Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

Define emissivity and absorptivity. What is Kirchhoff's law of radiation?

How do \((a)\) draft and \((b)\) cold floor surfaces cause discomfort for a room's occupants?

The heat generated in the circuitry on the surface of a silicon chip \((k=130 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is conducted to the ceramic substrate to which it is attached. The chip is \(6 \mathrm{~mm} \times 6 \mathrm{~mm}\) in size and \(0.5 \mathrm{~mm}\) thick and dissipates \(5 \mathrm{~W}\) of power. Disregarding any heat transfer through the \(0.5-\mathrm{mm}\) high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation.

The inner and outer surfaces of a \(0.5-\mathrm{cm}\) thick \(2-\mathrm{m} \times 2-\mathrm{m}\) window glass in winter are \(10^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. If the thermal conductivity of the glass is \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the amount of heat loss through the glass over a period of \(5 \mathrm{~h}\). What would your answer be if the glass were \(1 \mathrm{~cm}\) thick?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free