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Chapter 7: Problem 44

An array of 10 silicon chips, each of length \(L=10 \mathrm{~mm}\) on a side, is insulated on one surface and cooled on the opposite surface by atmospheric air in parallel flow with \(T_{\infty}=24^{\circ} \mathrm{C}\) and \(u_{\infty}=40 \mathrm{~m} / \mathrm{s}\). When in use, the same electrical power is dissipated in each chip, maintaining a uniform heat flux over the entire cooled surface. If the temperature of each chip may not exceed \(80^{\circ} \mathrm{C}\), what is the maximum allowable power per chip? What is the maximum allowable power if a turbulence promoter is used to trip the boundary layer at the leading edge? Would it be preferable to orient the array normal, instead of parallel, to the airflow?

Short Answer

Expert verified
The maximum allowable power per silicon chip with laminar flow is approximately \(0.301 \mathrm{W}\). If a turbulence promoter is used, the maximum allowable power increases to \(0.819 \mathrm{W}\). As for the orientation, changing it to normal to the airflow may increase the convection rate but also add flow disturbances. Deciding whether this option is preferable requires a more in-depth analysis weighing the trade-offs between convection rates and flow disturbances.

Step by step solution

01

Find properties of air

First, we need to find the properties of air at the mean temperature, which is the average of the chip's surface temperature and the free stream air temperature. For this exercise, we will consider air properties at \(\left(24+80\right)/2 = 52^{\circ}\mathrm{C}\) or \(325\mathrm{K}\). Using air property tables at the given temperature and around atmospheric pressure, we get: - Dynamic Viscosity \(\mu = 2.0 \times 10^{-5} \mathrm{~kg/m.s}\) - Thermal Conductivity, \(k = 0.029 \mathrm{~W/m.K}\) - Specific Heat, \(c_p = 1005 \mathrm{~J/kg.K}\) - Density, \(\rho = 1.10 \mathrm{~kg/m^3}\)
02

Calculate the Reynolds number

The Reynolds number (\(\mathrm{Re}\)) for airflow over the chip is: \(\mathrm{Re} = \frac{\rho u_{\infty} L}{\mu}\) Plugging in the known values, we get: \(\mathrm{Re} = \frac{1.10\times 40 \times 0.01}{2.0\times 10^{-5}} = 2.2 \times 10^5\)
03

Calculate the Nusselt number for laminar flow

Since the Reynolds number is greater than \(5\times 10^4\), the flow is turbulent. However, we will first calculate the Nusselt number for laminar flow as follows: \(Nu = 0.664 \times \mathrm{Re}^{0.5} \times Pr^{1/3}\) In case of air, the Prandtl number (\(Pr\)) is close to \(0.71\). Thus, we get: \(Nu = 0.664 \times \sqrt{2.2 \times 10^5} \times {0.71}^{1/3} \approx 184.6\)
04

Calculate the heat transfer coefficient for laminar flow

The heat transfer coefficient (\(h\)) can be found using the Nusselt number and thermal conductivity: \(h = \frac{k \times Nu}{L}\) Plugging in the values, we get: \(h = \frac{0.029 \times 184.6}{0.01} = 537.34 \mathrm{~W/m^2.K}\)
05

Calculate the maximum allowable power for laminar flow

The maximum allowable power can be found using the following formula: \(P = h \times A \times (T_{s} - T_{\infty})\) Here, \(A = L^2\) is the area of one chip. Rearranging the formula, we get: \(P = \frac{h \times L^2}{1} \times (80 - 24)\) Plugging in the values, we get: \(P = 537.34 \times 0.01^2 \times (56) \approx 0.301 \mathrm{W}\)
06

Check the maximum allowable power with turbulence promoter

If a turbulence promoter is used, the Nusselt number for turbulent flow needs to be considered. The Nusselt number formula for turbulent flow is: \(Nu = 0.037 \times \mathrm{Re}^{0.8} \times Pr^{1/3}\) Replacing the values, we get: \(Nu = 0.037 \times (2.2 \times 10^5)^{0.8} \times 0.71^{1/3} \approx 501.0\) Now calculate the heat transfer coefficient for turbulent flow: \(h = \frac{0.029 \times 501.0}{0.01} = 1452.9 \mathrm{~W/m^2.K}\) Finally, we calculate the maximum allowable power for turbulent flow: \(P = \frac{1452.9 \times 0.01^2}{1} \times (80 - 24) \approx 0.819 \mathrm{W}\) Thus, the maximum power per chip is \(0.819 \mathrm{W}\) if a turbulence promoter is used.
07

Discussion on the array orientation

Changing the orientation of the chips so that they are normal to the airflow would increase the exposed surface area and potentially lowering the temperature since a higher convection rate occurs. However, placing the chips normal to the airflow would create flow disturbance and increase the resistance to the flow, reducing the efficiency of the case. Determining if this arrangement is preferable would require a more in-depth analysis considering the trade-offs between higher convection rates and flow disturbances.

