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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 gasoline moves through the fuel line at approximately 2.55 m/s.

Step by step solution

01

Understand Given Data

The first step is to summarize the given information:- Mass flow rate of gasoline, \( \dot{m} = 5.88 \times 10^{-2} \mathrm{kg/s} \)- Density of gasoline, \( \rho = 735 \mathrm{kg/m^{3}} \)- Radius of the fuel line, \( r = 3.18 \times 10^{-3} \mathrm{m} \)Our task is to find the speed (velocity) of gasoline, \( v \).
02

Calculate the Cross-Sectional Area of the Pipe

The cross-sectional area, \( A \), of the fuel line is calculated using the formula for the area of a circle: \( A = \pi r^2 \). Substitute the radius: \[ A = \pi (3.18 \times 10^{-3})^2 \approx 3.18 \times 10^{-5} \mathrm{m^2} \]
03

Use Mass Flow Rate to Find Velocity

The mass flow rate \( \dot{m} \) relates to density \( \rho \), area \( A \), and velocity \( v \) by the formula: \[ \dot{m} = \rho A v \]Solve for velocity \( v \): \[ v = \frac{\dot{m}}{\rho A} \] Substitute the known values: \[ v = \frac{5.88 \times 10^{-2}}{735 \times 3.18 \times 10^{-5}} \approx 2.55 \mathrm{m/s} \]
04

Verify Units

Ensure the units are consistent:- Mass flow rate \( \mathrm{kg/s} \) - Density \( \mathrm{kg/m^3} \)- Area \( \mathrm{m^2} \)Velocity comes out in \( \mathrm{m/s} \), confirming the calculation.

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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 fundamental concept in fluid dynamics. It defines the amount of mass passing through a particular section of a system per unit of time. Think of it as a way to measure how much 'stuff' moves through a pipe or any opening over time.

Mathematically, mass flow rate is denoted by \( \dot{m} \) and is expressed in the units of kilograms per second (\( \mathrm{kg/s} \)).

In our example, the fuel pump sends gasoline to the engine at a mass flow rate of \( 5.88 \times 10^{-2} \mathrm{kg/s} \). This means that approximately \( 0.0588 \) kilograms of gasoline passes through the line every second.

To calculate or utilize mass flow rates, you generally need a few key variables:
  • Density \( \rho \)
  • The cross-sectional area \( A \) of the pipe or duct
  • The velocity \( v \) of the fluid
Understanding mass flow rate is crucial for analyzing and designing systems like engines, where consistent fuel delivery is important.
Density
Density is a crucial parameter in fluid dynamics. It tells us how much mass a certain volume of the fluid contains. In other words, it's the measure of compactness of a fluid, defining how tightly the fluid's mass is packed.

Density is represented by the symbol \( \rho \) and is typically measured in \( \mathrm{kg/m^3} \). For our gasoline example, the density is given as \( 735 \mathrm{kg/m^3} \). This means every cubic meter of gasoline holds \( 735 \) kilograms of mass.

This property is pivotal when working with fluids in motion, especially in calculations involving mass flow rate and pressure.
  • High density generally means a fluid is heavier and may flow differently under the same conditions compared to a fluid with lower density.
  • Understanding density helps in determining the force needed to move the fluid and how it's distributed in space.
Thus, in dynamic systems, density defines how other processes are computed, as it directly influences mass flow and other related quantities.
Velocity Calculation
Velocity calculation is often the key focus when analyzing how fluids move through systems. To find the velocity of a fluid, one utilizes the relationship between mass flow rate, density, and cross-sectional area.

The formula that connects these parameters is \( \dot{m} = \rho A v \), where:
  • \( \dot{m} \) is the mass flow rate.
  • \( \rho \) is the fluid density.
  • \( A \) is the cross-sectional area where the fluid flows.
  • \( v \) is the velocity we're solving for.
Rearranging this formula gives us the fluid's velocity: \[ v = \frac{\dot{m}}{\rho A} \]In our exercise, we substitute the values we've been given: \[ v = \frac{5.88 \times 10^{-2}}{735 \times 3.18 \times 10^{-5}} \approx 2.55 \mathrm{m/s} \]

This means the gasoline moves through the fuel line at a speed of approximately \( 2.55 \mathrm{m/s} \). By understanding how these pieces fit together, one can predict and control the movement of fluids in systems with greater precision.

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Most popular questions from this chapter

An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is 15.2 N. When completely submerged in water, its apparent weight is 13.7 N. What is the volume of the object?

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. The tube is contained inside a large 鈥渕edicine dropper,鈥 into which the liquid is drawn using the squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). The hydrometer has a weight of \(W=5.88 \times 10^{-2} \mathrm{N}\) and a cross-sectional area of tw \(A=7.85 \times 10^{-5} \mathrm{m}^{2} .\) How far from the bottom of the tube should the mark be put that denotes (a) battery acid \(\left(\rho=1280 \mathrm{kg} / \mathrm{m}^{3}\right)\) and (b) antifreeze \(\left(\rho=1073 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)

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Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of 1.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

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