/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 A square ( \(10 \mathrm{~mm} \ti... [FREE SOLUTION] | 91Ó°ÊÓ

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A square ( \(10 \mathrm{~mm} \times 10 \mathrm{~mm}\) ) silicon chip is insulated on one side and cooled on the opposite side by atmospheric air in parallel flow at \(u_{\infty}=20 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=\) \(24^{\circ} \mathrm{C}\). When in use, electrical power dissipation within the chip maintains a uniform heat flux at the cooled surface. If the chip temperature may not exceed \(80^{\circ} \mathrm{C}\) at any point on its surface, what is the maximum allowable power? What is the maximum allowable power if the chip is flush mounted in a substrate that provides for an unheated starting length of \(20 \mathrm{~mm}\) ?

Short Answer

Expert verified
In conclusion, the maximum allowable power for the square silicon chip in the given conditions without an unheated starting length is 3.37 W, and with an unheated starting length of 20 mm, it is 3.66 W.

Step by step solution

01

1. Calculate the film temperature

Calculate the film temperature: \(T_f = \dfrac{24^{\circ} \mathrm{C} + 80^{\circ} \mathrm{C}}{2} = 52^{\circ} \mathrm{C}\) The following properties of air at \(T_f = 52^{\circ} \mathrm{C}\) can be found in any standard air properties table: - Dynamic Viscosity, \(\mu = 2.06 \times 10^{-5} \mathrm{~kg/m.s}\) - Thermal Conductivity, \(k = 0.0285 \mathrm{~W/m.K}\) - Prandtl Number, \(Pr = 0.7\) Next, we need to calculate the Reynolds number of the flow to determine the flow regime and calculate the convective heat transfer coefficient accordingly.
02

2. Calculate the Reynolds Number

Calculate the Reynolds Number of the flow, using: \(Re = \dfrac{u_{\infty} L}{\nu}\) where \(u_{\infty}\) is the free-stream velocity, \(L\) is the length of the chip (unheated starting length is 0 in this case, and the square chip length is 10 mm), and \(\nu\) is the kinematic viscosity of air. First, we need to obtain the kinematic viscosity \(\nu\) as: \(\nu = \dfrac{\mu}{\rho}\) where \(\rho\) is the density of air, which can be determined using the Ideal Gas Law: \(\rho = \dfrac{p}{RT}\) Assuming air pressure \(p = 101.3 \mathrm{~kPa}\) and the gas constant for air \(R = 0.287 \mathrm{~kJ/kg.K}\), \(\rho = \dfrac{101.3 \mathrm{~kPa}}{(0.287 \mathrm{~kJ/kg.K})(52 + 273)} = 1.14 \mathrm{~kg/m^3}\) Now, we calculate the kinematic viscosity \(\nu\): \(\nu = \dfrac{2.06 \times 10^{-5} \mathrm{~kg/m.s}}{1.14 \mathrm{~kg/m^3}} = 1.81 \times 10^{-5} \mathrm{~m^2/s}\) Now, we can determine the Reynolds Number: \(Re = \dfrac{(20 \mathrm{~m/s})(0.01 \mathrm{~m})}{1.81 \times 10^{-5} \mathrm{~m^2/s}} = 1.10 \times 10^4\) Since \(Re > 4000\), the flow is turbulent, and we can use the Dittus-Boelter equation to determine the convective heat transfer coefficient, \(h\).
03

3. Calculate the convective heat transfer coefficient

Using the Dittus-Boelter equation for turbulent flow: \(Nu = 0.023 Re^{0.8} Pr^{0.4}\) where \(Nu\) is the Nusselt number. \(Nu = 0.023 (1.10 \times 10^4)^{0.8} (0.7)^{0.4} = 212.2\) To find the convective heat transfer coefficient, \(h\), use the formula: \(h = \dfrac{Nu \cdot k}{L}\) \(h = \dfrac{212.2 \cdot 0.0285 \mathrm{~W/m.K}}{0.01 \mathrm{~m}} = 601.7 \mathrm{~W/m^2.K}\) Now that we have the convective heat transfer coefficient, we can determine the heat transfer rate from the silicon chip to the air, \(q\).
04

