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Chapter 1: Problem 17

Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of water that will cover 1 acre of land to a depth of \(1 \mathrm{ft}\). A severe thunderstorm dumped \(2.0\) in. of rain in \(30 \mathrm{~min}\) on a town of area \(26 \mathrm{~km}^{2}\). What volume of water, in acre-feet, fell on the town?

Short Answer

Expert verified
1,073.33 acre-feet

Step by step solution

01

Convert area to acres

First convert the town's area from \(26 \ \text{km}^2\) to acres. Use the conversion factor \(1\ \text{km}^2 \approx 247.105 \ \text{acres}\). Multiply the town's area by this conversion factor: \[26 \ \text{km}^2 \times 247.105 \ \text{acres/km}^2 = 6,424.73 \ \text{acres}\]
02

Convert rainfall depth to feet

Convert the depth of rainfall from inches to feet. Use the conversion factor \(1 \ \text{inch} = \frac{1}{12} \ \text{feet}\). Multiply the depth of rainfall by this conversion factor: \[2.0 \ \text{inches} \times \ \frac{1}{12} \ \text{feet/inch} = \frac{2.0}{12} \ \text{feet} = 0.167 \ \text{feet}\]
03

Calculate volume in acre-feet

Now, calculate the volume of water that fell on the town by multiplying the area in acres by the depth of the rain in feet: \[6,424.73 \ \text{acres} \times 0.167 \ \text{feet} = 1,073.33 \ \text{acre-feet}\]

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Key Concepts

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

Introduction to Unit Conversion
Unit conversion is a critical tool in physics and engineering. It helps in transforming measurements from one unit to another, ensuring consistency and accuracy. For instance, in the given exercise, we convert kilometers squared (km虏) to acres and inches to feet. These conversions facilitate accurate rainfall volume calculations.
Remember, conversion factors are crucial here. For area, we use: 1 km虏 鈮� 247.105 acres. For length, we use: 1 inch = 1/12 feet. Engaging with unit conversions not only enhances your problem-solving skills but also aids in grasping multiple disciplines seamlessly.
Basics of Hydraulic Engineering
Hydraulic engineering focuses on the movement and storage of water. In this field, precise volume calculations are vital. Engineers may use units like acre-feet to describe large volumes of water, which makes managing resources more practical. An acre-foot is the volume needed to cover one acre to a depth of one foot.
Such units simplify the calculation of water volumes in large-scale applications. Like in our problem, determining how much rainfall collected over a large area can be efficiently solved using these principles. Understanding these basics helps in tackling real-world hydraulic problems effectively.
Understanding Volume Calculation
Volume calculation in physics involves determining the space occupied within a bounded area. In the exercise, this involves converting atmospheric precipitation's depth (in inches) over a defined area (in km虏) into acre-feet. Here's a step-by-step refresher:
1. **Convert Area**: By converting the area from km虏 to acres, we get units that are more manageable for water volume calculations.
2. **Convert Depth**: Transforming rainfall depth from inches to feet helps in utilizing easily convertible consistent units.
3. **Calculate Volume**: Finally, multiplying the area in acres and depth in feet yields the volume in acre-feet.
This logical approach ensures accurate results and is a valuable skill in various fields, from environmental science to urban planning.

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Most popular questions from this chapter

Assuming the length of the day uniformly increases by \(0.0010 \mathrm{~s}\) per century, calculate the cumulative effect on the measure of time over 20 centuries. (Such slowing of Earth's rotation is indicated by observations of the occurrences of solar eclipses during this period.)

A cubic centimeter in a typical cumulus cloud contains 50 to 500 water droplets, which have a typical radius of \(10 \mu \mathrm{m} .\) (a) How many cubic meters of water are in a cylindrical cumulus cloud of height \(3.0 \mathrm{~km}\) and radius \(1.0 \mathrm{~km} ?\) (b) How many 1-liter pop bottles would that water fill? (c) Water has a mass per unit volume (or density) of \(1000 \mathrm{~kg} / \mathrm{m}^{3}\). How much mass does the water in the cloud have?

Until 1883 , every city and town in the United States kept its own local time. Today, travelers reset their watches only when the time change equals \(1.0 \mathrm{~h}\). How far, on the average, must you travel in degrees of longitude until your watch must be reset by \(1.0 \mathrm{~h}\) ? (Hint: Earth rotates \(360^{\circ}\) in about 24 h.)

Fermi Physicist Enrico Fermi once pointed out that a standard lecture period (50 min) is close to 1 microcentury. (a) How long is a microcentury in minutes? (b) Using percentage difference \(=\left(\frac{\text { actual }-\text { approximation }}{\text { actual }}\right) 100\) find the percentage difference from Fermi's approximation.

Gold, which has a mass of \(19.32 \mathrm{~g}\) for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If \(1.000 \mathrm{oz}\) of gold, with a mass of \(27.63 \mathrm{~g}\), is pressed into a leaf of \(1.000 \mu \mathrm{m}\) thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius \(2.500 \mu \mathrm{m}\), what is the length of the fiber?
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