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A chip that is of length \(L=5 \mathrm{~mm}\) on a side and thick. ness \(t=I \mathrm{~mm}\) is encased in a ceramic substrate, and its exposed surface is convectively cooled by a diefectric liquid for which \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\mathrm{z}}=20^{\circ} \mathrm{C}\). In the off-mode the chip is in thermal equilibrium with the coolant \(\left(T_{i}=T_{*}\right)\). When the chip is energized, however, its Icmperafure increases until a new steady-state is established. For purpeises of analysis, the cnergized chip is characterized by uniform volumetric heating with \(q=9 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming an infinite contacl resistance between the chip and substrate and negligible conduction resistance within the chip. determine the steady-state chip temperature \(T_{f}\). Following activation of the chip, how long does it take to come within \(1^{-1} C\) of this temperature? The chip density and specific heat are \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

Step by step solution

01

Determine Heat Transfer

The first step is to determine the rate of heat transfer from the chip to the coolant by convection. Using the formula for convective heat transfer, we have:\[ q_{ ext{conv}} = hA(T_f - T_{ ext{z}}) \]where \(A\) is the chip's exposed surface area. Since the chip is a square with sides of length \(L=5\,\mathrm{mm}\), the area \(A = L^2 = (5\times10^{-3})^2 = 25\times10^{-6}\,\mathrm{m}^2\). Substituting into the convective heat transfer equation gives:\[ q_{ ext{conv}} = 150\times25\times10^{-6}(T_f - 20) \]

02

Apply Energy Balance

Since the rate of heat generation within the chip must equal the rate of heat removal at steady state, we have:\[ q_{ ext{gen}} = q_{ ext{conv}} \]The volumetric heat generation within the chip is given by:\[ q_{ ext{gen}} = qV = q(L^2 t) \]Substituting the values, we find:\[ q_{ ext{gen}} = 9 \times 10^6 \times 25\times10^{-6}\times 1\times10^{-3} = 225\times10^{3}\,\mathrm{W/m}^3 \]

03

Solve for Steady-State Temperature

Set the equations for generated and convected heat equal to find the steady-state temperature of the chip. From Step 1 and Step 2:\[ 225\times10^{3} = 150\times25\times10^{-6}(T_f - 20) \]Solving for \(T_f\):\[ 225 = 3.75(T_f - 20) \]\[ T_f - 20 = 60 \]\[ T_f = 80\,\,\mathrm{°C} \]

04

Calculate Time to Reach Steady-State

To find the time it takes for the chip to reach within \(1^{-1}\,^{ ext{C}}\) of its steady-state temperature, use the thermal system's response time constant, \(\tau\):\[ \tau = \frac{\rho ct}{h} \]Substitute the given values:\[ \tau = \frac{2000 \times 700 \times 1\times10^{-3}}{150} = 9.333\,\mathrm{s} \]The equation for the transient response near steady-state is:\[ T(t) = T_f + (T_i - T_f) e^{-t/\tau} \]Solve \(T(t) = 79\,\,\mathrm{°C}\), since the temperature drops by 1°C of steady state:\[ 79 = 80 - 60e^{-t/9.333} \]\[ e^{-t/9.333} = \frac{1}{60} \]\[ -\frac{t}{9.333} = \,\ln\left(\frac{1}{60}\right) \]\[ t = 9.333\,\ln(60) \approx 40\,\mathrm{s} \]

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Key Concepts

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

Convective Cooling

Convective cooling is a process where heat is removed from an object by the movement of a fluid, in this case, a dielectric liquid. The liquid absorbs heat from the electronic device and carries it away, aiding in temperature control. The amount of heat transferred depends on several factors: the heat transfer coefficient \(h\), the exposed surface area \(A\), and the temperature difference between the object and the fluid.

• The heat transfer coefficient \(h\) is a measure of a fluid's ability to conduct heat and is provided as \(150 \, \mathrm{W/m^2 \cdot K}\).
• The surface area, calculated for a square chip, is \(A = L^2 = 25 \times 10^{-6} \, \mathrm{m^2}\).
• The temperature difference drives the heat transfer; a greater difference results in faster cooling.

In this exercise, convective cooling helps maintain the chip temperature by removing the heat generated due to internal heating. The rate of convective heat transfer can be expressed as \(q_{\text{conv}} = hA(T_f - T_{z})\), showing how each variable impacts the cooling efficiency.

Volumetric Heating

Volumetric heating occurs when heat is uniformly distributed throughout the volume of an object, such as an electronic chip. In this scenario, the chip is powered, causing its temperature to rise as it generates heat internally.

• Given the volumetric heat generation rate \(q = 9 \times 10^6 \, \mathrm{W/m^3}\), the total heat generated is the product of this rate and the chip's volume, \(V\).
• The volume \(V\) can be calculated using \(V = L^2 t = 25 \times 10^{-6} \, \mathrm{m^2} \times 1 \times 10^{-3} \, \mathrm{m}\), which results in an overall heating magnitude of \(q_{\text{gen}} = 225 \times 10^3 \, \mathrm{W}\).

This uniform internal heating demands effective cooling mechanisms, like convective cooling, to ensure the chip reaches a new stable temperature and prevent overheating.

Steady-State Temperature

The steady-state temperature is the temperature at which an object remains constant over time because the heat generated internally matches the heat transferred away. For the chip, the steady-state temperature is achieved when the rate of heat generation from volumetric heating equals the rate of heat removal by convection.
Through the energy balance: \(q_{\text{gen}} = q_{\text{conv}}\).

• Using the equation \(225\times 10^3 = 150\times 25 \times 10^{-6} (T_f - 20)\), the steady-state temperature \(T_f\) can be solved.
• On resolving, this results in \(T_f = 80^\circ \mathrm{C}\).

Only when this balance is achieved does the chip operate safely and effectively, maintaining a consistent temperature even as it remains powered.

Energy Balance in Heat Transfer

Energy balance in heat transfer involves ensuring that the total energy entering a system equals the energy leaving it, plus any changes within. For the chip, the energy balance equation is critical to find how it maintains a stable temperature during operation.

• At steady-state, no accumulation of energy occurs, meaning the heat generated \(q_{\text{gen}}\) must sustain the heat convected away \(q_{\text{conv}}\).
• This balance is represented by setting the two heat terms equal, leading to \(q_{\text{gen}} = q_{\text{conv}}\).

Understanding this balance ensures engineers can design cooling systems that effectively manage the temperature of electronic devices. Furthermore, determining how quickly the chip approaches its final steady-state temperature can be done using the concept of thermal time constant \(\tau\). The calculation shows that significant temperature stabilization occurs in approximately 40 seconds, ensuring the chip quickly adapts to its operating conditions.