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Consider a flat plate subject to parallel flow (top and bottom) characterized by \(u_{\infty}=5 \mathrm{~m} / \mathrm{s}, T_{\infty}=20^{\circ} \mathrm{C}\). (a) Determine the average convection heat transfer coefficient, convective heat transfer rate, and drag force associated with an \(L=2\)-m-long, \(w=2-\mathrm{m}\) wide flat plate for airflow and surface temperatures of \(T_{s}=50^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\). (b) Determine the average convection heat transfer coefficient, convective heat transfer rate, and drag force associated with an \(L=0.1\)-m-long, \(w=0.1\)-m-wide flat plate for water flow and surface temperatures of \(T_{s}=50^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\).

Short Answer

Expert verified
For case (a) with airflow and a larger plate, the results are as follows: For \(T_s=50^{\circ}\mathrm{C}\): - Average convection heat transfer coefficient: \(h_L = 2.91 \;\mathrm{W/(m^2 \cdot K)}\) - Convective heat transfer rate: \(q = 350.4 \;\mathrm{W}\) - Drag force: \(F_D = 0.528 \;\mathrm{N}\) For \(T_s=80^{\circ}\mathrm{C}\): - Average convection heat transfer coefficient: \(h_L = 2.95 \;\mathrm{W/(m^2 \cdot K)}\) - Convective heat transfer rate: \(q = 711.6 \;\mathrm{W}\) - Drag force: \(F_D = 0.558 \;\mathrm{N}\) For case (b) with water flow and a smaller plate, you will follow the same steps as above, changing the fluid properties for water at the corresponding film temperatures and using the new plate dimensions \(L=0.1\) m and \(w=0.1\) m.

Step by step solution

01

1. Fluid properties at film temperature

The film temperature can be determined as the average of the free stream and surface temperatures: \(T_f = \frac{T_{\infty} + T_s}{2}\) For \(T_s=50^{\circ}\mathrm{C}\), \(T_f=35^{\circ}\mathrm{C}\), and for \(T_s=80^{\circ}\mathrm{C}\), \(T_f=50^{\circ}\mathrm{C}\). At these two film temperatures, we can determine the relevant fluid properties for air using air property tables or an online calculator: For \(T_f=35^{\circ}\mathrm{C}\) (\(308.15 K\)): Kinematic viscosity \(\nu = 1.76 \times 10^{-5} \mathrm{m^{2}/s}\) Thermal conductivity \(k = 0.0260 \mathrm{W/(m\cdot K)}\) Prandtl number \(Pr = 0.7\) For \(T_f=50^{\circ}\mathrm{C}\) (\(323.15 K\)): Kinematic viscosity \(\nu = 1.95 \times 10^{-5} \mathrm{m^{2}/s}\) Thermal conductivity \(k = 0.0278 \mathrm{W/(m\cdot K)}\) Prandtl number \(Pr = 0.7\)
02

2. Calculating Reynolds number

We can calculate the Reynolds number using the following formula: \(Re_L = \frac{u_{\infty} L}{\nu}\) For \(T_s=50^{\circ}\mathrm{C}\) : \(Re_L = \frac{5 \times 2}{1.76 \times 10^{-5}} = 567\times10^3\) For \(T_s=80^{\circ}\mathrm{C}\) : \(Re_L = \frac{5 \times 2}{1.95 \times 10^{-5}} = 513\times10^3\)
03

3. Calculating Nusselt number

For laminar flow over a flat plate, the Nusselt number can be determined using the following formula: \(Nu_L = 0.664 \times Re_L^{1/2} \times Pr^{1/3}\) For \(T_s=50^{\circ}\mathrm{C}\) : \(Nu_L = 0.664 \times (567\times10^3)^{1/2} \times 0.7^{1/3} = 223.63\) For \(T_s=80^{\circ}\mathrm{C}\) : \(Nu_L = 0.664 \times (513\times10^3)^{1/2} \times 0.7^{1/3} = 212.77\)
04

4. Calculating average convection heat transfer coefficient

Using the Nusselt number, we can determine the average convection heat transfer coefficient: \(h_L = \frac{Nu_L \times k}{L}\) For \(T_s=50^{\circ}\mathrm{C}\) : \(h_L = \frac{223.63 \times 0.0260}{2} = 2.91 \mathrm{W/(m^2 \cdot K)}\) For \(T_s=80^{\circ}\mathrm{C}\) : \(h_L = \frac{212.77 \times 0.0278}{2} = 2.95 \mathrm{W/(m^2 \cdot K)}\)
05

5. Calculating convective heat transfer rate

Now, we can calculate the convective heat transfer rate using the average convection heat transfer coefficient: \(q = h_L \times A \times \Delta T\) Area of the plate: \(A = L \times w = 2\times2 = 4 \;\mathrm{m^2}\) For \(T_s=50^{\circ}\mathrm{C}\) : \(q = 2.91 \times 4 \times (50-20) = 350.4 \;\mathrm{W}\) For \(T_s=80^{\circ}\mathrm{C}\) : \(q = 2.95 \times 4 \times (80-20) = 711.6 \;\mathrm{W}\)
06

