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While driving down a highway early in the evening, the air flow over an automobile establishes an overall heat transfer coefficient of \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The passenger cabin of this automobile exposes \(9 \mathrm{~m}^{2}\) of surface to the moving ambient air. On a day when the ambient temperature is \(33^{\circ} \mathrm{C}\), how much cooling must the air conditioning system supply to maintain a temperature of \(20^{\circ} \mathrm{C}\) in the passenger cabin? (a) \(670 \mathrm{~W}\) (b) \(1284 \mathrm{~W}\) (c) \(2106 \mathrm{~W}\) (d) \(2565 \mathrm{~W}\) (e) \(3210 \mathrm{~W}\)

Short Answer

Expert verified
Answer: The required cooling capacity is 2106 W.

Step by step solution

01

Calculate the temperature difference

To find the temperature difference between the cabin and the outside air, subtract the cabin temperature from the ambient temperature: \(\Delta T = T_{ambient} - T_{cabin} = 33^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 13 \mathrm{~K}\) (NOTE: We use Kelvins(K) for temperature differences in the formula.)
02

Calculate the heat transfer rate

Now that we have the temperature difference, we can use the convective heat transfer equation to find the cooling capacity required: \(Q = hA\Delta T\) where \(h = 18 \frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}}\), \(A = 9 \mathrm{~m}^{2}\), and \(\Delta T = 13 \mathrm{~K}\). Plugging in the given values, we get: \(Q = 18 \frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}} \cdot 9 \mathrm{~m}^{2} \cdot 13 \mathrm{~K} = 2106 \mathrm{~W}\) The cooling capacity required for the air conditioning system to maintain the temperature at \(20^{\circ} \mathrm{C}\) in the passenger cabin is \(2106 \mathrm{~W}\), which corresponds to the choice (c) in the possible answers.

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Key Concepts

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

Understanding Heat Transfer Coefficient
The heat transfer coefficient is a measure of a material's ability to allow heat to pass through it. In the context of our exercise, it represents the effectiveness of the air flow over the car in transferring heat away from the vehicle's passenger cabin. The larger the heat transfer coefficient, the more efficient the heat transfer process becomes. An understanding of this concept is critical for engineers and designers as it influences various decisions, from the selection of materials to the design of heating and cooling systems.

As stated in the problem, the heat transfer coefficient (\( h \) ) for the air flow over the automobile is given as 18 W/m²·K. This indicates that for each square meter of the car’s surface and for each degree Kelvin of temperature difference between the surface and the ambient air, 18 Watts of heat is transferred. It is important to note that the heat transfer coefficient is dependent on the properties of the fluid (air in this case), the flow conditions, and the surface geometry and properties of the material in question.
Temperature Difference in Heat Transfer
The temperature difference, often denoted as ∆T, is the driving force behind heat transfer by convection. It represents the difference in temperature between the surface of interest and the surrounding fluid; in this case, the car cabin and the ambient air. The greater the difference in temperature, the greater the heat transfer rate.

To calculate this, we subtract the interior temperature from the ambient temperature, which gives us a temperature gradient. If this temperature gradient did not exist, there would be no heat flow. In our exercise, the ambient air is warmer at 33°C while the desired cabin temperature is 20°C, resulting in a temperature difference of 13K (note that temperatures in such calculations are often converted to Kelvin to maintain consistency in units). Understanding this temperature difference is essential for designing thermal systems, such as the car's air conditioning system, to ensure adequate control over the environment.
Cooling Capacity: Maintaining Comfort
Cooling capacity is the amount of heat energy that an air conditioning system must remove to maintain a certain temperature level in a given space. In thermal management, this is a critical value as it determines the power that the system must have to ensure comfortable living or working conditions. The cooling capacity is often expressed in Watts (W) or British Thermal Units per hour (BTU/hr).

In our example, the air conditioning system must supply significant cooling capacity to counteract the heat transferred by convection through the car cabin’s walls. Using the heat transfer coefficient, surface area, and temperature difference, we calculated the necessary cooling capacity as 2106W. This tells us how powerful the air conditioning system needs to be to maintain the cabin temperature at a comfortable 20°C despite the higher outside temperature. Consequently, this information is pivotal for HVAC engineers when sizing and installing air conditioning systems.

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Most popular questions from this chapter

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of \(2 \mathrm{~m}^{2}\) with emissivity and absorptivity of \(0.9\). The surface temperature of the absorber is \(35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Net heat rate absorbed by the solar collector heats the water from an inlet temperature \(\left(T_{\text {in }}\right)\) to an outlet temperature \(\left(T_{\text {out }}\right)\). If the water flow rate is \(5 \mathrm{~g} / \mathrm{s}\) with a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the temperature rise of the water.

A 25 -cm-diameter black ball at \(130^{\circ} \mathrm{C}\) is suspended in air, and is losing heat to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat transfer from the black ball is (a) \(217 \mathrm{~W}\) (b) \(247 \mathrm{~W}\) (c) \(251 \mathrm{~W}\) (d) \(465 \mathrm{~W}\) (e) \(2365 \mathrm{~W}\)

Determine a positive real root of this equation using \(E E S\) : $$ 3.5 x^{3}-10 x^{0.5}-3 x=-4 $$

The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the \(U\)-factor. The value of the \(U\)-factor ranges from about \(1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(0.22 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a \(1.2-\mathrm{m} \times 1.8-\mathrm{m}\) window of a house that is maintained at \(20^{\circ} \mathrm{C}\) when the outdoor air temperature is \(-8^{\circ} \mathrm{C}\).

The roof of a house consists of a 22-cm-thick (st) concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The emissivity of the outer surface of the roof is \(0.9\), and the convection heat transfer coefficient on that surface is estimated to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner surface of the roof is maintained at \(15^{\circ} \mathrm{C}\). On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\) while the night sky temperature for radiation heat transfer is \(255 \mathrm{~K}\). Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is \(\$ 1.20\) / therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-hour period.

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