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Chapter 1: Problem 6

A glass window of width \(W=1 \mathrm{~m}\) and height \(H=2 \mathrm{~m}\) is \(5 \mathrm{~mm}\) thick and has a thermal conductivity of \(k_{e}=\) \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the inner and outer surface temperatures of the glass are \(15^{\circ} \mathrm{C}\) and \(-20^{\circ} \mathrm{C}\), respectively, on a cold winter day, what is the rate of heat loss through the glass? To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining. panes are separated by an air space if the spacing is \(10 \mathrm{~mm}\) and the glass surfaces in contact with the air have temperatures of \(10^{\circ} \mathrm{C}\) and \(-15^{\circ} \mathrm{C}\), what is the rate of heat loss from a \(I \mathrm{~m} \times 2 \mathrm{~m}\) window? The thermal conductivity of air is \(k_{t}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
Single pane heat loss: 19600 W; double pane with air space loss: 120 W.

Step by step solution

01

Calculate the Area of the Window

The area \(A\) of the window is calculated using the formula \(A = W \times H\). Given that the width \(W = 1 \space m\) and the height \(H = 2 \space m\), the area \(A\) is \(A = 1 \space m \times 2 \space m = 2 \space m^2\).
02

Determine the Temperature Difference

The temperature difference \(\Delta T\) across the single pane window is the difference between the inner and outer surface temperatures, which is \(15^{\circ}C - (-20^{\circ}C) = 35^{\circ}C\).
03

Calculate the Rate of Heat Loss for the Single Pane

To determine the rate of heat loss \(Q\) through the single pane window, use the formula: \[ Q = \frac{k_e \cdot A \cdot \Delta T}{d} \]where \(k_e = 1.4 \space W/m \cdot K\), \(d = 0.005 \space m\), \(A = 2 \space m^2\), and \(\Delta T = 35^{\circ}C\). Calculate this as:\[ Q = \frac{1.4 \cdot 2 \cdot 35}{0.005} = 19600 \space W \]
04

Calculate the Temperature Difference for Air Space

For the double pane with air space, the temperature difference \(\Delta T_1\) between the two air surfaces is \(10^{\circ}C - (-15^{\circ}C) = 25^{\circ}C\).
05

Calculate the Rate of Heat Loss for the Air Space

Use the formula for heat transfer rate through the air space:\[ Q_a = \frac{k_t \cdot A \cdot \Delta T_1}{d_a} \]where \(k_t = 0.024 \space W/m \cdot K\), \(d_a = 0.01 \space m\), \(A = 2 \space m^2\), and \(\Delta T_1 = 25^{\circ}C\). Calculating gives:\[ Q_a = \frac{0.024 \cdot 2 \cdot 25}{0.01} = 120 \space W \]

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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 an essential property of materials that helps us understand how they conduct heat. It is denoted by the symbol \(k\) and expresses how well a material allows heat to pass through it. Higher thermal conductivity means that the material is an excellent conductor of heat, and heat energy can move through it easily. In our exercise, the glass window has a thermal conductivity of \(1.4 \, \text{W/m}\cdot\text{K}\). This value indicates how effectively the glass material allows heat to traverse from the warmer side to the cooler side.
Materials with high thermal conductivity, like metals, are ideal for applications where efficient heat transfer is necessary. Conversely, materials with low thermal conductivity, like wood or air, are better suited for insulation, as they inhibit heat flow. Understanding and choosing the right materials based on their thermal conductivities can greatly affect the efficiency of heating and cooling systems.
Rate of Heat Loss
The rate of heat loss, often symbolized as \(Q\), is the amount of heat energy that transfers through a material over a given time. It is crucial in determining how much energy is needed to maintain a temperature difference across a barrier. The formula used to calculate the rate of heat loss through a material is:
  • \( Q = \frac{k \cdot A \cdot \Delta T}{d} \)
Where:
  • \(k\) is the thermal conductivity of the material,
  • \(A\) is the area through which heat is being transferred,
  • \(\Delta T\) is the temperature difference across the material,
  • \(d\) is the thickness of the material.
In the given problem, we applied this formula to find the heat loss through both a single and double pane of glass. For the single pane, it was calculated to be \(19,600 \, W\). By adding an air space and utilizing the double pane design, the heat loss reduced significantly. This shows how modifying the structure or the material of a barrier can enhance its insulating properties and reduce energy costs.
Temperature Difference
Temperature difference, often represented as \(\Delta T\), is the driving force behind heat transfer. It is simply the difference between the temperature on one side of a material and the temperature on the other side. In our scenario, the temperature difference across the single pane window is \(35^{\circ}C\), calculated as \(15^{\circ}C - (-20^{\circ}C)\).
Understanding temperature differences is crucial in heat transfer as it influences the rate of heat loss. Greater temperature differences result in a more significant driving force, leading to larger amounts of heat energy being transferred. Hence, in engineering designs, reducing the temperature difference can help minimize energy loss. In double-pane windows, the temperature difference across the inner air space was reduced to \(25^{\circ}C\), helping to limit the heat loss and maintaining a better-insulated environment.

