. (d) On \(q_{r}^{\prime \prime}-t\) coordinates, sketch the variation of the heat flux with time for locations at \(r=0\) and \(r=r_{o^{*}}\).

A steam pipe is wrapped with insulation of inner and outer radii \(r_{i}\) and \(r_{o}\), respectively. At a particular instant the temperature distribution in the insulation is known to be of the form $$ T(r)=C_{1} \ln \left(\frac{r}{r_{o}}\right)+C_{2} $$ Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?

A plane wall of thickness \(2 L=40 \mathrm{~mm}\) and thermal conductivity \(k=5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation at a rate \(\dot{q}\), while convection heat transfer occurs at both of its surfaces \((x=-L,+L)\), each of which is exposed to a fluid of temperature \(T_{\infty}=20^{\circ} \mathrm{C}\). Under steady-state conditions, the temperature distribution in the wall is of the form \(T(x)=a+b x+c x^{2}\) where \(a=82.0^{\circ} \mathrm{C}, b=-210^{\circ} \mathrm{C} / \mathrm{m}, c=-2 \times 10^{4 \circ} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. The origin of the \(x\)-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation \(\dot{q}\) in the wall? (c) Determine the surface heat fluxes, \(q_{x}^{\prime \prime}(-L)\) and \(q_{x}^{\prime \prime}(+L)\). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at \(x=-L\) and \(x=+L\) ? (e) Obtain an expression for the heat flux distribution \(q_{x}^{\prime \prime}(x)\). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated \((\dot{q}=0)\), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with \(\dot{q}=0\) ? How much energy must be removed by the fluid per unit area of the wall \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) to reach this state? The density and specific heat of the wall material are \(2600 \mathrm{~kg} / \mathrm{m}^{3}\) and \(800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

To determine the effect of the temperature dependence of the thermal conductivity on the temperature distribution in a solid, consider a material for which this dependence may be represented as $$ k=k_{o}+a T $$ where \(k_{o}\) is a positive constant and \(a\) is a coefficient that may be positive or negative. Sketch the steady-state temperature distribution associated with heat transfer in a plane wall for three cases corresponding to \(a>0\), \(a=0\), and \(a<0\).

A spherical particle of radius \(r_{1}\) experiences uniform thermal generation at a rate of \(\dot{q}\). The particle is encapsulated by a spherical shell of outside radius \(r_{2}\) that is cooled by ambient air. The thermal conductivities of the particle and shell are \(k_{1}\) and \(k_{2}\), respectively, where \(k_{1}=2 k_{2}\). (a) By applying the conservation of energy principle to spherical control volume \(A\), which is placed at an arbitrary location within the sphere, determine a relationship between the temperature gradient \(d T / d r\) and the local radius \(r\), for \(0 \leq r \leq r_{1}\). (b) By applying the conservation of energy principle to spherical control volume \(\mathrm{B}\), which is placed at an arbitrary location within the spherical shell, determine a relationship between the temperature gradient \(d T / d r\) and the local radius \(r\), for \(r_{1} \leq r \leq r_{2}\). (c) On \(T-r\) coordinates, sketch the temperature distribution over the range \(0 \leq r \leq r_{2}\).
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