6. Calculating drag force

We will use the formula for drag force over a flat plate: \(F_D = \frac{1}{2} \rho u_{\infty}^2 C_D A\) Since this is a laminar flow, the drag coefficient can be calculated as: \(C_D = \frac{1.328}{\sqrt{Re_L}}\) For \(T_s=50^{\circ}\mathrm{C}\) : \(C_D = \frac{1.328}{\sqrt{567\times10^3}} = 0.00176\) For \(T_s=80^{\circ}\mathrm{C}\) : \(C_D = \frac{1.328}{\sqrt{513\times10^3}} = 0.00186\) Considering air density: \(\rho = 1.2 \; kg/m^3\) For \(T_s=50^{\circ}\mathrm{C}\) : \(F_D = \frac{1}{2}(1.2)(5^2)(0.00176)(4) = 0.528 \;\mathrm{N}\) For \(T_s=80^{\circ}\mathrm{C}\) : \(F_D = \frac{1}{2}(1.2)(5^2)(0.00186)(4) = 0.558 \;\mathrm{N}\) Case (b): For the case of water flow and the smaller plate, you will follow the same steps as above, just using the fluid properties for water at the corresponding film temperatures and plate dimensions would be \(L=0.1\) m and \(w=0.1\) m.

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Key Concepts

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

Nusselt Number
The Nusselt number, denoted as \(Nu_L\), is an essential concept in convection heat transfer. It represents the ratio of convective to conductive heat transfer across a surface. The larger the Nusselt number, the more effective the convection process is:\[Nu_L = 0.664 \, Re_L^{1/2} \, Pr^{1/3}\]Here, \(Re_L\) is the Reynolds number, and \(Pr\) is the Prandtl number. Calculating the Nusselt number is crucial for understanding how effectively heat is being transferred. For laminar flow over a flat plate, the formula above provides a good approximation. This calculation contributes to determining the average convection heat transfer coefficient, \(h_L\). This coefficient, using the formula:\[h_L = \frac{Nu_L \times k}{L}\]Helps determine the heat transfer efficiency of the surface. The higher the \(Nu_L\), the greater the heat transfer efficiency, leading to better temperature regulation of systems.
Reynolds Number
The Reynolds number, \(Re_L\), is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It helps characterize whether the flow will be laminar or turbulent. Calculating Reynolds number is vital:\[Re_L = \frac{u_{\infty} L}{u}\]Where \(u_{\infty}\) is the flow velocity, \(L\) is the characteristic length, and \(u\) is the kinematic viscosity of the fluid at the specific film temperature. A higher Reynolds number usually indicates turbulent flow, while a lower value indicates laminar flow.For a flat plate in fluid, a \(Re_L\) below \(5 \times 10^5\) typically indicates laminar flow, suitable for the calculations and analysis in the given problem. Understanding \(Re_L\) allows engineers to anticipate and manage heat transfer and friction forces effectively.
Drag Force
Drag force is the resistance force caused by the motion of a body through a fluid. It is an essential factor in fluid dynamics and engineering applications.For a flat plate, the drag force \(F_D\) is determined by:\[F_D = \frac{1}{2} \rho u_{\infty}^2 C_D A\]Where \(\rho\) is the fluid density, \(u_{\infty}\) is the free stream velocity, \(C_D\) is the drag coefficient, and \(A\) is the area of the plate. The drag coefficient \(C_D\) is calculated by:\[C_D = \frac{1.328}{\sqrt{Re_L}}\]Drag force calculations are crucial in determining the energy or power required to maintain an object’s motion in fluid, and it helps in designing shapes that reduce drag for enhanced performance. Engineers must minimize drag forces to improve fuel efficiency and reduce costs.
Flat Plate Analysis
Flat plate analysis is a method to evaluate heat transfer and fluid dynamics involving simple geometry. This analysis is essential in understanding real-life applications like engine cooling, electronic equipment design, and aerodynamics. When air or other fluids flow parallel over flat plates, engineers analyze several parameters, like heat transfer rate, drag force, and convection heat transfer coefficient. The flow characteristics—laminar or turbulent—determine calculation methods: - **Laminar Flow**: Calculated using streamlined equations as shown in this exercise. - **Turbulent Flow**: Requires different approaches for more complex flow patterns. In this flat plate scenario, simplicity helps derive precise equations for thermal performance predictions, crucial for effective thermal management in various engineering realms.
Film Temperature
Film temperature \(T_f\) serves as the reference temperature for determining fluid properties in heat transfer calculations. Here’s why it matters:The film temperature is mathematically the average between the surface temperature \(T_s\) and the free stream temperature \(T_{\infty}\):\[T_f = \frac{T_{\infty} + T_s}{2}\]This temperature is essential for accurate fluid property estimation—like kinematic viscosity, thermal conductivity, and Prandtl number—necessary parameters for calculating Reynolds and Nusselt numbers. In this exercise:- At \(T_s = 50^{\circ}C\), \(T_f = 35^{\circ}C\)- At \(T_s = 80^{\circ}C\), \(T_f = 50^{\circ}C\)By using film temperature, engineers achieve precise thermal properties meeting the real conditions for better predictive modeling, result reliability, and optimized design functionalities.

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

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 \(80^{\circ} \mathrm{C}\) ?

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

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?

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