/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Clothing made of several thin la... [FREE SOLUTION] | 91Ó°ÊÓ

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Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold climates because it is light, fashionable, and a very effective thermal insulator. So it is no surprise that such clothing has largely replaced thick and heavy old-fashioned coats. Consider a jacket made of five layers of \(0.1-\mathrm{mm}\)-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(1.5\)-mm- thick air space \((k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the layers. Assuming the inner surface temperature of the jacket to be \(28^{\circ} \mathrm{C}\) and the surface area to be \(1.25 \mathrm{~m}^{2}\), determine the rate of heat loss through the jacket when the temperature of the outdoors is \(0^{\circ} \mathrm{C}\) and the heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What would your response be if the jacket is made of a single layer of \(0.5-\mathrm{mm}\)-thick synthetic fabric? What should be the thickness of a wool fabric ( \(k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) if the person is to achieve the same level of thermal comfort wearing a thick wool coat instead of a five-layer ski jacket?

Short Answer

Expert verified
To calculate the total thermal resistance of the ski jacket with multi-layer insulation, we first find the thermal resistance of one fabric layer and one air layer: 1. \(R_{fabric} = \frac{0.1\times10^{-3}}{0.13\times1.25} = \frac{0.0001}{0.1625} = 0.000615\,m^2\cdot K/W\) (approx.) 2. \(R_{air} = \frac{1.5\times10^{-3}}{0.026\times1.25} = \frac{0.0015}{0.0325} = 0.046154\,m^2\cdot K/W\) (approx.) Now, we can find the total thermal resistance: 3. \(R_{total} = 5 R_{fabric} + 4 R_{air} = 5 (0.000615) + 4 (0.046154) = 0.003075 + 0.184616 = 0.187691\,m^2\cdot K/W\) (approx.) The total thermal resistance of the ski jacket with multi-layer insulation is approximately 0.187691 m²⋅K/W.

Step by step solution

01

Calculate the total thermal resistance of the ski jacket with multi-layer insulation

First, let's consider the multi-layer jacket. We need to find the total thermal resistance (\(R_{total}\)) of the jacket. The thermal resistance of a layer can be calculated using the formula: \(R = \frac{L}{kA}\) where \(L\) = thickness of the layer, \(k\) = thermal conductivity of the material, and \(A\) = surface area. For the multi-layer jacket, there are 5 layers of fabric with air in between. Let's calculate their thermal resistance: 1. Calculate the thermal resistance of one fabric layer (\(R_{fabric}\)) by substituting the given values (\(L=0.1\ mm,\ k=0.13\ W/m\cdot K,\ A=1.25\ m^2)\): \(R_{fabric} = \frac{0.1\times10^{-3}}{0.13\times1.25}\) 2. Calculate the thermal resistance of one air layer (\(R_{air}\)) by substituting the given values (\(L=1.5\ mm,\ k=0.026\ W/m\cdot K,\ A=1.25\ m^2)\): \(R_{air} = \frac{1.5\times10^{-3}}{0.026\times1.25}\) 3. To find the total thermal resistance \(R_{total}\), consider that each air layer is between two fabric layers, so we have: \(R_{total} = 5 R_{fabric} + 4 R_{air}\) Now calculate the total thermal resistance.

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Key Concepts

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

Thermal Resistance
Thermal resistance, often denoted by the symbol 'R', is a measure of a material's ability to resist heat flow. It is analogous to electrical resistance in how it opposes the flow, but in this case, the flow is of thermal energy rather than electrical current. The higher the thermal resistance, the better the material insulates because it's more effective at preventing heat transfer.

The formula for calculating the thermal resistance of a given layer is:
\[R = \frac{L}{kA}\]
where:\[L\] represents the thickness of the material,
\[k\] is the thermal conductivity of the material, and
\[A\] is the surface area through which heat is being transferred.

In scenarios involving multiple layers, such as in the case of multi-layer insulation found in ski jackets, the total thermal resistance is the sum of the thermal resistances of all individual layers. For instance, with several alternating layers of fabric and air, you would add the resistance values of each layer together to get the overall thermal resistance. This cumulative resistance helps to determine how well the jacket will insulate against the cold environment outside.

