Question

A room at \(20^{\circ} \mathrm{C}\) air temperature is loosing heat to the outdoor air at \(0^{\circ} \mathrm{C}\) at a rate of \(1000 \mathrm{~W}\) through a \(2.5\)-m-high and 4 -m-long wall. Now the wall is insulated with \(2-\mathrm{cm}\) thick insulation with a conductivity of \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surface of the wall, the room air temperature, and the outdoor air temperature to remain unchanged. Also, disregard radiation. (a) \(20 \mathrm{~W}\) (b) \(561 \mathrm{~W}\) (c) \(388 \mathrm{~W}\) (d) \(167 \mathrm{~W}\) (e) \(200 \mathrm{~W}\)

Short answer

Answer: The new rate of heat loss after adding the insulation is approximately 20 W.

Step-by-step solution

Find the temperature difference between indoor and outdoor air

We are given the temperature of indoor and outdoor air. To find the temperature difference, subtract the outdoor air temperature from the indoor air temperature: \(\Delta T = T_{indoor} - T_{outdoor} = 20^{\circ} C - 0^{\circ} C = 20^{\circ} C\)

Calculate the area of the wall

To calculate the area of the wall, multiply the height and length of the wall: \(A = height \times length = 2.5 \mathrm{m} \times 4 \mathrm{m} = 10 \mathrm{m^2}\)

Determine the thermal resistance of the insulation

The thermal resistance of the insulation can be calculated using the formula: \(R_{insulation} = \frac{thickness}{k} = \frac{0.02 \mathrm{m}}{0.02 \mathrm{W/m \cdot K}} = 1 \mathrm{K/W}\)

Calculate the total thermal resistance

We're given that the rate of heat loss (without the insulation) is 1000 W. So, we can calculate the total thermal resistance of the wall without insulation: \(R_{total} = \frac{\Delta T}{Q} = \frac{20 \mathrm{K}}{1000 \mathrm{W}} = 0.02 \mathrm{K/W}\) Now that we've found the insulation's thermal resistance, we can add we add the thermal resistance value of the insulation to find the total thermal resistance of the wall with insulation: \(R_{total}^{'} = R_{total} + R_{insulation} = 0.02 \mathrm{K/W} + 1 \mathrm{K/W} = 1.02 \mathrm{K/W}\)

Calculate the new rate of heat loss

Now we can use the new total thermal resistance to calculate the new rate of heat loss: \(Q^{'} = \frac{\Delta T}{R_{total}^{'}} = \frac{20 \mathrm{K}}{1.02 \mathrm{K/W}} = 19.61 \mathrm{W}\) The result is approximately 20 W, which corresponds to the answer (a) \(20 \mathrm{W}\).

Key concepts

Understanding Thermal Resistance

Thermal resistance is a concept that measures the ability of a material to resist the flow of heat. It is especially useful in the context of insulation and building design, because it helps determine how effectively a material can prevent heat exchange between different areas. To put it simply, thermal resistance is to heat transfer what electrical resistance is to electrical current.

Let's use the example given in the exercise. In this scenario, the insulation added to the wall has a thermal resistance calculated by dividing its thickness by its thermal conductivity, resulting in a value of 1 K/W. This number represents how much temperature difference (in Kelvin) would cause one watt of heat to flow through the insulator per second. The higher the thermal resistance, the less heat is lost, as there is more resistance to the heat flow.

In a practical sense, when you add insulation to a wall, the total thermal resistance of the wall increases. It's like adding an extra layer of clothing on a cold day—it slows down the rate at which your body loses heat to the environment. Hence, to improve the heat efficiency of a room or any enclosed space, one should aim to increase the overall thermal resistance of the enclosure. This concept is why we often seek materials with high thermal resistance for insulation purposes.

The Role of Thermal Conductivity in Heat Loss

Thermal conductivity is a measure of a material's ability to conduct heat. It represents how easily heat can pass through a material, with high values indicating that a material is a good heat conductor, like metal, and lower values indicating a poor heat conductor, such as foam insulation.

The exercise provided gives us the thermal conductivity of the wall insulation (0.02 W/m·K), which is a crucial piece of data for calculating how much heat will be lost through the wall. A key point in understanding thermal conductivity is that it's an intrinsic property of the material itself, and not affected by its dimensions or shape. However, when it comes to a physical installation, the size, shape, and thickness of the material will determine how much heat is transferred – which brings us back to thermal resistance, which does take into account the material’s dimensions.

Generally, in building and home improvement, materials with low thermal conductivity are sought for insulation to minimize heat transfer and therefore energy usage. The challenge lies in balancing good insulation while also addressing factors like cost, durability, and other structural needs.

Heat Transfer Coefficient's Impact on Insulation

The heat transfer coefficient is an indicator of how well heat is transmitted from a material to its surroundings (like air or water). In technical terms, it quantifies the amount of heat that passes through a unit area of a material, for a given temperature difference between the material and its environment. Higher coefficients mean that heat is transferred more efficiently, and vice versa. Its role in our context is passive, yet it's an essential part of the whole system that determines how much energy is lost to the surroundings.

In the exercise, we are asked to assume the heat transfer coefficients on the inner and outer surface of the wall remain constant. The significance of this assumption can't be overstated. It allows a straightforward calculation by ignoring the complex nature of how different materials might alter these coefficients. For further heat loss calculations, understanding the heat transfer coefficient can help in selecting the right materials for a building's exterior that will interface effectively with the environment.

For instance, glass has a different heat transfer coefficient compared to brick. When designing or improving a building’s energy efficiency, this coefficient is pivotal. One could theoretically use materials that inherently have a lower heat transfer coefficient to reduce heat loss or gain without significantly altering the structure's thermal resistance.