A dormitory at a large university, built 50 years ago, has...

A dormitory at a large university, built 50 years ago, has exterior walls constructed of $L_{s}=25$-mm-thick sheathing with a thermal conductivity of $k_{s}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. To reduce heat losses in the winter, the university decides to encapsulate the entire dormitory by applying an $L_{i}=25$-mm-thick layer of extruded insulation characterized by $k_{i}=0.029 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ to the exterior of the original sheathing. The extruded insulation is, in turn, covered with an $L_{g}=5$-mm-thick architectural glass with $k_{g}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Determine the heat flux through the original and retrofitted walls when the interior and exterior air temperatures are $T_{\infty 0, i}=22^{\circ} \mathrm{C}$ and $T_{\infty, a}=$ $-20^{\circ} \mathrm{C}$, respectively. The inner and outer convection heat transfer coefficients are $h_{i}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and $h_{o}=$ $25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively.

Key Concepts

Thermal Conduction

Thermal conduction is the process by which heat is transferred through a material due to a temperature difference. It is governed by Fourier’s law, which states that the heat flux is proportional to the negative gradient of temperature and the thermal conductivity of the material. In the context of building envelopes, conduction determines how well materials such as insulation, glass, or sheathing transfer heat through their thicknesses.

Thermal Resistance

Thermal resistance is a measure of a material's ability to resist heat flow and is calculated by dividing the thickness of a material by its thermal conductivity. When multiple layers are present, such as in composite walls, the overall thermal resistance is the sum of the individual resistances. This concept is central to evaluating and comparing the insulating performance of different wall assemblies.

Convection Heat Transfer

Convection heat transfer involves the movement of heat between a surface and a fluid (such as air) in contact with it, characterized by convection heat transfer coefficients. In building energy analysis, interior and exterior convection coefficients are used to account for the resistance to heat flow at the surfaces of the building envelope, significantly impacting overall heat loss or gain.

Composite Wall Analysis

Composite wall analysis involves evaluating a wall system made up of several layers, each with its own material properties such as thickness and thermal conductivity. The overall heat transfer through such a system is determined by adding the resistances from conduction through each layer and the resistances due to convection on the surfaces. This integrated approach is essential for accurately predicting the thermal performance of both original and retrofitted building envelopes.

Overall Heat Transfer Coefficient (U-value)

The overall heat transfer coefficient, or U-value, quantifies the rate of heat transfer through a building element by combining the effects of conduction and convection. A lower U-value indicates better insulation and reduced heat loss. The U-value is determined by taking the inverse of the sum of all individual thermal resistances, and it is a key metric in assessing the thermal efficiency of wall assemblies and retrofit interventions.

Transcript

00:01 In this question we have given that there is a window fitted with a single glass of thickness 2 mm and we have to find the rate of heat flow through the window and it is given that inside temperature of room is 32 degrees celsius and outside temperature is 40 degrees celsius so this is our single glass window and we have given that the inside temperature of the room is that is 32 degrees celsius this is t2 is 32 degrees celsius and outside temperature is that is t1 is 40 degrees celsius we have given the thickness of this glass is 2mm so this is 2 mm so in the first part of the equation we have to find the heat flow that this is given by q by t so q by t is equal to this is k a into t1 minus t2 upon l so this is now put all the data in this equation so this is 1 into area is 2 and this is 40 minus 32 upon this is 2 into 10 to the power minus 3 so here we get the value that is 8 ,000 joule per second so this is the answer for our first part of the cushion now in the second part of the cushion it is told that this window is replaced by two glass panes and each having a thickness of 1 mm and separated by a distance of 1 mm so for second part it is given that this is the glass window it is separated by a distance of 1 meter and this there is a air in this gap so this is air and there is our glass window so first of all i'm finding the resistance of this glass window let is this is r1 for this this is r2 and this for this resistance is r3 so r1 can be written as l upon k a and this will equal to l1 upon k1a similarly r2 will be equal to l2 upon k2a and resistance r3 is given by l3 nat resistance is given by r is equal to r1 plus r2 plus r3 so this this is l1 upon k1a plus l2 upon k2a plus l3 upon k3a.

03:02 Now put all the data in this equation.

03:04 So this is 1 into 10 to the power minus 3 into 1 .05 upon 0 .05...

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