4. Calculate the heat transfer rate

Calculate the heat transfer rate using: \(q = h A (T_s - T_{\infty})\) where \(A\) is the area of the chip. \(A = (0.01 \mathrm{~m})^2 = 1 \times 10^{-4} \mathrm{~m^2}\) \(q = (601.7 \mathrm{~W/m^2.K})(1 \times 10^{-4} \mathrm{~m^2})(80^{\circ} \mathrm{C} - 24^{\circ} \mathrm{C}) = 3.37 \mathrm{~W}\) Now, we have the heat transfer rate from the silicon chip to the air, which is equal to the maximum allowable power dissipation within the chip when no unheated starting length is considered. For the case with an unheated starting length of 20 mm, we need to recalculate the Reynolds number, Nusselt number, and heat transfer coefficient, and then determine the maximum allowable power.
05

5. Recalculate parameters and find maximum allowable power for flush-mounted chip

To recompute the Reynolds number, use a total length of 30 mm (20 mm starting length + 10 mm chip length): \(Re = \dfrac{(20 \mathrm{~m/s})(0.03 \mathrm{~m})}{1.81 \times 10^{-5} \mathrm{~m^2/s}} = 3.31 \times 10^4\) Since this flow is still turbulent, we can use the Dittus-Boelter equation to find the new Nusselt number: \(Nu = 0.023 (3.31 \times 10^4)^{0.8} (0.7)^{0.4} = 682.4\) Now, find the new convective heat transfer coefficient, \(h\): \(h = \dfrac{682.4 \cdot 0.0285 \mathrm{~W/m.K}}{0.03 \mathrm{~m}} = 654.9 \mathrm{~W/m^2.K}\) Finally, calculate the new heat transfer rate and maximum allowable power: \(q = (654.9 \mathrm{~W/m^2.K})(1 \times 10^{-4} \mathrm{~m^2})(80^{\circ} \mathrm{C} - 24^{\circ} \mathrm{C}) = 3.66 \mathrm{~W}\) So, the maximum allowable power for the chip when flush mounted with an unheated starting length of 20 mm is 3.66 W. In conclusion, the maximum allowable power for the chip without an unheated starting length is 3.37 W, and with an unheated starting length of 20 mm it is 3.66 W.

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Key Concepts

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

Reynolds Number
The Reynolds number (Re) is a dimensionless quantity in fluid mechanics that helps predict flow patterns in different fluid flow situations. It compares the inertial forces to the viscous forces within a fluid, which affects the flow regime (laminar or turbulent) and consequently the heat transfer characteristics. The formula for calculating the Reynolds number is: \[\begin{equation}Re = \frac{u_{\infty} L}{u}\end{equation}\]where \(u_{\infty}\) is the free-stream velocity, \(L\) is the characteristic length (such as the side of the silicon chip in the given problem), and \(u\) is the kinematic viscosity of the fluid. In the original exercise, the transition to turbulent flow occurs when Re exceeds 4000. High Re values typically denote turbulent flow, which enhances mixing and potentially increases heat transfer due to the higher levels of fluid motion and mixing.

Understanding the role of the Reynolds number in predicting the flow regime is crucial in engineering applications like the cooling of electronic components. In the exercise, the calculation of Re helps to inform the appropriate convective heat transfer correlations to use for determining the cooling performance necessary to maintain the chip temperature below a certain threshold.
Nusselt Number
The Nusselt number (Nu) is another critical dimensionless parameter in convective heat transfer which relates the heat transfer at a surface to the heat transfer in the fluid away from the surface. It is expressed as:\[\begin{equation}Nu = \frac{hL}{k}\end{equation}\]where \(h\) is the convective heat transfer coefficient, \(L\) is the characteristic length, and \(k\) is the thermal conductivity of the fluid. The Nusselt number is a measure of the enhancement of heat transfer through a fluid as a result of convection, compared with heat transfer through conduction alone. For instance, a larger Nu signifies more effective convection, which may be due to increased turbulence or other factors.

In the provided exercise, the Nusselt number is calculated using the Dittus-Boelter equation, which is appropriate for turbulent flow conditions. This number is instrumental for determining the heat transfer coefficient, which in turn helps to calculate the heat flux between the silicon chip and the cooling air.
Heat Flux
Heat flux (\(q\)) is precisely the rate of heat energy transfer per unit area, expressed in watts per square meter (\(W/m^2\)). The heat flux is fundamental in understanding and calculating the heat transfer between surfaces and the environment, as in the case of the silicon chip in the exercise. The formula to find the heat flux from a surface is:\[\begin{equation}q = hA(T_s - T_{\infty})\end{equation}\]where \(h\) is the heat transfer coefficient, \(A\) is the area over which the heat transfer takes place, \(T_s\) is the surface temperature, and \(T_{\infty}\) is the fluid temperature far from the surface.

