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°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°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°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—either the 95% heat loss reduction or the specific exterior temperature condition for safety.

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Most popular questions from this chapter

Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually decrease heat transfer from a surface.

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to bery long, its efficiency is (a) \(0.60\) (b) \(0.67\) (c) \(0.72\) (d) \(0.77\) (e) \(0.88\)

What is a conduction shape factor? How is it related to the thermal resistance?

Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

Two 3-m-long and \(0.4-\mathrm{cm}\)-thick cast iron \((k=\) \(52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) steam pipes of outer diameter \(10 \mathrm{~cm}\) are connected to each other through two 1 -cm-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

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