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A \(200-\mathrm{ft}\)-long section of a steam pipe whose outer diameter is 4 in passes through an open space at \(50^{\circ} \mathrm{F}\). The average temperature of the outer surface of the pipe is measured to be $280^{\circ} \mathrm{F}$, and the average heat transfer coefficient on that surface is determined to be $6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F}\(. Determine \)(a)$ the rate of heat loss from the steam pipe and \((b)\) the annual cost of this energy loss if steam is generated in a natural gas furnace having an efficiency of 86 percent and the price of natural gas is \(\$ 1.10 /\) therm (1 therm \(=100,000\) Btu).

Short Answer

Expert verified
Answer: The rate of heat loss from the steam pipe is approximately 276000π Btu/h, and the annual cost of this energy loss is approximately $96,539.41.

Step by step solution

01

Calculate the pipe's surface area

First, we need to find the surface area (A) of the pipe. The formula for the surface area of a cylinder is \(A=2\pi rh\), where r is the radius of the pipe and h is the height (length) of the pipe. We are given the diameter (4 in), so we can easily find the radius by dividing the diameter by 2. \(r = \frac{4\,\text{in}}{2} = 2\,\text{in}\) or \(r = \frac{1}{6}\,\text{ft}\) We are also given the length of the pipe (200 ft), so the surface area is: \(A = 2\pi * \frac{1}{6}\,\text{ft} * 200\,\text{ft} = 200\pi\,\text{ft}^2\)
02

Calculate the rate of heat loss

Now, we can use the formula for heat transfer through convection to find the rate of heat loss (Q) from the pipe. The formula is \(Q = hA(T_s - T_\infty)\), where h is the heat transfer coefficient, A is the surface area, \(T_s\) is the outer surface temperature of the pipe, and \(T_\infty\) is the open space temperature. We are given the heat transfer coefficient, h (6 Btu / h·ft²·°F), the outer surface temperature, \(T_s\) (280°F), and the open space temperature, \(T_\infty\) (50°F). Plugging in the values, we get: \(Q = 6\,\text{Btu/h·ft}^2·\text{°F} * 200\pi\,\text{ft}^2 * (280\,\text{°F} - 50\,\text{°F}) = 276000\pi\,\text{Btu/h}\)
03

Calculate the annual energy loss

To calculate the annual energy loss, multiply the rate of heat loss (Q) by the number of hours in a year (8,760 h): \(\text{Annual energy loss} = 276000\pi\,\text{Btu/h} * 8760\,\text{h} = 2411500800\pi\,\text{Btu}\)
04

Calculate the annual cost of energy loss

To find the annual cost of this energy loss, we will first convert the energy loss to therms (1 therm = 100,000 Btu) and then use the cost of natural gas and the efficiency of the natural gas furnace (86% or 0.86). \(\text{Energy loss in therms} = \frac{2411500800\pi\,\text{Btu}}{100,000\,\text{Btu/therm}} = 24115.008\pi\,\text{therms}\) Since the furnace is 86% efficient, we need to divide the energy loss by the efficiency to find the amount of natural gas needed: \(\text{Natural gas needed} = \frac{24115.008\pi\,\text{therms}}{0.86} = 28017.2186\pi\,\text{therms}\) Now, we can multiply the amount of natural gas needed by the price of natural gas (\(\$1.10/\text{therm}\)) to find the annual cost of energy loss: \(\text{Annual cost of energy loss} = 28017.2186\pi\,\text{therms} * \$1.10/\text{therm} \approx \$96,539.41\) So, (a) the rate of heat loss from the steam pipe is approximately \(276000\pi\,\text{Btu/h}\); and (b) the annual cost of this energy loss is approximately \$96,539.41.

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Most popular questions from this chapter

The inner and outer surfaces of a 25 -cm-thick wall in summer are at \(27^{\circ} \mathrm{C}\) and \(44^{\circ} \mathrm{C}\), respectively. The outer surface of the wall exchanges heat by radiation with surrounding surfaces at \(40^{\circ} \mathrm{C}\) and by convection with ambient air also at $40^{\circ} \mathrm{C}\( with a convection heat transfer coefficient of \)8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Solar radiation is incident on the surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\). If both the emissivity and the solar absorptivity of the outer surface are \(0.8\), determine the effective thermal conductivity of the wall.

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