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One way to cool chips mounted on the circuit boards of a computer is to encapsulate the boards in metal frames that provide efficient pathways for conduction to supporting cold plates. Heat generated by the chips is then dissipated by transfer to water flowing through passages drilled in the plates. Because the plates are made from a metal of large thermal conductivity (typically aluminium or copper), they may be assumed to be at a temperature, \(T_{s, c p^{-}}\) (a) Consider circuit boards attached to cold plates of height \(H=750 \mathrm{~mm}\) and width \(L=600 \mathrm{~mm}\), each with \(N=10\) holes of diameter \(D=10 \mathrm{~mm}\). If operating conditions maintain plate temperatures of \(T_{\text {s.tp }}=32^{\circ} \mathrm{C}\) with water flow at \(\dot{m}_{1}=0.2 \mathrm{~kg} / \mathrm{s}\) per passage and \(T_{m, i}=7^{\circ} \mathrm{C}\), how much heat may be dissipated by the circuit boards? (b) To enhance cooling, thereby allowing increased power generation without an attendant increase in system temperatures, a hybrid cooling scheme may be used. The scheme involves forced airflow over the encapsulated circuit boards, as well as water flow through the cold plates. Consider conditions for which \(N_{\mathrm{cb}}=10\) circuit boards of width \(W=350 \mathrm{~mm}\) are attached to the cold plates and their average surface temperature is \(T_{s, \text { do }}=47^{\circ} \mathrm{C}\) when \(T_{s, \text { ep }}=32^{\circ} \mathrm{C}\). If air is in parallel flow over the plates with \(u_{\infty}=10 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=7^{\circ} \mathrm{C}\), how much of the heat generated by the circuit boards is transferred to the air?

Step by step solution

01

Determine the heat transfer rate due to water cooling

Use the given parameters and the relation for heat transfer rate, \(Q = \dot{m}_1 * c_p * \Delta T\), where \(c_p\) is the specific heat of water and \(\Delta T\) is the temperature difference between the inlet and outlet water temperatures. First, we need to find the outlet water temperature, \(T_{m, o}\). Since the total mass flow rate of water is \(\dot{m}_{1} * N\), \( T_{m, o} = T_{m, i} + \frac{Q}{\dot{m}_1 * N * c_p} \) The plate temperature, \(T_{s, c p^{-}} = T_{\text {s.tp }}\) (given). So, the temperature difference between the plate and water is 32 - 7 = 25°C. We can calculate the heat transfer rate by rearranging the equation for \(Q\): \( Q = \dot{m}_{1} * N * c_p * (T_{s, c p^{-}} - T_{m, i}) \)

02

Calculate the heat transfer rate due to water cooling

Now, substitute the given parameter values in the equation to find the heat transfer rate due to water cooling: \( Q = 0.2 * 10 * 4187 * (32 - 7) \) \( Q = 20.8 \times 10^3 \, \text{W} \) So, 20.8 kW of heat is dissipated by the circuit boards through water cooling.

03

Determine the heat transfer coefficient for forced airflow cooling

Next, we need to calculate the heat transfer coefficient, \(h\), for the parallel flow of air over the circuit boards. We will use the empirical formula for the average Nusselt number, \(Nu = \frac{h * W}{k}\), for parallel flow over flat plates, where \(k\) is the thermal conductivity of air. Since we know the plate temperature and the inlet air temperature, \(T_{p} = T_{s, \text{do }}\), and we have the free stream velocity, we can use the Reynolds number and Prandtl number to find the Nusselt number. Now, we can use the following correlations: \( Reynolds \, number,\, Re_x = \frac{u_\infty x}{v} \); \( Prandtl \, number,\, Pr = \frac{c_pf_{air} * \mu }{k} \) \( Nu = 0.664 * Re_x^{1/2} * Pr^{1/3} \) We have to utilize the properties of air at the film temperature (average of the plate and free stream temperatures). Once we have the Nusselt number, we can find the heat transfer coefficient, \(h=(k/W) * Nu\).

04

Compute the total heat transfer rate to the air

Now that we have obtained the value for the heat transfer coefficient, \(h\), we can calculate the total heat transfer rate, \(Q_{total}\), using the formula: \( Q_{total} = h * A * (T_{s, \text{do }} - T_\infty) \) Here, \(A\) is the total heat transfer area, which is equal to \(N_{\text{cb}} * W * L\).

