Question

A 10-m-long 5-cm-outer-radius cylindrical steam pipe is covered with \(3-\mathrm{cm}\) thick cylindrical insulation with a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the rate of heat loss from the pipe is \(1000 \mathrm{~W}\), the temperature drop across the insulation is (a) \(163^{\circ} \mathrm{C}\) (b) \(600^{\circ} \mathrm{C}\) (c) \(48^{\circ} \mathrm{C}\) (d) \(79^{\circ} \mathrm{C}\) (e) \(251^{\circ} \mathrm{C}\)

Short answer

Answer: (c) 48°C

Step-by-step solution

Identifying the given parameters

From the exercise, we have the following parameters: - Length of the pipe, L = 10m, - Outer radius of the pipe, r1 = 5cm = 0.05m, - Thickness of the insulation, t = 3cm = 0.03m, - Thermal conductivity of the insulation, k = 0.05 W/m.K, - Rate of heat loss from the pipe, Q = 1000 W. With this information, we can now proceed to find the temperature drop across the insulation.

Calculate the inner and outer radii of the insulation

The inner radius of the insulation (r2) is equal to the outer radius of the pipe (r1), which is 0.05m. The outer radius of the insulation (r3) will be the sum of the inner radius and the thickness of the insulation: r3 = r2 + t = 0.05m + 0.03m = 0.08m Now we have the inner and outer radii of the insulation, r2 = 0.05m and r3 = 0.08m.

Use the formula for heat transfer through a cylindrical wall

The formula for the rate of heat transfer through a cylindrical wall is: Q = 2 * pi * L * (T1 - T2) * k / ln(r3/r2) where Q is the rate of heat transfer, L is the length of the pipe, T1 is the temperature at the inner radius (r2), T2 is the temperature at the outer radius (r3), k is the thermal conductivity, r2 is the inner radius of the insulation, and r3 is the outer radius of the insulation. We are given Q, L, k, r2, and r3, and we need to find the temperature drop, which is (T1 - T2).

Rearrange the equation and solve for the temperature drop

Rearranging the formula to find (T1 - T2): (T1 - T2) = Q * ln(r3/r2) / (2 * pi * L * k) Now plug in the given values: (T1 - T2) = (1000 W) * ln(0.08m/0.05m) / [2 * pi * (10m) * (0.05 W/m.K)] (T1 - T2) ≈ 48.5 °C

Identify the correct answer among the given options

The calculated temperature drop across the insulation is approximately 48.5°C, so the correct option among the given choices is: (c) 48°C

Key concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. In simpler terms, it tells us how quickly or slowly heat can pass through a material. Materials with high thermal conductivity, like metals, allow heat to pass through quickly, while materials with low thermal conductivity, like wool or rubber, slow down the transfer of heat.

In this specific exercise, we are dealing with an insulating material wrapped around a steam pipe. The insulation has a thermal conductivity of 0.05 W/m.K, which is relatively low. This means the insulation is quite effective at reducing the heat loss from the pipe. It acts as a barrier that prevents heat from escaping quickly into the surrounding environment. This property is crucial when trying to minimize energy loss or maintain the temperature of a system.

Understanding thermal conductivity is vital in various applications, such as designing thermal insulation for homes, creating more efficient engines, or even designing spacesuits for astronauts.

Cylindrical Insulation

Cylindrical insulation is often used around pipes or cylindrical objects to control heat transfer. In this setup, insulating material, such as foam or fiberglass, is wrapped around a pipe to prevent heat from escaping or entering. The primary goal is to maintain the desired temperature inside the pipe with minimal energy loss.

Let's imagine our cylindrical steam pipe wrapped in insulation. The steam pipe has a surface from which heat can escape. By covering it with an insulating layer of thickness 3 cm, we create a barrier to heat loss. The effectiveness of this insulation depends on its material properties (like thermal conductivity) and its thickness.

• **Inner Radius (r2):** This is the radius of the pipe itself, the starting point for the insulation layer.
• **Outer Radius (r3):** This is the radius of the entire system, including the pipe and the thickness of the insulation layer.

The setup ensures that heat transfer occurs more slowly, thus effectively managing the system’s thermal efficiency.

Temperature Drop Calculation

In this exercise, we calculate the temperature drop across the insulation layer. This drop is the difference between the temperature on the outside surface of the pipe and the outer surface of the insulation.

To find this temperature difference, we use the formula for heat transfer through a cylindrical wall:\[ Q = \frac{2 \pi L (T1 - T2) k}{\ln\left(\frac{r3}{r2}\right)} \]Here, \(Q\) is the heat transfer rate, \(L\) is the length of the pipe, \(k\) is the thermal conductivity, \(r2\) and \(r3\) are the inner and outer radii, respectively, and \((T1 - T2)\) is the temperature difference we are looking for.

We rearrange the formula to solve for \((T1 - T2)\), resulting in:\[ (T1 - T2) = \frac{Q \ln\left(\frac{r3}{r2}\right)}{2 \pi L k} \]
By plugging in the given values, we find that the temperature drop across the insulation is approximately 48.5°C. This calculation is crucial because it tells us how effective our insulation setup is at reducing heat loss. Understanding these concepts allows for better design and optimization of thermal systems.