A house has a composite wall of wood, fiberglass insulation,...

A house has a composite wall of wood, fiberglass insulation, and plaster board, as indicated in the sketch. On a cold winter day, the convection heat transfer coefficients are $h_{o}=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and $h_{i}=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The total wall surface area is $350 \mathrm{~m}^{2}$. (a) Determine a symbolic expression for the total thermal resistance of the wall, including inside and outside convection effects for the prescribed conditions. (b) Determine the total heat loss through the wall. (c) If the wind were blowing violently, raising $h_{o}$ to $300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the percentage increase in the heat loss. (d) What is the controlling resistance that determines the amount of heat flow through the wall?

A house has a composite wall of wood, fiberglass insulation, and plaster board, as indicated in the sketch. On a cold winter day, the convection heat transfer coefficients are $h_{o}=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and $h_{i}=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The total wall surface area is $350 \mathrm{~m}^{2}$.
(a) Determine a symbolic expression for the total thermal resistance of the wall, including inside and outside convection effects for the prescribed conditions.
(b) Determine the total heat loss through the wall.
(c) If the wind were blowing violently, raising $h_{o}$ to $300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the percentage increase in the heat loss.
(d) What is the controlling resistance that determines the amount of heat flow through the wall?

Key Concepts

Thermal Resistance

Thermal resistance is the measure of a material’s opposition to heat flow. In a composite system, such as a wall with multiple layers or interfaces, the overall thermal resistance is determined by the sum of the individual resistances of each layer (conduction resistance) as well as the resistances at the interfaces due to convection. This concept is crucial for predicting temperature differences and rates of energy transfer in thermodynamic systems.

Composite Wall Construction

Composite wall construction refers to a structure made of several different material layers, each with its own thermal properties. When these layers are arranged in series, as in a wall comprising wood, insulation, and plaster board, the total thermal resistance is the sum of the individual resistances, with each layer contributing differently depending on its thickness and thermal conductivity. This concept is vital for understanding and designing building envelopes for energy efficiency.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient is a parameter that quantifies the convective heat transfer between a surface and a fluid (often air). It plays an essential role in calculating the convective thermal resistance at the boundaries of a material or structure. Higher coefficients indicate more efficient heat transfer due to convection, which can significantly affect the overall heat loss in systems exposed to environmental conditions such as wind.

Overall Heat Loss Calculation

Overall heat loss calculation involves determining the rate of heat transfer through a composite structure by using the total thermal resistance and the temperature difference between the two sides of the structure. This is typically done using an expression derived from Fourier’s law for conduction and Newton’s law of cooling for convection. It allows for the prediction of energy losses and is crucial for evaluating the thermal performance of buildings.

Controlling (Bottleneck) Resistance

The controlling or bottleneck resistance is the component within a series thermal resistance network that has the greatest value and, therefore, most significantly limits the overall heat flow. Identifying this resistance is essential because it dictates the thermal performance of the entire system. By understanding which layer or interface contributes the most to the resistance, one can target improvements to enhance energy efficiency.

Transcript

00:01 This equation, we'll need to calculate the heat transfer by conduction for the three areas of the house, the walls, the roof, and the windows.

00:10 So for that equation, we need to take the heat of conduction over time is equal to k times a, thermal conductivity times area, times length.

00:29 And this is for walls plus k times a.

00:37 This again thermal conductivity times the area over length for the roof plus k times a over length for the windows and we are going to multiply all that by the change in temperature so this is our equation we figured out by simply plugging in our numbers so we have 0 .023 joules times seconds times meters times degrees celsius is our units that's going to be times 410 meters squared and then for length we have 1 .75 times 10 to negative 1 meters then plus for our roof we have 0 .12 joules over second size of meters times degrees celsius times 280 meters squared and all that is over 6 .5 times 10 to the negative 2 centimeters.

02:21 And then for our windows we have plus 0 .84.

02:29 And again, this is going to be joules over second size of meters times degrees celsius times 33 meters squared over 6 .5 times 10 to the negative 3 meters.

02:52 And we're going to take all that and multiply it by 33 degrees.

03:00 And so if we do this math, we come up with an answer of 1 .596 times 10 to the 5th, which is equal to, in simpler terms, 1 .6 times 10 to the 5th, watts for the first part of our answer so the second part of our problem we need to figure out in 30 minutes how much energy we would so house so since the temperature is ranging from 20 degrees to 33 degrees celsius we will use 26 .5 as the average temperature so we have here the heat that is added is equal to the heat of the raised temperature plus the heat of conduction is equal to density of air times v times the specific heat of the air times change in temperature and then we were going to add to that plus heat of conduction over time times 1 ,800 seconds which is our 30 minutes.

04:48 So we can go ahead and just plug in our numbers.

04:52 So we will have 1 .29 kilograms per meter cubed times 750 meters cubed for our volume.

05:15 And then we're going to multiply that by 0 .24...

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