/*! 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 7 Free convection heat transfer is... [FREE SOLUTION] | 91影视

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Free convection heat transfer is sometimes quantified by writing Equation \(4.20\) as \(q_{\text {coov }}=S k_{\text {eff }} \Delta T_{1-2}\), where \(k_{\text {eff }}\) is an effective thermal conductivity. The ratio \(k_{\text {eff }} / k\) is greater than unity because of fluid motion driven by buoyancy forces, as represented by the dashed streamlines. An experiment for the configuration shown yields a heat transfer rate per unit length of \(q_{\text {conv }}^{\prime}=110 \mathrm{~W} / \mathrm{m}\) for surface temperatures of \(T_{1}=53^{\circ} \mathrm{C}\) and \(T_{2}=15^{\circ} \mathrm{C}\), respectively. For inner and outer cylinders of diameters \(d=20 \mathrm{~mm}\) and \(D=60 \mathrm{~mm}\), and an eccentricity factor of \(z=10 \mathrm{~mm}\), determine the value of \(k_{\text {eff. }}\). The actual thermal conductivity of the fluid is \(k=0.255 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The effective thermal conductivity (\(k_{\text{eff}}\)) for this free convection heat transfer experiment is approximately \(29.71\, \mathrm{W/m\cdot K}\).

Step by step solution

01

Compute the surface area S

The inner cylinder has a diameter of \(d=20 \mathrm{~mm}\), and the outer cylinder has a diameter of \(D=60 \mathrm{~mm}\). Since the eccentricity factor (\(z=10 \mathrm{~mm}\)), the heights of the cylindrical surfaces are equal and can be calculated by subtracting the radius difference and the eccentricity factor: \(H = \frac{D}{2} - \frac{d}{2} - z = \frac{60}{2} - \frac{20}{2} - 10 = 10\,\mathrm{mm}\). Now, we can calculate the surface area \(S\) of the annular space between the concentric cylinders using the formula: \(S=2 \pi L H\) Assuming the unit length is 1 m, we have \(L = 1\, \mathrm{m}\). \(S = 2 \pi (1)(0.01\, \mathrm{m}) = 0.02\pi\, \mathrm{m}^{2}\)
02

Calculate the temperature difference 螖T鈧佲倐

We know the surface temperatures \(T_1\) and \(T_2\): \(T_{1}=53^{\circ} \mathrm{C}\) and \(T_{2}=15^{\circ} \mathrm{C}\). To find the temperature difference \(\Delta T_{1-2}\), we subtract \(T_2\) from \(T_1\): \(\Delta T_{1-2} = T_1 - T_2 = 53 - 15 = 38\, ^{\circ}\mathrm{C}\).
03

Solve for effective thermal conductivity k_eff

Now we can use the given equation \(q_{\text{coov}} = S k_{\text{eff}} \Delta T_{1-2}\) to solve for \(k_{\text{eff}}\). We are given the heat transfer rate per unit length (\(q_{\text{conv}}'\)), which is equal to \(q_{\text{coov}}\). Therefore, \(k_{\text{eff}} = \frac{q_{\text{coov}}}{S \Delta T_{1-2}}\) We can now plug in the values: \(k_{\text{eff}} = \frac{110\, \mathrm{W/m}}{0.02\pi\, \mathrm{m}^{2} \times 38\, ^{\circ}\mathrm{C}}\) \(k_{\text{eff}} = 29.71\, \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) So, the effective thermal conductivity (\(k_{\text{eff}}\)) for this free convection heat transfer experiment is approximately \(29.71\, \mathrm{W/m\cdot K}\).

