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Chapter 3: Problem 179

Steam at \(235^{\circ} \mathrm{C}\) is flowing inside a steel pipe \((k=\) \(61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(10 \mathrm{~cm}\) and \(12 \mathrm{~cm}\), respectively, in an environment at \(20^{\circ} \mathrm{C}\). The heat transfer coefficients inside and outside the pipe are \(105 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine ( \(a\) ) the thickness of the insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) needed to reduce the heat loss by 95 percent and \((b)\) the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to \(40^{\circ} \mathrm{C}\) for safety reasons.

Short Answer

Expert verified
To determine this, follow the steps below: 1. Calculate the current heat transfer rate (Q) using the given parameters and the overall heat transfer coefficient (U). 2. Determine the insulation thickness that reduces heat loss by 95%, considering the new heat transfer rate (Q_reduced) and a redefined thermal resistance (R'_{total}). 3. Calculate the insulation thickness required for an exposed surface temperature of 40掳C using a redefined thermal resistance (R''_{total}). After performing the calculations mentioned in the given solution, you will obtain the insulation thickness required for both conditions.

Step by step solution

01

Calculate the current heat transfer

First, let's convert the diameters to meters: \(d_{in}=0.1\,\text{m}\) and \(d_{out}=0.12\,\text{m}\). Now, we can use the thermal circuit analogy to determine the overall heat transfer coefficient \(U\). The total thermal resistance to heat transfer from inside to outside is given by: \(R_{total} = \frac{1}{h_{in}A_{in}} + \frac{\ln{(r_{out}/r_{in})}}{2\pi k_{steel}L} + \frac{1}{h_{out}A_{out}}\) Next, we need to calculate the heat transfer rate inside and outside the pipe: \(A_{in}=2\pi r_{in}\,L\) \(A_{out}=2\pi r_{out}\,L\) \(r_{in} = \frac{d_{in}}{2} =0.05\,\text{m}\) \(r_{out} = \frac{d_{out}}{2} =0.06\,\text{m}\) Now plug the values into \(R_{total}\) and calculate the overall heat transfer coefficient, U: \(R_{total} = \frac{1}{h_{in}2\pi r_{in} L} + \frac{\ln{(r_{out}/r_{in})}}{2\pi k_{steel} L} + \frac{1}{h_{out}2\pi r_{out} L}\) \(U = \frac{1}{R_{total}}\) Heat transfer rate, \(Q = UA(T_{steam} - T_{env})\)
02

Determine insulation thickness to reduce heat loss by 95%

To reduce the heat loss by 95%, we want the heat transfer rate to be: \(Q_{reduced} =Q(1-0.95)=0.05Q\) Redefine the thermal resistance to include the insulation: \(R'_{total} = \frac{1}{h_{in}A_{in}} + \frac{\ln{(r_{out}/r_{in})}}{2\pi k_{steel}L} + \frac{\ln{(r_{insulation}/r_{out})}}{2\pi k_{insulation}L} + \frac{1}{h_{out}A_{insulation}}\) \(Q_{reduced} = U'A(T_{steam}-T_{env}) \implies U' = \frac{Q_{reduced}}{A(T_{steam}-T_{env})}\) Now we can solve for the insulation thickness using the equation: \(\frac{\ln{(r_{insulation}/r_{out})}}{2\pi k_{insulation} L} = \frac{1}{U'} - \frac{1}{h_{in}A_{in}} - \frac{\ln{(r_{out}/r_{in})}}{2\pi k_{steel} L} - \frac{1}{h_{out}A_{insulation}}\) Calculate the thickness of the insulation, \(t_{insulation} = r_{insulation} - r_{out}\).
03

