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You are in charge of energy conservation at your plant. There is a \(300 \mathrm{~m}\) run of 6 in. O.D. pipe carrying steam at \(250^{\circ} \mathrm{C}\). The company requires that any insulation must pay for itself in one year. The thermal resistances are such that the surface of the pipe will stay close to \(250^{\circ} \mathrm{C}\) in air at \(25^{\circ} \mathrm{C}\) when \(\bar{h}=10\) \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\). Calculate the annual energy savings in \(\mathrm{kW} \cdot \mathrm{h}\) that will result if a 1 in layer of \(85 \%\) magnesia insulation is added. If energy is worth 6 cents per \(\mathrm{kW} \cdot \mathrm{h}\) and insulation costs \(\$ 75\) per installed linear meter, will the insulation pay for itself in one year?

Short Answer

Expert verified
The insulation will pay for itself in less than one year.

Step by step solution

01

Calculate Initial Heat Loss

Calculate the initial heat loss from the uninsulated pipe using the formula: \[ q = (T_s - T_a) \cdot \bar{h} \cdot \pi \cdot D \cdot L \]where:- \(T_s = 250^{\circ} \mathrm{C}\) is the surface temperature of the pipe,- \(T_a = 25^{\circ} \mathrm{C}\) is the ambient temperature,- \(\bar{h} = 10 \, \mathrm{W}/\mathrm{m}^2 \, \mathrm{K}\) is the heat transfer coefficient,- \(D = 6 \, \text{inches} \times 0.0254 \, \mathrm{m}/\mathrm{inch} = 0.1524 \, \mathrm{m}\) is the pipe diameter,- \(L = 300 \, \mathrm{m}\) is the length of the pipe.Substitute in these values and calculate \(q\):\[ q = (250 - 25) \cdot 10 \cdot \pi \cdot 0.1524 \cdot 300 \] \[ q = 335,320 \, \mathrm{W} \]
02

Calculate Heat Loss with Insulation

Use the formula for heat loss with insulation:\[ q_{\text{insulated}} = \frac{2 \pi L (T_s - T_a)}{\ln\left(\frac{D_2}{D_1}\right)/k + 1/\bar{h}} \] where:- \(D_1 = 0.1524 \, \mathrm{m}\) is the outer diameter of the pipe,- \(D_2 = D_1 + 2(1 \, \text{inch} \times 0.0254 \, \mathrm{m/inch}) = 0.2032 \, \mathrm{m}\) is the diameter with insulation,- \(k = 0.049 \, \mathrm{W}/\mathrm{m}\,\mathrm{K}\) is the thermal conductivity of 85% magnesia.Substitute these into the formula:\[ q_{\text{insulated}} = \frac{2 \pi \cdot 300 \cdot (250 - 25)}{\ln(0.2032/0.1524) / 0.049 + 1/10} \]\[ q_{\text{insulated}} = 49,292 \, \mathrm{W} \]
03

Calculate Energy Savings

Calculate the energy savings due to insulation:\[ \text{Energy Savings} = (q - q_{\text{insulated}}) \cdot \text{seconds in a year} \]where seconds in a year is \(365 \times 24 \times 3600\).\[ \text{Energy Savings} = (335,320 - 49,292) \cdot 365 \times 24 \times 3600 \]Convert to \(\mathrm{kW} \cdot \mathrm{h}\) by dividing by 3,600:\[ \text{Energy Savings} = 2,503,402 \, \mathrm{kW} \cdot \mathrm{h} \]
04

Determine Cost Savings and Insulation Cost

Calculate the cost savings by multiplying the energy savings by the cost per \(\mathrm{kW} \cdot \mathrm{h}\), which is 0.06 dollars.\[ \text{Cost Savings} = 2,503,402 \times 0.06 = 150,204 \, \text{dollars} \]Calculate the cost of insulation:\[ \text{Insulation Cost} = 75 \times 300 = 22,500 \, \text{dollars} \]
05

Conclusion

Compare the cost savings with the insulation cost.Since the annual cost savings (\\(150,204) is greater than the insulation cost (\\)22,500), the insulation will pay for itself in less than one year.

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Key Concepts

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

Heat Transfer
When discussing heat transfer, it's important to understand how thermal energy moves from one substance to another. In simple terms, heat transfer is the movement of thermal energy from a hotter object to a cooler one. This process can occur in three different ways:
  • Conduction: Heat is transferred through direct contact. Think of how a metal spoon gets hot when placed in a pot of boiling water.
  • Convection: The transfer occurs through fluid motion. This is common in gases and liquids, such as warm air rising from a heater.
  • Radiation: Heat is transferred through electromagnetic waves. An example is the warmth felt from sunlight.
In a steel pipe carrying steam, heat is primarily lost through conduction. The heat from the steam is conducted through the pipe material, eventually being transferred to the surrounding air. This unregulated transfer means the pipe loses a significant amount of energy, which is where insulation can play an important role in minimizing heat loss.
Energy Conservation
Energy conservation is all about using energy efficiently and minimizing waste. When managing a system like a steam pipe, conserving energy means minimizing the amount of heat lost to the environment. This savings is not only beneficial for reducing energy costs but also for lessening the environmental impact.

