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Chapter 3: Problem 78

A 50 -m-long section of a steam pipe whose outer (鈧�) 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 an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not?

One wall of a refrigerated warehouse is \(10.0\)-m-high and \(5.0\)-m-wide. The wall is made of three layers: \(1.0\)-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fibreglass \((k=\) \(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0-\mathrm{cm}\) thick gypsum board \((k=\) \(0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts \((k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), each \(2.0 \mathrm{~cm}\) in diameter and \(12.0 \mathrm{~cm}\) long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?

The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?

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.

A turbine blade made of a metal alloy \((k=\) \(17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a length of \(5.3 \mathrm{~cm}\), a perimeter of \(11 \mathrm{~cm}\), and a cross-sectional area of \(5.13 \mathrm{~cm}^{2}\). The turbine blade is exposed to hot gas from the combustion chamber at \(973^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(538 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The base of the turbine blade maintains a constant temperature of \(450^{\circ} \mathrm{C}\) and the tip is adiabatic. Determine the heat transfer rate to the turbine blade and temperature at the tip.
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