The maximization of the heat flux is often a target in the design of thermal systems, especially to prevent overheating, as in the case with the silicon chip needing to stay below \(80°C\). Calculating the heat flux allows engineers to determine the maximum allowable power in electronic components without exceeding the temperature limit, further highlighting the practical importance of heat transfer in system design and operation.

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Most popular questions from this chapter

The roof of a refrigerated truck compartment is of composite construction, consisting of a layer of foamed urethane insulation \(\left(t_{2}=50 \mathrm{~mm}, k_{i}=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) sandwiched between aluminum alloy panels \(\left(t_{1}=5 \mathrm{~mm}\right.\), \(\left.k_{p}=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The length and width of the roof are \(L=10 \mathrm{~m}\) and W \(=3.5 \mathrm{~m}\), respectively, and the temperature of the inner surface is \(T_{s, i}=-10^{\circ} \mathrm{C}\). Consider conditions for which the truck is moving at a speed of \(V=105 \mathrm{~km} / \mathrm{h}\), the air temperature is \(T_{\infty}=32^{\circ} \mathrm{C}\), and the solar irradiation is \(G_{S}=750 \mathrm{~W} / \mathrm{m}^{2}\). Turbulent flow may be assumed over the entire length of the roof. (a) For equivalent values of the solar absorptivity and the emissivity of the outer surface \(\left(\alpha_{S}=\varepsilon=0.5\right)\), estimate the average temperature \(T_{s, o}\) of the outer surface. What is the corresponding heat load imposed on the refrigeration system? (b) A special finish \(\left(\alpha_{S}=0.15, \varepsilon=0.8\right)\) may be applied to the outer surface. What effect would such an application have on the surface temperature and the heat load? (c) If, with \(\alpha_{S}=\varepsilon=0.5\), the roof is not insulated \(\left(t_{2}=0\right)\), what are the corresponding values of the surface temperature and the heat load?

Consider the velocity boundary layer profile for flow over a flat plate to be of the form \(u=C_{1}+C_{2} y\). Applying appropriate boundary conditions, obtain an expression for the velocity profile in terms of the boundary layer thickness \(\delta\) and the free stream velocity \(u_{\infty}\). Using the integral form of the boundary layer momentum equation (Appendix G), obtain expressions for the boundary layer thickness and the local friction coefficient, expressing your result in terms of the local Reynolds number. Compare your results with those obtained from the exact solution (Section 7.2.1) and the integral solution with a cubic profile (Appendix \(G\) ).

Air at a pressure of 1 atm and a temperature of \(50^{\circ} \mathrm{C}\) is in parallel flow over the top surface of a flat plate that is heated to a uniform temperature of \(100^{\circ} \mathrm{C}\). The plate has a length of \(0.20 \mathrm{~m}\) (in the flow direction) and a width of \(0.10 \mathrm{~m}\). The Reynolds number based on the plate length is 40,000 . What is the rate of heat transfer from the plate to the air? If the free stream velocity of the air is doubled and the pressure is increased to \(10 \mathrm{~atm}\), what is the rate of heat transfer?

Determine the convection heat loss from both the top and the bottom of a flat plate at \(T_{s}=80^{\circ} \mathrm{C}\) with air in parallel flow at \(T_{\infty}=25^{\circ} \mathrm{C}, u_{\infty}=3 \mathrm{~m} / \mathrm{s}\). The plate is \(t=1 \mathrm{~mm}\) thick, \(L=25 \mathrm{~mm}\) long, and of depth \(w=50 \mathrm{~mm}\). Neglect the heat loss from the edges of the plate. Compare the convection heat loss from the plate to the convection heat loss from an \(L_{c}=50\)-mm-long cylinder of the same volume as that of the plate. The convective conditions associated with the cylinder are the same as those associated with the plate.

Steel (AISI 1010) plates of thickness \(\delta=6 \mathrm{~mm}\) and length \(L=1 \mathrm{~m}\) on a side are conveyed from a heat treatment process and are concurrently cooled by atmospheric air of velocity \(u_{\infty}=10 \mathrm{~m} / \mathrm{s}\) and \(T_{x}=20^{\circ} \mathrm{C}\) in parallel flow over the plates. For an initial plate temperature of \(T_{i}=300^{\circ} \mathrm{C}\), what is the rate of heat transfer from the plate? What is the corresponding rate of change of the plate temperature? The velocity of the air is much larger than that of the plate.

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