91Ó°ÊÓ

Cooling of outdoor electronic equipment such as in telecommunications towers is difficult due to seasonal and diurnal variations of the air temperature, and potential fouling of heat exchange surfaces due to dust accumulation or insect nesting. A concept to provide a nearly constant sink temperature in a hermetically sealed environment is shown below. The cool surface is maintained at nearly constant groundwater temperature \(\left(T_{1}=5^{\circ} \mathrm{C}\right)\) while the hot surface is subjected to a constant heat load from the electronic equipment \(\left(q_{2}=50 \mathrm{~W}, T_{2}\right)\). Connecting the surfaces is a concentric tube of length \(L=10 \mathrm{~m}\) with \(D_{i}=100 \mathrm{~mm}\) and \(D_{o}=150 \mathrm{~mm}\). A fan moves air at a mass flow rate of \(m=0.0325 \mathrm{~kg} / \mathrm{s}\) and dissipates \(P=10 \mathrm{~W}\) of thermal energy. Heat transfer to the cool surface is described by \(q_{1}^{N}=\bar{h}_{1}\left(T_{h_{1} o}-T_{1}\right)\) while heat transfer from the hot surface is described by \(q_{2}^{\prime \prime}=\bar{h}_{2}\left(T_{2}-T_{f_{0}}\right)\) where \(T_{f_{0}}\) is the fan outlet temperature. The values of \(\bar{h}_{1}\) and \(h_{2}\) are 40 and \(60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. To isolate the electronics from ambient temperature variations, the entire device is insulated at its outer surfaces. The design engineer is concerned that conduction through the wall of the inner tube may adversely affect the device performance. Determine the value of \(T_{2}\) for the limiting cases of (i) no conduction resistance in the inner tube wall and (ii) infinite conduction resistance in the inner tube wall. Does the proposed device maintain maximum temperatures below \(80^{\circ} \mathrm{C}\) ?

Step by step solution

01

Determine \(q_1^N\) for the cool surface

We will use the following equation: \(q_1^N = \bar{h}_1 \left(T_{h_{1} o} - T_{1}\right)\) Since \(T_{1}\) is maintained at a constant temperature, \(T_{h_{1} o} = T_{1}\), and \(q_1^N = 0\).

02

Calculate \(q_2'\) for the hot surface

Using the equation: \(q_2'=q_{2}^{\prime\prime}S=\bar{h}_{2}\left(T_{2}-T_{f_{0}}\right)S\) Since \(q_2 = 50 \mathrm{~W}\), the total surface area of the inner tube is: \(S = \pi D_i L = \pi (0.1)(10) = \pi \mathrm{~m}^2\) Now, we can solve for \(T_2\): \(q_2' = h_2 (T_2 - T_{f_0})S\)

03

Determine the total heat load and fan outlet temperature (\(T_{f_0}\))

Considering the heat load from the fan, we have \(q_{t} = q_2 + P = 60 \mathrm{~W}\) The total heat load \(q_{t}\) is equal to the sum of heat removed by the fan and heat exchanged at the cool surface: \(q_t = m c(T_{f_0} - T_{1}) + q_1^N\) Solving for the fan outlet temperature \(T_{f_0}\) and using the specific heat of air \(c = 1005 \mathrm{~J} / \mathrm{kg}\cdot \mathrm{K}\), we get: \(T_{f_0} = T_1 + \frac{q_t}{mc} = 5 + \frac{60}{(0.0325)(1005)} = 5 + 7.61 = 12.61^{\circ}\mathrm{C}\)

04

Determine \(T_2\) based on the hot surface temperature

Now, we can substitute \(T_{f_0}\) back into the equation for \(q_2'\) and solve for \(T_2\): \(q_2' = h_2 (T_2 - T_{f_0})S\) \(T_2 = \frac{q_2'}{h_2 S} + T_{f_0} = \frac{50}{(60)(\pi)} + 12.61 = 15.80^{\circ}\mathrm{C}\) ##Case (ii) - Infinite conduction resistance in the inner tube wall##

05

Calculate \(q_1^N\) for the cool surface with infinite conduction resistance

Since conduction resistance is infinite, no heat is transferred to the cool surface, so \(q_1^N=0\).

