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Chapter 9: Problem 161

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 (c) down (d) recirculation around the sphere (e) no motion

Short Answer

Expert verified
a) towards the dry ice sphere b) away from the dry ice sphere c) parallel to the surface of the dry ice sphere d) recirculation around the sphere Answer: d) recirculation around the sphere

Step by step solution

01

Determine the motion of air due to temperature difference

When air is at a different temperature than the surface it is in contact with, heat transfer occurs between the air and the surface. This temperature difference will cause a change in air density. Cold air is denser than warm air, which means that it will naturally sink while warm air rises.
02

Analyze the behavior of air around the dry ice sphere

The dry ice sphere is at -79°C and the atmospheric air is at 30°C. The air around the dry ice will cool down and becomes denser than the surrounding air. This denser air will sink due to gravity, causing the less dense (warm) air to rise.
03

Identify the general direction of air movement

The denser cold air around the dry ice sphere will sink, forcing the warm air above it to rise. This will create a circular motion, with cold air near the dry ice moving downward, and warm air rising above it. Eventually, this motion will lead to a recirculation pattern around the sphere.
04

Choose the correct answer

Given the above analysis on air movement due to temperature difference and the circular motion of air around the dry ice sphere, the correct answer is: (d) recirculation around the sphere

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Most popular questions from this chapter

A hot liquid is poured into a spherical tank with an inner diameter of $3 \mathrm{~m}\( and a wall thickness of \)3 \mathrm{~cm}$. The tank wall is made of a material with a thermal conductivity of $0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The hot liquid in the tank causes the inner surface temperature to be \(100^{\circ} \mathrm{C}\), while the tank outer surface is exposed to air at \(20^{\circ} \mathrm{C}\) and has an emissivity of \(0.35\). Determine the outer surface temperature of the tank. Assume that the properties of air can be evaluated at \(40^{\circ} \mathrm{C}\) and 1 atm pressure. Is this a good assumption?

A hot liquid $\left(c_{p}=1000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( flows at a flow rate of \)0.05 \mathrm{~kg} / \mathrm{s}$ inside a copper pipe with an inner diameter of \(45 \mathrm{~mm}\) and a wall thickness of \(5 \mathrm{~mm}\). At the pipe exit, the liquid temperature decreases by \(10^{\circ} \mathrm{C}\) from its temperature at the inlet. The outer surface of the \(5-\mathrm{m}\)-long copper pipe is black oxidized, which subjects the outer surface to radiation heat transfer. The air temperature surrounding the pipe is \(10^{\circ} \mathrm{C}\). Assuming that the properties of air can be evaluated at \(35^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure, determine the outer surface temperature of the pipe. Is $35^{\circ} \mathrm{C}$ an appropriate film temperature for evaluation of the air properties?

Consider a heat sink with optimum fin spacing. Explain how heat transfer from this heat sink will be affected by \((a)\) removing some of the fins on the heat sink and (b) doubling the number of fins on the heat sink by reducing the fin spacing. The base area of the heat sink remains unchanged at all times.

A can of engine oil with a length of \(150 \mathrm{~mm}\) and a diameter of $100 \mathrm{~mm}$ is placed vertically in the trunk of a car. On a hot summer day, the temperature in the trunk is \(43^{\circ} \mathrm{C}\). If the surface temperature of the can is \(17^{\circ} \mathrm{C}\), determine heat transfer rate from the can surface. Neglect the heat transfer from the ends of the can.

Consider a \(0.2-\mathrm{m}\)-diameter and \(1.8-\mathrm{m}\)-long horizontal cylinder in a room at \(20^{\circ} \mathrm{C}\). If the outer surface temperature of the cylinder is \(40^{\circ} \mathrm{C}\), the natural convection heat transfer coefficient is (a) \(2.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(4.1 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(5.2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(6.1 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
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