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During a winter day, the window of a patio door with a height of \(1.8 \mathrm{~m}\) and width of \(1.0 \mathrm{~m}\) shows a frost line near its base. The room wall and air temperatures are \(15^{\circ} \mathrm{C}\). (a) Explain why the window would show a frost layer at the base rather than at the top. (b) Estimate the heat loss through the window due to free convection and radiation. Assume the window has a uniform temperature of \(0^{\circ} \mathrm{C}\) and the emissivity of the glass surface is \(0.94\). If the room has electric baseboard heating, estimate the corresponding daily cost of the window heat loss for a utility rate of \(0.18 \mathrm{\$} / \mathrm{kW} \cdot \mathrm{h}\).

Short Answer

Expert verified
The frost layer forms at the base of the window because cold air sinks due to its higher density than warm air. The daily cost of the window heat loss using electric baseboard heating at a utility rate of \(0.18\ \mathrm{\$} / \mathrm{kWh}\) is approximately \(0.43\).

Step by step solution

01

Explain the frost layer formation at the base

The frost layer forms at the base of the window because cold air sinks. Since cold air is denser than warm air, it settles at the bottom, causing the temperature at the base of the window to be cooler than the temperature at the top. This temperature difference leads to the formation of frost at the bottom of the window.
02

Calculate free convection heat loss

To estimate the heat loss through the window due to free convection, we can use the following formula: \( q_c = h_c \cdot A \cdot \Delta T \) Where: \(q_c\) = heat loss due to free convection [W], \(h_c\) = convection heat transfer coefficient [W/m虏路K], \(A\) = surface area of the window [m虏], \(\Delta T\) = temperature difference between the window and the room air [K]. Let's first calculate the surface area of the window: \(A = 1.8 \times 1.0 = 1.8\ \mathrm{m^2}\) Next, we need to find \(h_c\) for free convection between the window and air. According to various references, a reasonable value for the convection heat transfer coefficient for a vertical glass surface is about \(h_c \approx 2.5 \ \mathrm{W/m^2 \cdot K}\). The temperature difference \(\Delta T\) is given as: \(\Delta T = T_{air} - T_{window} = 15 - 0 = 15\) Now, we can calculate the free convection heat loss: \( q_c = 2.5 \times 1.8 \times 15 = 67.5\ \mathrm{W} \)
03

Calculate radiation heat loss

The heat loss due to radiation can be calculated using the Stefan-Boltzmann law: \( q_r = \epsilon \cdot \sigma \cdot A \cdot (T_{window}^4 - T_{air}^4) \) Where: \(q_r\) = heat loss due to radiation [W], \(\epsilon\) = emissivity of the glass surface (given as 0.94), \(\sigma\) = Stefan-Boltzmann constant (\(5.6703\times10^{-8} \ \mathrm{W/m^2 \cdot K^4}\)), \(T_{window}\) and \(T_{air}\) are given in Kelvin. First, let's convert the Celsius temperatures to Kelvin: \(T_{window} = 0 + 273.15 = 273.15\ \mathrm{K}\), \(T_{air} = 15 + 273.15 = 288.15\ \mathrm{K}\). Now, we can calculate the radiation heat loss: \( q_r = 0.94 \times 5.6703 \times 10^{-8} \times 1.8 \times (273.15^4 - 288.15^4) \approx 31.72\ \mathrm{W} \)
04

Calculate total heat loss and daily cost

The total heat loss is the sum of the free convection and radiation heat losses: \( q_{total} = q_c + q_r = 67.5 + 31.72 = 99.22\ \mathrm{W} \) To calculate the daily cost of the window heat loss, first, find the daily heat loss in kilowatt-hours: \( \frac{99.22 \ \mathrm{W} \times 24\ \mathrm{h}}{1000} = 2.381\ \mathrm{kWh} \) Now, multiply the daily heat loss by the utility rate to find the daily cost: \( 2.381\ \mathrm{kWh} \times 0.18\ \mathrm{\$} / \mathrm{kWh} = 0.4286\ \mathrm{\$} \)
05