In the context of the exercise, energy conservation is achieved by insulating the pipe. Adding insulation reduces the rate of heat transfer from the pipe to the outside air, which helps to retain more of the steam’s energy within the system. To determine whether the insulation is worth the cost, you need to consider:
  • The initial heat loss from the uninsulated pipe.
  • The decrease in heat loss due to the insulation.
  • The annual energy savings achieved by reducing this heat loss.
The exercise shows that with the addition of magnesia insulation, significant savings are achieved, thus demonstrating the financial and practical benefits of energy conservation.
Thermal Conductivity
Thermal conductivity is a material property that indicates how well a material can conduct heat. High thermal conductivity means heat passes through easily, while low thermal conductivity means the material is good at insulating against heat transfer.

In the exercise, 85% magnesia is used as the insulating material. Its thermal conductivity value is relatively low (\(0.049 \, \text{W/mK}\)), making it effective for slowing down the heat transfer from the steam pipe to the outside environment. This low thermal conductivity of the magnesia ensures that less heat is lost, aiding significantly in energy savings.
  • Lower thermal conductivity means better insulation.
  • Materials like magnesia help retain heat, thereby conserving energy.
  • Choosing the right insulating material is crucial for achieving effective energy savings.
By understanding thermal conductivity, one can not only select the appropriate material for insulating pipes but also optimize energy efficiency in various applications involving heat transfer.

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Most popular questions from this chapter

An outside pipe is insulated and we measure its temperature with a thermocouple. The pipe serves as an electrical resistance heater, and \(\dot{q}\) is known from resistance and current measurements. The inside of the pipe is cooled by the flow of liquid with a known bulk temperature. Evaluate the heat transfer coefficient, \(\bar{h}\), in terms of known information. The pipe dimensions and properties are known. [Hint: Remember that \(\bar{h}\) is not known and we cannot use a boundary condition of the third kind at the inner wall to get \(T(r) .]\)

A \(1 \mathrm{~kW}\) commercial electric heating rod, \(8 \mathrm{~mm}\) in diameter and \(0.3 \mathrm{~m}\) long, is to be used in a highly corrosive gaseous environment. Therefore, it has to be provided with a cylindrical sheath of fireclay. The gas flows by at \(120^{\circ} \mathrm{C}\), and \(\bar{h}\) is \(230 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) outside the sheath. The surface of the heating rod cannot exceed \(800^{\circ} \mathrm{C}\). Set the maximum sheath thickness and the outer temperature of the fireclay. [Hint: use heat flux and temperature boundary conditions to get the temperature distribution. Then use the additional convective boundary condition to obtain the sheath thickness.]

Heat is generated at \(54,000 \mathrm{~W} / \mathrm{m}^{3}\) in a \(0.16 \mathrm{~m}\) diameter sphere. The sphere is cooled by natural convection with fluid at \(0{ }^{\circ} \mathrm{C}\), and \(\bar{h}=\left[2+6\left(T_{\text {surface }}-T_{\infty}\right)^{1 / 4}\right] \mathrm{W} / \mathrm{m}^{2} \mathrm{~K}, k_{\text {sphere }}=9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Find the surface temperature and center temperature of the sphere.

A student working heat transfer problems late at night needs a cup of hot cocoa to stay awake. She puts milk in a pan on an electric stove and seeks to heat it as rapidly as she can, without burning the milk, by turning the stove on high and stirring the milk continuously. Explain how this works using an analogous electric circuit. Is it possible to bring the entire bulk of the milk up to the burn temperature without burning part of it?

A slab of thickness \(L\) is subjected to a constant heat flux, \(q_{1}\), on the left side. The right-hand side if cooled convectively by an environment at \(T_{\infty}\). (a) Develop a dimensionless equation for the temperature of the slab. (b) Present dimensionless equation for the left- and right-hand wall temperatures as well. (c) If the wall is firebrick, \(10 \mathrm{~cm}\) thick, \(q_{1}\) is \(400 \mathrm{~W} / \mathrm{m}^{2}, \bar{h}=20\) \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\), and \(T_{\infty}=20^{\circ} \mathrm{C}\), compute the lefthand and righthand temperatures.

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