/*! 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 36 Air at \(27^{\circ} \mathrm{C}\)... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 36

Air at \(27^{\circ} \mathrm{C}\) with a free stream velocity of \(10 \mathrm{~m} / \mathrm{s}\) is used to cool electronic devices mounted on a printed circuit board. Each device, \(4 \mathrm{~mm} \times 4 \mathrm{~mm}\), dissipates \(40 \mathrm{~mW}\), which is removed from the top surface. A turbulator is located at the leading edge of the board, causing the boundary layer to be turbulent. (a) Estimate the surface temperature of the fourth device located \(15 \mathrm{~mm}\) from the leading edge of the board. (b) Generate a plot of the surface temperature of the first four devices as a function of the free stream velocity for \(5 \leq u_{s} \leq 15 \mathrm{~m} / \mathrm{s}\). (c) What is the minimum free stream velocity if the surface temperature of the hottest device is not to exceed \(80^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
(a) The surface temperature of the fourth device is approximately \(34.72^{\circ} \mathrm{C}\) or \(307.72\:K\). (b) To generate the plot of surface temperatures, calculate the surface temperature for different free-stream velocities and plot the results accordingly. (c) To find the minimum free-stream velocity, express h in terms of u_s using the given formulas and solve for the required conditions.

Step by step solution

01

(a) Estimate the surface temperature of the fourth device:

First, we need to calculate the heat transfer coefficient (h) using the given information. As the flow is turbulent, we can use the Dittus-Boelter equation to find the heat transfer coefficient: \( Nu = 0.023 \times Re^{0.8} \times Pr^{0.4} \) where Nu is the Nusselt number, Re is the Reynolds number, and Pr is the Prandtl number. We have: Free stream velocity, \( u_s = 10\:m/s \) Temperature, \( T_{\infty} =27^{\circ}C = 300\:K \) Device dimensions: \( L = 0.004\:m, W = 0.004\:m \) Power dissipated, \( P = 40\:mW = 0.04\:W \) Boundary layer thickness, \( \delta = 15\:mm = 0.015\:m \) First, we need to calculate the Reynolds number (Re): \( Re = \frac{u_s \delta}{\nu} \) where \(\nu\) is the kinematic viscosity of air, which is given by: \( \nu = 15.11 \times 10^{-6} \:m^2/s \) So, \( Re = \frac{10 \times 0.015}{15.11 \times 10^{-6}} = 9945 \) Now we need to find the Prandtl number (Pr): \( Pr = \frac{\mu c_p}{k} \) using the properties of air at \( T_\infty = 300K \): Dynamic viscosity, \( \mu = 1.846 \times 10^{-5} \:kg/(m \cdot s) \) Specific heat capacity at constant pressure, \(c_p = 1005 \:J/(kg \cdot K) \) Thermal conductivity, \(k = 0.0263 \:W/(m \cdot K) \) So, \( Pr = \frac{1.846 \times 10^{-5} \cdot 1005}{0.0263} = 0.708 \) Now we can find the Nusselt number (Nu): \( Nu = 0.023 \times 9945^{0.8} \times 0.708^{0.4} = 185.15 \) And, finally, we can find the heat transfer coefficient (h): \( h = \frac{Nu \cdot k}{\delta} = \frac{185.15 \cdot 0.0263}{0.015} = 322.78 \:W/(m^2 \cdot K) \) Now, to calculate the surface temperature of the 4th device, we use the convection heat transfer formula: \( P = h \times A \times (T_{surface} - T_{\infty}) \) where A is the surface area of the device. For the 4th device, \( A = L \times W = 0.004 \times 0.004 = 16 \times 10^{-6}\: m^2 \) Let \( \Delta T = T_{surface} - T_{\infty} \) So, \( \Delta T = \frac{P}{h \times A} = \frac{0.04}{322.78 \times 16 \times 10^{-6}} = 7.72 K \) Thus, the surface temperature of the 4th device is: \( T_{surface} = T_{\infty} + \Delta T = 300 + 7.72 = 307.72\:K \) or \( 34.72^{\circ} \mathrm{C} \).
02