To improve understanding, it's important to picture thermal resistance as a barrier to heat flow - the thicker and less conductive the material, the more formidable the barrier.
Thermal Conductivity
Thermal conductivity, represented by the variable \(k\), is a property of a material that indicates its ability to conduct heat. It is defined as the amount of heat (in watts) that can pass through a one-meter thickness of the material with a one-square-meter cross-sectional area for a temperature difference of one degree Celsius. The unit of thermal conductivity is \(W/m\cdot K\).

A higher value of \(k\) means that the material is a good conductor of heat and allows heat to pass through it more readily. Conversely, a lower value of \(k\) means that the material is a poor conductor and therefore a good insulator. Thermal conductivity is a critical factor in designing insulation systems, like those used in cold-weather clothing. For example, in our exercise, the synthetic fabric with a \(k\) value of \(0.13 W/m\cdot K\) is less conductive than many metals but is more conductive than the trapped air with a \(k\) value of \(0.026 W/m\cdot K\), making the combination effective for insulation purposes.

Everyday Examples

In everyday life, materials with high thermal conductivity, like copper or aluminum, are used in cooking utensils and radiators for their ability to transfer heat quickly. In contrast, materials with low thermal conductivity, like rubber, wood, or air, are used in thermal insulating applications to slow down the heat flow, such as in thermal flasks, building insulation, or indeed, ski jackets.
Multilayer Insulation
Multilayer insulation (MLI) is an advanced insulation technique that uses multiple thin layers of materials, often separated by spacers like air gaps, to greatly reduce heat transfer. Each layer serves as a barrier for heat that is trying to escape or enter, and the air gaps further reduce the overall thermal conductivity, as air is a poor conductor of heat.

MLI is effective because it tackles heat transfer through all three mechanisms: conduction, convection, and radiation. By alternating layers of conductive material with non-conductive spacers, MLI reduces conduction. Spaces between layers minimize convection, and reflective surfaces address radiative heat transfer.

In the context of our exercise, the ski jacket made of several thin layers of fabric with trapped air in between is an excellent example of MLI. The air gaps between the synthetic fabric layers act as thermal barriers, due to air's low thermal conductivity. This structure makes the jacket lightweight while providing superior insulation compared to a single thicker layer of material. For optimal comprehension, visualize MLI as a series of hurdles that heat must jump over to get through, with each hurdle making it progressively harder for the heat to pass, thereby keeping the wearer warm.

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Most popular questions from this chapter

The \(700 \mathrm{~m}^{2}\) ceiling of a building has a thermal resistance of \(0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is \(-10^{\circ} \mathrm{C}\) and the interior is at \(20^{\circ} \mathrm{C}\) is (a) \(23.1 \mathrm{~kW} \quad\) (b) \(40.4 \mathrm{~kW}\) (c) \(55.6 \mathrm{~kW}\) (d) \(68.1 \mathrm{~kW}\) (e) \(88.6 \mathrm{~kW}\)

Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3 -mm- thick layers of glass \((k=\) \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) separated by a \(12-\mathrm{mm}\)-wide stagnant air space \((k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at \(24^{\circ} \mathrm{C}\) while the temperature of the outdoors is \(-5^{\circ} \mathrm{C}\). Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be \(h_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{2}=\) \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat transfer by radiation.

How does the thermal resistance network associated with a single-layer plane wall differ from the one associated with a five-layer composite wall?

A plate consists of two thin metal layers pressed against each other. Do we need to be concerned about the thermal contact resistance at the interface in a heat transfer analysis or can we just ignore it?

Consider a house with a flat roof whose outer dimensions are \(12 \mathrm{~m} \times 12 \mathrm{~m}\). The outer walls of the house are \(6 \mathrm{~m}\) high. The walls and the roof of the house are made of \(20-\mathrm{cm}-\) thick concrete \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the inner and outer surfaces of the house are \(15^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. Accounting for the effects of the edges of adjoining surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ignoring the effects of the edges and corners and treating the roof as a \(12 \mathrm{~m} \times 12 \mathrm{~m}\) surface and the walls as \(6 \mathrm{~m} \times 12 \mathrm{~m}\) surfaces for simplicity?

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