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Chapter 9: Problem 95

A vertical, double-pane window, which is \(1 \mathrm{~m}\) on a side and has a \(25-\mathrm{mm}\) gap filled with atmospheric air, separates quiescent room air at \(T_{\infty, i}=20^{\circ} \mathrm{C}\) from quiescent ambient air at \(T_{\infty, o}=-20^{\circ} \mathrm{C}\). Radiation exchange between the window panes, as well as between each pane and its surroundings, may be neglected. (a) Neglecting the thermal resistance associated with conduction heat transfer across each pane, determine the corresponding temperature of each pane and the rate of heat transfer through the window. (b) Comment on the validity of neglecting the conduction resistance of the panes if each is of thickness \(L_{p}=6 \mathrm{~mm}\).

Short Answer

Expert verified
The temperatures of the inner and outer panes are \(10^{\circ}\mathrm{C}\) and \(-30^{\circ}\mathrm{C}\), respectively, and the rate of heat transfer through the window is \(38.4\,\mathrm{W}\). Neglecting the conduction resistance of the panes is a valid assumption, as the thermal resistance of the glass panes is more than 100 times smaller than the thermal resistance of the air gap.

Step by step solution

01

Determine the average air temperature in the gap

First, we need to determine the average temperature of the air in the gap between the panes: \(T_{avg} = \frac{T_{\infty,i} + T_{\infty,o}}{2} = \frac{20^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C}}{2} = 0^{\circ}\mathrm{C}\)
02

Calculate the thermal conductivity of the air

Next, we will calculate the thermal conductivity of the air between the panes at the average temperature. We can refer to the air properties table and at \(0^{\circ}\mathrm{C}\), the thermal conductivity of air is: \(k_{air} \approx 0.024\,\mathrm{W/(m \cdot K)}\)
03

Calculate the total heat transfer rate

We can use Fourier's law of heat conduction to find the total heat transfer rate, \(Q\), as follows: \(Q = \frac{k_{air} \cdot A \cdot \Delta T}{L}\) Where: \(A = 1\,\mathrm{m^2}\) (Area of the window) \(\Delta T = T_{\infty,i} - T_{\infty,o} = 20^{\circ}\mathrm{C} - (-20^{\circ}\mathrm{C}) = 40\,\mathrm{K}\) (Temperature difference) \(L = 0.025\,\mathrm{m}\) (thickness of the air gap) Now, plugging the values into the equation, we get: \(Q = \frac{0.024\,\mathrm{W/(m \cdot K)} \cdot 1\,\mathrm{m^2} \cdot 40\,\mathrm{K}}{0.025\,\mathrm{m}} = 38.4\,\mathrm{W}\)
04

Calculate the temperature of each pane

To determine the temperature of each pane, we need to consider the heat transfer through the air gap only. Since the temperature drop across the air gap is linear, we can use the following formula: \(T_{pane, i} = T_{\infty, i} - \frac{Q \cdot L}{2 \cdot k_{air} \cdot A}\) \(T_{pane, i} = 20^{\circ}\mathrm{C} - \frac{38.4\,\mathrm{W} \cdot 0.025\,\mathrm{m}}{2 \cdot 0.024\,\mathrm{W/(m \cdot K)} \cdot 1\,\mathrm{m^2}} = 10^{\circ} \mathrm{C}\) \(T_{pane, o} = T_{\infty, o} + \frac{Q \cdot L}{2 \cdot k_{air} \cdot A}\) \(T_{pane, o} = -20^{\circ}\mathrm{C} + \frac{38.4\,\mathrm{W} \cdot 0.025\,\mathrm{m}}{2 \cdot 0.024\,\mathrm{W/(m \cdot K)} \cdot 1\,\mathrm{m^2}} = -30^{\circ} \mathrm{C}\)
05

Part (b) Commenting on the validity of neglecting conduction resistance of the panes

