/*! 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 29 An array of electronic chips is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An array of electronic chips is mounted within a sealed rectangular enclosure, and cooling is implemented by attaching an aluminum heat \(\operatorname{sink}(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The base of the heat sink has dimensions of \(w_{1}=w_{2}=\) \(100 \mathrm{~mm}\), while the 6 fins are of thickness \(t=10 \mathrm{~mm}\) and pitch \(S=18 \mathrm{~mm}\). The fin length is \(L_{f}=50 \mathrm{~mm}\), and the base of the heat sink has a thickness of \(L_{b}=10 \mathrm{~mm}\). If cooling is implemented by water flow through the heat sink, with \(u_{\infty}=3 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=17^{\circ} \mathrm{C}\), what is the base temperature \(T_{b}\) of the heat sink when power dissipation by the chips is \(P_{\text {elec }}=1800 \mathrm{~W}\) ? The average convection coefficient for surfaces of the fins and the exposed base may be estimated by assuming parallel flow over a flat plate. Properties of the water may be approximated as \(k=0.62 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=995 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \nu=7.73 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), and \(\operatorname{Pr}=5.2\).

Short Answer

Expert verified
The base temperature \(T_{b}\) of the heat sink when the power dissipation of the chips is \(1800 W\) is determined by following a series of steps involving calculating the fin efficiency, total heat transfer area, and total heat transfer. First, estimate the average convection coefficient and use it to calculate the fin efficiency. Next, determine the overall heat transfer area of the fins and base, then calculate the total heat transfer. Finally, solve for the base temperature \(T_{b}\) based on the given power dissipation.

Step by step solution

01

Calculate the Fin Efficiency

We start by calculating the fin efficiency. We can use the formula for the fin efficiency of a rectangular fin as follows: \( \eta = \frac{\tanh \left(\frac{h SK}{2 k_{al} t^2}\right)}{\frac{h SK}{2 k_{al} t^2}} \) where \(h\) is the convection coefficient, \(S\) is the fin spacing, \(K_{al}\) is the thermal conductivity of the aluminum heat sink, and \(t\) is the fin thickness. But before calculating the fin efficiency, we need to first estimate the average convection coefficient.
02

Estimate the Average Convection Coefficient

To estimate the average convection coefficient, we can use the correlation for flow over a flat plate: \( h = \frac{k_{w}}{L_{b}} \cdot 0.037 Re^{4 / 5} Pr^{1/3} \) where, \(k_{w}\) is the thermal conductivity of water, \(L_{b}\) is the base length of the heat sink, and \(Re\) and \(Pr\) are the Reynolds and Prandtl numbers for water flow, respectively. The Reynolds number is calculated as: \( Re = \frac{u_{\infty} L_{b}}{\nu} \) Substitute the given values to calculate the Reynolds number and convection coefficient.
03

Calculate Fin Efficiency

Now that we have estimated the average convection coefficient, we can calculate the fin efficiency using the formula from Step 1. Substitute the given values and the calculated convection coefficient to find the fin efficiency:
04

Determine the Overall Heat Transfer Area

With the fin efficiency, we can determine the overall heat transfer area of the fins and the exposed base: \( A_{total} = A_{base} + (N \cdot \eta \cdot A_{fin}) \) where \(N\) is the number of fins and \(A_{base}\) and \(A_{fin}\) represent the exposed base area and the single fin surface area, respectively.
05

Calculate Total Heat Transfer

The total heat transfer from the heat sink can be calculated using the formula: \( Q = h A_{total} (T_{b} - T_{\infty}) \) Substitute the given values, calculated convection coefficient, and overall heat transfer area into the equation.
06

Determine Base Temperature

Now, we can determine the base temperature \(T_{b}\) based on the given power dissipation: \( P_{elec} = Q \) Solve for the base temperature \(T_{b}\) using the total heat transfer value. This will give you the base temperature of the heat sink when the power dissipation of the chips is \(1800 W\).

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.

Fin Efficiency
Fin efficiency plays a critical role in the design and analysis of heat sinks in thermal management systems. It represents the ratio of actual heat transfer to the ideal heat transfer from a fin, assuming that the entire fin is at the base temperature. The efficiency depends on several factors, including the material's thermal conductivity, the fin's geometry, and the convection coefficient.

The mathematical expression to calculate this efficiency considers the physical and thermal properties, allowing us to evaluate the effectiveness of the fins in dissipating heat. In practice, higher fin efficiency means that the fin material and design are highly effective in transferring heat away from the component such as electronic chips.
Convection Coefficient
The convection coefficient, represented as 'h' in thermal analysis, measures the convective heat transfer per unit area and per degree of temperature difference between the object's surface and the surrounding fluid. It's a crucial variable when assessing the cooling performance of heat sinks.

