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Chapter 3: Problem 78

A 50 -m-long section of a steam pipe whose outer (€) diameter is \(10 \mathrm{~cm}\) passes through an open space at \(15^{\circ} \mathrm{C}\). The average temperature of the outer surface of the pipe is measured to be \(150^{\circ} \mathrm{C}\). If the combined heat transfer coefficient on the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine (a) the rate of heat loss from the steam pipe; \((b)\) the annual cost of this energy lost if steam is generated in a natural gas furnace that has an efficiency of 75 percent and the price of natural gas is $$\$ 0.52 /$$ therm ( 1 therm \(=105,500 \mathrm{~kJ})\); and \((c)\) the thickness of fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) needed in order to save 90 percent of the heat lost. Assume the pipe temperature to remain constant at \(150^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Based on the given information and formulas, we can solve step by step as follows: Step 1: Calculate the rate of heat loss from the steam pipe \(A = \pi DL = \pi(0.1)(50) = 15.71 \mathrm{m}^{2}\) \(q = (20)(\pi(0.1)(50))(135) = 42,411 \mathrm{W} = 42.411 \mathrm{kW}\) Step 2: Calculate the annual cost of energy lost \(q_\mathrm{kJ/year} = q \cdot \frac{3600 \mathrm{~kJ}}{1 \mathrm{~kW} \cdot \mathrm{h}} \cdot \frac{8760 \mathrm{~h}}{1 \mathrm{~year}} = 42.411 \cdot 3600 \cdot 8760 = 1,334,317,456 \mathrm{~kJ/year}\) \(q_\mathrm{therm/year} = \frac{q_\mathrm{kJ/year}}{105,500 \mathrm{~kJ/therm}} = 12,655 \mathrm{~therm/year}\) \(C = \frac{q_\mathrm{therm/year}}{\eta} \cdot P = \frac{12,655}{0.75} \cdot 0.52 = \$8,767.67\) Step 3: Calculate the required insulation thickness \(q_\mathrm{target} = \frac{q}{10} = 4,241.1 \mathrm{W}\) \(x = \frac{(0.035)(\pi(0.1)(50))(135)}{q_\mathrm{target}} = \frac{(0.035)(15.71)(135)}{4,241.1} = 0.219 \mathrm{m}\) Thus, the rate of heat loss from the steam pipe is 42.411 kW, the annual cost of energy lost is \$8,767.67, and the required insulation thickness to save 90 percent of the heat lost is 0.219 m.

Step by step solution

01

Calculate the rate of heat loss from the steam pipe

To find the rate of heat loss, we will use the formula for the rate of heat transfer, which is: \(q = hA \Delta T\) where \(q\) is the rate of heat loss, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the pipe, and \(\Delta T\) is the temperature difference between the pipe and the surrounding air. Given the outer diameter (\(D = 0.1 \mathrm{m}\)), length (\(L = 50 \mathrm{m}\)), heat transfer coefficient (\(h = 20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)), and temperature difference (\(\Delta T = 150^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 135 \mathrm{K}\)), we can calculate the surface area and the rate of heat loss as follows: \(A = \pi DL = \pi(0.1)(50)\) \(q = (20)(\pi(0.1)(50))(135)\)
02

Calculate the annual cost of energy lost

Given the efficiency of the natural gas furnace (\(\eta = 0.75\)) and the price of natural gas (\(P = \$0.52/\mathrm{therm}\)), we can find the annual cost of the energy lost as follows: First, convert the rate of heat loss to kJ/year: \(q_\mathrm{kJ/year} = q \cdot \frac{3600 \mathrm{~kJ}}{1 \mathrm{~kW} \cdot \mathrm{h}} \cdot \frac{8760 \mathrm{~h}}{1 \mathrm{~year}}\) Next, find the energy lost in terms of therms per year: \(q_\mathrm{therm/year} = \frac{q_\mathrm{kJ/year}}{105,500 \mathrm{~kJ/therm}} \) Finally, find the annual cost of energy lost: \(C = \frac{q_\mathrm{therm/year}}{\eta} \cdot P\)
03

