A 10-in-thick, 30-ft-long, and 10-ft-high wall is to be constructed using 9 -in-long solid bricks \(\left(k=0.40 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) of cross section 7 in \(\times 7\) in, or identical size bricks with nine square air holes \(\left(k=0.015 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) that are 9 in long and have a cross section of \(1.5\) in \(\times 1.5 \mathrm{in}\). There is a \(0.5\)-in-thick plaster layer \(\left(k=0.10 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}\right)\) between two adjacent bricks on all four sides and on both sides of the wall. The house is maintained at \(80^{\circ} \mathrm{F}\) and the ambient temperature outside is \(30^{\circ} \mathrm{F}\). Taking the heat transfer coefficients at the inner and outer surfaces of the wall to be \(1.5\) and \(4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat transfer through the wall constructed of \((a)\) solid bricks and (b) bricks with air holes.

Step by step solution

01

Determine the thermal resistance of the plaster layer

First, we must calculate the thermal resistance of the plaster layer, with a thickness of 0.5 in and a thermal conductivity of k_plaster = 0.10 Btu/h·ft·°F. The surface area (A) of the plaster is the same as the surface area of the bricks. A = 30 ft × 10 ft = 300 ft^2. The equation for thermal resistance in a plane wall is R = t / (k·A). Therefore, the thermal resistance of the plaster layer (R_plaster) can be calculated as: R_plaster = (0.5 in/12 ft/in) / (0.10 Btu/h·ft·°F × 300 ft^2)

02

Determine the thermal resistance of the solid bricks

Next, we must calculate the thermal resistance of the solid bricks, with a thickness of 10 in and a thermal conductivity of k_brick = 0.40 Btu/h·ft·°F: R_brick = (10 in / 12 ft/in) / (0.40 Btu/h·ft·°F × 300 ft^2)

03

Calculate the overall thermal resistance of the wall

Now, we must find the overall thermal resistance of the wall, considering the two plaster layers and the solid brick layer: R_total = R_plaster + R_brick + R_plaster

04

Calculate the heat transfer rate for solid bricks

With the overall thermal resistance found, we can calculate the heat transfer rate (Q) through the wall using the equation Q = ΔT / R_total, where ΔT is the temperature difference between inside and outside: ΔT = T_inside - T_outside = 80°F - 30°F = 50°F Q_solid = ΔT / R_total #b) For bricks with air holes#

05

Determine the thermal resistance of the bricks with air holes

Now we will calculate the thermal resistance for bricks with air holes. The thermal conductivity of the air holes is given as k_hole = 0.015 Btu/h·ft·°F and the cross-sectional area of the air holes is 1.5 in × 1.5 in. There are nine air holes in each brick and the thermal resistance of the air holes will be in parallel with the solid brick material's thermal resistance. R_hole = (9 in) / (0.015 Btu/h·ft·°F × 9 × (1.5 in * 1.5 in / 144 ft²/in²))

06

Calculate the equivalent thermal resistance R_equiv

The bricks with air holes have solid brick material and air holes in parallel, so we need to calculate the equivalent thermal resistance for both (R_equiv): 1/R_equiv = (1/R_brick) + (1/R_hole)

07

Calculate the overall thermal resistance for bricks with air holes

The overall thermal resistance for bricks with air holes is the sum of the plaster layers' and the equivalent thermal resistances: R_total_holes = R_plaster + R_equiv + R_plaster

08

Calculate the heat transfer rate for bricks with air holes

Using the equation Q = ΔT / R_total_holes, we can determine the heat transfer rate through the wall made with bricks containing air holes: Q_holes = ΔT / R_total_holes Now you have calculated the rate of heat transfer through the wall made of solid bricks (Q_solid) and bricks with air holes (Q_holes).

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Key Concepts

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

Thermal Resistance

Understanding thermal resistance is crucial when optimizing the insulation of structures. Think of it as a measure of how well a material resists heat flow. The higher the thermal resistance, the better the material insulates. For instance, in the calculation for our wall's plaster layer, the thermal resistance is given by the formula
\(R = \frac{t}{k \times A}\)
where \(t\) represents the thickness of the material, \(k\) the thermal conductivity, and \(A\) the surface area over which heat transfer occurs.

In the provided exercise, we saw that both the plaster and bricks contribute to the wall's total thermal resistance. The thermal resistance for each layer is summed up because heat must pass through each sequentially. Building constructions often layer materials to create a higher thermal resistance and therefore better insulate the interior from exterior temperatures.

Thermal Conductivity

Thermal conductivity is indicative of a material's ability to conduct heat. Given in units of \(\frac{Btu}{h \times ft \times{ }^{\circ}F}\), it quantifies how rapidly heat is transferred through a material due to a temperature difference.

In the context of our wall exercise, solid bricks have higher thermal conductivity compared to bricks with air holes. This explains why solid bricks, with a \(k\)-value of 0.40, would transfer heat faster than bricks with air holes, which have a lower \(k\)-value of 0.015. Lower thermal conductivity is advantageous in insulative materials because it slows down heat transfer, helping maintain desired temperatures within a space.

Heat Transfer Rate

The rate of heat transfer through a wall is directly related to both the thermal resistance of the materials in the wall and the temperature difference across it. The basic formula used in the exercise to calculate this rate is
\(Q = \frac{\Delta T}{R_{total}}\)
where \(Q\) is the heat transfer rate, \(\Delta T\) represents the temperature difference between the indoors and outdoors, and \(R_{total}\) is the sum of the thermal resistances of each layer in the wall.

A practical application of this concept is seen in the comparison between solid bricks and bricks with air holes. Bricks with air holes exhibit a lower heat transfer rate due to their increased thermal resistance, a desirable trait for energy efficiency in a building. By comprehensively understanding how to manipulate these variables, one can precisely calculate and optimize the thermal performance of building envelopes.