/*! 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

Copper spheres of \(20-\mathrm{mm}\) diameter are quenched by being dropped into a tank of water that is maintained at \(280 \mathrm{~K}\). The spheres may be assumed to reach the terminal velocity on impact and to drop freely through the water. Estimate the terminal velocity by equating the drag and gravitational forces acting on the sphere. What is the approximate height of the water tank needed to cool the spheres from an initial temperature of \(360 \mathrm{~K}\) to a center temperature of \(320 \mathrm{~K}\) ?

Dry air at atmospheric pressure and \(350 \mathrm{~K}\), with a free stream velocity of \(25 \mathrm{~m} / \mathrm{s}\), flows over a smooth, porous plate \(1 \mathrm{~m}\) long. (a) Assuming the plate to be saturated with liquid water at \(350 \mathrm{~K}\), estimate the mass rate of evaporation per unit width of the plate, \(n_{\mathrm{A}}^{\prime}(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\). (b) For air and liquid water temperatures of 300,325 , and \(350 \mathrm{~K}\), generate plots of \(n_{\mathrm{A}}^{\prime}\) as a function of velocity for the range from 1 to \(25 \mathrm{~m} / \mathrm{s}\).

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 \(20 \mathrm{~mm}\) ?

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 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?
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