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Consider a fluid whose volume does not change with temperature at constant pressure. What can you say about natural convection heat transfer in this medium?

Short Answer

Expert verified
Answer: Natural convection heat transfer is driven by density differences due to temperature gradients within the fluid, leading to buoyancy-driven flow. In this case, although the fluid's volume remains constant, density differences still arise due to temperature changes. Thus, natural convection heat transfer will still occur in this medium, with the driving force being solely due to density differences resulting from temperature changes rather than volume expansion. Heat will be transferred through the fluid, driven by buoyancy forces.

Step by step solution

01

Understanding natural convection heat transfer

Natural convection heat transfer occurs when a fluid is subjected to temperature gradients, leading to density differences within the fluid. These density differences cause the fluid to move due to buoyancy forces, resulting in heat being transferred through the fluid. In most cases, heating a fluid causes it to expand, decreasing its density, and generating buoyancy-driven flow. However, in this case, the fluid's volume remains constant, and we need to investigate the consequences of this property for natural convection heat transfer.
02

Analyzing the impact of constant volume on density

Since the fluid has a constant volume at constant pressure, regardless of temperature changes, we can use the ideal gas law to understand the relation between its temperature and density. The ideal gas law is: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. Since the volume V and the pressure P are constant for our fluid, the following relation holds true: rho = n/V = P/(RT) where rho is the density. In this scenario, since the volume does not change with temperature, the density of the fluid is only a function of its temperature.
03

Investigating the buoyancy-driven flow

Buoyancy-driven flow is a direct result of density differences in a fluid. When a fluid element with a lower density is surrounded by fluid elements with higher density, it experiences a net upward force due to buoyancy. In our case, since the density of the fluid is dependent on temperature, the buoyancy-driven flow will still occur when there is a temperature gradient within the fluid.
04

Identifying the effect on natural convection heat transfer

Since the fluid's volume remains constant, and density differences still arise due to temperature gradients, we can conclude that natural convection heat transfer will still occur in this medium. The difference lies in the fact that the driving force for natural convection in this case is solely due to density differences resulting from temperature changes, rather than due to volume expansion as well. Nonetheless, heat will still be transferred through the fluid, driven by buoyancy forces.

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Key Concepts

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

Buoyancy Forces
Buoyancy forces are the upward forces that keep objects afloat in a fluid, and they play a fundamental role in natural convection heat transfer. These forces arise due to differences in fluid density: a region of fluid that is less dense than the surrounding fluid will experience a net upward force. This is essentially Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. In the context of natural convection, when part of a fluid is heated, it expands and becomes less dense. This portion of fluid then rises through the cooler, denser fluid. However, as noted in our exercise, if the fluid's volume does not change with temperature, the expansion does not occur. Despite this, temperature gradients can still cause density changes, and thus, buoyancy forces are present to create motion within the fluid, leading to heat transfer.
Ideal Gas Law
The ideal gas law is a fundamental equation that helps us comprehend the behavior of gases under various conditions. It is represented by the equation \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles, \(R\) for the universal gas constant, and \(T\) for temperature. From this equation, we see the interrelation between these variables. When considering a scenario where a fluid's volume is constant, as the temperature (\(T\)) increases, the pressure (\(P\)) or the number of moles (\(n\)) must change to maintain the balance if the gas follows this law closely. This implies a direct connection between density (\(\rho = \frac{n}{V}\)) and temperature, which is crucial for predicting the behavior of natural convection heat transfer in gaseous mediums.
Density Differences in Fluids
Density differences in fluids are the reason for buoyancy and the driving factor behind natural convection. The density (\(\rho\)) of a fluid is its mass per unit volume, and it can change with temperature variations if the fluid is compressible or due to thermal expansion in the case of liquids. In the exercise, the fluid's volume does not change with temperature at a constant pressure, unlike most fluids that expand upon heating. This unique property still leads to density variations solely based on temperature changes, since hotter fluid becomes lighter. These changes in density mean that different parts of the fluid will have varying buoyancy forces acting upon them, ultimately causing convection currents and enabling heat transfer.
Temperature Gradients
Temperature gradients are variations in temperature over a distance within a material. These gradients are the driving force of natural convection, as they give rise to density gradients in the fluid. When a part of the fluid is warmer than its surroundings, a temperature gradient is established. This gradient can lead to the movement of fluid particles from a hotter region to a cooler one, creating a convective flow. In the case of the fluid in our exercise, even though the volume does not change, the temperature gradients will still result in density variations that can cause convection currents. This means that despite the fluid's unique behavior, as long as there are temperature gradients, natural convection can facilitate heat transfer, reinforcing the importance of understanding both the thermal and physical properties of fluids in thermal systems.

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Most popular questions from this chapter

Consider a double-pane window whose air space width is \(20 \mathrm{~mm}\). Now a thin polyester film is used to divide the air space into two 10-mm-wide layers. How will the film affect \((a)\) convection and (b) radiation heat transfer through the window?

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 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 \(k=0.02401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735, v=\) \(1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) )

A 4-m-diameter spherical tank contains iced water at \(0^{\circ} \mathrm{C}\). The tank is thin-shelled and thus its outer surface temperature may be assumed to be same as the temperature of the iced water inside. Now the tank is placed in a large lake at \(20^{\circ} \mathrm{C}\). The rate at which the ice melts is (a) \(0.42 \mathrm{~kg} / \mathrm{s}\) (b) \(0.58 \mathrm{~kg} / \mathrm{s}\) (c) \(0.70 \mathrm{~kg} / \mathrm{s}\) (d) \(0.83 \mathrm{~kg} / \mathrm{s}\) (e) \(0.98 \mathrm{~kg} / \mathrm{s}\) (For lake water, use \(k=0.580 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=9.45, v=\) \(0.1307 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \beta=0.138 \times 10^{-3} \mathrm{~K}^{-1}\) )

Consider a \(15-\mathrm{cm} \times 20\)-cm printed circuit board \((\mathrm{PCB})\) that has electronic components on one side. The board is placed in a room at \(20^{\circ} \mathrm{C}\). The heat loss from the back surface of the board is negligible. If the circuit board is dissipating \(8 \mathrm{~W}\) of power in steady operation, determine the average temperature of the hot surface of the board, assuming the board is \((a)\) vertical, \((b)\) horizontal with hot surface facing up, and (c) horizontal with hot surface facing down. Take the emissivity of the surface of the board to be \(0.8\) and assume the surrounding surfaces to be at the same temperature as the air in the room. Evaluate air properties at a film temperature of \(32.5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

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