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Chapter 4: Problem 181

Citrus trees are very susceptible to cold weather, and extended exposure to subfreezing temperatures can destroy the crop. In order to protect the trees from occasional cold fronts with subfreezing temperatures, tree growers in Florida usually install water sprinklers on the trees. When the temperature drops below a certain level, the sprinklers spray water on the trees and their fruits to protect them against the damage the subfreezing temperatures can cause. Explain the basic mechanism behind this protection measure and write an essay on how the system works in practice.

Short Answer

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Question: Explain how water sprinklers help protect citrus trees from subfreezing temperatures. Answer: Water sprinklers help protect citrus trees from subfreezing temperatures by utilizing the latent heat of fusion released during the phase change of water when it freezes. When the water sprayed on the trees starts to freeze and turn into ice, it releases this latent heat, helping to maintain a temperature above freezing point around the tree and its fruits. This prevents damages to the tree due to cold temperatures. This method is most effective when combined with an automatic temperature monitoring system and an adequate water supply. However, it is important to use additional preventive measures for long-term and more severe cold weather conditions.

Step by step solution

01

Introduction

Citrus trees can be severely affected by cold weather, and extended exposure to subfreezing temperatures can cause significant damage to the crop. To protect the trees from extreme cold temperatures, a common countermeasure adopted by citrus tree growers in Florida is to install water sprinklers on the trees. This essay will explain how this protection measure works, the basic mechanism behind it, and how it is implemented in practice.
02

The Science Behind Water Sprinklers

The fundamental principle behind using water sprinklers to protect citrus trees from cold temperature damage is based on the heat released during the phase change of water. When water freezes, it undergoes a phase change from liquid to solid. During this process, it releases a certain amount of heat, known as the latent heat of fusion, which helps to keep the surrounding environment warmer.
03

Water pHase Change

When the temperature drops below the freezing point, the water sprayed on the trees by the sprinklers will start to freeze. As the water freezes and turns into ice, it releases the latent heat of fusion, which is approximately 334 joules per gram. This release of heat will help to maintain a temperature above the freezing point around the tree and its fruits, preventing any damage from the cold.
04

The Sprinkler System

The sprinkler system consists of a network of sprinkler heads mounted on poles or directly on the trees, connected to a water source. Depending on the specific requirements of the tree grove, these sprinkler heads can be designed to provide full coverage to the trees and their fruits. The system ensures that the water is evenly distributed and covers all exposed surfaces of the trees.
05

Activating the Sprinkler System

To prevent unnecessary water usage, the system can be equipped with temperature sensors that will monitor the ambient temperature. When the temperature drops below a predetermined level, typically a few degrees above the freezing point, the sprinkler system will automatically activate, spraying water on the trees and their fruits.
06

Precautions and Limitations

It is essential to maintain an adequate water supply during the protection process to ensure the continuous release of heat around the tree; otherwise, the ice may form without releasing enough heat, and the tree could still be damaged. Moreover, this method is most effective for short-term freezing events, as prolonged exposure to freezing temperatures can still cause damage to the tree structure and the roots.
07

Conclusion

In summary, the water sprinkler system used for protecting citrus trees from cold damage works by utilizing the latent heat of fusion released during the phase change of water. This heat helps to keep the tree and its fruit above freezing temperature, thereby preventing any damage from subfreezing temperatures. It is an effective protection method when properly designed and operated with adequate water supply and monitoring systems. However, this technique should be coupled with other preventive measures for long-term and more severe cold weather conditions.

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Most popular questions from this chapter

A long 35-cm-diameter cylindrical shaft made of stainless steel \(304\left(k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=477 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) comes out of an oven at a uniform temperature of \(400^{\circ} \mathrm{C}\). The shaft is then allowed to cool slowly in a chamber at \(150^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the shaft \(20 \mathrm{~min}\) after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Solve this problem using analytical one-term approximation method (not the Heisler charts).

How can the contamination of foods with microorganisms be prevented or minimized? How can the growth of microorganisms in foods be retarded? How can the microorganisms in foods be destroyed?

When water, as in a pond or lake, is heated by warm air above it, it remains stable, does not move, and forms a warm layer of water on top of a cold layer. Consider a deep lake \(\left(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that is initially at a uniform temperature of \(2^{\circ} \mathrm{C}\) and has its surface temperature suddenly increased to \(20^{\circ} \mathrm{C}\) by a spring weather front. The temperature of the water \(1 \mathrm{~m}\) below the surface 400 hours after this change is (a) \(2.1^{\circ} \mathrm{C}\) (b) \(4.2^{\circ} \mathrm{C}\) (c) \(6.3^{\circ} \mathrm{C}\) (d) \(8.4^{\circ} \mathrm{C}\) (e) \(10.2^{\circ} \mathrm{C}\)

Consider a curing kiln whose walls are made of \(30-\mathrm{cm}-\) thick concrete with a thermal diffusivity of \(\alpha=0.23 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Initially, the kiln and its walls are in equilibrium with the surroundings at \(6^{\circ} \mathrm{C}\). Then all the doors are closed and the kiln is heated by steam so that the temperature of the inner surface of the walls is raised to \(42^{\circ} \mathrm{C}\) and the temperature is maintained at that level for \(2.5 \mathrm{~h}\). The curing kiln is then opened and exposed to the atmospheric air after the steam flow is turned off. If the outer surfaces of the walls of the kiln were insulated, would it save any energy that day during the period the kiln was used for curing for \(2.5 \mathrm{~h}\) only, or would it make no difference? Base your answer on calculations.

The water main in the cities must be placed at sufficient depth below the earth's surface to avoid freezing during extended periods of subfreezing temperatures. Determine the minimum depth at which the water main must be placed at a location where the soil is initially at \(15^{\circ} \mathrm{C}\) and the earth's surface temperature under the worst conditions is expected to remain at \(-10^{\circ} \mathrm{C}\) for a period of 75 days. Take the properties of soil at that location to be \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.4 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Answer: \(7.05 \mathrm{~m}\)
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