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Chapter 9: Problem 6

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: In a fluid that doesn't change volume with temperature at constant pressure, the density remains nearly constant even when the temperature changes. As natural convection heat transfer relies on density variations to create convection currents, a fluid with this behavior will have significantly reduced or even nonexistent natural convection heat transfer.

Step by step solution

01

Understand the behavior of natural convection heat transfer in fluids

Natural convection heat transfer occurs when a fluid experiences a temperature difference, which results in density variations throughout the fluid. The density differences cause the fluid to move and form convection currents, leading to heat transfer within the fluid medium.
02

Analyze the fluid's properties provided in the exercise

In this case, the volume of the fluid does not change with temperature at constant pressure. This implies that the fluid's density remains nearly constant even when its temperature changes.
03

Relate volume change to density variation and its effect on natural convection heat transfer

Since fluid volume doesn't change with temperature variations under constant pressure, the fluid's density doesn't change much either. In other words, the change in the buoyancy force caused by temperature fluctuations is negligible. Because natural convection relies on buoyancy forces to create convection currents and promote heat transfer, we can conclude that natural convection heat transfer will be very limited, or even negligible, in this fluid medium.
04

Conclusion

For a fluid whose volume doesn't change with temperature at constant pressure, natural convection heat transfer will be significantly reduced or even nonexistent. This is because the absence of a significant density difference in the fluid, which is necessary for driving natural convection currents and thus, heat transfer.

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Most popular questions from this chapter

Why are the windows considered in three regions when analyzing heat transfer through them? Name those regions, and explain how the overall \(U\)-value of the window is determined when the heat transfer coefficients for all three regions are known.

A vertical 4-ft-high and 6-ft-wide double-pane window consists of two sheets of glass separated by a 1 -in air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be $65^{\circ} \mathrm{F}\( and \)40^{\circ} \mathrm{F}$, determine the rate of heat transfer through the window by \((a)\) natural convection and (b) radiation. Also, determine the \(R\)-value of insulation of this window such that multiplying the inverse of the \(R\)-value by the surface area and the temperature difference gives the total rate of heat transfer through the window. The effective emissivity for use in radiation calculations between two large parallel glass plates can be taken to be \(0.82\).

Consider a \(3-\mathrm{m}\)-high rectangular enclosure consisting of two surfaces separated by a \(0.1-\mathrm{m}\) air gap at \(1 \mathrm{~atm}\). If the surface temperatures across the air gap are \(30^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\), determine the ratio of the heat transfer rate for the horizontal orientation (with hotter surface at the bottom) to that for vertical orientation.

Write a computer program to evaluate the variation of temperature with time of thin square metal plates that are removed from an oven at a specified temperature and placed vertically in a large room. The thickness, the size, the initial temperature, the emissivity, and the thermophysical properties of the plate as well as the room temperature are to be specified by the user. The program should evaluate the temperature of the plate at specified intervals and tabulate the results against time. The computer should list the assumptions made during calculations before printing the results. For each step or time interval, assume the surface temperature to be constant and evaluate the heat loss during that time interval and the temperature drop of the plate as a result of this heat loss. This gives the temperature of the plate at the end of a time interval, which is to serve as the initial temperature of the plate for the beginning of the next time interval. Try your program for \(0.2-\mathrm{cm}\)-thick vertical copper plates of $40 \mathrm{~cm} \times 40 \mathrm{~cm}\( in size initially at \)300^{\circ} \mathrm{C}\( cooled in a room at \)25^{\circ} \mathrm{C}$. Take the surface emissivity to be \(0.9\). Use a time interval of \(1 \mathrm{~s}\) in calculations, but print the results at \(10-\mathrm{s}\) intervals for a total cooling period of \(15 \mathrm{~min}\).

A vertical \(0.9\)-m-high and \(1.5\)-m-wide double-pane window consists of two sheets of glass separated by a \(2.0-\mathrm{cm}\) air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be \(20^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), the rate of heat transfer through the window is (a) \(16.3 \mathrm{~W}\) (b) \(21.7 \mathrm{~W}\) (c) \(24.0 \mathrm{~W}\) $\begin{array}{ll}\text { (d) } 31.3 \mathrm{~W} & \text { (e) } 44.6 \mathrm{~W}\end{array}$ (For air, use $k=0.02551 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \operatorname{Pr}=0.7296\(, \)\nu=1.562 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\(. Also, the applicable correlation is \)\left.\mathrm{Nu}=0.42 \mathrm{Ra}^{1 / 4} \operatorname{Pr}^{0.012}(H / L)^{-0.3} .\right)$
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