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Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of \(300-\mathrm{mm}\) inner diameter and \(30-\mathrm{mm}\) wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation \((k=0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tubes, while degradation of the insulation is reduced by wrapping it in a thin sheet of aluminum having an emissivity of \(\varepsilon=0.20\). The air and wall temperatures of the power plant are \(27^{\circ} \mathrm{C}\). (a) Assuming that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to ensure that the temperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss(b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length. per meter of tube length?

Step by step solution

01

Calculate the thermal resistance of the steel tube

We can calculate the thermal resistance of the steel tube (\(R_{st}\)) using the following equation: \(R_{st}=\frac{\ln(r_{2}/r_{1})}{2 \pi k_{st} L}\) Where: - \(r_1 =\) inner radius of the tube = \(300 \times 10^{-3} / 2 = 0.15\, \mathrm{m}\) - \(r_2 =\) outer radius of the tube = \((300 + 30) \times 10^{-3} / 2 = 0.165\, \mathrm{m}\) - \(k_{st} =\) thermal conductivity of the steel tube = $35\, \mathrm{W / m \cdot K}\) - L = length of the tube (we are considering 1 meter for the exercise) Now calculate the \(R_{st}\).

02

Calculate the thermal resistance of the insulation

Next, we will calculate the thermal resistance of insulation (\(R_{ins}\)). We are not given the insulation thickness, so let the outer radius of the insulation be \(r_3\). Then we can find the resistance as: \(R_{ins}=\frac{\ln(r_{3}/r_{2})}{2 \pi k_{ins} L}\) Where: - \(k_{ins} =\) thermal conductivity of the insulation = $0.10\, \mathrm{W / m \cdot K}\)

03

Calculate the convection resistance between the aluminum and air

We are given the convection coefficient outside the aluminum sheet (\(h_c = 6\, \mathrm{W / m^2 \cdot K}\)). We can calculate the convection resistance (\(R_c\)) as follows: \(R_{c}=\frac{1}{h_c A}\) Where: - A = area of heat transfer = \(2 \pi r_3 L\)

04

Calculate the total resistance of the system

Now that we have the individual resistance values, we can calculate the total resistance (\(R_{total}\)) of the system by summing them up: \(R_{total}=R_{st}+R_{ins}+R_{c}\)

05

Apply energy balance and solve for insulation thickness

We will apply the energy balance equation considering the temperature difference between the steam and air, and the temperature of the aluminum: \(Q=\frac{T_{steam}-T_{air}}{R_{total}}=\frac{T_{steam}-T_{aluminum}}{R_{st}+R_{ins}}\) We are given: - \(T_{steam} = 575^{\circ} \mathrm{C}\) - \(T_{air} = 27^{\circ} \mathrm{C}\) - \(T_{aluminum} = 50^{\circ} \mathrm{C}\) By plugging these values and solving for \(r_3\), we can determine the minimum insulation thickness.

06

Calculate heat loss per meter of tube length

Once we have the minimum thickness of insulation, we can use the energy balance equation to find the heat loss in \(W/m\): \(Q=\frac{T_{steam}-T_{air}}{R_{total}}\)

07

(b) Effect of insulation thickness on the temperature of the aluminum and the heat loss per unit tube length

To explore the effect of insulation thickness, we can vary the outer radius of the insulation, \(r_3\), and calculate corresponding aluminum temperature and heat loss values using the energy balance equation and resistance formulas. This will help us understand how increasing or decreasing insulation thickness affects the aluminum's temperature and the overall heat loss from the system.

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Key Concepts

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

Heat Transfer

Understanding how heat moves from one place to another is essential for managing energy efficiently. Heat transfer occurs through three main mechanisms: conduction, convection, and radiation. In solids, heat transfer mainly occurs via conduction, where kinetic energy is passed from atom to atom or through free electrons. Convection is the transfer of heat by the physical movement of a fluid (liquid or gas). Radiation involves heat transfer through electromagnetic waves and can occur in a vacuum.

Consider a hot cup of coffee left on the table; over time, it cools down as heat is transferred from the coffee to the surrounding air. This is an example of heat transfer in action. The cup transfers heat to the air mainly through conduction at the point of contact, and then the air carries it away through convection. Moreover, the coffee emits heat via radiation, which doesn't require a medium to travel through.

Thermal Resistance

Thermal resistance is a measure of a material's ability to resist the flow of heat. It is determined by the material's intrinsic properties, such as its thermal conductivity, as well as its dimensions. Thermal resistance is analogous to electrical resistance and follows similar principles. The higher the thermal resistance, the better the material is at insulating.

When it comes to insulating pipes, like in the given exercise, choosing a material with high thermal resistance effectively slows down heat loss from the pipe's contents, keeping the fluid at the desired temperature and preventing heat from escaping into the surroundings. Stepping through the solution, we calculate thermal resistances for different layers (steel tube, insulation) to understand how they contribute to the total heat transfer and determine the optimal insulation thickness.

Conduction in Solids

Heat conduction in solids is a way of transfer that happens due to temperature differences within the material. When one part of a solid is heated, the particles in that area gain kinetic energy and begin to vibrate more rapidly. These vibrations spread, transferring energy to neighboring particles. The process continues, resulting in heat moving from the warmer area to the cooler one.

In the context of the exercise, conduction is what allows heat to move through the steel tube walls and insulation. The thermal conductivity (\( k \)) of the material plays a significant role here: the higher it is, the more easily heat can pass through. As the exercise suggests, we measure this property to determine how much insulation is required to prevent excessive heat from being conducted to the external environment.