(b) Generate a plot of the surface temperature of the first four devices:

To generate a plot of the surface temperature of the first four devices as a function of the free stream velocity for \( 5 \leq u_{s} \leq 15\:m/s \), you can use the above steps with different free stream velocities to obtain the surface temperature values. Then, plot the surface temperature against the free stream velocity.
03

(c) Find the minimum free-stream velocity:

To find the minimum free-stream velocity for the hottest device surface temperature not to exceed \(80^{\circ} \mathrm{C}\), we can use the heat transfer formula: \( P = h \times A \times (T_{max} - T_{\infty}) \) where \( T_{max} = 80^{\circ}C\) or \( 353.15\:K \). We already have the values of P and A. Now, we have to find h in terms of the free stream velocity (u_s) and substitute it into the equation: \( h = \frac{Nu \cdot k}{\delta} = \frac{ 0.023 \times Pr^{0.4} \cdot Re^{0.8} \cdot k}{\delta} \) Now, consider: \( Pr = \frac{\mu c_p}{k} \) \( Re = \frac{u_s \delta}{\nu} \) So, \( h = 0.023 \times (\frac{\mu c_p}{k})^{0.4} \times (\frac{u_s \delta}{\nu})^{0.8} \times \frac{k}{\delta} \) Now, use the above formula to express \( h \) in terms of \( u_s \). Then, calculate the minimum free stream velocity for the hottest device surface temperature not to exceed \(80^{\circ} \mathrm{C}\) using the heat transfer formula: \( P = h \times A \times (T_{max} - T_{\infty}) \)

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Key Concepts

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

Dittus-Boelter Equation
In the realm of heat transfer, the Dittus-Boelter equation is a powerful tool used to estimate the convective heat transfer coefficient for turbulent flows. This equation is particularly applicable when dealing with fluids moving through pipes or along flat surfaces, where the flow becomes turbulent. The formula is given as: \[ Nu = 0.023 imes Re^{0.8} imes Pr^{0.4} \] where \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number, and \(Pr\) is the Prandtl number. This equation makes it easy to calculate the heat transfer rates in engineering applications. However, it's critical to ensure the flow is, in fact, turbulent as the equation is derived under that assumption. Given its simplicity, it's a staple in many engineering curricula and practical applications.
  • Applies primarily to turbulent flow
  • Relies on Reynolds and Prandtl numbers
Understanding this equation is key to mastering concepts of convective heat transfer.
Nusselt Number
The Nusselt number, often symbolized as \(Nu\), plays an essential role in the analysis of convective heat transfer. It is a dimensionless number and is a measure of convective heat transfer relative to conductive heat transfer across a boundary. Essentially, the Nusselt number tells us how effective convection is in comparison to conduction:- High \(Nu\) implies efficient convective transfer- Depends on flow conditions and surface characteristics
The Nusselt number allows engineers to determine whether the heat transfer is primarily dominated by conduction or by convection. In simple terms, higher Nusselt numbers typically indicate more efficient heat transfer in situations where convection is the primary heat transfer mode. This makes \(Nu\) a crucial factor in designing systems for heating, cooling, or managing thermal conditions.
Reynolds Number
The Reynolds number (Re) is a dimensionless value that describes the nature of fluid flow. It is primarily a ratio of inertial forces to viscous forces and helps predict flow patterns in different fluid flow situations. The Reynolds number is defined by:\[ Re = \frac{u L}{u} \] where \(u\) is the velocity of the fluid, \(L\) is a characteristic length, and \(u\) is the kinematic viscosity. A higher Reynolds number indicates turbulent flow, while a lower number suggests laminar flow. This distinction is critical in many applications because the flow regime significantly affects heat and mass transfer rates as well as the energy efficiency of processes. In engineering, accurately computing the Reynolds number helps to ensure safe and efficient fluid handling systems.
Prandtl Number
The Prandtl number (Pr) is another dimensionless number that is vital in the study of heat transfer. It represents the ratio of momentum diffusivity to thermal diffusivity. Simply put, it gives us an idea of the relative thickness of the velocity boundary layer to the thermal boundary layer:\[ Pr = \frac{c_p \mu}{k} \] where \(c_p\) is the specific heat capacity at constant pressure, \(\mu\) is the dynamic viscosity, and \(k\) is the thermal conductivity. In practical terms, the Prandtl number helps determine how quickly heat diffuses compared to the velocity of the fluid.
  • A smaller Prandtl number means heat diffuses quickly, causing thicker thermal boundary layers
  • A larger Prandtl number indicates slower heat diffusion relative to velocity
This number is particularly useful in comparing convective heat transfer performances among different fluids.
Heat Transfer Coefficient
The heat transfer coefficient, often denoted by \(h\), is a measure of the heat transfer capability of a fluid in motion. It describes how well heat is transferred between a solid and a fluid in contact with its surface. You can calculate it using the Nusselt number from the Dittus-Boelter equation:\[ h = \frac{Nu \cdot k}{L} \] where \(k\) is the thermal conductivity of the fluid and \(L\) is the characteristic length. Higher values of \(h\) mean more effective heat transfer from the surface to the fluid or vice-versa.
  • Critical for calculating heat exchange in various systems
  • Affect design and efficiency of thermal management systems
Understanding the heat transfer coefficient is crucial for engineers to ensure their systems maintain the required thermal conditions efficiently.

