/*! 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 137 Air at \(20^{\circ} \mathrm{C}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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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 surface is (a) \(7383 \mathrm{~W}\) (b) \(8985 \mathrm{~W}\) (c) \(11,231 \mathrm{~W}\) (d) 14,672 W (e) \(20,402 \mathrm{~W}\) (For air, use \(k=0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7228, \nu=1.798 \times\) \(\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

Short Answer

Expert verified
Answer: The rate of heat transfer from the surface is 8985 W.

Step by step solution

01

Calculate Reynolds Number

First, we need to calculate the Reynolds number for the airflow over the plate. The Reynolds number is given by the formula: \(Re = \frac{V \cdot L}{\nu}\) where \(V\) is the fluid velocity, \(L\) is the length of the plate, and \(\nu\) is the kinematic viscosity of the fluid. For this exercise, we have \(V = 5 \mathrm{~m} / \mathrm{s}\), \(L = 4\,\text{m}\), and \(\nu = 1.798 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). So, we can find the Reynolds number as: \(Re = \frac{5\,\text{m/s} \cdot 4\,\text{m}}{1.798 \times 10^{-5}\,\text{m}^2/\text{s}} = 1,113,460.51 \)
02

Calculate Nusselt Number

The Nusselt number can be determined with the following empirical correlation: \(Nu = 0.664\,Re^{1/2}\,Pr^{1/3}\) Here, \(Nu\) represents Nusselt number, and \(Pr\) refers to the Prandtl number. With \(Re = 1,113,460.51\) and \(Pr = 0.7228\), we can calculate the Nusselt number as: \(Nu = 0.664 \cdot (1,113,460.51)^{1/2} \cdot (0.7228)^{1/3} = 375.27\)
03

Calculate Convective Heat Transfer Coefficient

Now, we can find the convective heat transfer coefficient, \(h\), by using the formula: \(h = \frac{k \cdot Nu}{L}\) where \(k\) is the thermal conductivity of air. For this problem, \(k = 0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). So, the convective heat transfer coefficient is: \(h = \frac{0.02735\,\text{W/mK} \cdot 375.27}{4\,\text{m}} = 2.58\,\text{W/m}^{2}\cdot \mathrm{K}\)
04

Calculate Rate of Heat Transfer

Finally, we can calculate the rate of heat transfer using the convective heat transfer coefficient and given surface temperature and dimensions. The rate of heat transfer, \(Q\), is given by: \(Q = h \cdot A \cdot \Delta T\) where \(A\) is the surface area of the plate and \(\Delta T\) is the temperature difference between the surface and air. For this problem, \(A = 4\,\text{m} \cdot 3\,\text{m} = 12\,m^2\), and \(\Delta T = (80-20)^{\circ}C = 60^{\circ} \mathrm{C}\). Now we can find the rate of heat transfer: \(Q = 2.58\,\text{W/m}^{2}\mathrm{K} \cdot 12\,\text{m}^2 \cdot 60^{\circ} \mathrm{C} = 8985\,\text{W}\) Based on our calculations, the rate of heat transfer from the surface is 8985 W, which corresponds to option (b).

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Key Concepts

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

Reynolds Number
In the realm of fluid dynamics, the Reynolds number is pivotal in determining the flow characteristics over surfaces. It is a dimensionless quantity that helps predict whether the flow will be laminar or turbulent. The formula for calculating the Reynolds number is \(Re = \frac{V \cdot L}{u}\). Here, \(V\) stands for fluid velocity, \(L\) represents the characteristic length (in this exercise, it's the length of the plate), and \(u\) is the kinematic viscosity of the fluid.
A higher Reynolds number signifies a more turbulent flow, while a lower value indicates laminar flow. For example, in the provided exercise, the Reynolds number calculated was \(1,113,460.51\), suggesting a turbulent flow over the plate. It's essential to compute this number accurately since it aids in determining other factors like drag coefficient and heat transfer rates.
Nusselt Number
The Nusselt number is another important dimensionless parameter in convective heat transfer. It expresses the ratio between the convective and conductive heat transfer across a boundary. The formula utilized in the exercise is an empirical one: \(Nu = 0.664\,Re^{1/2}\,Pr^{1/3}\). This calculation involves the Reynolds number, further connecting it to the flow dynamics, alongside the Prandtl number (\(Pr\)), which describes fluid flow properties.
In this context, a higher Nusselt number means more effective convective heat transfer across the plate. It indicates the enhancement of thermal energy transfer due to the flow. For our case, \(Nu = 375.27\) was calculated, showcasing efficient heat transfer from the surface, thanks to turbulent flow and favorable thermal properties of the air.
Kinematic Viscosity
Kinematic viscosity \((u)\) is a measure of a fluid's resistance to flow and shear under gravitational forces. It is essentially dynamic viscosity divided by the fluid's density, providing insight into the fluid's pouring and spreading behavior. In the equation \(Re = \frac{V \cdot L}{u}\), kinematic viscosity plays a crucial role in determining the Reynolds number.
Specifically, in air at \(20^{\circ}C\), the kinematic viscosity is given as \(1.798 \times 10^{-5} \,\text{m}^{2}/\text{s}\). This low value suggests that air flows relatively easily, resulting in lower internal resistance to flow. Kinematic viscosity is vital in determining flow regimes and is a constant factor in planning and predicting heat transfer scenarios.
Thermal Conductivity of Air
Thermal conductivity \((k)\) refers to the ability of a material, in this case air, to conduct heat. It is measured in \(\text{W/mK}\) and is crucial for designing and analyzing heat transfer applications. In the given exercise, air's thermal conductivity is \(0.02735 \,\text{W/mK}\).
Thermal conductivity plays a key role when calculating the convective heat transfer coefficient \(h\), using the formula \(h = \frac{k \cdot Nu}{L}\). This coefficient helps quantify the heat transfer rate from the plate surface to air flowing over it. Air's relatively low thermal conductivity denotes that it is not the best conductor of heat; however, when combined with proper surface area and temperature gradient, it can facilitate significant convective heat transfer, as seen in the exercise's outcome.

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Most popular questions from this chapter

What is drag? What causes it? Why do we usually try to minimize it?

Air at \(20^{\circ} \mathrm{C}(1 \mathrm{~atm})\) is flowing over a \(5-\mathrm{cm}\) diameter sphere with a velocity of \(3.5 \mathrm{~m} / \mathrm{s}\). If the surface temperature of the sphere is constant at \(80^{\circ} \mathrm{C}\), determine \((a)\) the average drag coefficient on the sphere and \((b)\) the heat transfer rate from the sphere.

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: \(k=\) \(0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=\) \(\left.1.872 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{\mathrm{s}}=1.729 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

Why is flow separation in flow over cylinders delayed in turbulent flow?

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?

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