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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

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 \(G\) ).

Dry air at \(35^{\circ} \mathrm{C}\) and a velocity of \(15 \mathrm{~m} / \mathrm{s}\) flows over a long cylinder of \(20-\mathrm{mm}\) diameter. The cylinder is covered with a thin porous coating saturated with water, and an embedded electrical heater supplies power to maintain the coating surface temperature at \(20^{\circ} \mathrm{C}\). (a) What is the evaporation rate of water from the cylinder per unit length \((\mathrm{kg} / \mathrm{h}+\mathrm{m})\) ? What electrical power per unit length of the cylinder \((\mathrm{W} / \mathrm{m})\) is required to maintain steady-state conditions? (b) After a long period of operation, all the water is evaporated from the coating and its surface is dry. For the same free stream conditions and heater power of part (a), estimate the temperature of the surface.

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.

Consider atmospheric air at \(u_{\infty}=2 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=300 \mathrm{~K}\) in parallel flow over an isothermal flat plate of length \(L=1 \mathrm{~m}\) and temperature \(T_{s}=350 \mathrm{~K}\). (a) Compute the local convection coefficient at the leading and trailing edges of the heated plate with and without an unheated starting length of \(\xi=1 \mathrm{~m}\). (b) Compute the average convection coefficient for the plate for the same conditions as part (a). (c) Plot the variation of the local convection coefficient over the plate with and without an unheated starting length.

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?

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