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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 (\( \mathrm{Re} \)) is key in determining the flow characteristics of a fluid. It is defined as the ratio between the inertial forces and the viscous forces within a fluid flow. It's important because it helps predict whether the flow will be laminar or turbulent.
  • A higher Reynolds number indicates that inertial forces dominate, which is usually associated with turbulent flow.
  • A lower Reynolds number suggests that viscous forces dominate, indicating a laminar flow.
The formula to calculate the Reynolds number in a flow situation, such as airflow over a chip, is given by:\[\mathrm{Re} = \frac{\rho u_{\infty} L}{\mu}\]Where:
  • \( \rho \) is the fluid density.
  • \( u_{\infty} \) is the velocity of the fluid.
  • \( L \) is the characteristic length, often the length of the object the fluid flows over.
  • \( \mu \) is the dynamic viscosity of the fluid.
For our exercise, the calculation confirmed a Reynolds number of \( 2.2 \times 10^5 \), indicating a turbulent flow. Understanding the Reynolds number is crucial for assessing flow behaviors such as the transition from laminar to turbulent flow.
Nusselt number
The Nusselt number (\( Nu \)) is an essential dimensionless number in heat transfer. It correlates the convective heat transfer to the conductive heat transfer across a boundary. Essentially, it helps determine the efficiency of heat transfer in a fluid flow over a surface.
  • A higher Nusselt number suggests more efficient convective heat transfer, which is desirable in cooling applications.
  • For laminar flow, the Nusselt number is typically calculated with empirical correlations like \( Nu = 0.664 \times \mathrm{Re}^{0.5} \times Pr^{1/3} \)
  • For turbulent flow, a different correlation is used, such as \( Nu = 0.037 \times \mathrm{Re}^{0.8} \times Pr^{1/3} \)
The Nusselt number allows the determination of the heat transfer coefficient (\( h \)) using the formula:\[h = \frac{k \times Nu}{L}\]where \( k \)is the thermal conductivity of the fluid. In our example, the Nusselt number for laminar flow was found to be approximately 184.6, but with turbulent flow, it increased to approximately 501.0, indicating a substantial improvement in heat transfer.
Turbulent flow
Turbulent flow refers to a type of fluid flow characterized by chaotic and irregular fluctuations. Unlike laminar flow, turbulent flow does not follow a streamlined path and is more complex and less predictable.
  • This flow type occurs when the Reynolds number exceeds a critical threshold, typically around \( 5 \times 10^4 \).
  • In turbulent flow, eddies, vortices, and rapid changes in pressure and flow velocity dominate.
In the context of the problem, when the inflow surpassed \( \mathrm{Re} = 2.2 \times 10^5 \), turbulent flow was determined. This increased the heat transfer coefficient significantly. Using a turbulence promoter can trigger turbulent flow at lower Reynolds numbers or earlier in the flow over a surface. This property was leveraged to achieve a more substantial cooling effect on the chips by raising the Nusselt number, hence enhancing the power dissipation capability of each chip.
Laminar flow
Laminar flow is often described as smooth, streamlined, or regular flow. In laminar flow, fluid particles move in parallel layers and there is minimal mixing between them. It occurs at lower Reynolds numbers, typically less than \( 2 \times 10^3 \).
  • Laminar flow is more stable and predictable than turbulent flow.
  • It has lower momentum transfer across layers compared to turbulent flow, which results in lower heat transfer efficiency.
In our specific setting, we initially calculated parameters as if the flow might be laminar, even though the problem context indicated turbulent flow. Despite the smooth nature of laminar flow, in many engineering applications, especially cooling or heating, turbulent flow is preferred due to its higher heat transfer capabilities. However, laminar flow can be advantageous in situations where low friction and high accuracy are necessary.

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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?

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 \(D=4 \mathrm{~mm} ?\)

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.

The cylindrical chamber of a pebble bed nuclear reactor is of length \(L=10 \mathrm{~m}\), and diameter \(D=3 \mathrm{~m}\). The chamber is filled with spherical uranium oxide pellets of core diameter \(D_{p}=50 \mathrm{~mm}\). Each pellet generates thermal energy in its core at a rate of \(\dot{E}_{g}\) and is coated with a layer of non-heat-generating graphite, which is of uniform thickness \(\delta=5 \mathrm{~mm}\), to form a pebble. The uranium oxide and graphite each have a thermal conductivity of \(2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The packed bed has a porosity of \(\varepsilon=0.4\). Pressurized helium at 40 bars is used to absorb the thermal energy from the pebbles. The helium enters the packed bed at \(T_{i}=450^{\circ} \mathrm{C}\) with a velocity of \(3.2 \mathrm{~m} / \mathrm{s}\). The properties of the helium may be assumed to be \(c_{p}=5193 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.3355 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2.1676 \mathrm{~kg} / \mathrm{m}^{3}, \mu=4.214 \times\) \(10^{-5} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}, \operatorname{Pr}=0.654\). (a) For a desired overall thermal energy transfer rate of \(q=125 \mathrm{MW}\), determine the mean outlet temperature of the helium leaving the bed, \(T_{o}\), and the amount of thermal energy generated by each pellet, \(\dot{E}_{g^{*}}\) (b) The amount of energy generated by the fuel decreases if a maximum operating temperature of approximately \(2100^{\circ} \mathrm{C}\) is exceeded. Determine the maximum internal temperature of the hottest pellet in the packed bed. For Reynolds numbers in the range \(4000 \leq R e_{D} \leq 10,000\), Equation \(7.81\) may be replaced by \(\varepsilon \bar{j}_{H}=2.876 R e_{D}^{-1}+0.3023 R e_{D}^{-0.35}\).

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 \(8.42 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).
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