Determine insulation thickness to reduce exposed surface temperature to 40掳C

To achieve an exposed surface temperature of \(40\,^{\circ}\mathrm{C}\), we can use the following relationship: \(T_{insulation} - T_{env} = U'R'_{total} (T_{steam} - T_{insulation})\) In this case, \(T_{insulation} = 40\,^{\circ}\mathrm{C}\) and \(T_{env}=20\,^{\circ}\mathrm{C}\). Redefine the thermal resistance as before: \(R''_{total} = \frac{1}{h_{in}A_{in}} + \frac{\ln{(r_{out}/r_{in})}}{2\pi k_{steel}L} + \frac{\ln{(r_{insulation}/r_{out})}}{2\pi k_{insulation}L} + \frac{1}{h_{out}A_{insulation}}\) Now we can solve for the insulation thickness using the equation: \(\frac{\ln{(r_{insulation}/r_{out})}}{2\pi k_{insulation}L} = \frac{1}{U''} - \frac{1}{h_{in}A_{in}} - \frac{\ln{(r_{out}/r_{in})}}{2\pi k_{steel}L} - \frac{1}{h_{out}A_{insulation}}\) Calculate the thickness of the insulation, \(t_{insulation} = r_{insulation} - r_{out}\).

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics that involves the movement of thermal energy from one place to another. It is crucial in analyzing how well heating or cooling systems perform and in calculating the insulation thickness for pipes or buildings. In the given exercise, we are examining a scenario where steam is flowing inside a steel pipe, and our goal is to minimize heat loss.

There are three modes of heat transfer: conduction, convection, and radiation. In our scenario, conduction occurs through the metal of the pipe, and convection is happening on both the inside and outside surfaces of the pipe due to the difference in temperature between the steam, the pipe material, and the surrounding environment. The exercise focuses on calculating the necessary insulation to achieve a significant reduction in heat loss by 95% and ensure safety by keeping the exterior pipe temperature at 40掳C.
Thermal Resistance
Thermal resistance refers to the ability of a material or system to resist the flow of heat. It's an analogous concept to electrical resistance, but instead of impeding electrical current, it hinders the transfer of heat. The higher the thermal resistance, the less heat is lost over time, making it a critical factor in insulation.

When performing insulation thickness calculation, as in our textbook exercise, we must understand the various resistances heat encounters as it moves from the hot steam inside the pipe, through the pipe's steel wall, the insulation, and eventually to the cooler environment outside. These resistances are then combined to form a total system resistance, which helps us to quantify the efficiency of the insulation. In the solution steps, we are provided with a formula that incorporates the thermal resistances of the pipe and insulation to solve for the heat transfer rate reduction and the necessary insulation thickness.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, symbolized by 'U', is a measure that incorporates all forms of heat transfer (conduction, convection, and radiation) in a system. It is an indicator of how well a series of materials can transfer heat. The overall heat transfer coefficient is inversely proportional to the total thermal resistance of the system.

Let's apply this to our exercise, where calculating the 'U' factor allows us to understand the heat transfer characteristics of the pipe with and without insulation. When insulation is added, we calculate a new 'U' to reflect the additional resistance. By comparing the original and the new 'U' values, we can determine the effectiveness of different insulation thicknesses and choose the one that meets our goal鈥攅ither the 95% heat loss reduction or the specific exterior temperature condition for safety.

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Most popular questions from this chapter

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Consider a flat ceiling that is built around \(38-\mathrm{mm} \times\) \(90-\mathrm{mm}\) wood studs with a center-to-center distance of \(400 \mathrm{~mm}\). The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a \(13-\mathrm{mm}\) plywood, a layer of felt \(\left(R=0.011 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), and linoleum \(\left(R=0.009 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\). Both sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide air space between the studs ( \(a\) ) does not have any reflective surface, (b) has a reflective surface with \(\varepsilon=0.05\) on one side, and ( ) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the air space.

Consider a wall that consists of two layers, \(A\) and \(B\), with the following values: \(k_{A}=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{A}=8 \mathrm{~cm}, k_{B}=\) \(0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{B}=5 \mathrm{~cm}\). If the temperature drop across the wall is \(18^{\circ} \mathrm{C}\), the rate of heat transfer through the wall per unit area of the wall is (a) \(180 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(153 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(89.6 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(72 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(51.4 \mathrm{~W} / \mathrm{m}^{2}\)

A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

A \(5-\mathrm{mm}\)-diameter spherical ball at \(50^{\circ} \mathrm{C}\) is covered by a \(1-\mathrm{mm}\)-thick plastic insulation \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The ball is exposed to a medium at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball.
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