Hot- and cold-water pipes \(8 \mathrm{~m}\) long run parallel to each other in a thick concrete layer. The diameters of both pipes are \(5 \mathrm{~cm}\), and the distance between the centerlines of the pipes is \(40 \mathrm{~cm}\). The surface temperatures of the hot and cold pipes are \(60^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. Taking the thermal conductivity of the concrete to be \(k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the rate of heat transfer between the pipes.

Short Answer

Expert verified
Answer: The rate of heat transfer between the hot and cold water pipes is approximately \(270\,\mathrm{W}\).

Step by step solution

01

Determine the dimensions of the problem

The distance between the centers of the two pipes is given as \(40 \mathrm{~cm}\). However, we will need the distance between their surfaces to get the thickness of the concrete layer where the heat transfer takes place. The diameter of both pipes is \(5 \mathrm{~cm}\), so their radius is \(2.5 \mathrm{~cm}\). Therefore, we can calculate the thickness of the concrete layer between them: \(thickness = 40\,\text{cm} - 2\cdot 2.5\,\text{cm}\)
02

Calculate the thickness of the concrete layer

Now, we will calculate the concrete thickness: \(thickness = 40\,\text{cm} - 5\,\text{cm} = 35\,\text{cm}\) We will now convert the thickness to meters: \(thickness = 35\,\text{cm} \cdot \frac{1\,\text{m}}{100\,\text{cm}} = 0.35\,\text{m}\)
03

Calculate the temperature difference

The temperature difference between the hot pipe's surface and the cold pipe's surface is given as \(ΔT = T_{hot} - T_{cold}\). So, let's plug in the values: \(ΔT = 60^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 45^{\circ} \mathrm{C}\)
04

Apply the formula for heat transfer through a solid medium

Now we can use the heat transfer formula: \(Q = \frac{kAΔT}{d}\) where: - \(Q\) is the heat transfer rate, - \(k\) is the thermal conductivity of the concrete (\(0.75\,\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\)), - \(A\) is the area through which heat is being transferred (we will assume the heat transfer occurs along the entire \(8\,\text{m}\) length of the pipe), - \(ΔT\) is the temperature difference (\(45^{\circ} \mathrm{C}\)), and - \(d\) is the thickness of the concrete layer (\(0.35\,\text{m}\)). First, we need to determine the area \(A\): \(A = 8\,\text{m} \cdot 0.35\,\text{m}\)
05

Calculate the area

Now, we will calculate the area: \(A = 8\,\text{m} \cdot 0.35\,\text{m} = 2.8\,\text{m}^2\)
06

Calculate the heat transfer rate

Finally, we can plug all the values into the heat transfer formula and calculate the heat transfer rate: \(Q = \frac{0.75\,\mathrm{W} / \mathrm{m} \cdot \mathrm{K} \cdot 2.8\,\text{m}^2 \cdot 45^{\circ} \mathrm{C}}{0.35\,\text{m}}\) \(Q = \frac{94.5\,\mathrm{W}}{0.35\,\text{m}}\) \(Q \approx 270\,\mathrm{W}\) So, the rate of heat transfer between the hot and cold water pipes is approximately \(270\,\mathrm{W}\).

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Key Concepts

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

Thermal Conductivity
Thermal conductivity, represented by the symbol \(k\), is a material property that quantifies how well a substance conducts heat. In the context of heat transfer, it is used to determine the rate at which heat energy is transferred through a material due to a temperature gradient. Materials with high thermal conductivity, like metals, are effective at transferring heat, while those with low thermal conductivity, such as wood or concrete, transfer heat more slowly.

In the exercise provided, the thermal conductivity of concrete is \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). This value is crucial in calculating the heat transfer rate between two pipes encased in concrete. To optimize the student's learning process, it's important to stress that thermal conductivity is not dependent on the amount of material or its temperature but is an inherent property of the material itself. Understanding this concept will help when dealing with exercises involving different materials and varying thermal conductivities.
Conduction in Solids
Conduction in solids is the process whereby heat is transferred through a material without any movement of the substance itself. This is because the heat energy causes particles within the solid to vibrate and transfer energy to neighboring particles. The efficiency of heat conduction is highly dependent on the material's thermal conductivity.

In our exercise, heat conduction occurs through the concrete layer separating the hot and cold water pipes. The concrete's particles which are closer to the hot pipe absorb heat and vibrate more vigorously, transferring energy to adjacent particles towards the cooler regions near the cold pipe. This microscopic view helps students visualize heat transfer at a particle level, an important aspect to grasp for a solid understanding of conduction in solids.
Temperature Difference
Temperature difference, often denoted as \(\Delta T\), is the driving force behind heat transfer. The greater the temperature difference between two regions, the higher the rate of heat transfer. This concept is exemplified in the exercise, where there's a temperature difference of \(45^\circ \mathrm{C}\) between the hot and cold pipes. It's essential for students to recognize that heat always flows from the hotter to the cooler region to reach thermal equilibrium.

The exercise illustrates the direct proportionality between temperature difference and heat transfer rate: the larger the \(\Delta T\), the higher the heat transfer rate, assuming all other factors remain constant. This relationship underpins many thermal systems and is critical for students to understand when solving real-world heat transfer problems.

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Most popular questions from this chapter

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.04 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-cm-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.

Consider a 25-m-long thick-walled concrete duct \((k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section. The outer dimensions of the duct are \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\), and the thickness of the duct wall is \(2 \mathrm{~cm}\). If the inner and outer surfaces of the duct are at \(100^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer through the walls of the duct. Answer: \(47.1 \mathrm{~kW}\)

Two flow passages with different cross-sectional shapes, one circular another square, are each centered in a square solid bar of the same dimension and thermal conductivity. Both configurations have the same length, \(T_{1}\), and \(T_{2}\). Determine which configuration has the higher rate of heat transfer through the square solid bar for \((a) a=1.2 b\) and \((b) a=2 b\).

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