/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 (a) Estimate the time it would t... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take if you could divert a moderate size river, flowing at \(5000 \mathrm{m}^{3} / \mathrm{s}\) into the pool?

Short Answer

Expert verified
(a) It would take approximately 22.22 hours to fill the swimming pool using a garden hose delivering 60 L/min. (b) It would take 0.016 seconds to fill the swimming pool by diverting a moderate size river flowing at \(5000 \mathrm{m}^{3} / \mathrm{s}\).

Step by step solution

01

Understand the problem

Read the problem carefully to understand that the capacity of the swimming pool is 80,000 L, and we have to calculate the time it would take to fill it up using two different methods.
02

Convert capacity to a uniform unit

In part (a), the garden hose delivers water at a rate of 60 L/min, and in part (b), the river flows at a rate of \(5000 \mathrm{m}^{3} / \mathrm{s}\). To proceed with the calculation, first, we need to convert the capacity of the swimming pool to a uniform unit. In this case, we will convert the capacity from liters to cubic meters. For that, use the following conversion: 1 L = 0.001 m³. So, \(80,000 \times 0.001 = 80 \mathrm{m}^{3}\).
03

Calculate time to fill the pool using a garden hose

Now, we have the capacity of the pool in cubic meters (80 m³), and the flow rate of the garden hose in liters per minute (60 L/min). First, let's convert the flow rate of the hose to cubic meters per minute. Use the same conversion: 1 L = 0.001 m³. So, \(60 \times 0.001 = 0.06 \mathrm{m}^{3}\)/min. Next, to calculate the time, divide the pool capacity by the flow rate of the garden hose: Time_to_fill = Pool_capacity / Flow_rate_of_garden_hose Substitute the values: Time_to_fill = \(80 \mathrm{m}^{3} / 0.06 \mathrm{m}^{3}\)/min Calculate the time: Time_to_fill = 1333.33 min Finally, convert minutes to hours: Time_to_fill = \(\frac{1333.33 \mathrm{min}}{60} = 22.22 \mathrm{hours}\) (approximately)
04

Calculate time to fill the pool using a river flow

In part (b), the river flows at a rate of \(5000 \mathrm{m}^{3} / \mathrm{s}\). To calculate the time, divide the pool capacity by the flow rate of the river: Time_to_fill = Pool_capacity / Flow_rate_of_river Substitute the values: Time_to_fill = \(80 \mathrm{m}^{3} / 5000 \mathrm{m}^{3}\)/s Calculate the time: Time_to_fill = 0.016 s
05

Present the final answer

(a) It would take approximately 22.22 hours to fill the swimming pool using a garden hose delivering 60 L/min. (b) It would take 0.016 seconds to fill the swimming pool by diverting a moderate size river flowing at \(5000 \mathrm{m}^{3} / \mathrm{s}\).

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Key Concepts

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

Volume Conversion
Understanding volume conversion is crucial when dealing with measurements in varying units. In our example, the pool's volume is given in liters, but the flow rate of the river is in cubic meters per second. To make accurate calculations, we need these measurements in the same unit.

Volume conversion follows a simple but essential principle: the conversion factor. For instance, to convert liters to cubic meters, the conversion factor is 0.001 cubic meters per liter, because there are 1,000 liters in one cubic meter. By multiplying the volume in liters by the conversion factor, you obtain the volume in cubic meters. This step allows for direct comparison and computation, which is vital for evaluations like determining how long it would take to fill a pool.
Rate of Flow
The rate of flow, often measured in volume per unit of time, tells us how quickly a fluid moves through a given cross-section. In fluid dynamics, this is a fundamental concept because it helps understand how fast systems like pipes or rivers can deliver water.

Calculating the rate of flow is straightforward: divide the volume by the time it took for that volume to flow. In our example with the garden hose and river, the flow rates are given, but when only the total volume and time are known, the rate of flow can be calculated by inverting the operation used to find time.

Understanding flow rates is essential in fields ranging from engineering to environmental science, as they directly affect how systems are designed and how they operate. In a practical scenario, like filling our hypothetical swimming pool, it demonstrates the vast difference in time required when using a garden hose versus a river, illustrating the power of flow rates in real-world applications.
Unit Conversion
Unit conversion is a necessary skill in many scientific and engineering fields. Any time you are presented with quantities in different units, as in the swimming pool problem, unit conversion ensures that you can compare 'apples to apples.' To convert units, use a conversion factor that relates the two units, like converting from minutes to seconds or liters to cubic meters.

Within the SI unit system, multiples and submultiples are often in powers of ten, making conversion relatively straightforward. However, when converting between SI and non-SI units or between units within either system that aren't simply related by powers of ten, it's crucial to know or look up the correct conversion factor.

Incorrect unit conversion can lead to a miscalculation which in practical situations can cause significant mistakes or inefficiencies, hence why accuracy and attention to detail are paramount. For students, mastering unit conversion is not only about finding the right answers but also about developing a rigorous approach to problem-solving.

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Most popular questions from this chapter

Fluid originally flows through a tube at a rate of \(100 \mathrm{cm}^{3} / \mathrm{s} .\) To illustrate the sensitivity of flow rate to various factors, calculate the new flow rate for the following changes with all other factors remaining the same as in the original conditions. (a) Pressure difference increases by a factor of \(1.50 .\) (b) A new fluid with 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube is used with a radius 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half the length, and the pressure difference is increased by a factor of 1.50 .

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