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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

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

How does the thermal resistance network associated with a single-layer plane wall differ from the one associated with a five-layer composite wall?

Two 5-cm-diameter, 15-cm-long aluminum bars \((k=\) \(176 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with ground surfaces are pressed against each other with a pressure of \(20 \mathrm{~atm}\). The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the twobar system are maintained at temperatures of \(150^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine \((a)\) the rate of heat transfer along the cylinders under steady conditions and (b) the temperature drop at the interface. Answers: (a) \(142.4 \mathrm{~W}\), (b) \(6.4^{\circ} \mathrm{C}\)

A pipe is insulated to reduce the heat loss from it. However, measurements indicate that the rate of heat loss has increased instead of decreasing. Can the measurements be right?

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

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