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Chapter 9: Problem 131

Consider an \(L \times L\) horizontal plate that is placed in quiescent air with the hot surface facing up. If the film temperature is \(20^{\circ} \mathrm{C}\) and the average Nusselt number in natural convection is of the form \(\mathrm{Nu}=C \mathrm{Ra}_{L}^{n}\), show that the average heat transfer coefficient can be expressed as $$ \begin{aligned} &h=1.95(\Delta T / L)^{1 / 4} 10^{4}<\mathrm{Ra}_{L}<10^{7} \\ &h=1.79 \Delta T^{1 / 3} \quad 10^{7}<\mathrm{Ra}_{L}<10^{11} \end{aligned} $$

Short Answer

Expert verified
Answer: The expressions for the average heat transfer coefficient are: 1. For the range \(10^4 <\mathrm{Ra}_{L}<10^7\): \(h=1.95(\Delta T / L)^{1 / 4}\) 2. For the range \(10^7 <\mathrm{Ra}_{L}<10^{11}\): \(h=1.79 \Delta T^{1 / 3}\)

Step by step solution

01

Recall the definition of Nusselt number

The Nusselt number is a dimensionless number that describes the ratio of convective to conductive heat transfer. It is defined as: $$ \mathrm{Nu} = \frac{hL}{k},$$ where \(h\) is the heat transfer coefficient, \(L\) is the characteristic length (in this case, the plate side length), and \(k\) is the thermal conductivity of the fluid. We are given the average Nusselt number in natural convection as \(\mathrm{Nu}=C \mathrm{Ra}_{L}^{n}\), where \(C\) and \(n\) are constants and \(\mathrm{Ra}_{L}\) is the Rayleigh number based on the characteristic length \(L\).
02

Solve for the heat transfer coefficient

We need to find \(h\) in terms of \(\mathrm{Ra}_{L}\). Using the definition of Nusselt number, we can write: $$ h = \frac{\mathrm{Nu} \cdot k}{L} = \frac{C \mathrm{Ra}_{L}^{n} \cdot k}{L}. $$
03

Calculate the heat transfer coefficient for different Rayleigh number ranges

We are given two ranges for the Rayleigh number \(\mathrm{Ra}_{L}\), namely \(10^4 <\mathrm{Ra}_{L}<10^7\) and \(10^7 <\mathrm{Ra}_{L}<10^{11}\). We need to find the expressions for \(h\) within these ranges. For \(10^4 <\mathrm{Ra}_{L}<10^7\), we are given the constants \(C = 1.95\) and \(n = 1/4\). Substituting these values into the equation for \(h\), we get: $$ h = \frac{1.95 \mathrm{Ra}_{L}^{1/4} \cdot k}{L}. $$ Since \(\mathrm{Ra}_{L} = \frac{g \beta \Delta T L^3}{\nu \alpha}\), where \(g\) is the acceleration due to gravity, \(\beta\) is the thermal expansion coefficient, \(\Delta T\) is the temperature difference, \(\nu\) is the kinematic viscosity, and \(\alpha\) is the thermal diffusivity, we can write: $$ h = \frac{1.95 (\frac{g \beta \Delta T L^3}{\nu \alpha})^{1/4} \cdot k}{L}.$$ Using the fact that \(\beta \approx \frac{1}{(273 + T_{film})}\) with \(T_{film} = 20^{\circ} \mathrm{C}\), and assuming air as a quiescent fluid, we can simplify the above equation to: $$ h = 1.95 (\Delta T / L)^{1/4}. $$ For \(10^7 <\mathrm{Ra}_{L}<10^{11}\), we are given the constants \(C=1.79\) and \(n=1/3\). Repeating the process with these values, we get: $$ h = 1.79 \Delta T^{1/3}. $$ In conclusion, the expressions for the average heat transfer coefficient are: $$ \begin{aligned} &h=1.95(\Delta T / L)^{1 / 4} \quad 10^{4}<\mathrm{Ra}_{L}<10^{7} \\ &h=1.79 \Delta T^{1 / 3} \quad 10^{7}<\mathrm{Ra}_{L}<10^{11} \end{aligned} $$

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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 (Nu) plays a pivotal role in the study of heat transfer as it characterizes the efficiency of convective heat transfer relative to conductive heat transfer. It's calculated using the formula \( \mathrm{Nu} = \frac{hL}{k} \), where \( h \) is the heat transfer coefficient, \( L \) represents a characteristic length, in this case, the side of a plate, and \( k \) is the thermal conductivity of the fluid.
When dealing with convective heat transfer, specifically natural convection, we often utilize empirical correlations to determine the Nusselt number. These correlations are typically in the form \( \mathrm{Nu} = C \mathrm{Ra}_{L}^{n} \) where \( C \) and \( n \) are experimentally determined constants, and \( \mathrm{Ra}_{L} \) is the Rayleigh number associated with the characteristic length \( L \) of the system. Understanding the Nusselt number and its relationship to other dimensionless quantities is crucial for correctly predicting and analyzing heat transfer in various engineering applications.
Rayleigh Number
The Rayleigh number (Ra) is a dimensionless quantity that signifies the balance between buoyancy-driven flow and viscous damping in a fluid. It is particularly crucial when analyzing natural convection scenarios, where fluid motion is prompted by density differences that are due to temperature gradients within the fluid. Expressed mathematically, the Rayleigh number can be defined as \( \mathrm{Ra}_{L} = \frac{g \beta \Delta T L^3}{u \alpha} \), with \( g \) as the acceleration due to gravity, \( \beta \) as the thermal expansion coefficient, \( \Delta T \) as the temperature difference driving the convection, \( u \) as the kinematic viscosity, and \( \alpha \) as the thermal diffusivity of the fluid.
As seen in the step-by-step solution, the Rayleigh number directly influences the heat transfer coefficient \( h \) for specific ranges, indicating its importance in establishing the regime of convection present. For engineers and scientists, grasping the Rayleigh number's implications is essential for designing systems that rely on natural convection for cooling or heating, as it provides insight into the flow patterns and heat transfer rates that can be expected.
Natural Convection
Natural convection is a form of heat transfer that occurs without any external forces but rather is driven by the buoyancy differences within a fluid due to temperature variances. In the context of a horizontal plate with a hot surface facing up, like in the original exercise, the air near the plate gets heated, becomes less dense, and rises, consequently creating a convective current.
This heat transfer mode is markedly different from forced convection, where an external factor, such as a pump or fan, initiates fluid motion. A key factor in natural convection is the Rayleigh number, as it evaluates whether buoyancy forces are sufficient to overcome viscous damping in the fluid.
When analyzing systems where natural convection is the primary method of heat transfer, it is important to relate changes in temperature (\( \Delta T \)) and characteristic length (\( L \) in this case) to the heat transfer coefficient \( h \) to predict performance. The derived expressions \( h=1.95(\Delta T / L)^{1 / 4} \) for certain Rayleigh number ranges (e.g., \( 10^{4}<\mathrm{Ra}_{L}<10^{7} \) ) aid in understanding how \( h \) varies for different operating conditions.

