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Chapter 4: Problem 16

A tube of diameter \(50 \mathrm{~mm}\) having a surface temperature of \(85^{\circ} \mathrm{C}\) is embedded in the center plane of a concrete slab \(0.1 \mathrm{~m}\) thick with upper and lower surfaces at \(20^{\circ} \mathrm{C}\). Using the appropriate tabulated relation for this configuration, find the shape factor. Determine the heat transfer rate per unit length of the tube.

Short Answer

Expert verified
The shape factor (S) for the given configuration is \(\frac{2 \pi L \ln{\frac{0.125}{0.025}}}{1.5}\), where L is the length of the tube. Using the temperature difference of 65 K, the heat transfer rate per unit length of the tube is found to be \(807.7 L \ \frac{\mathrm{W}}{\mathrm{m}}\).

Step by step solution

01

Gather the given information and find the required parameters

We are given: - Diameter of the tube, \(D = 50 \ \mathrm{mm}\) - Surface temperature of the tube, \(T_s = 85^{\circ} \mathrm{C}\) - Thickness of the concrete slab, \(t = 0.1 \ \mathrm{m}\) - Upper and lower surface temperature of the concrete slab, \(T_c = 20^{\circ} \mathrm{C}\) We need to find two parameters: the shape factor (S) and the heat transfer rate per unit length of the tube (Q).
02

Find the shape factor (S)

Since the tube is embedded in a cylinder of concrete material, and we have a symmetrical configuration, we can assume that there will be uniform heat loss from the outer surface of the tube into the surrounding material. The shape factor (S) for steady-state conductive heat transfer between coaxial cylinders is given by the formula: \[S = \frac{2 \pi L \ln{\frac{R_2}{R_1}}}{k}\] Where: - \(L\) = length of the tube - \(R_1\) = inner radius (radius of the tube) - \(R_2\) = outer radius (including the thickness of the concrete slab) - \(k\) = thermal conductivity of the concrete First, let's convert the diameter of the tube to the radius in meters: \[R_1 = \frac{D}{2} = \frac{50 \ \mathrm{mm}}{2} \cdot \frac{1\ \mathrm{m}}{1000 \ \mathrm{mm}} = 0.025 \ \mathrm{m}\] Next, let's determine the outer radius: \[R_2 = R_1 + t = 0.025 + 0.1 = 0.125 \ \mathrm{m}\] Assuming the thermal conductivity of the concrete to be \(k = 1.5 \ \frac{\mathrm{W}}{\mathrm{m \cdot K}}\), we can now calculate the shape factor: \[S = \frac{2 \pi L \ln{\frac{0.125}{0.025}}}{1.5}\] Since the length of the tube is not given, we will leave it as \(L\) in the equation for now. The expression for the shape factor becomes: \[S = \frac{2 \pi L \ln{\frac{0.125}{0.025}}}{1.5}\]
03

Calculate the temperature difference

Now, let's find the temperature difference between the surface of the tube and the concrete slab: \[\Delta T = T_s - T_c = 85 - 20 = 65 \ \mathrm{K}\]
04

Find the heat transfer rate per unit length of the tube (Q)

We can use the shape factor (S) found in step 2 and the temperature difference found in step 3 to determine the heat transfer rate per unit length of the tube: \[Q = S \cdot k \cdot \Delta T = \left(\frac{2 \pi L \ln{\frac{0.125}{0.025}}}{1.5}\right) \cdot 1.5 \cdot 65\] \[Q = 2 \pi L \ln{\frac{0.125}{0.025}} \cdot 65\] Solving for Q, we get: \[Q = 807.7 L \ \frac{\mathrm{W}}{\mathrm{m}}\] So, the heat transfer rate per unit length of the tube is \(807.7 \ \frac{\mathrm{W}}{\mathrm{m}}\).

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Key Concepts

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

Shape Factor
The shape factor in heat transfer serves as a crucial element in calculating how effectively heat can move between different bodies or surfaces.
When discussing conductors like cylinders embedded in other materials, the shape factor determines how the configuration of these objects influences heat flow.
It is used to simplify complex systems into manageable equations, primarily allowing us to understand the heat exchange without diving deep into advanced math.
  • The shape factor is particularly useful as it allows us to account for geometry and orientation in heat transfer equations.
  • For cylindrical systems, like the problem we see here, the shape factor is derived using the natural logarithm to relate the radii of coaxial cylinders.
  • This relationship reflects how the heat spreads out from the tube into the surrounding concrete.
By understanding the shape factor, we develop insights into the thermal interaction across different shapes and configurations, allowing us to calculate the heat transfer quantity effectively.
Conductive Heat Transfer
Conductive heat transfer is the process where heat flows through a material due to a temperature difference. In our problem, this is how heat moves from the hot surface of the embedded tube into the surrounding concrete slab.
This type of heat transfer occurs without any movement of the material itself鈥攊t's all about vibrations, or microscopic movements, of particles within the substance.
  • Conductive heat transfer is governed by Fourier's law, which states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the area through which the heat is flowing.
  • In practice, this means the heat flows from areas of higher temperature to areas of lower temperature till equilibrium is reached.
  • In cylindrical objects, the geometry heavily impacts how effectively conduction occurs, thus the importance of using the shape factor.
Grasping this concept helps illustrate why understanding the physical arrangement of objects is vital when evaluating heat transfer.
Thermal Conductivity
Thermal conductivity represents a material's ability to conduct heat. Basically, it's how well heat can move through a specific material. In our case, the thermal conductivity of concrete impacts how quickly the heat from the tube dissipates into the slab.
Materials with high thermal conductivity, like metals, allow heat to flow quickly, while materials with low conductivity, like rubber or air, slow down the flow.
  • In our problem, the provided thermal conductivity of concrete is essential to ensure our calculations reflect the actual rate of heat transfer.
  • The concept proceeds as a constant in the calculations, showing the intrinsic property of a material to transfer heat.
  • Knowing the thermal conductivity coupled with a shape factor helps evaluate how much more heat one material will conduct than another.
This understanding is vital when deciding what materials to use in construction or designing systems for temperature regulation.
Temperature Difference
Temperature difference is the driving force behind heat transfer. It indicates how much heat energy will move from a warmer area to a cooler one. In this specific exercise, we calculate it as the difference between the surface temperature of the tube and the surface temperature of the concrete slab.
The greater the temperature difference, the faster the heat transfer process occurs.
  • This concept is a straightforward yet critical component of all heat transfer modes, acting as the initial push for heat flow.
  • In calculations like those undertaken, temperature difference pairs with thermal properties to determine the heat transfer rate.
  • The knowledge of this difference allows one to anticipate how quickly equilibrium might be reached within a system.
Ultimately, temperature difference is essential to understanding not just the rate of heat movement, but also the resulting energy distribution across materials.