Calculate the required insulation thickness

To save 90 percent of the heat lost, we need to reduce the heat transfer rate by a factor of 10. We will use the following formula for the heat transfer rate with insulation: \(q_\mathrm{insulated} = \frac{kA\Delta T}{x}\) where \(q_\mathrm{insulated}\) is the rate of heat transfer with insulation, \(k\) is the thermal conductivity of the insulation, and \(x\) is the insulation thickness. First, find the target heat transfer rate with insulation: \(q_\mathrm{target} = \frac{q}{10}\) Next, rearrange the formula to solve for insulation thickness \(x\): \(x = \frac{kA\Delta T}{q_\mathrm{target}}\) Finally, substitute the given values and calculate the required insulation thickness: \(x = \frac{(0.035)(\pi(0.1)(50))(135)}{q_\mathrm{target}}\)

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It quantifies the rate at which heat passes through a material and is denoted by the symbol 'k'. This property is crucial in understanding how materials act as insulators or conductors of heat. In the context of the given exercise, the thermal conductivity of fiberglass insulation is provided as \(k = 0.035 \frac{W}{m \times K}\), indicating that it's a good insulator with low thermal conductivity.

When calculating the thickness of insulation required to prevent heat loss, a material's thermal conductivity serves as a foundational value. The lower the thermal conductivity, the less heat is transferred through the material, and thus, the better it is for insulation purposes. This characterizes the efficiency of the material in hindering heat flow and is essential for calculating how much insulation is necessary to achieve a certain level of heat conservation, as seen in step 3 of the solution.
Energy Conservation in Thermal Systems
Energy conservation in thermal systems is based on the principle that energy cannot be created or destroyed, only transferred or converted from one form to another. This fundamental concept, also known as the first law of thermodynamics, is applied when calculating the heat loss from a steam pipe, and subsequently, the cost of the energy lost due to this heat transfer.

In the provided exercise, heat loss is determined by the temperature difference between the steam pipe and its surroundings, the surface area of the pipe, and the heat transfer coefficient. The aim is to minimize the loss of heat, which is energy taking the form of heat from the hot steam, to the cooler environment. By knowing the amount of energy lost, one can calculate the economic impact and devise strategies, such as insulation, to conserve energy and reduce cost, as detailed in steps 1 and 2 of the solution. Understanding the thermal system's energy conservation helps in designing measures that can effectively reduce energy waste and improve efficiency.
Insulation Thickness
The insulation thickness is a key factor in the efficacy of insulation material in reducing heat loss from a system. In our exercise, the steam pipe is losing heat to its surroundings, and one way to minimize this loss is by wrapping the pipe with insulation. The necessary thickness of this insulation significantly impacts the rate of heat transfer, and calculating it requires insights into energy conservation and knowledge of the thermal conductivity of the insulation material.

The aim is to have enough insulation to create a substantial barrier against heat flow without being excessively thick. This balance is critical for cost-effectiveness and practicality. In step 3 of the provided solution, the insulation thickness calculation involves the determination of the desired rate of heat loss (which in this case is 90% less than the current rate), the thermal conductivity of the insulation, the temperature gradient, and the geometry of the pipe. It illustrates how an adequate thickness of the insulation material can lead to significant energy savings by retaining heat within the thermal system more effectively.

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Most popular questions from this chapter

What is the reason for the widespread use of fins on surfaces?

A 10-m-long 5-cm-outer-radius cylindrical steam pipe is covered with \(3-\mathrm{cm}\) thick cylindrical insulation with a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the rate of heat loss from the pipe is \(1000 \mathrm{~W}\), the temperature drop across the insulation is (a) \(163^{\circ} \mathrm{C}\) (b) \(600^{\circ} \mathrm{C}\) (c) \(48^{\circ} \mathrm{C}\) (d) \(79^{\circ} \mathrm{C}\) (e) \(251^{\circ} \mathrm{C}\)

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.

Two flow passages with different cross-sectional shapes, one circular another square, are each centered in a square solid bar of the same dimension and thermal conductivity. Both configurations have the same length, \(T_{1}\), and \(T_{2}\). Determine which configuration has the higher rate of heat transfer through the square solid bar for \((a) a=1.2 b\) and \((b) a=2 b\).

Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one? (a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness
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