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Most popular questions from this chapter

To enhance heat transfer from a silicon chip of width \(W=4 \mathrm{~mm}\) on a side, a copper pin fin is brazed to the surface of the chip. The pin length and diameter are \(L=12 \mathrm{~mm}\) and \(D=2 \mathrm{~mm}\), respectively, and atmospheric air at \(V=10 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=300 \mathrm{~K}\) is in cross flow over the pin. The surface of the chip, and hence the base of the pin, are maintained at a temperature of \(T_{b}=350 \mathrm{~K}\). (a) Assuming the chip to have a negligible effect on flow over the pin, what is the average convection coefficient for the surface of the pin? (b) Neglecting radiation and assuming the convection coefficient at the pin tip to equal that calculated in part (a), determine the pin heat transfer rate. (c) Neglecting radiation and assuming the convection coefficient at the exposed chip surface to equal that calculated in part (a), determine the total rate of heat transfer from the chip. (d) Independently determine and plot the effect of increasing velocity \((10 \leq V \leq 40 \mathrm{~m} / \mathrm{s})\) and pin diameter \((2 \leq D \leq 4 \mathrm{~mm})\) on the total rate of heat transfer from the chip. What is the heat rate for \(V=40 \mathrm{~m} / \mathrm{s}\) and \(D=4 \mathrm{~mm} ?\)

The use of rock pile thermal energy storage systems has been considered for solar energy and industrial process heat applications. A particular system involves a cylindrical container, \(2 \mathrm{~m}\) long by \(1 \mathrm{~m}\) in diameter, in which nearly spherical rocks of \(0.03-\mathrm{m}\) diameter are packed. The bed has a void space of \(0.42\), and the density and specific heat of the rock are \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=879 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. Consider conditions for which atmospheric air is supplied to the rock pile at a steady flow rate of \(1 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(90^{\circ} \mathrm{C}\). The air flows in the axial direction through the container. If the rock is at a temperature of \(25^{\circ} \mathrm{C}\), what is the total rate of heat transfer from the air to the rock pile?

Consider a liquid metal \((\operatorname{Pr} \leqslant 1)\), with free stream conditions \(u_{\infty}\) and \(T_{\infty}\), in parallel flow over an isothermal flat plate at \(T_{s}\). Assuming that \(u=u_{\infty}\) throughout the thermal boundary layer, write the corresponding form of the boundary layer energy equation. Applying appropriate initial \((x=0)\) and boundary conditions, solve this equation for the boundary layer temperature field, \(T(x, y)\). Use the result to obtain an expression for the local Nusselt number \(\mathrm{Nu}_{x}\). Hint: This problem is analogous to one-dimensional heat transfer in a semiinfinite medium with a sudden change in surface temperature.

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 \(G\) ).

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 \(\operatorname{Pr}=5.2\).
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