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Most popular questions from this chapter

A vertical slab of Woods metal is joined to a substrate on one surface and is melted as it is uniformly irradialed by a laser souree on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\). and the melt rans off by gravity as soon as it is formed. The abserptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{y}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantanecus rate of melting in \(\mathrm{kg} / \mathrm{s}-\mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\mathrm{w}}=20 r \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \(\left(\varepsilon=0.4, T_{\text {max }}=\right.\) \(20^{\circ} \mathrm{C}\), determine the instantaneous rate of melting during imadiation.

The concrete slab of a basement is \(11 \mathrm{~m}\) long. \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basemient is heated by a gas furnace operating at an cfficiency of \(\eta_{r}=0.90\) and natural gas is priced at \(C_{r}=50.01 \mathrm{MJ}\), what is the daily cost of the heat loss?

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), while the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) Neglecting radiation exchange with the surruundings, calculate the temperiture of the roof under seady. state conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the sume ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airfiow conditions over the roof, increasing with increasing air specd. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

An electric resistance heater is embedded in a lony cylinder of diameter \(30 \mathrm{~mm}\). When water with a temperature of \(25^{\circ} \mathrm{C}\) and velocity of \(1 \mathrm{~m} / \mathrm{s}\) flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of \(90^{\circ} \mathrm{C}\) is \(28 \mathrm{~kW} / \mathrm{m}\). When air, also at \(25^{\circ} \mathrm{C}\), but with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) is flowing, the power per unit length required to maintain the same surface temperature is \(400 \mathrm{~W} / \mathrm{m}\). Calculate and compare the convection coefficients for the flows of water and air.

A computer consists of an array of five printed circuit boards (PCBs), each dissipating \(P_{h}=20 \mathrm{~W}\) of power. Cooling of the electronic components on a board is provided by the forced flow of air, equally distributed in passages formed by adjoining boards, and the convection coefficient associated with heat transfer from the components to the air is approximately \(h=200 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). Air enters the computer console at a temperiture of \(T_{6}=20^{\circ} \mathrm{C}\), and flow is driven by a fin whese power consumption is \(P_{f}=25 \mathrm{~W}\). Iniet ain \(\forall_{1} T_{i}\) (a) If the tempensture rise of the air flow. \(\left(T_{e}-T_{i}\right)_{\text {is }}\) is ot to exceed \(15^{\circ} \mathrm{C}\), what is the minimum allowable volumetric flow rate \(\forall\) of the air? 'The density and specific heat of the air may be approximated is \(\rho=1.161\) \(\mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) The component that is most suncejtible to thermal failure dissipates I W/cm \({ }^{2}\) of surface area. To minimive the potential for thermal failure, where should the component be installed on a PCB? What is its surface lemperature at this location?
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