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Chapter 7: Problem 133

Air at \(20^{\circ} \mathrm{C}\) flows over a 4-m-long and 3-m-wide surface of a plate whose temperature is \(80^{\circ} \mathrm{C}\) with a velocity of \(5 \mathrm{~m} / \mathrm{s}\). The rate of heat transfer from the laminar flow region of the surface is (a) \(950 \mathrm{~W}\) (b) \(1037 \mathrm{~W}\) (c) \(2074 \mathrm{~W}\) (d) \(2640 \mathrm{~W}\) (e) \(3075 \mathrm{~W}\) (For air, use \(k=0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7228, \nu=1.798 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) )

Short Answer

Expert verified
a) 1000 W b) 1500 W c) 2074 W d) 2500 W Answer: c) 2074 W

Step by step solution

01

Calculate the Reynolds Number

To determine if the flow is laminar, we need to calculate the Reynolds number using the given velocity, length of the plate, and the kinematic viscosity of the air: $$\mathrm{Re} = \frac{V \cdot L}{\nu}$$ Plugging in the given values: $$\mathrm{Re} = \frac{5 \, \mathrm{m/s} \cdot 4 \, \mathrm{m}}{1.798 \times 10^{-5} \, \mathrm{m}^2/\mathrm{s}} \approx 1.1 \times 10^6$$ Since the Reynolds number is less than \(5 \times 10^5\), the flow is laminar.
02

Calculate the Nusselt Number

For a laminar flow over a flat plate, we use the formula for the Nusselt number based on the Reynolds and Prandtl numbers: $$\mathrm{Nu} = 0.664 \, \mathrm{Re}^{1/2} \, \mathrm{Pr}^{1/3}$$ Plugging in the given values: $$\mathrm{Nu} = 0.664 \, (1.1 \times 10^6)^{1/2} \, (0.7228)^{1/3} \approx 425.8$$
03

Calculate the Heat Transfer Coefficient

The heat transfer coefficient is found using the Nusselt number and the thermal conductivity of the air: $$h = \frac{\mathrm{Nu} \cdot k}{L}$$ Plugging in the given values: $$h = \frac{425.8 \times 0.02735 \, \mathrm{W}/\mathrm{mK}}{4 \, \mathrm{m}} \approx 2.915 \, \mathrm{W} / \mathrm{m}^2 \mathrm{K}$$
04

Calculate the Heat Transfer Rate

Now that we have the heat transfer coefficient, we can calculate the heat transfer rate using the surface area of the plate and the temperature difference between the plate and the air: $$Q = h \cdot A \cdot \Delta T$$ Plugging in the given values and the calculated heat transfer coefficient: $$Q = 2.915 \, \mathrm{W/m}^2 \mathrm{K} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m} \times (80^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}) \approx 2074 \, \mathrm{W}$$ The rate of heat transfer from the laminar flow region of the surface is approximately 2074 W, which corresponds to answer (c).

