Question

Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3 -mm- thick layers of glass $(k=$ $0.78\mathrm{W}/\mathrm{m}\cdot \mathrm{K}$ ) separated by a $12-\mathrm{mm}$-wide stagnant air space $(k=0.026\mathrm{W}/\mathrm{m}\cdot \mathrm{K})$. Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at ${24}^{\circ }\mathrm{C}$ while the temperature of the outdoors is $-5^{\circ }\mathrm{C}$. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be $h_1=10\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$ and $h_2=$ $25\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$, and disregard any heat transfer by radiation.

Short answer

Answer: The steady rate of heat transfer through the double-pane window is approximately 358.73 W, and the temperature of its inner surface is approximately ${9.0}^{\circ }\mathrm{C}$.

Step-by-step solution

Calculate thermal resistances

First, we need to calculate the thermal resistances for convection and conduction across the window system. These thermal resistances are R1 for convection on inner glass surface, R2 for conduction in the inner glass layer, R3 for conduction in the air space, R4 for conduction in the outer glass layer, and R5 for convection on outer glass surface. We can calculate the convection and conduction resistances using the following equations: $$R_{conv}=\frac{1}{hA}$$ $$R_{cond}=\frac{L}{kA}$$ where h is the convection heat transfer coefficient, k is the thermal conductivity, A is the area, and L is the thickness.

Determine the total thermal resistance

Next, we will find the total thermal resistance by adding all the thermal resistances in the window system. $$R_{total}=R1+R2+R3+R4+R5$$

Calculate the steady rate of heat transfer

Now, we can calculate the steady rate of heat transfer (Q) by using the following equation: $$Q=\frac{T_{in}-T_{out}}{R_{total}}$$ where $T_{in}$ is the temperature inside the room and $T_{out}$ is the temperature outside.

Determine the temperature of the inner surface

Finally, we can find the temperature of the inner surface (T1) using the following equation: $$T1=T_{in}-Q\cdot R1$$ Now, let's calculate the values.

(Calculations)

The area of the double-pane window is 1.2 m * 2 m = 2.4 m². All the resistances are in series. Thus, they will all have the same area, A = 2.4 m². $$R1=\frac{1}{h_{1}A}=\frac{1}{(10\frac{\text{W}}{{\text{m}}^2\text{K}})(2.4{\text{m}}^2)}=0.04167\text{ K/W}$$ $$R2=\frac{L}{kA}=\frac{0.003\text{m}}{(0.78\frac{\text{W}}{\text{m K}})(2.4{\text{m}}^2)}=0.00161\text{ K/W}$$ $$R3=\frac{L}{kA}=\frac{0.012\text{m}}{(0.026\frac{\text{W}}{\text{m K}})(2.4{\text{m}}^2)}=0.01923\text{ K/W}$$ $$R4=\frac{L}{kA}=\frac{0.003\text{m}}{(0.78\frac{\text{W}}{\text{m K}})(2.4{\text{m}}^2)}=0.00161\text{ K/W}$$ $$R5=\frac{1}{h_{2}A}=\frac{1}{(25\frac{\text{W}}{{\text{m}}^2\text{K}})(2.4{\text{m}}^2)}=0.01667\text{ K/W}$$

(Calculations)

Now, let's determine the total thermal resistance. $$R_{total}=R1+R2+R3+R4+R5=0.04167+0.00161+0.01923+0.00161+0.01667=0.08079\text{ K/W}$$

(Calculations)

We can now calculate the steady rate of heat transfer (Q). $$Q=\frac{T_{in}-T_{out}}{R_{total}}=\frac{24°C-(-5°C)}{0.08079\text{ K/W}}=358.73\text{ W}$$

(Calculations)

Finally, let's find the temperature of the inner surface (T1). $$T1=T_{in}-Q\cdot R1=24°C-(358.73\text{ W})(0.04167\text{ K/W})=9.05°C$$ ≈ ${9.0}^{\circ }\mathrm{C}$. The steady rate of heat transfer through the double-pane window is approximately 358.73 W, and the temperature of its inner surface is approximately ${9.0}^{\circ }\mathrm{C}$.

Key concepts

Thermal Resistance

Thermal resistance is a measure of how well an object resists the flow of heat. Imagine it as insulation in a house; the more insulation, the harder it is for heat to flow through the walls. Similarly, in a material, thermal resistance determines how difficult it is for heat to transfer from one side to another.
In thermal systems, we often treat thermal resistance like electrical resistance in circuits. It's calculated using specific formulas depending on the type of heat transfer involved. If it's conduction, we use the formula:

• $R_{cond}=\frac{L}{kA}$

Here, $R_{cond}$ is thermal resistance due to conduction, $L$ is the thickness, $k$ is the thermal conductivity, and $A$ is the area.
For convection, we use:

• $R_{conv}=\frac{1}{hA}$

Where $h$ is the heat transfer coefficient, unique to the nature of the surface and fluid in contact. In the exercise, these resistances collectively form a sort of barrier to heat transfer across the window.

Conduction

Conduction is the process where heat energy is transmitted through a material. This happens when two areas of a material are at different temperatures. Heat flows from the warmer area to the cooler area. Think about touching a hot stove. The heat moves from the stove to your hand, that's conduction at work.

• Materials with high thermal conductivity, like metals, transfer heat easily.
• Materials with low thermal conductivity, like wood and foam, are effective insulators.

For our window example, the heat mainly conducts through glass layers and the air space. The resistances, $R2$ and $R4$, pertain to the glass, while $R3$ relates to the air space. Each material's thickness and thermal conductivity factor into calculating these resistances. The low thermal conductivity value of air highlights it as a poor conductor, which means it serves as a good insulator in this case.

Convection

Convection is the transfer of heat by the movement of a fluid, either a liquid or a gas. If you boil water, you'll notice the warm water rising to the top while cooler water sinks. That's convection. It happens naturally, as in that boiling pot, or it can be forced, like with fans or pumps.In windows, convection occurs at the boundary where the glass meets the air, both inside and out. The heat transfer coefficient $h$ determines how quickly heat is transferred between the glass and the moving air. In the exercise, values $h_1$ and $h_2$ dictate the speed of heat transfer on the inside and outside window surfaces, respectively. The convection resistances $R1$ and $R5$ are calculated taking these coefficients into account. These resistances tell us how well the surfaces resist the transfer of heat due to the fluid flow.

Heat Transfer Coefficient

The heat transfer coefficient, denoted as $h$, is an essential parameter in any calculation involving convection heat transfer. It measures the ability of a fluid to carry heat away from a surface it touches. A higher coefficient means the fluid is good at removing or adding heat.

• This coefficient depends on the nature of the flow (like turbulent or laminar) and the physical properties of the fluid.
• In the exercise, $h_1=10\mathrm{W}/{\mathrm{m}}^2\mathrm{K}$ is used for the inside window surface (often less agitated indoor air), while $h_2=25\mathrm{W}/{\mathrm{m}}^2\mathrm{K}$ fits the often windier outdoor surface.

By including these coefficients in their calculations, engineers ensure heat transfer efficiency is appropriately modeled in real-world scenarios. The heat transfer coefficients combine with surface areas to give the resistances $R1$ and $R5$, showing how effective convection is at transferring heat through the window.