Question

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

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

Step-by-step solution

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: \(thickness = 40\,\text{cm} - 2\cdot 2.5\,\text{cm}\)

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: \(thickness = 35\,\text{cm} \cdot \frac{1\,\text{m}}{100\,\text{cm}} = 0.35\,\text{m}\)

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: \(ΔT = 60^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 45^{\circ} \mathrm{C}\)

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 \(A\): \(A = 8\,\text{m} \cdot 0.35\,\text{m}\)

Calculate the area

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

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 \(270\,\mathrm{W}\).

Key concepts

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.