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Suprathane, manufactured by Rubicon Chemicals Inc., is a wall insulation consisting of a sandwich of urethane foam covered with a protective layer on either side. Each protective layer is a composite of aluminum foil over kraft paper interlaced with high-strength glass fiber. For a \(11 / 4\) in-thick sandwich, the \(\mathrm{R}\) value is \(13.2\left[\mathrm{Btu} / \mathrm{hr} \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\right]^{-1}\). When used in conjunction with a conventional \(31 / 2\) in-thick mineral wool blanket, a wall with a combined \(R\) value of \(22.7\) is claimed. (i) What is the thermal resistance per unit area of the sandwich in SI units? (ii) If inside and outside heat transfer coefficients are estimated to be \(7 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively, what is the rate of heat loss through a \(5 \mathrm{~m}\)-long, \(3 \mathrm{~m}\)-high wall when the inside temperature is \(20^{\circ} \mathrm{C}\) and the outside temperature is \(-20^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The sandwich's thermal resistance is 2.32452 m²K/W in SI units. The rate of heat loss through the wall is 237.07 W.

Step by step solution

01

Convert R Value to SI Units

The given thermal resistance (R value) in imperial units is \(13.2 \text{ Btu/hr ft}^2{}^{\circ} \text{F}^{-1}\). To convert to SI units, we use the conversion factor: \(1 \text{ Btu/hr ft}^2{}^{\circ} \text{F}^{-1} = 0.1761 \text{ m}^2 \text{ K/W}\). Thus, the R value in SI units is \(13.2 \times 0.1761 = 2.32452 \text{ m}^2 \text{ K/W}\).
02

Setup for Heat Transfer Calculation

We need to calculate the rate of heat loss through a wall. The wall size is given as 5 m × 3 m, hence its area \(A\) is \(5 \times 3 = 15 \text{ m}^2\). The temperature difference between inside and outside is \(20 - (-20) = 40^{\circ} \text{C}\).
03

Calculate Total Thermal Resistance

The total thermal resistance \(R_{\text{total}}\) of the wall setup includes the resistance of the inside surface, wall materials (the suprathane sandwich and mineral wool blanket), and the outside surface. The resistance due to the surface films is \(\frac{1}{7} + \frac{1}{20}\). Thus, \(R_{\text{total}} = \frac{1}{7} + 2.32452 + \frac{1}{20} = 2.52452\text{ m}^2 \text{ K/W} \).
04

Calculate Heat Loss Through the Wall

Using the formula for heat loss, \(Q = \frac{\Delta T}{R_{\text{total}}} \times A\), substitute the values: \(Q = \frac{40}{2.52452} \times 15 = 237.07 \text{ W}\). This is the rate of heat loss through the 5 m long, 3 m high wall.

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Key Concepts

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

Heat Transfer
Heat transfer is the movement of thermal energy from one object or material to another, and it can occur in three main ways: conduction, convection, and radiation. In this exercise, we're primarily dealing with conduction, which happens when heat moves through materials, like in our wall consisting of the Suprathane insulation and mineral wool blanket.

Conduction happens when molecules in a warmer area transfer their energy to molecules in a cooler area. In simple terms, when one side of a Suprathane sandwich gets warm, heat travels through the material to the cooler side. Convection, on the other hand, involves the movement of heat between a solid surface and a fluid (like air) in motion, but our main focus is on conduction in this scenario.

The thermal resistance of a material, which we calculated using the R value, helps us understand how well the material resists heat flow. Higher R values mean better insulation and less heat transfer. Using the formula, we understand that increasing the wall's thermal resistance helps in reducing the rate of heat loss, keeping the inside warm and the outside cold even in contrasting temperature conditions.
Insulation Materials
Insulation materials are crucial in controlling the energy efficiency of buildings. The primary purpose of insulation like the Suprathane sandwich in this exercise is to reduce heat flow, helping maintain a desired temperature inside a building even if it varies drastically outside.

Different materials have different insulating properties and can be combined for improved effect. The Suprathane in our example is made of urethane foam with a protective outer layer that includes aluminum foil, kraft paper, and a high-strength glass fiber mesh. These combined layers help create a barrier to heat loss.