The thickness of each pane is given as \(L_p = 6\,\mathrm{mm}\) To investigate the validity of neglecting conduction resistance, we need to compare the thermal resistance of the air gap with that of the glass panes. The thermal resistance of the air gap is: \(R_{air} = \frac{L}{k_{air} \cdot A} = \frac{0.025\,\mathrm{m}}{0.024\,\mathrm{W/(m \cdot K)} \cdot 1\,\mathrm{m^2}} = 1.042\,\mathrm{K/W}\) Now, let's find the thermal conductivity of the glass: At room temperature, the thermal conductivity of glass is approximately \(1\,\mathrm{W/(m \cdot K)}\). The thermal resistance of each glass pane is: \(R_{glass} = \frac{L_p}{k_{glass} \cdot A} = \frac{0.006\,\mathrm{m}}{1\,\mathrm{W/(m \cdot K)} \cdot 1\,\mathrm{m^2}} = 0.006\,\mathrm{K/W}\) The thermal resistance of the glass panes is more than 100 times smaller than the thermal resistance of the air gap. This means that neglecting the conduction resistance of the glass panes is a valid assumption for this problem.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a property of materials that indicates how well they conduct heat. It's a measure of the rate at which heat flows through a material per unit area and temperature gradient. In our exercise, we calculate the thermal conductivity of air at an average temperature of 0掳C. We find that at this temperature, the thermal conductivity of air is approximately 0.024 W/(m路K).

Thermal conductivity can vary based on the material and its temperature. That's why it's essential to refer to property tables or charts for accurate values when solving heat transfer problems. Here are some key points about thermal conductivity:
  • Higher thermal conductivity means a material is a better heat conductor.
  • Metals typically have high thermal conductivity, making them good conductors of heat.
  • Insulators like air and wood have low thermal conductivity, indicating they do not conduct heat as well, which is ideal for thermal resistance purposes.
Understanding thermal conductivity is crucial for evaluating materials' heat transfer capabilities in various engineering applications.
Fourier's Law
Fourier's Law is a fundamental principle that describes the conduction of heat through materials. It states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the area perpendicular to that gradient. The equation for Fourier's Law is given by:
\[ Q = -k rac{ ext{d}T}{ ext{d}x}A \] Where:
  • \(Q\) is the heat transfer rate (Watts)
  • \(k\) is the thermal conductivity (W/(m路K))
  • \(\text{d}T/\text{d}x\) is the temperature gradient (K/m)
  • \(A\) is the cross-sectional area (m2)
In our problem, we use a simplified version of Fourier's Law to calculate the total heat transfer through the window by substituting the temperature difference and thickness of the air gap. This simplification leads to:\[ Q = \frac{k_{air} \cdot A \cdot \Delta T}{L} \]It's essential for solving practical heat transfer problems since it allows us to determine how much heat is moving through a given material, providing insight into thermal insulation efficiency and energy consumption.
Thermal Resistance
Thermal resistance quantifies how well a material or assembly resists the flow of heat. It's similar to electrical resistance but applies to thermal systems. The thermal resistance of a material is calculated by dividing the material thickness by its thermal conductivity times the area through which heat is transferred. For the air gap and glass panes in our problem, thermal resistance is calculated separately.

The formula for thermal resistance \(R\) is given by:
\[ R = \frac{L}{k \cdot A} \]Where:
  • \(L\) is the thickness of the material (m)
  • \(k\) is the thermal conductivity (W/(m路K))
  • \(A\) is the cross-sectional area (m2)
In this problem, the air gap has a much higher thermal resistance than the glass panes, which is why the panes' thermal resistance can be neglected. Key takeaways are:
  • Higher thermal resistance indicates better insulation.
  • It helps determine the efficiency of thermal barriers.
  • Comparing thermal resistances helps prioritize areas for improved insulation in heat transfer problems.
By understanding thermal resistance, engineers can design systems that either enhance or reduce heat flow depending on the specific application's needs.

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Most popular questions from this chapter

A long, uninsulated steam line with a diameter of \(89 \mathrm{~mm}\) and a surface emissivity of \(0.8\) transports steam at \(200^{\circ} \mathrm{C}\) and is exposed to atmospheric air and large surroundings at an equivalent temperature of \(20^{\circ} \mathrm{C}\). (a) Calculate the heat loss per unit length for a calm day. (b) Calculate the heat loss on a breezy day when the wind speed is \(8 \mathrm{~m} / \mathrm{s}\). (c) For the conditions of part (a), calculate the heat loss with a 20 -mm- thick layer of insulation ( \(k=\) \(0.08 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Would the heat loss change significantly with an appreciable wind speed?

