Question

Consider a 1.2-m-high and 2-m-wide glass window with a thickness of \(6 \mathrm{~mm}\), thermal conductivity \(k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and emissivity \(\varepsilon=0.9\). The room and the walls that face the window are maintained at \(25^{\circ} \mathrm{C}\), and the average temperature of the inner surface of the window is measured to be \(5^{\circ} \mathrm{C}\). If the temperature of the outdoors is \(-5^{\circ} \mathrm{C}\), determine \((a)\) the convection heat transfer coefficient on the inner surface of the window, \((b)\) the rate of total heat transfer through the window, and \((c)\) the combined natural convection and radiation heat transfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermal resistance of the glass in this case?

Short answer

The convection heat transfer coefficient on the inner surface of the window (h_in) is \(13\,\mathrm{W/m^2 \cdot K}\). (b) What is the rate of total heat transfer through the window? The rate of total heat transfer through the window (Q_total) is \(936\,\mathrm{W}\). (c) What is the combined natural convection and radiation heat transfer coefficient on the outer surface of the window? The combined natural convection and radiation heat transfer coefficient on the outer surface of the window (h_out) is \(26\,\mathrm{W/m^2 \cdot K}\). Is it reasonable to neglect the thermal resistance of the glass in this case? Yes, it is reasonable to neglect the thermal resistance of the glass in this case, since the glass thickness is only \(6\mathrm{~mm}\) and the heat transfer coefficients (h_in and h_out) have much larger values, making the contribution of the glass to the overall heat transfer negligible.

Step-by-step solution

Calculate the conduction heat transfer (Q_cond) through the glass window

Using the Fourier's law of heat conduction, we can find the conduction heat transfer through the glass window. The formula for heat conduction is: \(Q_{cond} = k \cdot A \cdot \frac{ΔT}{d}\) Where: - k is the thermal conductivity of the glass (\(0.78 \,\mathrm{W/m \cdot K}\)) - A is the area of the window (\(1.2\,\mathrm{m} \times 2\,\mathrm{m} = 2.4\, \mathrm{m^2}\)) - ΔT is the temperature difference between the room and the outer surface of the window (\(25^\circ \mathrm{C} - (-5^\circ \mathrm{C}) = 30\,\mathrm{K}\)) - d is the thickness of the glass window (\(6\,\mathrm{mm} = 0.006\,\mathrm{m}\)) Now, calculate the conduction heat transfer: \(Q_{cond} = 0.78 \times 2.4 \times \frac{30}{0.006} = 936\,\mathrm{W}\)

Calculate the convection heat transfer coefficient (h_in) on the inner surface of the window

The formula for convection heat transfer is: \(Q_{conv} = h_{in} \cdot A \cdot (T_{room} - T_{avg_{in}})\) Since we have already calculated the conduction heat transfer (Q_cond), we can assume that Q_cond ≈ Q_conv. Therefore, we can solve for the convection heat transfer coefficient (h_in): \(h_{in} = \frac{Q_{conv}}{A \cdot (T_{room} - T_{avg_{in}})} = \frac{936\,\mathrm{W}}{2.4\,\mathrm{m^2} \times (25^\circ \mathrm{C} - 5^\circ \mathrm{C})} = 13\,\mathrm{W/m^2 \cdot K}\)

Calculate the total heat transfer (Q_total) through the window

As we mentioned before, in this case, we can assume that Q_cond ≈ Q_conv. Therefore, the total heat transfer (Q_total) through the window is: \(Q_{total} = Q_{conv} = 936\,\mathrm{W}\)

Calculate the combined natural convection and radiation heat transfer coefficient on the outer surface of the window (h_out)

