Steam at \(450^{\circ} \mathrm{F}\) is flowing through a steel pipe \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) whose inner and outer diameters are \(3.5\) in and \(4.0\) in, respectively, in an environment at \(55^{\circ} \mathrm{F}\). The pipe is insulated with 2 -in-thick fiberglass insulation \((k=\) \(\left.0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and \(5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?

Short Answer

Expert verified
Answer: The rate of heat loss from the steam per foot length of the pipe is 1920.52 Btu/h, and the error involved in neglecting the steel pipe's thermal resistance is 0.65%.

Step by step solution

01

Convert units to consistent values

To ensure consistency in our calculations, let's first convert inches to feet: Inner diameter: \(3.5 \,\text{in} = 3.5/12 \,\text{ft} = 0.29167 \,\text{ft}\) Outer diameter: \(4.0 \,\text{in} = 4/12 \,\text{ft} = 0.33333 \,\text{ft}\) Insulation thickness: \(2 \,\text{in} = 2/12 \,\text{ft} = 0.16667 \,\text{ft}\)
02

Calculate thermal resistances and the overall heat transfer coefficient (U)

There are three types of thermal resistances: 1. Convection inside the pipe (R1): \(\frac{1}{h_1 \cdot A_1} = \frac{1}{30 \cdot (2\pi\cdot0.29167/2) } = 0.01131 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) 2. Conduction through the steel pipe (R2): \(\frac{\ln(D_2/D_1)}{2\pi k_1 L} = \frac{\ln(0.33333/0.29167)}{2\pi \cdot 8.7} = 0.00149 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) 3. Conduction through the insulation (R3): \(\frac{\ln(D_3/D_2)}{2\pi k_2 L} = \frac{\ln((0.33333+0.16667)/0.33333)}{2\pi \cdot 0.020} = 0.25671 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) 4. Convection outside the pipe (R4): \(\frac{1}{h_2 \cdot A_2} = \frac{1}{5 \cdot (2\pi\cdot0.5) } = 0.05305 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) Total thermal resistance (R_total) = R1 + R2 + R3 + R4 = 0.01131 + 0.00149 + 0.25671 + 0.05305 = 0.32257 \,\text{h} \cdot \text{ft\(^2\)} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}$ Overall heat transfer coefficient (U) = \(\frac{1}{R_\text{total}} = \frac{1}{0.32257} = 3.099 \,\text{Btu} \cdot \text{h}^{-1} \cdot \text{ft\)^{-2}\(} \cdot { }^{\circ}\text{F}^{-1}\)
03

Calculate heat loss per foot length of the pipe (Q)

To find the heat loss, we can use the formula: \(Q = U \cdot A \cdot \Delta T\) Area of the outer surface (A): \(2\pi(\frac{D_3}{2})\cdot L = 2\pi(\frac{0.33333+0.16667}{2}) = 1.5708 \,\text{ft\)^2\(}\) Temperature difference (\(\Delta T\)): \(450 \,^{\circ}\text{F} - 55 \,^{\circ}\text{F} = 395 \,^{\circ}\text{F}\) Heat loss (Q): \(3.099 \cdot 1.5708 \cdot 395 = 1920.52 \,\text{Btu/h}\)
04

Calculate heat loss without the thermal resistance of the steel pipe (Q_without_R2)

To calculate this, we use the same formula to find the heat transfer rate without considering R2: R_total_without_R2 = R1 + R3 + R4 = 0.01131 + 0.25671 + 0.05305 = 0.32108 \,\text{h} \cdot \text{ft\(^2\)} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}$ Overall heat transfer coefficient (U_without_R2) = \(\frac{1}{R_\text{total without R2}} = \frac{1}{0.32108} = 3.114 \,\text{Btu} \cdot \text{h}^{-1} \cdot \text{ft\)^{-2}\(} \cdot { }^{\circ}\text{F}^{-1}\) Heat loss without considering the steel pipe's resistance (Q_without_R2): \(3.114 \cdot 1.5708 \cdot 395 = 1932.99 \,\text{Btu/h}\)
05

Calculate the error involved due to neglecting R2

To find the error, we can use the formula: Error = \(\frac{Q_\text{without R2} - Q}{Q} \times 100\%\) Error = \(\frac{1932.99 - 1920.52}{1920.52} \times 100\% = 0.65\%\) The error involved in neglecting the thermal resistance of the steel pipe is 0.65%.

