Question

A \(1.0 \mathrm{~m} \times 1.5 \mathrm{~m}\) double-pane window consists of two 4-mm-thick layers of glass \((k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that are separated by a \(5-\mathrm{mm}\) air gap \(\left(k_{\text {air }}=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The heat flow through the air gap is assumed to be by conduction. The inside and outside air temperatures are \(20^{\circ} \mathrm{C}\) and \(-20^{\circ} \mathrm{C}\), respectively, and the inside and outside heat transfer coefficients are 40 and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine \((a)\) the daily rate of heat loss through the window in steady operation and \((b)\) the temperature difference across the largest thermal resistence.

Short answer

Question: Calculate the daily heat loss through a double-pane window and the temperature difference across the largest thermal resistance, given the inside heat transfer coefficient 𝑥, the outside heat transfer coefficient 𝑦, the area of the window 𝐴, the thickness and thermal conductivity of the glass layers 𝐿1 and 𝑘1, the thickness and thermal conductivity of the air gap 𝐿2 and 𝑘2, and the inside and outside air temperatures 𝑇1 and 𝑇2. Answer: To calculate the daily heat loss and temperature difference across the largest thermal resistance, follow these steps: 1. Calculate the individual thermal resistances: convection resistance of the inner surface, conduction resistance of the first glass layer, conduction resistance of the air gap, conduction resistance of the second glass layer, and convection resistance of the outer surface, using the provided parameters. 2. Calculate the total thermal resistance by adding all the individual resistances. 3. Calculate the heat rate using the total thermal resistance and the given temperature differences. 4. Calculate the daily heat loss by multiplying the heat rate by the number of seconds in a day. 5. Find the largest thermal resistance and calculate the temperature difference across it using the heat rate. Using the given parameters, you can substitute the values into the formulas provided in the solution to find the daily heat loss and the temperature difference across the largest thermal resistance.

Step-by-step solution

Identify and calculate individual thermal resistances

We have the following thermal resistances in series: 1. Convection resistance of the inner surface 2. Conduction resistance of the first glass layer 3. Conduction resistance of the air gap 4. Conduction resistance of the second glass layer 5. Convection resistance of the outer surface Let's calculate these individual thermal resistances: 1. Convection resistance of the inner surface \(R_{\text{conv,1}} = \frac{1}{h_{1} \cdot A}\) , where \(h_1\) is the inside heat transfer coefficient and A is the area of the window. 2. Conduction resistance of the first glass layer \(R_{\text{cond,1}} = \frac{L_1}{k_1\cdot A}\), where L1 is the thickness of the first glass layer, \(k_1\) is the thermal conductivity of the glass. 3. Conduction resistance of the air gap \(R_{\text{cond,2}} = \frac{L_2}{k_2\cdot A}\) , where L2 is the thickness of the air gap, \(k_2\) is the thermal conductivity of the air gap. 4. Conduction resistance of the second glass layer \(R_{\text{cond,3}} = \frac{L_1}{k_1\cdot A}\) (The same formula as for the first glass layer) 5. Convection resistance of the outer surface \(R_{\text{conv,2}} = \frac{1}{h_{2} \cdot A}\) , where \(h_2\) is the outside heat transfer coefficient.

Calculate the total thermal resistance

Now, we will calculate the total thermal resistance which is the sum of all individual thermal resistances: \(R_\text{total} = R_{\text{conv,1}} + R_{\text{cond,1}} + R_{\text{cond,2}} + R_{\text{cond,3}} + R_{\text{conv,2}}\)

Calculate Heat Rate

Next, we will calculate the heat rate using total thermal resistance and the given temperature difference: \(\dot{Q} = \frac{T_1 - T_2}{R_\text{total}}\), where \(T_1\) is the inside air temperature and \(T_2\) is the outside air temperature.

Calculate the Daily Heat Loss

To find the daily heat loss, we will multiply the heat rate by the number of seconds in a day: \(Q_\text{daily} = \dot{Q} \cdot \text{Seconds in a day}\)

Find the largest thermal resistance and calculate temperature difference

Now, we have to find the largest thermal resistance among all the individual thermal resistances. Once we find that, we will calculate the temperature difference across it using the heat rate and the following formula: \(\Delta T_\text{largest} = \dot{Q} \cdot R_\text{largest}\)

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in the study of thermodynamics and plays a crucial role in various engineering fields. It can be described as the movement of heat energy from one point to another due to a difference in temperature.

There are three main modes of heat transfer: conduction, convection, and radiation. Conduction involves the transfer of heat through a solid medium without any physical movement of the material itself. It's the process we consider when heat flows through the layers of a double-pane window or a wall. Convection, on the other hand, involves the transfer of heat by the physical movement of fluid, which can be a liquid or gas. When you feel the warmth of a heater through the air, that's convection at work. Lastly, radiation is heat transfer through electromagnetic waves, like the heat from the sun.

In the context of the provided exercise, both conduction and convection are at play. The double-pane window allows heat to conduct through its glass and air layers, and convection is responsible for heat transfer at both the inner and outer surfaces of the window.

Conduction in Solids

Conduction in solids is a key concept when analyzing heat loss in structures like buildings or in designing devices that require thermal management. Within a solid, heat conduction occurs as kinetic energy is transferred from molecule to molecule or through free electrons in the case of metals. This mode of heat transfer is quantified by Fourier's law of heat conduction, which states that the heat transfer rate through a material is directly proportional to the negative gradient of temperature and the area through which heat is transferred, and it is inversely proportional to the material's thickness.

The mathematical representation of Fourier's law is:
\[\begin{equation}\begin{aligned} \text{Q} = -k \times A \times \frac{dT}{dx}\text{Q} is the heat transfer per unit time (W),\ k is the thermal conductivity of the material (W/(m·K)),\ A is the area through which heat is being transferred (m^2),\ \frac{dT}{dx} is the temperature gradient through the thickness of the material (K/m).\end{aligned}\end{equation}\]

For the double-pane window in the exercise, thermal conduction occurs through the glass layers and the air gap. The thermal conductivity value 'k' provided for glass and air is crucial to calculate the rate of heat transfer.

Convection Heat Transfer

Convection heat transfer involves the movement of heat by fluid motion, which can be a result of natural buoyancy forces or forced by a fan or pump, for instance. It's an important mode of heat transfer in many engineering applications like heating, ventilation, and air conditioning (HVAC) systems.

The rate of convective heat transfer is governed by Newton's law of cooling, which states: \[\begin{equation}\begin{aligned}\text{Q} = h \times A \times (T_s - T_\text{fluid})\text{Q} is the heat transfer rate (W),\ h is the convective heat transfer coefficient (W/(m^2·K)),\ A is the surface area (m^2),\ T_s is the surface temperature (K),\ T_\text{fluid} is the fluid temperature (K).\end{aligned}\end{equation}\]

In the scenario of the double-pane window, both sides of the glass are exposed to air, with one side being the warmer indoor air and the other being the colder outdoor air. The heat transfer coefficients given for the inside and outside air (40 and 20 W/m²·K, respectively) are vital for calculating the convective resistance. Therefore, understanding convective heat transfer helps us predict how effective a surface will be at exchanging heat with surrounding air.