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Most popular questions from this chapter

The elemental unit of an air heater consists of a long circular rod of diameter \(D\), which is encapsulated by a finned sleeve and in which thermal energy is generated by ohmic heating. The \(N\) fins of thickness \(t\) and length \(L\) are integrally fabricated with the square sleeve of width \(w\). Under steady-state operating conditions, the rate of thermal energy generation corresponds to the rate of heat transfer to airflow over the sleeve. (a) Under conditions for which a uniform surface temperature \(T_{s}\) is maintained around the circumference of the heater and the temperature \(T_{\infty}\) and convection coefficient \(h\) of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the heat rate for \(T_{s}=300^{\circ} \mathrm{C}, D=20 \mathrm{~mm}\), an aluminum sleeve \(\left(k_{s}=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right), w=40 \mathrm{~mm}\), \(N=16, t=4 \mathrm{~mm}, L=20 \mathrm{~mm}, T_{\infty}=50^{\circ} \mathrm{C}\), and \(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) For the foregoing heat rate and a copper heater of thermal conductivity \(k_{h}=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the required volumetric heat generation within the heater and its corresponding centerline temperature? (c) With all other quantities unchanged, explore the effect of variations in the fin parameters \((N, L, t)\) on the heat rate, subject to the constraint that the fin thickness and the spacing between fins cannot be less than \(2 \mathrm{~mm}\).

Derive the nodal finite-difference equations for the following configurations. (a) Node \((m, n)\) on a diagonal boundary subjected to convection with a fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\). (b) Node \((m, n)\) at the tip of a cutting tool with the upper surface exposed to a constant heat flux \(q_{o}^{\prime \prime}\), and the diagonal surface exposed to a convection cooling process with the fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\).

Consider an aluminum heat sink \((k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), such as that shown schematically in Problem 4.28. The inner and outer widths of the square channel are \(w=20 \mathrm{~mm}\) and \(W=40 \mathrm{~mm}\), respectively, and an outer surface temperature of \(T_{s}=50^{\circ} \mathrm{C}\) is maintained by the array of electronic chips. In this case, it is not the inner surface temperature that is known, but conditions \(\left(T_{\infty}, h\right)\) associated with coolant flow through the channel, and we wish to determine the rate of heat transfer to the coolant per unit length of channel. For this purpose, consider a symmetrical section of the channel and a two-dimensional grid with \(\Delta x=\Delta y=5 \mathrm{~mm}\). (a) For \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the unknown temperatures, \(T_{1}, \ldots, T_{7}\), and the rate of heat transfer per unit length of channel, \(q^{\prime}\). (b) Assess the effect of variations in \(h\) on the unknown temperatures and the heat rate.

An igloo is built in the shape of a hemisphere, with an inner radius of \(1.8 \mathrm{~m}\) and walls of compacted snow that are \(0.5 \mathrm{~m}\) thick. On the inside of the igloo, the surface heat transfer coefficient is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\); on the outside, under normal wind conditions, it is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal conductivity of compacted snow is \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature of the ice cap on which the igloo sits is \(-20^{\circ} \mathrm{C}\) and has the same thermal conductivity as the compacted snow. (a) Assuming that the occupants' body heat provides a continuous source of \(320 \mathrm{~W}\) within the igloo, calculate the inside air temperature when the outside air temperature is \(T_{\infty}=-40^{\circ} \mathrm{C}\). Be sure to consider heat losses through the floor of the igloo. (b) Using the thermal circuit of part (a), perform a parameter sensitivity analysis to determine which variables have a significant effect on the inside air temperature. For instance, for very high wind conditions, the outside convection coefficient could double or even triple. Does it make sense to construct the igloo with walls half or twice as thick?

Using the thermal resistance relations developed in Chapter 3, determine shape factor expressions for the following geometries: (a) Plane wall, cylindrical shell, and spherical shell. (b) Isothermal sphere of diameter \(D\) buried in an infinite medium.
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