Estimation of the average convection coefficient for a heat sink involves correlations derived for particular flow conditions, like those over a flat plate. The value of 'h' contrasts across different parts of the heat sink, influenced by both the velocity of the fluid flow and its properties.
Reynolds Number
Reynolds number is a dimensionless value used to predict the flow regime—laminar, transitional, or turbulent flow—in fluid mechanics. It's a ratio that compares inertial forces to viscous forces within the fluid flow. For heat sinks cooled by fluid flow, like water, a higher Reynolds number generally indicates a turbulent flow, which enhances the convection coefficient and thereby improves the cooling performance.

In our case, using the provided flow velocity, fluid properties, and characteristic length, the Reynolds number was calculated to determine the average convection coefficient. Understanding and calculating the Reynolds number helps in predicting the nature of the flow and, thus, the performance of the heat sink.
Heat Transfer
Heat transfer is a fundamental concept in thermal systems, encompassing the movement of thermal energy from one place to another. It occurs through three main mechanisms: conduction, convection, and radiation. In the context of a heat sink, conduction occurs within the solid material, while convection is responsible for heat removal from the heat sink to the surrounding fluid.

To determine the thermal performance of a heat sink, we calculate the total heat transfer, which considers the area for heat dissipation (taking into account the fin efficiency and the number of fins) along with the temperature difference between the base of the heat sink and the flowing fluid. The given power dissipation by the electronic chips must match the heat transferred away by the heat sink to maintain the components at a safe operating temperature.
Thermal Conductivity
Thermal conductivity, denoted as 'k', is a material property that indicates how well a material can conduct heat. High thermal conductivity materials are often chosen for heat sink construction as they can quickly spread heat away from heat sources like electronic components.

In our exercise, the aluminum's high thermal conductivity is essential for distributing the heat efficiently through the fins and into the surrounding water. It's a key parameter when calculating both the fin efficiency and the convection coefficient, ultimately affecting how well the heat sink can keep the electronic chips at the desired temperature.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A spherical droplet of alcohol, \(0.5 \mathrm{~mm}\) in diameter, is falling freely through quiescent air at a velocity of \(1.8 \mathrm{~m} / \mathrm{s}\). The concentration of alcohol vapor at the surface of the droplet is \(0.0573 \mathrm{~kg} / \mathrm{m}^{3}\), and the diffusion coefficient for alcohol in air is \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Neglecting radiation and assuming steady-state conditions, calculate the surface temperature of the droplet if the ambient air temperature is \(300 \mathrm{~K}\). The latent heat of vaporization is \(8.42 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).

Motile bacteria are equipped with flagella that are rotated by tiny, biological electrochemical engines which, in turn, propel the bacteria through a host liquid. Consider a nominally spherical Escherichia coli bacterium that is of diameter \(D=2 \mu \mathrm{m}\). The bacterium is in a water-based solution at \(37^{\circ} \mathrm{C}\) containing a nutrient which is characterized by a binary diffusion coefficient of \(D_{\mathrm{AB}}=0.7 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\) and a food energy value of \(\mathcal{N}=16,000 \mathrm{~kJ} / \mathrm{kg}\). There is a nutrient density difference between the fluid and the shell of the bacterium of \(\Delta \rho_{\mathrm{A}}=860 \times 10^{-12} \mathrm{~kg} / \mathrm{m}^{3}\). Assuming a propulsion efficiency of \(\eta=0.5\), determine the maximum speed of the E. coli. Report your answer in body diameters per second.

Consider steady, parallel flow of atmospheric air over a flat plate. The air has a temperature and free stream velocity of \(300 \mathrm{~K}\) and \(25 \mathrm{~m} / \mathrm{s}\). (a) Evaluate the boundary layer thickness at distances of \(x=1,10\), and \(100 \mathrm{~mm}\) from the leading edge. If a second plate were installed parallel to and at a distance of \(3 \mathrm{~mm}\) from the first plate, what is the distance from the leading edge at which boundary layer merger would occur? (b) Evaluate the surface shear stress and the \(y\)-velocity component at the outer edge of the boundary layer for the single plate at \(x=1,10\), and \(100 \mathrm{~mm}\). (c) Comment on the validity of the boundary layer approximations.

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.

An \(L=1\)-m-long vertical copper tube of inner diameter \(D_{i}=20 \mathrm{~mm}\) and wall thickness \(t=2 \mathrm{~mm}\) contains liquid water at \(T_{w}=0^{\circ} \mathrm{C}\). On a winter day, air at \(V=3 \mathrm{~m} / \mathrm{s}, T_{\infty}=-20^{\circ} \mathrm{C}\) is in cross flow over the tube. (a) Determine the heat loss per unit mass from the water (W/kg) when the tube is full of water. (b) Determine the heat loss from the water (W/kg) when the tube is half full.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.