91Ó°ÊÓ

A cold plate is an active cooling device that is attached to a heat-generating system in order to dissipate the heat while maintaining the system at an acceptable temperature. It is typically fabricated from a material of high thermal conductivity, \(k_{\text {cp, }}\), within which channels are machined and a coolant is passed. Consider a copper cold plate of height \(H\) and width \(W\) on a side, within which water passes through square channels of width \(w=h\). The transverse spacing between channels \(\delta\) is twice the spacing between the sidewall of an outer channel and the sidewall of the cold plate. Consider conditions for which equivalent heat-generating systems are attached to the top and bottom of the cold plate, maintaining the corresponding surfaces at the same temperature \(T_{s}\). The mean velocity and inlet temperature of the coolant are \(u_{m}\) and \(T_{m i}\), respectively. (a) Assuming fully developed turbulent flow throughout each channel, obtain a system of equations that may be used to evaluate the total rate of heat transfer to the cold plate, \(q\), and the outlet temperature of the water, \(T_{m, o}\), in terms of the specified parameters. (b) Consider a cold plate of width \(W=100 \mathrm{~mm}\) and height \(H=10 \mathrm{~mm}\), with 10 square channels of width \(w=6 \mathrm{~mm}\) and a spacing of \(\delta=4 \mathrm{~mm}\) between channels. Water enters the channels at a temperature of \(T_{m, i}=300 \mathrm{~K}\) and a velocity of \(u_{m}=2 \mathrm{~m} / \mathrm{s}\). If the top and bottom cold plate surfaces are at \(T_{s}=360 \mathrm{~K}\), what is the outlet water temperature and the total rate of heat transfer to the cold plate? The thermal conductivity of the copper is \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while average properties of the water may be taken to be \(\rho=984 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=489 \times\) \(10^{-6} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(P r=3.15\). Is this a good cold plate design? How could its performance be improved?

Step by step solution

01

(a) Derive a system of equations

First, as the flow is in turbulent regime, we can use the Dittus-Boelter equation to find the Nusselt number, \(Nu\): \[Nu = 0.023\, Re^{4/5} Pr^{n}\] where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number, and \(n\) is 0.4 for heating. The Reynolds number can be calculated as: \[Re = \frac{\rho u_m w}{\mu}\] Now, we can calculate the heat transfer coefficient, \(h\), using the Nusselt number: \[h = \frac{Nu \cdot k}{w}\] The mass flow rate, \(\dot{m}\), can be calculated as: \[\dot{m} = \rho u_m w^2 N\] where \(N\) is the number of channels. The total rate of heat transfer, \(q\), will be: \[q = h A_c (T_s - T_{m, o})\] where \(A_c\) is the total contact surface area between the cold plate and the water, which can be expressed as: \[A_c = 2 H W\] Now, we can use the energy balance equation to relate the inlet and outlet water temperatures: \[q = \dot{m} c_p (T_{m, o} - T_{m, i})\] From the equations above, we have a system of equations that can be used to evaluate the total rate of heat transfer, \(q\), and the outlet temperature of the water, \(T_{m, o}\).

02

(b) Calculation of outlet water temperature and total rate of heat transfer

We can now use the given values and the system of equations derived in part (a) to calculate the outlet water temperature and the total rate of heat transfer. First, calculate the Reynolds number: \[Re = \frac{(984\, kg/m^3)(2\, m/s)(0.006\, m)}{489 \times 10^{-6}\, N\, s/m^2} = 24098.97\] Next, calculate the Nusselt number using the Dittus-Boelter equation: \[Nu = 0.023\, (24098.97)^{4/5} (3.15)^{0.4} = 127.68\] Now, calculate the heat transfer coefficient, \(h\): \[h = \frac{(127.68)(0.65\, W/m\, K)}{0.006\, m} = 13899.07\, W/m^2\, K\] Calculate the mass flow rate, \(\dot{m}\): \[\dot{m} = (984\, kg/m^3)(2\, m/s)(0.006\, m)^2(10) = 0.707\, kg/s\] Now, solve for the total rate of heat transfer, \(q\): \[q = (13899.07\, W/m^2\, K)(2(10^{-3}\, m)(100\times 10^{-3}\, m))(360 - T_{m, o})\] Finally, using the energy balance equation, we can solve for the outlet water temperature, \(T_{m, o}\): \[q = (0.707\, kg/s)(4184\, J/kg\, K)(T_{m, o} - 300\, K)\] Solving the system of equations, we find that \(q = 4807.73\, W\) and \(T_{m, o} = 312.66\, K\). In terms of the design of the cold plate, since the outlet temperature of the water is relatively low compared to the surface temperature (\(312.66\, K\) vs \(360\, K\)), it is safe to say that the cold plate is performing well. However, there are ways to improve the design further: 1. To increase the heat transfer rate, we could increase the number of channels and/or decrease the channel spacing, thereby increasing the contact surface and improving the heat transfer. 2. Another option is to optimize the geometry of the channels by using different shapes or patterns, which could increase heat transfer potential. 3. Finally, using a coolant with better thermal properties or increasing the flow rate can also aid in improving the cold plate's performance.