Conclusion

This exercise required explaining the frost layer formation at the base of a window and estimating the heat loss through the window due to free convection and radiation. The frost forms at the base of a window because cold air sinks as it is denser than warm air. The daily cost of the window heat loss using electric baseboard heating at a utility rate of \(0.18\ \mathrm{\$} / \mathrm{kWh} is approximately \)0.43.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Free Convection
When we talk about free convection, we're addressing a natural phenomenon where heat is transferred due to the bulk movement of molecules within fluids such as air or water. This occurs without any external source, hence 'free'. It's driven by differences in density caused by temperature variances within the fluid. Imagine a hot cup of tea steaming in a cool room; the heat from the tea naturally rises, while cooler air descends to take its place.

In our window example, colder air close to the frosty window gets denser and sinks, displacing the warmer air upward, creating a circulation loop. This loop continuously moves heat away from the window and explains why the frost layer forms at the base. The rate of heat transfer due to free convection can be estimated with a formula:
\[ q_c = h_c \cdot A \cdot \Delta T \]
Where \( q_c \) is the heat transfer due to convection in watts (W), \( h_c \) represents the convection heat transfer coefficient in watts per square meter-kelvin (W/m虏路K), \( A \) is the area through which heat is being transferred in square meters (m虏), and \( \Delta T \) is the temperature difference between the surface and the fluid in kelvin (K).

It's essential to accurately estimate \( h_c \) for specific conditions to determine the heat loss. If \( h_c \) is overestimated, we could end up assuming more heat loss than what actually occurs, while underestimating it could lead to insufficient heating solutions. The balance of understanding and calculating free convection is critical for efficient thermal management in various applications, including heating systems and thermal insulation.
Radiation Heat Transfer Explored
Radiation heat transfer differs from convection as it doesn't require a medium to occur. This type of heat transfer happens through electromagnetic waves and can take place in a vacuum, such as space. Consider how the Sun heats the Earth; despite the vast emptiness of space, the warmth still reaches us.

The amount of heat transferred by radiation is influenced by several factors, including the surface's temperature, its ability to emit radiation (emissivity), and its area. In the context of our winter window, it is important to quantify the heat loss due to radiation to understand the window鈥檚 overall thermal performance. To do this, we use the Stefan-Boltzmann law framed in the equation:
\[ q_r = \epsilon \cdot \sigma \cdot A \cdot (T_{window}^4 - T_{air}^4) \]
Here, \( q_r \) is the heat transfer by radiation in watts (W), \( \epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant (5.6703 x 10^-8 W/m虏路K鈦), and \( T \) represents temperature in kelvins (K). The fourth power dependency on temperature implies that even small differences in temperature can lead to significant changes in heat transfer rates.

Improper estimation of either the surface鈥檚 emissivity or omitting the temperature differences raised to the fourth power could lead to incorrect predictions of the heat loss, which explains why our understanding of radiation heat transfer is imperative for successful thermal management in buildings and many other engineering applications.
Applying the Stefan-Boltzmann Law
The Stefan-Boltzmann law is a cornerstone of thermodynamics that relates the heat radiated from a black body to the fourth power of its temperature. This law is vital for working out radiation heat transfer problems like the one in our textbook exercise.

To apply it, we must acknowledge that real materials are not perfect black bodies, which is why we introduce the term emissivity (\( \epsilon \)), a measure of how effectively a material radiates energy. A perfect black body has an emissivity of 1, while all real materials have values less than 1.

In our exercise, we used the given emissivity of the window glass (0.94) to account for its real-world radiation properties. Combined with the Stefan-Boltzmann law, we derived an equation that models the radiation heat loss through the window:
\[ q_r = \epsilon \cdot \sigma \cdot A \cdot (T_{window}^4 - T_{air}^4) \]
It's crucial to keep in mind temperature in this equation must be in absolute units (kelvin), as it involves temperature differences to the fourth power. Overlooking this could result in a substantial miscalculation of heat loss. As thermal properties such as emissivity can vary between different materials and surface conditions, selecting the appropriate value is key to producing an accurate calculation 鈥 a concept central to thermal design in both nature and industry.