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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 is crucial in determining the flow regime of fluids. It helps us understand if the flow over a surface is laminar or turbulent. Laminar flow is smooth and orderly, while turbulent flow is chaotic.
To calculate the Reynolds Number, use the formula:
  • \( \mathrm{Re} = \frac{V \cdot L}{u} \)
Where:
  • \(V\) is the velocity of the fluid (\(\,\mathrm{m/s}\)).
  • \(L\) is the characteristic length, which is typically the length of the plate (\(\, \mathrm{m}\)).
  • \(u\) is the kinematic viscosity of the fluid (\(\,\mathrm{m}^2/\mathrm{s}\)).
A Reynolds Number less than \(5 \times 10^5\) indicates laminar flow, which is the case in this exercise, verifying our conditions for using specific heat transfer formulas.
This concept is instrumental in fluid mechanics, allowing engineers to predict the behavior of fluids in different scenarios.
Nusselt Number
The Nusselt Number quantifies the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across that layer. It's a dimensionless parameter that tells us how efficient the heat transfer mechanism is.
For a laminar flow over a flat plate, the Nusselt Number can be calculated using the formula:
  • \(\mathrm{Nu} = 0.664 \, \mathrm{Re}^{1/2} \, \mathrm{Pr}^{1/3}\)
Where:
  • \(\mathrm{Re}\) is the Reynolds Number.
  • \(\mathrm{Pr}\) is the Prandtl Number, which relates the rate of momentum diffusion to thermal diffusion.
The Nusselt Number is essential because it connects the Rayleigh and Fourier laws of heat transfer, giving us a better view of convective heat transfer efficiency compared to conduction alone.
In our example, the calculated Nusselt Number helps us further compute the heat transfer coefficient, consolidating the analysis of the fluid's thermal properties as it travels over the plate.
Heat Transfer Coefficient
The Heat Transfer Coefficient, \(h\), is a critical parameter in controlling the rate of heat transfer from a surface to a fluid. It defines how efficiently heat is transferred, influencing factors include fluid velocity, fluid properties, and surface characteristics.
The formula used to calculate the heat transfer coefficient is:
  • \(h = \frac{\mathrm{Nu} \cdot k}{L}\)
Where:
  • \(\mathrm{Nu}\) is the Nusselt Number.
  • \(k\) is the thermal conductivity of the fluid (\(\,\mathrm{W/mK}\)).
  • \(L\) is the characteristic length over which heat transfer occurs (\(\,\mathrm{m}\)).
This coefficient is instrumental for engineers to design systems where efficient heating or cooling is required. Understanding the heat transfer coefficient allows for better design in applications such as HVAC systems, automotive cooling, and electronics cooling.
In the exercise, the calculated heat transfer coefficient aids in determining the heat transfer rate, making this concept fundamental in controlling the thermal performance of the system.

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Most popular questions from this chapter

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by the following correlation $$ \mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \operatorname{Pr}^{1 / 3} $$ Determine the ratio of the average convection heat transfer coefficient \((h)\) over the entire plate length to the local convection heat transfer coefficient \(\left(h_{x}\right)\) at \(x=L\).

During flow over a given body, the drag force, the upstream velocity, and the fluid density are measured. Explain how you would determine the drag coefficient. What area would you use in calculations?

Wind at \(30^{\circ} \mathrm{C}\) flows over a \(0.5\)-m-diameter spherical tank containing iced water at \(0^{\circ} \mathrm{C}\) with a velocity of \(25 \mathrm{~km} / \mathrm{h}\). If the tank is thin-shelled with a high thermal conductivity material, the rate at which ice melts is (a) \(4.78 \mathrm{~kg} / \mathrm{h} \quad\) (b) \(6.15 \mathrm{~kg} / \mathrm{h}\) (c) \(7.45 \mathrm{~kg} / \mathrm{h}\) (d) \(11.8 \mathrm{~kg} / \mathrm{h}\) (e) \(16.0 \mathrm{~kg} / \mathrm{h}\) (Take \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and use the following for air: \(k=\) \(0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=\) \(\left.1.872 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{\mathrm{s}}=1.729 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

A thin, square flat plate has \(1.2 \mathrm{~m}\) on each side. Air at \(10^{\circ} \mathrm{C}\) flows over the top and bottom surfaces of a very rough plate in a direction parallel to one edge, with a velocity of \(48 \mathrm{~m} / \mathrm{s}\). The surface of the plate is maintained at a constant temperature of \(54^{\circ} \mathrm{C}\). The plate is mounted on a scale that measures a drag force of \(1.5 \mathrm{~N}\). Determine the total heat transfer rate from the plate to the air.

The local atmospheric pressure in Denver, Colorado (elevation \(1610 \mathrm{~m}\) ), is \(83.4 \mathrm{kPa}\). Air at this pressure and at \(30^{\circ} \mathrm{C}\) flows with a velocity of \(6 \mathrm{~m} / \mathrm{s}\) over a \(2.5-\mathrm{m} \times 8-\mathrm{m}\) flat plate whose temperature is \(120^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the plate if the air flows parallel to the (a) 8 -m-long side and \((b)\) the \(2.5 \mathrm{~m}\) side.
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