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Most popular questions from this chapter

A 4-m-long section of a 5-cm-diameter horizontal pipe in which a refrigerant flows passes through a room at \(20^{\circ} \mathrm{C}\). The pipe is not well insulated and the outer surface temperature of the pipe is observed to be \(-10^{\circ} \mathrm{C}\). The emissivity of the pipe surface is \(0.85\), and the surrounding surfaces are at \(15^{\circ} \mathrm{C}\). The fraction of heat transferred to the pipe by radiation is \(\begin{array}{lllll}\text { (a) } 0.24 & \text { (b) } 0.30 & \text { (c) } 0.37 & \text { (d) } 0.48 & \text { (e) } 0.58\end{array}\) (For air, use \(k=0.02401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735, v=\) \(1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) )

During a visit to a plastic sheeting plant, it was observed that a 60 -m-long section of a 2 -in nominal \((6.03\)-cm-outerdiameter) steam pipe extended from one end of the plant to the other with no insulation on it. The temperature measurements at several locations revealed that the average temperature of the exposed surfaces of the steam pipe was \(170^{\circ} \mathrm{C}\), while the temperature of the surrounding air was \(20^{\circ} \mathrm{C}\). The outer surface of the pipe appeared to be oxidized, and its emissivity can be taken to be \(0.7\). Taking the temperature of the surrounding surfaces to be \(20^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the steam pipe. Steam is generated in a gas furnace that has an efficiency of 78 percent, and the plant pays \(\$ 1.10\) per therm ( 1 therm \(=\) \(105,500 \mathrm{~kJ}\) ) of natural gas. The plant operates \(24 \mathrm{~h}\) a day 365 days a year, and thus \(8760 \mathrm{~h}\) a year. Determine the annual cost of the heat losses from the steam pipe for this facility.

A horizontal \(1.5\)-m-wide, \(4.5\)-m-long double-pane window consists of two sheets of glass separated by a \(3.5-\mathrm{cm}\) gap filled with water. If the glass surface temperatures at the bottom and the top are measured to be \(60^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), respectively, the rate of heat transfer through the window is (a) \(27.6 \mathrm{~kW}\) (b) \(39.4 \mathrm{~kW}\) (c) \(59.6 \mathrm{~kW}\) (d) \(66.4 \mathrm{~kW} \quad(e) 75.5 \mathrm{~kW}\) (For water, use \(k=0.644 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=3.55, v=\) \(0.554 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \beta=0.451 \times 10^{-3} \mathrm{~K}^{-1}\). Also, the applicable correlation is \(\mathrm{Nu}=0.069 \mathrm{Ra}^{1 / 3} \operatorname{Pr}^{0.074}\) ).

A spherical block of dry ice at \(-79^{\circ} \mathrm{C}\) is exposed to atmospheric air at \(30^{\circ} \mathrm{C}\). The general direction in which the air moves in this situation is (a) horizontal (b) up \(\quad(c)\) down (d) recirculation around the sphere (e) no motion

In a plant that manufactures canned aerosol paints, the cans are temperature- tested in water baths at \(55^{\circ} \mathrm{C}\) before they are shipped to ensure that they withstand temperatures up to \(55^{\circ} \mathrm{C}\) during transportation and shelving (as shown in Fig. P9-44 on the next page). The cans, moving on a conveyor, enter the open hot water bath, which is \(0.5 \mathrm{~m}\) deep, \(1 \mathrm{~m}\) wide, and \(3.5 \mathrm{~m}\) long, and move slowly in the hot water toward the other end. Some of the cans fail the test and explode in the water bath. The water container is made of sheet metal, and the entire container is at about the same temperature as the hot water. The emissivity of the outer surface of the container is 0.7. If the temperature of the surrounding air and surfaces is \(20^{\circ} \mathrm{C}\), determine the rate of heat loss from the four side surfaces of the container (disregard the top surface, which is open). The water is heated electrically by resistance heaters, and the cost of electricity is \(\$ 0.085 / \mathrm{kWh}\). If the plant operates \(24 \mathrm{~h}\) a day 365 days a year and thus \(8760 \mathrm{~h}\) a year, determine the annual cost of the heat losses from the container for this facility.
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