06

Determine \(T_2\) based on the fan outlet temperature

Since no heat is transferred to the cool surface, the total heat load equals only the heat removed by the fan: \(q_t = m c (T_{f_0} - T_{1})\) Solving for the fan outlet temperature \(T_{f_0}\), we get: \(T_{f_0} = T_1 + \frac{q_t}{mc} = 5 + \frac{60}{(0.0325)(1005)} = 12.61^{\circ}\mathrm{C}\) Now, we can substitute \(T_{f_0}\) back into the equation for \(q_2'\) and solve for \(T_2\): \(q_2' = h_2 (T_2 - T_{f_0})S\) \(T_2 = \frac{q_2'}{h_2 S} + T_{f_0} = \frac{50}{(60)(\pi)} + 12.61 = 15.80^{\circ}\mathrm{C}\)

07

Check if the maximum temperature stays below \(80^{\circ}\mathrm{C}\)

Finally, we compare the calculated values of \(T_2\) for both cases against the limit of \(80^{\circ}\mathrm{C}\). For case (i), \(T_2 = 15.80^{\circ}\mathrm{C} < 80^{\circ}\mathrm{C}\), and for case (ii), \(T_2 = 15.80^{\circ}\mathrm{C} < 80^{\circ}\mathrm{C}\), so the proposed device does maintain maximum temperatures below \(80^{\circ}\mathrm{C}\).

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 91Ó°ÊÓ!

Key Concepts

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

Heat Transfer in Thermal Systems

Understanding heat transfer is crucial when it comes to the design and operation of thermal systems, including the cooling of electronics as seen in the exercise. Heat transfer is the movement of thermal energy from one object or material to another. There are three modes of heat transfer: conduction, convection, and radiation. Conduction is the heat transfer through a solid material, convection is the transfer of heat by the physical movement of fluid, and radiation is the transfer of energy by electromagnetic waves.

In the given problem, we see examples of convection as the primary mode of heat transfer, where a fan is used to move air to dissipate thermal energy. The exercise also considers conduction through the wall of an inner tube that affects the device's performance. High conduction resistance would mean less heat transfer through the tube wall, while low resistance would indicate more heat dissipation. It's important for students to understand that altering conduction resistance impacts the overall efficiency and thermal regulation of the cooling system.

The Role of Thermal Energy in Heat Exchangers

Thermal energy refers to the kinetic energy of the molecules within a substance and is directly associated with the temperature of the material. In our exercise, the electronic equipment is generating heat that needs to be removed to prevent overheating. This dissipated heat is the thermal energy that we are aiming to transfer away from the electronics to maintain a safe operating temperature.

In a heat exchanger design, managing the thermal energy is key. The ground acts as a heat sink in the given setup, with groundwater temperature providing a near-constant temperature. This helps sustain a desired temperature difference, which is the driving force for heat transfer. Students should recognize how maintaining this temperature difference is vital and how it influences the effectiveness of the heat exchanger.

Importance of Heat Exchanger Design

The design of heat exchangers, like the concentric tube setup in the exercise, is a sophisticated task that entails understanding thermal dynamics, fluid mechanics, and material science. The exercise simulates a scenario where an engineer must evaluate the impact of conduction resistance on device performance within a heat exchanger.

The key to efficient heat exchanger design lies in maximizing heat transfer while minimizing energy loss. This is why the conduction resistance in tube walls is such an important factor. If the resistance is too high, necessary heat exchange might be impeded. On the other hand, if the resistance is too low, there may be unwanted heat loss or gain. The exercise calls upon students to understand how varying conduction resistance affects the temperature (\(T_2\)) of the electronics and to ensure this temperature remains below the critical threshold for safe operations.