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

Expert verified
(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

01

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.
02

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.
03

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: $$q \approx 98.0 \mathrm{~W} / \mathrm{m}$$
04

4. Find the correct answer

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

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Key Concepts

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

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.

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Most popular questions from this chapter

What is a variable? How do you distinguish a dependent variable from an independent one in a problem?

Liquid ethanol is a flammable fluid that has a flashpoint at \(16.6^{\circ} \mathrm{C}\). At temperatures above the flashpoint, ethanol can release vapors that form explosive mixtures, which could ignite when source of ignition is present. In a chemical plant, liquid ethanol is being transported in a pipe \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inside diameter of \(3 \mathrm{~cm}\) and a wall thickness of \(3 \mathrm{~mm}\). The pipe passes through areas where occasional presence of ignition source can occur, and the pipe's outer surface is subjected to a heat flux of \(1 \mathrm{~kW} / \mathrm{m}^{2}\). The ethanol flowing in the pipe has an average temperature of \(10^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Your task as an engineer is to ensure that the ethanol is transported safely and prevent fire hazard. Determine the variation of temperature in the pipe wall and the temperatures of the inner and outer surfaces of the pipe. Are both surface temperatures safely below the flashpoint of liquid ethanol?

Consider a large plane wall of thickness \(L=0.8 \mathrm{ft}\) and thermal conductivity \(k=1.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). The wall is covered with a material that has an emissivity of \(\varepsilon=0.80\) and a solar absorptivity of \(\alpha=0.60\). The inner surface of the wall is maintained at \(T_{1}=520 \mathrm{R}\) at all times, while the outer surface is exposed to solar radiation that is incident at a rate of \(\dot{q}_{\text {solar }}=300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\). The outer surface is also losing heat by radiation to deep space at \(0 \mathrm{~K}\). Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached.

Consider a water pipe of length \(L=17 \mathrm{~m}\), inner radius \(r_{1}=15 \mathrm{~cm}\), outer radius \(r_{2}=20 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the pipe material uniformly by a \(25-\mathrm{kW}\) electric resistance heater. The inner and outer surfaces of the pipe are at \(T_{1}=60^{\circ} \mathrm{C}\) and \(T_{2}=80^{\circ} \mathrm{C}\), respectively. Obtain a general relation for temperature distribution inside the pipe under steady conditions and determine the temperature at the center plane of the pipe.

Consider a long rectangular bar of length \(a\) in the \(x-\) direction and width \(b\) in the \(y\)-direction that is initially at a uniform temperature of \(T_{i}\). The surfaces of the bar at \(x=0\) and \(y=0\) are insulated, while heat is lost from the other two surfaces by convection to the surrounding medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient two-dimensional heat transfer with no heat generation, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

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