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Chapter 11: Problem 54

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of \(5.88 \times 10^{-2} \mathrm{kg} / \mathrm{s} .\) The density of the gasoline is \(735 \mathrm{kg} / \mathrm{m}^{3},\) and the radius of the fuel line is \(3.18 \times 10^{-3} \mathrm{m} .\) What is the speed at which the gasoline moves through the fuel line?

Short Answer

Expert verified
The speed of gasoline through the fuel line is approximately 2.52 m/s.

Step by step solution

01

Understand the Problem

We need to find the velocity of gasoline as it moves through a fuel line, given the mass flow rate, the density of gasoline, and the radius of the fuel line.
02

Use the Formula for Mass Flow Rate

The mass flow rate \( \dot{m} \) is given by the formula \( \dot{m} = \rho v A \), where \( \rho \) is the density, \( v \) is the velocity, and \( A \) is the cross-sectional area of the pipe.
03

Calculate the Cross-Sectional Area

The cross-sectional area \( A \) of a circle is \( \pi r^2 \) where \( r \) is the radius. So, substitute \( r = 3.18 \times 10^{-3} \text{ m} \) into the formula: \[ A = \pi (3.18 \times 10^{-3})^2 \approx 3.18 \times 10^{-5} \text{ m}^2 \].
04

Rearrange the Equation for Velocity

Rearrange the equation \( \dot{m} = \rho v A \) to solve for velocity \( v \): \[ v = \frac{\dot{m}}{\rho A} \].
05

Substitute the Known Values

Use the values \( \dot{m} = 5.88 \times 10^{-2} \text{ kg/s} \), \( \rho = 735 \text{ kg/m}^3 \), and \( A = 3.18 \times 10^{-5} \text{ m}^2 \) in the equation:\[ v = \frac{5.88 \times 10^{-2}}{735 \times 3.18 \times 10^{-5}} \].
06

Calculate the Speed

Perform the calculation to find the speed:\[ v \approx \frac{5.88 \times 10^{-2}}{0.023373} \approx 2.52 \text{ m/s} \].

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Key Concepts

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

Mass Flow Rate
Mass flow rate is a key parameter in fluid dynamics that describes how much mass of a fluid passes through a given point over a specific time interval. It is denoted by the symbol \( \dot{m} \). In essence, it tells us how quickly a fluid, such as gasoline, is moving through a system, like the fuel line in a car.
  • Units: Mass flow rate is measured in kilograms per second (\( \text{kg/s} \)).
  • Calculation: In fluid dynamics, it can be calculated if you know the density of the fluid, the velocity of the fluid, and the cross-sectional area through which the fluid moves. The formula is: \( \dot{m} = \rho v A \), where \( \rho \) is the density, \( v \) is the velocity, and \( A \) is the cross-sectional area.
It acts like a bridge between other important aspects, such as the density and velocity, making it crucial for determining other characteristics of the flow.
Density of Gasoline
Density is an important physical property in fluid dynamics that refers to the mass per unit volume of a substance. For gasoline, the density tells us how much mass of gasoline occupies a certain volume.
  • Units: The density of gasoline is typically measured in kilograms per cubic meter (\( \text{kg/m}^3 \)).
  • Relevance: In the context of our exercise, the density \( \rho = 735 \text{ kg/m}^3 \) is used to determine other properties such as mass flow rate and velocity of gasoline in the fuel line.
Understanding the density helps us relate the mass of the gasoline to the volume it occupies, which is vital when designing fuel systems for performance and efficiency.
Cross-Sectional Area
The cross-sectional area is a geometrical property that represents the size of the opening through which fluid flows. For a cylindrical fuel line, it is the area of a circle with the given radius.
  • Formula: The cross-sectional area \( A \) for a circular pipe is calculated using the formula \( A = \pi r^2 \), where \( r \) is the radius of the pipe.
  • Units: It is expressed in square meters (\( \text{m}^2 \)).
  • Importance: Knowing the cross-sectional area helps to link the mass flow rate and velocity of the fluid. In our case, this was key to calculating the velocity of the gasoline.
Understanding this concept is crucial for solving problems involving fluid flow through pipes, as it directly influences how fast or slow the fluid moves through the system.
Velocity Calculation
In fluid dynamics, velocity refers to the speed and direction at which a fluid, like gasoline, moves through a pipe. Calculating this speed is essential for understanding how efficiently a system operates.
  • Relation to Other Quantities: Velocity \( v \) can be determined if the mass flow rate \( \dot{m} \), density \( \rho \), and cross-sectional area \( A \) are known, using the rearranged equation: \( v = \frac{\dot{m}}{\rho A} \).
  • Units: Velocity is measured in meters per second (\( \text{m/s} \)).
  • Example Calculation: In our exercise, using the given values, the velocity was found to be approximately \( 2.52 \text{ m/s} \).
Calculating velocity is fundamental for analyzing and optimizing any system where fluid flow is involved, ensuring that the fluid is delivered promptly and efficiently to its destination.

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Most popular questions from this chapter

An airplane wing is designed so that the speed of the air across the top of the wing is \(251 \mathrm{m} / \mathrm{s}\) when the speed of the air below the wing is \(225 \mathrm{m} / \mathrm{s} .\) The density of the air is \(1.29 \mathrm{kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area \(24.0 \mathrm{m}^{2} ?\)

A 1.00-\(\mathrm{m}\) -tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to the atmosphere. What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure?

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of \(45 \mathrm{N}\) to the input piston, which has a radius \(r_{1} .\) As a result, the output plunger, which has a radius \(r_{2},\) applies a force to the car. The ratio \(r_{2} / r_{1}\) has a value of \(8.3 .\) Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

A solid concrete block weighs 169 N and is resting on the ground. Its dimensions are \(0.400 \mathrm{m} \times 0.200 \mathrm{m} \times 0.100 \mathrm{m} .\) A number of identical blocks are stacked on top of this one. What is the smallest number of whole blocks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block?

A liquid is flowing through a horizontal pipe whose radius is \(0.0200 \mathrm{m}\). The pipe bends straight upward through a height of \(10.0 \mathrm{m}\) and joins another horizontal pipe whose radius is \(0.0400 \mathrm{m} .\) What volume flow rate will keep the pressures in the two horizontal pipes the same?
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