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Mercury at \(25^{\circ} \mathrm{C}\) flows over a 3 -m-long and \(2-\mathrm{m}\)-wide flat plate maintained at \(75^{\circ} \mathrm{C}\) with a velocity of \(0.8 \mathrm{~m} / \mathrm{s}\). Determine the rate of heat transfer from the entire plate.

Short Answer

Expert verified
Answer: The rate of heat transfer from the flat plate to the mercury flow is 210,000 W.

Step by step solution

01

Calculate the temperature difference (ΔT)

Given the temperature of the mercury \(T_{\text{mercury}}=25^{\circ} \mathrm{C}\) and the temperature of the plate \(T_{\text{plate}}=75^{\circ} \mathrm{C}\), we first need to calculate the temperature difference as follows: $$ \Delta T = T_{\text{plate}} - T_{\text{mercury}} = 75 - 25 = 50 ^{\circ} \mathrm{C} $$
02

Estimate the convective heat transfer coefficient (h)

To estimate the convective heat transfer coefficient (h) for an external flow over a flat plate, one would usually calculate the Reynolds number (Re) to determine the flow conditions, use empirical correlations to calculate the Nusselt number (Nu), and then use the relation between h, Nu, and the thermal conductivity (k) of the fluid to find the value of h. However, for simplicity, we will use a previous published value of \(h \approx 700 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) for mercury flow over a flat plate.
03

Calculate the heat transfer rate per unit area (q')

Using the convective heat transfer equation, we can calculate the heat transfer rate per unit area (q'): $$ q' = h \cdot \Delta T = 700 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \cdot 50 ^{\circ} \mathrm{C} \\ q' = 35000 \frac{\mathrm{W}}{\mathrm{m}^2} $$
04

Calculate the surface area (A) of the flat plate

The dimensions of the flat plate are given as length \(L = 3\,\mathrm{m}\) and width \(W = 2\,\mathrm{m}\). The surface area (A) can be calculated as follows: $$ A = L \cdot W = 3 \, \mathrm{m} \cdot 2\, \mathrm{m} = 6\, \mathrm{m}^2 $$
05

Determine the total rate of heat transfer (Q)

Once we have calculated q' and the surface area of the flat plate, we can calculate the total rate of heat transfer (Q) from the plate as follows: $$ Q = q' \cdot A = 35000 \frac{\mathrm{W}}{\mathrm{m}^2} \cdot 6\, \mathrm{m}^2 = 210000\, \mathrm{W} $$ The rate of heat transfer from the entire plate to the mercury flow is \(210,000\, \mathrm{W}\).

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Key Concepts

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

Convection
Convection occurs when heat is transferred due to the actual movement of fluid molecules from one place to another.
This is a fundamental process observed in fluids, such as liquids and gases, where warmer parts of the fluid rise and cooler parts sink, creating a circulation pattern.
In the given problem, mercury at a lower temperature flows over a heated plate.
As mercury absorbs heat from the surface of the plate, convection currents are established. These currents facilitate the transfer of thermal energy from the plate to the mercury flow.

The nature of convection can be either forced, like with fans and pumps, or natural/generalized, driven by differential buoyancy forces due to temperature variations.
In this exercise, the mercury flow is a form of forced convection, where an imposed velocity causes the movement of fluid over the plate.
Thermal Conductivity
Thermal conductivity is a material property that indicates how well a material can conduct heat.
It is represented by the symbol \( k \) and has units of \( \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \).
A material with high thermal conductivity can transfer heat efficiently, while materials with low thermal conductivity act as insulators.

In the context of the given problem, mercury has a specific thermal conductivity value that influences how heat is transferred when it flows over the heated plate.
This property is crucial when calculating the Nusselt Number, which eventually helps to determine the convective heat transfer coefficient.
While we used a previously published value for the heat transfer coefficient in the exercise, understanding thermal conductivity is essential for empirical assessments in similar scenarios.
Reynolds Number
Reynolds Number, denoted as \( \mathrm{Re} \), is a dimensionless number used to predict flow regimes in fluid dynamics, including laminar or turbulent flow.
It is calculated as \( \mathrm{Re} = \frac{\rho VD}{\mu} \), where \( \rho \) is the fluid density, \( V \) is the flow velocity, \( D \) is a characteristic length (often diameter), and \( \mu \) is the dynamic viscosity of the fluid.

