Question

Hot water flows through a PVC \((k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) pipe whose inner diameter is \(2 \mathrm{~cm}\) and outer diameter is \(2.5 \mathrm{~cm}\). The temperature of the interior surface of this pipe is \(50^{\circ} \mathrm{C}\) and the temperature of the exterior surface is \(20^{\circ} \mathrm{C}\). The rate of heat transfer per unit of pipe length is (a) \(77.7 \mathrm{~W} / \mathrm{m}\) (b) \(89.5 \mathrm{~W} / \mathrm{m}\) (c) \(98.0 \mathrm{~W} / \mathrm{m}\) (d) \(112 \mathrm{~W} / \mathrm{m}\) (e) \(168 \mathrm{~W} / \mathrm{m}\)

Short answer

(a) 80.0 W/m (b) 96.0 W/m (c) 98.0 W/m (d) 100.0 W/m Solution: Using the formula for heat transfer in cylindrical systems and substituting the given values: $$q=\frac{2 \pi (0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}) (50^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C})}{ln[(2.5 \mathrm{~cm}) / (2 \mathrm{~cm})]}$$ Converting the values to the appropriate units and performing the calculations: $$q \approx 98.0 \mathrm{~W} / \mathrm{m}$$ Comparing our calculated value with the given choices, we find that the correct answer is (c) \(98.0 \mathrm{~W} / \mathrm{m}\).

Step-by-step solution

1. Recall the formula for heat transfer in cylindrical systems

To determine the rate of heat transfer per unit of pipe length, we will use the following formula for heat transfer in a cylindrical system: $$q=\frac{2 \pi k (T_i - T_o)}{ln(r_o / r_i)}$$ where \(q\) is the rate of heat transfer per unit length, \(k\) is the thermal conductivity of the pipe material, \(T_i\) and \(T_o\) are the temperatures of the inner and outer surfaces, and \(r_i\) and \(r_o\) are the radii of the inner and outer surfaces.

2. Plug given values into the formula

Now, substitute the given values of thermal conductivity, inner and outer temperatures, inner and outer diameters into the formula: $$q=\frac{2 \pi (0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}) (50^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C})}{ln[(2.5 \mathrm{~cm}) / (2 \mathrm{~cm})]}$$ Make sure to convert the diameters to meters and calculate the inner and outer radii by dividing the diameters by 2.

3. Perform the calculations

Once all the given values are plugged into the formula, convert the values to the appropriate units, and perform the calculations: $$q=\frac{2 \pi (0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}) (30 \mathrm{~K})}{ln[(0.0125 \mathrm{~m}) / (0.01 \mathrm{~m})]}$$ Calculating the above expression, we get the rate of heat transfer per unit of pipe length to be: $$q \approx 98.0 \mathrm{~W} / \mathrm{m}$$

4. Find the correct answer

Comparing our calculated value with the given choices, we find that the correct answer is: (c) \(98.0 \mathrm{~W} / \mathrm{m}\)

Key concepts

Thermal Conductivity

To understand how heat moves through materials, the concept of thermal conductivity is essential. Thermal conductivity, denoted as 'k' in scientific terms, is a measure of a material's ability to conduct heat. It quantifies the rate at which heat energy is transferred through a material due to a temperature difference. This property varies among different materials; for instance, metals typically have high thermal conductivity, which allows them to transfer heat efficiently, while insulators like PVC have lower values of thermal conductivity, causing them to transfer heat less effectively. In the context of the given exercise, the PVC pipe has a thermal conductivity value of \(0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), which tells us that for every meter in length and every degree of temperature difference (Kelvin), the pipe will transfer 0.092 Watts of heat energy. Understanding this concept is critical when calculating the rate of heat transfer through materials.

Temperature Gradient

The temperature gradient is a term that refers to the rate of change of temperature with distance. It's analogous to a slope that you might calculate when looking at a hill—the steeper the hill (or the greater the temperature difference), the larger the gradient. In heat transfer, a higher temperature gradient means a quicker transfer of heat. The gradient is driven by the difference in temperature between two points; in the case of the cylindrical pipe problem, the inner surface is at \(50^{\text{o}} \mathrm{C}\) and the outer surface is at \(20^{\text{o}} \mathrm{C}\), resulting in a gradient that causes heat to flow from the hotter interior to the cooler exterior. An important aspect to note here is that the temperature gradient is directly proportional to the rate of heat transfer—meaning, the larger the difference in temperature, the more heat will flow per unit of time.

Conduction in Cylindrical Coordinates

Conduction is the process of heat transfer through a material without the movement of the material itself. It occurs at a molecular level where vibrating molecules transfer energy to their neighbors. For cylindrical objects, such as pipes, conduction can be more complex than in flat materials due to the geometry. The mathematical treatment of conduction in cylindrical coordinates involves considering the radial flow of heat and using cylindrical-specific formulae, like the one given in our textbook exercise. The formula \(q=\frac{2 \pi k (T_i - T_o)}{\ln(r_o / r_i)}\) integrates thermal conductivity, the temperature difference between inner and outer surfaces, and the radii of these surfaces to calculate the rate of heat transfer per unit length of the cylinder. This equation acknowledges that the area through which heat is conducted changes with radius, which is a distinctive factor when compared to heat conduction in flat, rectangular materials. Utilizing the correct form of heat transfer equations for the shape of the material is crucial for an accurate calculation in applied thermodynamics.