/*! 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 50 Oil is flowing with a speed of \... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 50

Oil is flowing with a speed of \(1.22 \mathrm{~m} / \mathrm{s}\) through a pipeline with a radius of \(0.305 \mathrm{~m} .\) How many gallons of oil \(\left(1 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{~m}^{3}\right)\) flow in one day?

Short Answer

Expert verified
About 81,424.85 gallons of oil flow in one day.

Step by step solution

01

Calculate the Cross-Sectional Area

To find the cross-sectional area, use the formula for the area of a circle: \[ A = \pi r^2 \]where \( r \) is the radius of the pipeline. Here, \( r = 0.305 \) meters.So, \[ A = \pi (0.305)^2 \approx 0.2927 \text{ m}^2 \]
02

Calculate the Volume of Oil Flowing per Second

The volume flow rate can be calculated using the cross-sectional area and the flow speed. The formula is:\[ Q = A \cdot v \]where \( A \) is the area from Step 1, and \( v \) is the velocity of the oil, \( 1.22 \text{ m/s} \).So, \[ Q = 0.2927 \cdot 1.22 \approx 0.3571 \text{ m}^3/\text{s} \]
03

Calculate the Volume of Oil Flowing in One Day

To find the daily volume flow, multiply the volume per second by the number of seconds in a day:\[ V_{day} = Q \cdot 86400 \]where 86400 is the number of seconds in a day (24 hours \(\times\) 60 minutes/hour \(\times\) 60 seconds/minute).So,\[ V_{day} = 0.3571 \times 86400 \approx 30844.44 \text{ m}^3 \]
04

Convert the Volume to Gallons

Now, convert the daily volume from cubic meters to gallons using the conversion: \(1 \text{ gal} = 379 \times 10^{-3} \text{ m}^3\).\[ \text{Gallons} = \frac{30844.44}{379 \times 10^{-3}} \approx 81424.85 \text{ gal} \]

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Key Concepts

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

Volume Flow Rate
Volume flow rate is a key concept in fluid dynamics. It tells us the amount of fluid flowing through a cross-section per unit of time.
The formula to calculate volume flow rate, denoted by \( Q \), is:
  • \( Q = A \cdot v \)
where:
  • \( A \) is the cross-sectional area of the pipe
  • \( v \) is the velocity of the fluid
The resulting unit for volume flow rate is cubic meters per second \( (\text{m}^3/\text{s}) \).
By knowing the volume flow rate, you can easily scale up or down to find out how much fluid passes in minutes, hours, or even days by multiplying with the number of seconds in those periods.
In our exercise, oil is flowing through the pipeline at a constant speed, so knowing the volume flow rate helps in calculating how much oil flows in a day.
Cross-Sectional Area
To understand the flow of fluid through pipes, it's crucial to calculate the cross-sectional area, especially for circular pipes. The area determines how much fluid can pass through at a given speed.
The formula for the area \( A \) of a circle is:
  • \( A = \pi r^2 \)
where:
  • \( \pi \) is approximately 3.14159
  • \( r \) is the radius of the circular cross-section
In our exercise, with a pipeline radius of 0.305 meters, the cross-sectional area comes out to around 0.2927 \( \text{m}^2 \).
Knowing this, we can better understand the pipe's capacity to carry oil and directly use it to find the volume flow rate when paired with the fluid velocity.
Unit Conversion
Unit conversion is an essential step in many calculations, especially when dealing with different measurement systems. In fluid dynamics, you often need to convert between cubic meters, liters, gallons, etc.
In our problem, after finding the volume of oil in cubic meters, we need to convert it to gallons for easier understanding or because gallons may be the desired unit.
Given:
  • 1 gallon = 379 \( \times 10^{-3} \text{ m}^3 \)
To convert from cubic meters \( \text{m}^3 \) to gallons:
  • Divide the volume in cubic meters by 379 \( \times 10^{-3} \)
Our calculated daily oil volume in cubic meters converted to gallons allows for practical application, such as logistical planning. This highlights the importance of comfortably navigating different unit systems in fluid-related problems.

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Most popular questions from this chapter

Interactive Solution \(11.73\) at illustrates a model for solving this problem. A pressure difference of \(1.8 \times 10^{3} \mathrm{~Pa}\) is needed to drive water \(\left(\eta=1.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) through a pipe whose radius is \(5.1 \times 10^{-3} \mathrm{~m}\). The volume flow rate of the water is \(2.8 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}\). What is the length of the pipe?

In the human body, blood vessels can dilate, or increase their radii, in response to various stimuli, so that the volume flow rate of the blood increases. Assume that the pressure at either end of a blood vessel, the length of the vessel, and the viscosity of the blood remain the same, and determine the factor \(R_{\text {dilated }} / R_{\text {normal }}\) by which the radius of a vessel must change in order to double the volume flow rate of the blood through the vessel.

A Venturi meter is a device that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed \(v_{2}\) through a horizontal section of pipe whose cross-sectional area is \(A_{2}=0.0700 \mathrm{~m}^{2} .\) The gas has a density of \(\rho=1.30 \mathrm{~kg} / \mathrm{m}^{3} .\) The Venturi meter has a cross-sectional area of \(A_{1}=0.0500 \mathrm{~m}^{2}\) and has been substituted for a section of the larger pipe. The pressure difference between the two sections is \(P_{2}-P_{1}=120 \mathrm{~Pa}\). Find (a) the speed \(v_{2}\) of the gas in the larger original pipe and (b) the volume flow rate \(Q\) of the gas.

Accomplished silver workers in India can pound silver into incredibly thin sheets, as thin as \(3.00 \times 10^{-7} \mathrm{~m}\) (about onehundredth of the thickness of this sheet of paper). Find the area of such a sheet that can be formed from \(1.00 \mathrm{~kg}\) of silver.

Some researchers believe that the dinosaur Barosaurus held its head erect on a long neck, much as a giraffe does. If so, fossil remains indicate that its heart would have been about \(12 \mathrm{~m}\) below its brain. Assume that the blood has the density of water, and calculate the amount by which the blood pressure in the heart would have exceeded that in the brain. Size estimates for the single heart needed to withstand such a pressure range up to two tons. Alternatively, Barosaurus may have had a number of smaller hearts.
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