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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

A heat sink for cooling computer chips is fabricated from copper \(\left(k_{s}=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), with machined microchannels passing a cooling fluid for which \(T=25^{\circ} \mathrm{C}\) and \(h=30,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The lower side of the sink experiences no heat removal, and a preliminary heat sink design calls for dimensions of \(a=b=w_{s}=w_{f}=200 \mu \mathrm{m}\). A symmetrical element of the heat path from the chip to the fluid is shown in the inset. (a) Using the symmetrical element with a square nodal network of \(\Delta x=\Delta y=100 \mu \mathrm{m}\), determine the corresponding temperature field and the heat rate \(q^{\prime}\) to the coolant per unit channel length \((\mathrm{W} / \mathrm{m})\) for a maximum allowable chip temperature \(T_{c, \max }=75^{\circ} \mathrm{C}\). Estimate the corresponding thermal resistance between the chip surface and the fluid, \(R_{t, c-f}^{\prime}(\mathrm{m} \cdot \mathrm{K} / \mathrm{W})\). What is the maximum allowable heat dissipation for a chip that measures \(10 \mathrm{~mm} \times 10 \mathrm{~mm}\) on a side? (b) The grid spacing used in the foregoing finite-difference solution is coarse, resulting in poor precision for the temperature distribution and heat removal rate. Investigate the effect of grid spacing by considering spatial increments of 50 and \(25 \mu \mathrm{m}\). (c) Consistent with the requirement that \(a+b=400 \mu \mathrm{m}\), can the heat sink dimensions be altered in a manner that reduces the overall thermal resistance?

Consider nodal configuration 4 of Table \(4.2\). Derive the finite-difference equations under steady-state conditions for the following situations. (a) The upper boundary of the external corner is perfectly insulated and the side boundary is subjected to the convection process \(\left(T_{\infty}, h\right)\). (b) Both boundaries of the external corner are perfectly insulated. How does this result compare with Equation 4.43?

A plate \((k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is stiffened by a series of longitudinal ribs having a rectangular cross section with length \(L=8 \mathrm{~mm}\) and width \(w=4 \mathrm{~mm}\). The base of the plate is maintained at a uniform temperature \(T_{b}=\) \(45^{\circ} \mathrm{C}\), while the rib surfaces are exposed to air at a temperature of \(T_{\infty}=25^{\circ} \mathrm{C}\) and a convection coeffient of \(h=600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Using a finite-difference method with \(\Delta x=\Delta y=\) \(2 \mathrm{~mm}\) and a total of \(5 \times 3\) nodal points and regions, estimate the temperature distribution and the heat rate from the base. Compare these results with those obtained by assuming that heat transfer in the rib is one- dimensional, thereby approximating the behavior of a fin. (b) The grid spacing used in the foregoing finitedifference solution is coarse, resulting in poor precision for estimates of temperatures and the heat rate. Investigate the effect of grid refinement by reducing the nodal spacing to \(\Delta x=\Delta y=1 \mathrm{~mm}\) (a \(9 \times 3\) grid) considering symmetry of the center line. (c) Investigate the nature of two-dimensional conduction in the rib and determine a criterion for which the one-dimensional approximation is reasonable. Do so by extending your finite-difference analysis to determine the heat rate from the base as a function of the length of the rib for the range \(1.5 \leq L w \leq 10\), keeping the length \(L\) constant. Compare your results with those determined by approximating the rib as a fin.

Consider the nodal point 0 located on the boundary between materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\). Derive the finite-difference equation, assuming no internal generation.

Consider heat transfer in a one-dimensional (radial) cylindrical coordinate system under steady-state conditions with volumetric heat generation. (a) Derive the finite-difference equation for any interior node \(m\). (b) Derive the finite-difference equation for the node \(n\) located at the external boundary subjected to the convection process \(\left(T_{\infty 0}, h\right)\).
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