05

Determine the heat transfer rate to the air

Finally, we can compute the heat transfer rate to the air by subtracting the heat transfer rate due to water cooling from the total heat transfer rate: \( Q_{air} = Q_{total} - Q \) With all the required parameter values given in the problem statement, we can calculate the heat transfer rate to the air.

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Key Concepts

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

Water Cooling Heat Dissipation

When it comes to cooling computer components, water cooling is an efficient method due to its high capacity for heat dissipation. Water has a high specific heat capacity, meaning it can absorb a lot of heat energy before its temperature rises significantly. This principle is applied in computer cooling where water flows through metal plates or blocks that are in contact with the heat source. The water absorbs the heat, and as it moves, it carries the heat away from the components.

In the exercise scenario, the water flows through drilled passages in metal plates attached to circuit boards, effectively transferring heat. Using the given formula, the amount of heat dissipated can be determined. With a flow rate of 0.2 kg/s and a temperature rise of 25°C, a significant amount of heat—20.8 kW—is removed. This showcases the powerful impact water cooling has in maintaining computer components at optimal operating temperatures.

Forced Airflow Cooling

Forced airflow cooling is another common heat management technique where air is blown over the surfaces of components to transfer heat away. Air has a lower thermal conductivity compared to water, but when air is moved with sufficient velocity, it can effectively remove heat through convection.

In our exercise, the hybrid cooling scenario elaborates on this method, where both water and forced air are used for cooling. By introducing a steady stream of air at 10 m/s over the encapsulated circuit boards, we increase the system's capability to dissipate heat. The impact of this forced airflow is calculated using principles of heat transfer, factoring in air's properties through the correlation of Nusselt, Reynolds, and Prandtl numbers. As a result, a part of the heat generated by the circuit boards is effectively transferred to the air.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. In the context of computer cooling, materials with high thermal conductivity, such as copper and aluminum used in cold plates, are preferred as they enable rapid heat transfer from high-heat components to the cooling medium, be it water or air.

The exercise demonstrates the significance of selecting materials with high thermal conductivity for the cold plates, which ensures that the metal stays uniformly at the same temperature. This uniform temperature simplifies the analysis and enhances the efficiency of heat dissipation from the circuit boards to the cooling water flowing through the plates.

Nusselt Number

The Nusselt number (Nu) is a dimensionless parameter that represents the ratio of convective to conductive heat transfer across a boundary. In cases where forced airflow is used as a cooling mechanism, the Nusselt number can be crucial in determining the efficiency of heat transfer from a surface to the air flowing over it.

The exercise uses the Nusselt number to relate the heat transfer coefficient to the thermal conductivity of the air and the specific geometry of the surface. A high Nusselt number indicates efficient convection, which is desirable in the context of cooling computer components. By applying the empirical formula involving the Nusselt number, students can calculate the heat transfer coefficient needed for determining the total heat transfer rate to the air.

Reynolds Number

The Reynolds number (Re) is an essential dimensionless quantity used to predict flow patterns in different fluid flow situations. It compares the inertial forces to the viscous forces within the fluid, and it helps in determining whether the flow will be laminar or turbulent.

In forced airflow cooling, knowledge of the Reynolds number allows for the prediction of airflow characteristics over the circuit boards. In the provided exercise, it's used in the formula to calculate the Nusselt number, which then informs the heat transfer coefficient. A high Reynolds number typically indicates turbulent flow, which is associated with a higher rate of convective heat transfer—a desirable condition in cooling applications.

Prandtl Number

The Prandtl number (Pr) is another dimensionless number that plays a pivotal role in the analysis of fluid flow and heat transfer. It relates the momentum diffusivity (kinematic viscosity) to the thermal diffusivity of the fluid. In essence, it gives an indication of the relative thickness of the velocity and thermal boundary layers.

In the exercise, the Prandtl number is a critical factor when calculating the Nusselt number for forced airflow cooling. The Prandtl number provides a means to compare the rate at which momentum and heat are diffused within the air. For many fluids, including air, the Prandtl number is roughly around 0.7. This similarity in momentum and thermal diffusivity simplifies the determination of heat transfer in forced convection scenarios.