For flat plate scenarios, the length \( L \) can be used as the characteristic length.
Calculating the Reynolds Number helps determine which correlations to use when computing the Nusselt Number, which subsequently allows determining the convective heat transfer coefficient.
In our case, although we utilized an estimated value for the heat transfer coefficient directly, understanding \( \mathrm{Re} \) is crucial for more detailed analyses in fluid systems.
Nusselt Number
The Nusselt Number, symbolized as \( \mathrm{Nu} \), is a dimensionless number that provides a measure of the convective heat transfer relative to conductive heat transfer across a boundary.
It is given by \( \mathrm{Nu} = \frac{hL}{k} \), where \( h \) is the convective heat transfer coefficient, \( L \) is the characteristic length, and \( k \) is the thermal conductivity of the fluid.

A larger value of the Nusselt Number indicates that convection is dominating over conduction, suggesting efficient heat transfer through fluid motion relative to heat transfer by molecular diffusion.
In the problem, calculating the Nusselt Number using empirical correlations that factor in the Reynolds Number typically helps obtain the heat transfer coefficient \( h \).
This process is fundamental in applying theoretical understanding to practical heat transfer tasks like the one presented.
Heat Transfer Coefficient
The heat transfer coefficient, \( h \), is a parameter that quantifies the heat transfer rate per unit area per unit temperature difference across a boundary layer.
This coefficient is crucial in engineering contexts to design and analyze systems involving thermal transport, such as exchangers and electronic cooling.
In this exercise, the heat transfer coefficient for mercury flowing over the flat plate was given as \( 700 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K} \).

The value of \( h \) can be influenced by several factors, including fluid velocity, surface roughness, and the temperature difference between the fluid and the surface.
The calculation of \( h \) is linked directly to understanding the nature of fluid flow characterized by the Reynolds and Nusselt numbers.
Having an accurate value for \( h \) allows the determination of the total rate of heat transfer, which is crucial for ensuring system efficiency and performance in real-world applications.

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Most popular questions from this chapter

A \(0.2 \mathrm{~m} \times 0.2 \mathrm{~m}\) street sign surface has an absorptivity of \(0.6\) and an emissivity of \(0.7\), while the street sign is subjected to a cross flow wind at \(20^{\circ} \mathrm{C}\) with a velocity of \(1 \mathrm{~m} / \mathrm{s}\). Solar radiation is incident on the street sign at a rate of \(1100 \mathrm{~W} / \mathrm{m}^{2}\), and the surrounding temperature is \(20^{\circ} \mathrm{C}\). Determine the surface temperature of the street sign. Evaluate the air properties at \(30^{\circ} \mathrm{C}\). Treat the sign surface as a vertical plate in cross flow.

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

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, 1.7-m-long smooth pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. The drag force exerted on the pipe by the air is (a) \(0.4 \mathrm{~N}\) (b) \(1.1 \mathrm{~N}\) (c) \(8.5 \mathrm{~N}\) (d) \(13 \mathrm{~N}\) (e) \(18 \mathrm{~N}\) (For air, use \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=1.269 \mathrm{~kg} / \mathrm{m}^{3}\) )

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=0.3 \mathrm{~m}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, W/m² \(\mathbf{K}\) (f) Average convection heat transfer coefficient, W/m² \(\mathrm{K}\) (g) Rate of convective heat transfer, W

Air at 1 atm and \(20^{\circ} \mathrm{C}\) is flowing over the top surface of a \(0.5-\mathrm{m}\)-long thin flat plate. The air stream velocity is \(50 \mathrm{~m} / \mathrm{s}\) and the plate is maintained at a constant surface temperature of \(180^{\circ} \mathrm{C}\). Determine \((a)\) the average friction coefficient, \((b)\) the average convection heat transfer coefficient, and (c) repeat part (b) using the modified Reynolds analogy.

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