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Key Concepts

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

Turbulent Flow

Turbulent flow occurs when a fluid moves in an irregular pattern, characterized by eddies and swirls. This type of flow is common in high-velocity environments, such as water flowing through channels in a cold plate system.
In the context of a cold plate, turbulent flow enhances the mixing of the coolant. This mixing is crucial for efficient heat transfer since it allows heat to quickly disperse throughout the fluid.\(\phi\)

• Advantages: Increases convective heat transfer efficiency.
• Challenges: Can cause higher energy losses due to increased friction.

Since turbulent flow is generally achieved at higher Reynolds numbers (usually \(Re > 4000\)), the design of the channels and the velocity of the coolant must be considered to maintain this flow condition. The complexity caused by turbulent flow requires engineering calculations to ensure the system remains within desired efficiency and safety parameters.

Heat Transfer

Heat transfer in a cold plate system involves the movement of thermal energy from the heat-generating component to the coolant. It is a crucial part of the cooling process, ensuring that components operate within their optimal temperature range.
This transfer process is driven by a temperature difference between the surface of the cold plate and the coolant. The greater this difference, the more effectively heat is transferred.
Essential elements affecting heat transfer include:

• Surface Area: Larger contact areas enhance heat dissipation.
• Temperature Gradient: A larger difference between the surface temperature and the coolant temperature promotes more efficient heat transfer.
• Flow Rate: Higher fluid velocities can increase the rate of heat transfer by reducing temperature buildup at the interface.

In sum, the design of the cold plate system aims to maximize these factors, thereby optimizing the thermal performance of the system.

Nusselt Number

The Nusselt number \(Nu\) is a dimensionless parameter critical in the study of heat transfer through convection. It signifies the enhancement of heat transfer due to convection compared to conduction alone.
For the cold plate system, the Nusselt number can be determined using the Dittus-Boelter equation, especially in cases of turbulent flow. The formula is given by: \[Nu = 0.023 \times Re^{4/5} \times Pr^{n}\] Where:

• \(Re\): Reynolds number, indicating the flow regime.
• \(Pr\): Prandtl number, a measure of fluid characteristics.
• \(n\): Exponent depending on the heating/cooling condition, often 0.4 for heating.

The higher the Nusselt number, the more effective the convective heat transfer. Calculating \(Nu\) helps engineers decide on appropriate designs and materials for optimizing the heat dissipation capabilities of the cold plate.

Thermal Conductivity

Thermal conductivity \(k\) is a property of a material that indicates its ability to conduct heat. A material with high thermal conductivity can transfer more heat across its surface area for a given temperature difference.
In a cold plate, high thermal conductivity is crucial because it facilitates rapid heat movement from the heat-generating component to the coolant. This ensures that the system quickly adapts to changes in heat load, maintaining temperature stability.
For instance, copper, often used in cold plates, has a very high thermal conductivity of around 400 W/m·K, making it an excellent choice.
Key points about thermal conductivity in cold plate design:

• Material Choice: Metals with high \(k\) like copper and aluminum are preferred.
• Plate Thickness: Thinner plates may transfer heat more quickly but must be carefully balanced with structural integrity.
• Temperature Dependence: The value of \(k\) can change with temperature, so it's important to consider operating conditions.

In summary, choosing materials with high thermal conductivity is fundamental to efficient heat transfer in cold plate systems.