Question

A 4-m-high and 6-m-long wall is constructed of two large \(0.8\)-cm-thick steel plates \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separated by 1 -cm- thick and \(22-\mathrm{cm}\)-wide steel bars placed \(99 \mathrm{~cm}\) apart. The remaining space between the steel plates is filled with fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the temperature difference between the inner and the outer surfaces of the walls is \(22^{\circ} \mathrm{C}\), determine the rate of heat transfer through the wall. Can we ignore the steel bars between the plates in heat transfer analysis since they occupy only 1 percent of the heat transfer surface area?

Short answer

Answer: Yes, it is significant to consider the steel bars in the heat transfer analysis, as they contribute about 99.12% of the total heat transfer. Ignoring their effect would lead to incorrect results.

Step-by-step solution

Calculate the area of the wall

The wall is 4 meters high and 6 meters long, so the total area of the wall (A_total) will be: \(A_{total} = 4~\text{m} * 6~\text{m} = 24~\text{m}^2\)

Calculate the area of the steel bars in the wall

There are steel bars separated at a distance of 0.99 m placed inside the wall between the plates. The width of each steel bar (A_bar_width) is 0.22 m. To calculate the total number of steel bars, first, let's calculate how many bars fit in the 6-meter horizontal length: \(\text{Number of bars} = \frac{6~\text{m}}{0.99~\text{m}} \approx 6\) So, the total area occupied by the steel bars inside the wall (A_bars) is: \(A_{bars} = 6 * 0.22~\text{m} * 4~\text{m} = 5.28~\text{m}^2\)

Calculate the area of the fiberglass insulation

The fiberglass insulation fills the remaining space in the wall between the steel plates, except the steel bars. So, the area of the fiberglass insulation (A_fiberglass) is: \(A_{fiberglass} = A_{total} - A_{bars} = 24~\text{m}^2 - 5.28~\text{m}^2 = 18.72~\text{m}^2\)

Calculate the temperature difference

The temperature difference (∆T) between the inner and the outer surfaces of the wall is given as: \(\Delta T = 22^{\circ}\text{C}\)

Calculate the rate of heat transfer through the steel bars

Now, let's calculate the rate of heat transfer (Q) for the steel bars using the formula: \(Q_{bars} = \frac{k \cdot A_{bars} \cdot \Delta T}{\text{thickness}}\) where k = 15 W/m·K, A_bars = 5.28 m² and thickness = 0.008 m: \(Q_{bars} = \frac{15~\text{W/m} \cdot \text{K} * 5.28~\text{m}^2 * 22^{\circ}\text{C}}{0.008~\text{m}} = 23198.4~\text{W}\)

Calculate the rate of heat transfer through the fiberglass insulation

Similarly, let's calculate the rate of heat transfer (Q) for the fiberglass insulation using the formula: \(Q_{fiberglass} = \frac{k \cdot A_{fiberglass} \cdot \Delta T}{\text{thickness}}\) where k = 0.035 W/m·K, A_fiberglass = 18.72 m² and thickness = 0.008 m: \(Q_{fiberglass} = \frac{0.035~\text{W/m} \cdot \text{K} * 18.72~\text{m}^2 * 22^{\circ}\text{C}}{0.008~\text{m}} = 206.196~\text{W}\)

Calculate the total rate of heat transfer through the wall

To find the total rate of heat transfer through the wall, we sum the heat transfer through the steel bars and the fiberglass insulation: \(Q_{total} = Q_{bars} + Q_{fiberglass} = 23198.4~\text{W} + 206.196~\text{W} = 23404.596~\text{W}\)

Determine the significance of the steel bars in heat transfer analysis

To determine whether ignoring the steel bars has a significant effect on the heat transfer analysis, we can compare the heat transfer rates with and without the steel bars: Percentage of heat transfer through steel bars: \(\frac{Q_{bars}}{Q_{total}} * 100 = \frac{23198.4~\text{W}}{23404.596~\text{W}} * 100 \approx 99.12\%\) Thus, the steel bars contribute about 99.12% of the total heat transfer, and their effect cannot be ignored in the heat transfer analysis.