The formula for the combined natural convection and radiation heat transfer is: \(Q_{conv+rad} = h_{out} \cdot A \cdot (T_{avg_{in}} - T_{out})\) Again, we can assume that Q_conv+rad ≈ Q_total. Therefore, we can solve for the combined natural convection and radiation heat transfer coefficient (h_out): \(h_{out} = \frac{Q_{total}}{A \cdot (T_{avg_{in}} - T_{out})} = \frac{936\,\mathrm{W}}{2.4\,\mathrm{m^2} \times (5^\circ \mathrm{C} - (-5^\circ \mathrm{C})} = 26\,\mathrm{W/m^2 \cdot K}\) In conclusion: (a) The convection heat transfer coefficient on the inner surface of the window (h_in) is \(13\,\mathrm{W/m^2 \cdot K}\). (b) The rate of total heat transfer through the window (Q_total) is \(936\,\mathrm{W}\). (c) The combined natural convection and radiation heat transfer coefficient on the outer surface of the window (h_out) is \(26\,\mathrm{W/m^2 \cdot K}\). Considering that the glass thickness is only \(6\mathrm{~mm}\) and the heat transfer coefficients (h_in and h_out) have much larger values, it can be reasonable to neglect the thermal resistance of the glass in this case. This is because the relatively thin glass and high heat transfer coefficients make the contribution of the glass to the overall heat transfer negligible.

Key concepts

Conduction Heat Transfer

Conduction heat transfer refers to the transfer of heat through a material due to the movement of thermal energy from a region of higher temperature to a region of lower temperature. In the exercise, we consider a glass window where heat is conducted through the thickness of the glass.
To calculate conduction heat transfer, we use Fourier's law, which states: \[ Q_{cond} = k \cdot A \cdot \frac{\Delta T}{d} \]This formula tells us several important aspects:

• Thermal Conductivity \(k\): This is a measure of a material's ability to conduct heat, given in watts per meter-kelvin (W/m·K). For our glass window, it is 0.78 W/m·K.
• Area \(A\): The surface area through which heat transfer occurs. For the window, \(A = 1.2\,\mathrm{m} \times 2\,\mathrm{m} = 2.4\,\mathrm{m^2}\).
• Temperature Difference \(\Delta T\): The driving force for heat conduction. It is the difference in temperatures across the glass, calculated as \(30\,\mathrm{K}\) in our example.
• Thickness \(d\): The thickness of the material through which heat is conducted, noted as 6 mm or 0.006 m for our window.

Using these parameters, the calculated conduction heat transfer through the window is 936 W.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient \(h\) is a measure of the heat transfer between a solid surface and a fluid in motion adjacent to the surface. In this exercise, we consider the inner surface of the window exposed to air.
The convection heat transfer can be calculated using:\[ Q_{conv} = h_{in} \cdot A \cdot (T_{room} - T_{avg_{in}}) \]Here’s how it works out:

• Convection Coefficient \(h_{in}\): This value denotes the efficiency of heat transfer via convection across the boundary layer of the window's inner surface.
• Surface Area \(A\): Similar to conduction, it's the area interacting with the fluid. Here it remains 2.4 m².
• Temperature Difference: This involves the difference between the room temperature and the surface temperature, which in this case is \(20\,\mathrm{K}\).

Given the assumption that \(Q_{conv} \approx Q_{cond}\), the \(h_{in}\) is determined to be 13 W/m²·K.
This approximation plays a key role, as it reflects the balance of heat flowing into and out of the window.

Radiation Heat Transfer

Radiation heat transfer involves the emission and absorption of electromagnetic waves, primarily in the infrared spectrum. In our example, the outer surface also experiences this type of heat transfer alongside convection.
The combined effect is expressed with a combined convection and radiation heat transfer coefficient \(h_{out}\):\[ Q_{conv+rad} = h_{out} \cdot A \cdot (T_{avg_{in}} - T_{out}) \]Let's break it down:

• Emissivity \(\varepsilon\): A measure of the window's efficiency in emitting thermal radiation, given as 0.9 for the glass.
• Convection and Radiation Coefficient \(h_{out}\): It captures both processes together, calculated to be 26 W/m²·K.
• Temperature Difference: The transfer occurs between the inner surface temperature and the outdoor temperature, resulting in a difference of 10 K.

Radiation can contribute significantly to total heat transfer; however, in this exercise, approximations neatly synchronize convection and radiation, simplifying calculations.