Mineral wool blankets, which are used in conjunction with the Suprathane, add another layer of thermal resistance. These blankets often consist of fibers made from rock or metal slag, providing excellent insulation properties. By stacking different types of insulation materials, we can maximize the wall's overall R value, enhancing its ability to resist heat flow effectively.
Thermal Conductivity
Thermal conductivity is a material property that describes how well a material can conduct heat. It is a fundamental concept affecting thermal resistance, heat transfer, and insulation effectiveness.

High thermal conductivity indicates that a material is a good conductor of heat and will transfer heat more readily. Metals, for instance, tend to have high thermal conductivity, meaning they can quickly transfer heat. On the other hand, materials like the urethane foam found in Suprathane have low thermal conductivity, making them effective insulators by resisting heat flow through them.

When building an efficient insulation system, choosing materials with low thermal conductivity is essential. It ensures that even with temperature differences inside and outside of a wall, heat does not easily pass through. This is why materials with lower thermal conductivity, like the Suprathane sandwich, are used as part of effective insulation systems, optimizing energy use and comfort inside a building.

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Most popular questions from this chapter

A reactor vessel's contents are initially at \(290 \mathrm{~K}\) when a reactant is added, leading to an exothermic chemical reaction that releases heat at a rate of \(4 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). The volume and exterior surface area of the vessel are \(0.008 \mathrm{~m}^{3}\) and \(0.24 \mathrm{~m}^{2}\), respectively, and the overall heat transfer coefficient between the vessel contents and the ambient air at \(300 \mathrm{~K}\) is \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the reactants are well stirred, estimate their temperature after (i) I minute. (ii) 10 minutes. Take \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=3000 \mathrm{~J} / \mathrm{kg} \mathrm{K}\) for the reactants.

A hot-water cylinder contains 150 liters of water. It is insulated, and its outer surface has an area of \(3.5 \mathrm{~m}^{2}\). It is located in an area where the ambient air is \(25^{\circ} \mathrm{C}\), and the overall heat transfer coefficient between the water and the surroundings is \(1.0 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), based on outer surface area. If there is a power failure, how long will it take the water to cool from \(65^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) ? Take the density of water as \(980 \mathrm{~kg} / \mathrm{m}^{3}\) and its specific heat as \(4180 \mathrm{~J} / \mathrm{kg} \mathrm{K}\).

Two small blackened spheres of identical size-one of aluminum, the other of an unknown alloy of high conductivity-are suspended by thin wires inside a large cavity in a block of melting ice. It is found that it takes \(4.8\) minutes for the temperature of the aluminum sphere to drop from \(3^{\circ} \mathrm{C}\) to \(1^{\circ} \mathrm{C}\), and \(9.6\) minutes for the alloy sphere to undergo the same change. If the specific gravities of the aluminum and alloy are \(2.7\) and \(5.4\), respectively, and the specific heat of the aluminum is \(900 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), what is the specific heat of the alloy?

A thermistor is used to measure the temperature of an air stream leaving an air heater. It is located in a \(30 \mathrm{~cm}\) square duct and records a temperature of \(42.6^{\circ} \mathrm{C}\) when the walls of the duct are at \(38.1^{\circ} \mathrm{C}\). What is the true temperature of the air? The thermistor can be modeled as a \(3 \mathrm{~mm}\)-diameter sphere of emittance \(0.7\). The convective heat transfer coefficient from the air stream to the thermistor is estimated to be \(31 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

Solve the following ordinary differential equations: (i) \(\frac{d y}{d x}+\beta y=0\) (ii) \(\quad \frac{d y}{d x}+\beta y+\alpha=0\) (iii) \(\frac{d^{2} y}{d x^{2}}-\lambda^{2} y=0\) (iv) \(\frac{d^{2} y}{d x^{2}}+\lambda^{2} y=0\) (v) \(\frac{d^{2} y}{d x^{2}}-\lambda^{2} y+\alpha=0\) where \(\alpha, \beta\), and \(\lambda\) are constants.

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