Problem 50 - Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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, \[ A = \pi (0.305)^2 \approx 0.2927 \text{ m}^2 \]

Step 2: 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, \[ Q = 0.2927 \cdot 1.22 \approx 0.3571 \text{ m}^3/\text{s} \]

Step 3: 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,\[ V_{day} = 0.3571 \times 86400 \approx 30844.44 \text{ m}^3 \]

Step 4: Convert the Volume to Gallons

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

Key Concepts

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.