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Most popular questions from this chapter

Consider a horizontal 6-mm-thick, 100 -mm-long straight fin fabricated from plain carbon steel \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\varepsilon=0.5\) ). The base of the fin is maintained at \(150^{\circ} \mathrm{C}\), while the quiescent ambient air and the surroundings are at \(25^{\circ} \mathrm{C}\). Assume the fin tip is adiabatic. (a) Estimate the fin heat rate per unit width, \(q_{f}^{\prime}\). Use an average fin surface temperature of \(125^{\circ} \mathrm{C}\) to estimate the free convection coefficient and the linearized radiation coefficient. How sensitive is your estimate to the choice of the average fin surface temperature? (b) Generate a plot of \(q_{f}^{\prime}\) as a function of the fin emissivity for \(0.05 \leq \varepsilon \leq 0.95\). On the same coordinates, show the fraction of the total heat rate due to radiation exchange.

The front door of a dishwasher of width \(580 \mathrm{~mm}\) has a vertical air vent that is \(500 \mathrm{~mm}\) in height with a \(20-\mathrm{mm}\) spacing between the inner tub operating at \(52^{\circ} \mathrm{C}\) and an outer plate that is thermally insulated. (a) Determine the heat loss from the tub surface when the ambient air is \(27^{\circ} \mathrm{C}\). (b) A change in the design of the door provides the opportunity to increase or decrease the \(20-\mathrm{mm}\) spacing by \(10 \mathrm{~mm}\). What recommendations would you offer with regard to how the change in spacing will alter heat losses?

Many laptop computers are equipped with thermal management systems that involve liquid cooling of the central processing unit (CPU), transfer of the heated liquid to the back of the laptop screen assembly, and dissipation of heat from the back of the screen assembly by way of a flat, isothermal heat spreader. The cooled liquid is recirculated to the CPU and the process continues. Consider an aluminum heat spreader that is of width \(w=275 \mathrm{~mm}\) and height \(L=175 \mathrm{~mm}\). The screen assembly is oriented at an angle \(\theta=30^{\circ}\) from the vertical direction, and the heat spreader is attached to the \(t=3\)-mm-thick plastic housing with a thermally conducting adhesive. The plastic housing has a thermal conductivity of \(k=0.21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and emissivity of \(\varepsilon=0.85\). The contact resistance associated with the heat spreaderhousing interface is \(R_{t, c}^{\prime \prime}=2.0 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If the CPU generates, on average, \(15 \mathrm{~W}\) of thermal energy, what is the temperature of the heat spreader when \(T_{\infty}=T_{\text {sur }}=23^{\circ} \mathrm{C}\) ? Which thermal resistance (contact, conduction, radiation, or free convection) is the largest?

Certain wood stove designs rely exclusively on heat transfer by radiation and natural convection to the surroundings. Consider a stove that forms a cubical enclosure, \(L_{s}=1 \mathrm{~m}\) on a side, in a large room. The exterior walls of the stove have an emissivity of \(\varepsilon=0.8\) and are at an operating temperature of \(T_{s s s}=500 \mathrm{~K}\). The stove pipe, which may be assumed to be isothermal at an operating temperature of \(T_{s, p}=400 \mathrm{~K}\), has a diameter of \(D_{p}=0.25 \mathrm{~m}\) and a height of \(L_{p}=2 \mathrm{~m}\), extending from stove to ceiling. The stove is in a large room whose air and walls are at \(T_{\infty}=T_{\text {sur }}=300 \mathrm{~K}\). Neglecting heat transfer from the small horizontal section of the pipe and radiation exchange between the pipe and stove, estimate the rate at which heat is transferred from the stove and pipe to the surroundings.

According to experimental results for parallel airflow over a uniform temperature, heated vertical plate, the effect of free convection on the heat transfer convection coefficient will be \(5 \%\) when \(G r_{L} / R e_{L}^{2}=0.08\). Consider a heated vertical plate \(0.3 \mathrm{~m}\) long, maintained at a surface temperature of \(60^{\circ} \mathrm{C}\) in atmospheric air at \(25^{\circ} \mathrm{C}\). What is the minimum vertical velocity required of the airflow such that free convection effects will be less than \(5 \%\) of the heat transfer rate?

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