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Key Concepts

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

Thermal Resistance
Thermal resistance is a concept in thermodynamics that measures the opposition to heat flow through a material. It's analogous to electrical resistance, which measures how much a material resists the flow of electricity. In our context, thermal resistance is crucial for calculating how well a material insulates, or in other words, slows down the transfer of heat.

For instance, in the step-by-step solution provided, we see individual thermal resistances for various layers around the steam pipe: convection inside the pipe, conduction through the steel and the fiberglass insulation, and finally convection outside the pipe. By calculating the thermal resistance of each layer and summing them up, we can determine the total resistance to heat loss.

In practice, the lower the thermal resistance, the greater the heat flow. Insulation materials, such as fiberglass, are designed with high thermal resistance to minimize heat loss in thermal applications. Conversely, materials with low thermal resistance, like metals, are excellent heat conductors. Knowing the thermal resistance values for different materials is essential for engineers and architects to design energy-efficient systems and structures, such as the steam pipe in the exercise.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, commonly represented by the symbol U, is a measure that combines the thermal resistances of different layers and interfaces to quantify how easily heat can pass through a system. It's defined as the amount of heat that passes per unit area, per unit temperature difference, per unit time.

In the solution, the overall heat transfer coefficient (U) takes into account all the resistances - including those caused by conduction through the steel pipe and insulation, as well as convection inside and outside the pipe. It's calculated by taking the inverse of the total thermal resistance of the system. The U-factor is particularly important when designing environmental control systems, as it helps to predict the heat loss or gain, which in turn affects energy consumption and efficiency.

Note the step involving the calculation of U and its importance: the lower the U-value, the better the composite material is at insulating. This step is critical to properly assess the performance of the insulation in preventing heat loss, thereby guiding decisions regarding material selection and system design.
Conduction and Convection
Conduction and convection are two primary ways that heat transfers through materials and fluids. Conduction is the process of heat transfer through a material without any movement of the material itself. It occurs at the microscopic level as vibrating, heat-carrying particles bump into their neighbors, transferring energy. This principle is seen at work in the exercise when heat conducts through the steel pipe wall.

Convection, on the other hand, is the heat transfer due to the bulk movement of molecules within fluids (gases and liquids), carrying heat along with them. This is especially pertinent in systems where a fluid, like steam or water, is involved. In the provided exercise, convection is at play both inside and outside the steam pipe, where it carries heat away from the pipe's surface, contributing to the system's thermal resistance.

Understanding both conduction and convection is necessary not only to calculate heat loss effectively but also to design systems with optimal temperature control. These principles help us determine the efficiency of heaters, coolers, and insulators in real-world applications, which is directly linked to energy conservation and operational cost.

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Most popular questions from this chapter

Hot- and cold-water pipes \(8 \mathrm{~m}\) long run parallel to each other in a thick concrete layer. The diameters of both pipes are \(5 \mathrm{~cm}\), and the distance between the centerlines of the pipes is \(40 \mathrm{~cm}\). The surface temperatures of the hot and cold pipes are \(60^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. Taking the thermal conductivity of the concrete to be \(k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the rate of heat transfer between the pipes.

The plumbing system of a house involves a \(0.5-\mathrm{m}\) section of a plastic pipe \((k=0.16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner diameter \(2 \mathrm{~cm}\) and outer diameter \(2.4 \mathrm{~cm}\) exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about \(-5^{\circ} \mathrm{C}\) for a period of \(14 \mathrm{~h}\). The combined convection and radiation heat transfer coefficient on the outer surface of the pipe is estimated to be \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Assuming the pipe to contain stationary water initially at \(0^{\circ} \mathrm{C}\), determine if the water in that section of the pipe will completely freeze that night.

A 3-cm-long, 2-mm \(\times 2-\mathrm{mm}\) rectangular crosssection aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the fin efficiency is 65 percent, the effectiveness of this single fin is (a) 39 (b) 30 (c) 24 (d) \(18 \quad(e) 7\)

A total of 10 rectangular aluminum fins \((k=\) \(203 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) are placed on the outside flat surface of an electronic device. Each fin is \(100 \mathrm{~mm}\) wide, \(20 \mathrm{~mm}\) high and \(4 \mathrm{~mm}\) thick. The fins are located parallel to each other at a center- tocenter distance of \(8 \mathrm{~mm}\). The temperature at the outside surface of the electronic device is \(60^{\circ} \mathrm{C}\). The air is at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine \((a)\) the rate of heat loss from the electronic device to the surrounding air and \((b)\) the fin effectiveness.

Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain.

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