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Key Concepts

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

Effective Thermal Conductivity
In the context of free convection heat transfer, effective thermal conductivity (\(k_{\text{eff}}\)) serves as a crucial parameter. It incorporates not just the thermal properties of the fluid, but also the enhanced heat transfer due to fluid movement. When fluids move, they transfer heat more effectively than in a stagnant state, leading to an effective thermal conductivity that is higher than the fluid's static thermal conductivity (\(k\)).For instance, in the problem we are solving, the effective thermal conductivity is calculated to be significantly higher than the static conductivity due to the buoyancy-driven convection currents between cylinders. This increased heat transfer capability is expressed in terms like \(k_{\text{eff}}\), where the effect of motion allows better heat dissipation.
  • An experiment with concentric cylinders in free convection conditions showed that the calculated effective thermal conductivity was \(29.71\, \mathrm{W/m \cdot K}\)
  • This is substantially greater than the static conductivity of \(0.255\, \mathrm{W/m \cdot K}\)
Understanding effective thermal conductivity helps engineers design systems for optimal heat dissipation, ensuring efficiency and safety in thermal management applications.
Buoyancy Forces
Buoyancy forces are a fundamental concept in fluid dynamics and play a critical role in free convection heat transfer. When a fluid experiences a temperature gradient, it tends to move because of differences in density. Warmer parts of the fluid are less dense and rise, while cooler parts sink, creating a circulation pattern known as convection currents. These buoyancy forces are the driving mechanism behind the enhanced heat transfer seen in free convection. In the cylindrical configuration discussed in the exercise:
  • Buoyancy forces cause the fluid to move within the annular space between two cylinders
  • This motion increases the heat transfer rate compared to stagnant fluid
  • The effective thermal conductivity is greater than the static value due to these forces
The principle is akin to how a hot air balloon works, where heated air inside the balloon rises, creating lift. In thermal systems, utilizing buoyancy forces effectively can enhance cooling efficiency, making it an essential principle for engineers and designers.
Temperature Difference
The temperature difference (\(\Delta T\)) is a straightforward, yet profound concept crucial for understanding heat transfer processes. In essence, it is the driving factor behind heat flow and dictates the rate and direction of transfer. Heat naturally flows from an area of higher temperature to an area of lower temperature.In solving the exercise, the given temperatures:
  • Surface temperatures \(T_{1} = 53^{\circ} \mathrm{C}\) and \(T_{2} = 15^{\circ} \mathrm{C}\) were used
  • The temperature difference \(\Delta T_{1-2} = 38^{\circ} \mathrm{C}\)
This difference is crucial in calculations as it directly impacts the rate of convective heat transfer.A higher difference means greater energy transfer between the surfaces, influenced further by the fluid's movement and properties. In practical applications, controlling temperature differences can help manage system efficiency, as well as prevent thermal shock or overheating.
Cylindrical Geometry
Cylindrical geometry is often used in the design of various industrial equipment such as heat exchangers and insulation systems. In this specific problem, we have a smaller cylinder placed concentrically within a larger cylinder, creating an annular space. This setup is common for studying heat transfer due to its symmetrical properties that simplify mathematical analysis.Key aspects of cylindrical geometry in the context of heat transfer include:
  • The inner and outer cylinders described have diameters of\(d = 20\, \mathrm{mm}\) and \(D = 60\, \mathrm{mm}\) respectively
  • The eccentricity factor, \(z = 10\, \mathrm{mm}\), slightly offsets the cylinders from being perfectly concentric
Such arrangements allow for controlled experiments that can test theoretical models of conduction and convection.Cylindrical systems are efficient for maximizing surface area relative to volume, enhancing heat transfer rates. Understanding how geometry impacts heat transfer processes is vital in designing efficient heating or cooling systems that effectively utilize space within constraints.

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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?

An electrical heater \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Estimate the temperature reached by the heater when dissipating \(50 \mathrm{~W}\) with the surface of the block at a temperature of \(25^{\circ} \mathrm{C}\).

In Chapter 3 we assumed that, whenever fins are attached to a base material, the base temperature is unchanged. What in fact happens is that, if the temperature of the base material exceeds the fluid temperature, attachment of a fin depresses the junction temperature \(T_{j}\) below the original temperature of the base, and heat flow from the base material to the fin is two-dimensional. Consider conditions for which a long aluminum pin fin of diameter \(D=5 \mathrm{~mm}\) is attached to a base material whose temperature far from the junction is maintained at \(T_{b}=100^{\circ} \mathrm{C}\). Fin convection conditions correspond to \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) What are the fin heat rate and junction temperature when the base material is (i) aluminum ( \(k=\) \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and (ii) stainless steel \((k=15\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) ? (b) Repeat the foregoing calculations if a thermal contact resistance of \(R_{t, j}^{\prime \prime}=3 \times 10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) is associated with the method of joining the pin fin to the base material. (c) Considering the thermal contact resistance, plot the heat rate as a function of the convection coefficient over the range \(10 \leq h \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for each of the two materials.

Hot water is transported from a cogeneration power station to commercial and industrial users through steel pipes of diameter \(D=150 \mathrm{~mm}\), with each pipe centered in concrete \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section \((w=300 \mathrm{~mm})\). The outer surfaces of the concrete are exposed to ambient air for which \(T_{\infty}=0^{\circ} \mathrm{C}\) and \(h=\) \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the inlet temperature of water flowing through the pipe is \(T_{i}=90^{\circ} \mathrm{C}\), what is the heat loss per unit length of pipe in proximity to the inlet? The temperature of the pipe \(T_{1}\) may be assumed to be that of the inlet water. (b) If the difference between the inlet and outlet temperatures of water flowing through a 100 -m-long pipe is not to exceed \(5^{\circ} \mathrm{C}\), estimate the minimum allowable flow rate \(\dot{m}\). A value of \(c=4207 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) may be used for the specific heat of the water.

A long constantan wire of \(1-\mathrm{mm}\) diameter is butt welded to the surface of a large copper block, forming a thermocouple junction. The wire behaves as a fin, permitting heat to flow from the surface, thereby depressing the sensing junction temperature \(T_{j}\) below that of the block \(T_{\sigma}\). Copper block, \(T_{o}\) (a) If the wire is in air at \(25^{\circ} \mathrm{C}\) with a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the measurement error \(\left(T_{j}-T_{o}\right)\) for the thermocouple when the block is at \(125^{\circ} \mathrm{C}\). (b) For convection coefficients of 5,10 , and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the measurement error as a function of the thermal conductivity of the block material over the range 15 to \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Under what circumstances is it advantageous to use smaller diameter wire?

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