9.27 The vertical rear window of an automobile is of thickness \(L=8 \mathrm{~mm}\) and height \(H=0.5 \mathrm{~m}\) and contains fine-meshed heating wires that can induce nearly uniform volumetric heating, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). (a) Consider steady-state conditions for which the interior surface of the window is exposed to quiescent air at \(10^{\circ} \mathrm{C}\), while the exterior surface is exposed to air at \(-10^{\circ} \mathrm{C}\) moving in parallel flow over the surface with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). Determine the volumetric heating rate needed to maintain the interior window surface at \(T_{s, i}=15^{\circ} \mathrm{C}\). (b) The interior and exterior window temperatures, \(T_{s, i}\) and \(T_{s, \rho}\), depend on the compartment and ambient temperatures, \(T_{\infty, i}\) and \(T_{\infty, p}\), as well as on the velocity \(u_{\infty}\) of air flowing over the exterior surface and the volumetric heating rate \(\dot{q}\). Subject to the constraint that \(T_{s, i}\) is to be maintained at \(15^{\circ} \mathrm{C}\), we wish to develop guidelines for varying the heating rate in response to changes in \(T_{\infty,}, T_{\infty \rho,}\), and/or \(u_{\infty}\). If \(T_{\infty, i}\) is maintained at \(10^{\circ} \mathrm{C}\), how will \(\dot{q}\) and \(T_{s, \rho}\) vary with \(T_{\infty, o}\) for \(-25 \leq T_{\infty,,} \leq 5^{\circ} \mathrm{C}\) and \(u_{\infty}=10,20\), and \(30 \mathrm{~m} / \mathrm{s}\) ? If a constant vehicle speed is maintained, such that \(u_{\infty}=30 \mathrm{~m} / \mathrm{s}\), how will \(\dot{q}\) and \(T_{s, o}\) vary with \(T_{\infty, i}\) for \(5 \leq T_{\infty, i} \leq 20^{\circ} \mathrm{C}\) and \(T_{\infty, o}=-25,-10\), and \(5^{\circ} \mathrm{C}\) ?

A 200-mm-square, 10-mm-thick tile has the thermophysical properties of Pyrex \((\varepsilon=0.80)\) and emerges from a curing process at an initial temperature of \(T_{i}=140^{\circ} \mathrm{C}\). The backside of the tile is insulated while the upper surface is exposed to ambient air and surroundings at \(25^{\circ} \mathrm{C}\).

Consider window blinds that are installed in the air space between the two panes of a vertical double-pane window. The window is \(H=0.5 \mathrm{~m}\) high and \(w=0.5 \mathrm{~m}\) wide, and includes \(N=19\) individual blinds that are each \(L=25 \mathrm{~mm}\) wide. When the blinds are open, 20 smaller, square enclosures are formed along the height of the window. In the closed position, the blinds form a nearly continuous sheet with two \(t=12.5 \mathrm{~mm}\) open gaps at the top and bottom of the enclosure. Determine the convection heat transfer rate between the inner pane, which is held at \(T_{s, i}=20^{\circ} \mathrm{C}\), and the outer pane, which is at \(T_{s, o}=-20^{\circ} \mathrm{C}\), when the blinds are in the open and closed positions, respectively. Explain why the closed blinds have little effect on the convection heat transfer rate across the cavity.

Consider an experiment to investigate the transition to turbulent flow in a free convection boundary layer that develops along a vertical plate suspended in a large room. The plate is constructed of a thin heater that is sandwiched between two aluminum plates and may be assumed to be isothermal. The heated plate is \(1 \mathrm{~m}\) high and \(2 \mathrm{~m}\) wide. The quiescent air and the surroundings are both at \(25^{\circ} \mathrm{C}\). (a) The exposed surfaces of the aluminum plate are painted with a very thin coating of high emissivity \((\varepsilon=0.95)\) paint. Determine the electrical power that must be supplied to the heater to sustain the plate at a temperature of \(T_{s}=35^{\circ} \mathrm{C}\). How much of the plate is exposed to turbulent conditions in the free convection boundary layer? (b) The experimentalist speculates that the roughness of the paint is affecting the transition to turbulence in the boundary layer and decides to remove the paint and polish the aluminum surface ( \(\varepsilon=0.05\) ). If the same power is supplied to the plate as in part (a), what is the steady- state plate temperature? How much of the plate is exposed to turbulent conditions in the free convection boundary layer?
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