A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 7.53x102 kg/s. The density of the gasoline is 735 kg/m, and the radius of the fuel line is 2.56x103 m. What is the speed at which gasoline moves through the fuel line?

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter15: Fluids
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### Fuel Flow Rate Calculation

**Current Attempt in Progress**

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of \(7.53 \times 10^{-2}\) kg/s. The density of the gasoline is 735 kg/m\(^3\), and the radius of the fuel line is \(2.56 \times 10^{-3}\) m. What is the speed at which gasoline moves through the fuel line?

#### Calculation Input

- **Mass flow rate (\(\dot{m}\)):** \(7.53 \times 10^{-2}\) kg/s
- **Density of gasoline (\(\rho\)):** 735 kg/m\(^3\)
- **Radius of the fuel line (r):** \(2.56 \times 10^{-3}\) m

#### Solution

To find the speed at which the gasoline moves through the fuel line, we can use the relationship between mass flow rate, density, and velocity.

The formula to use is:

\[
\dot{m} = \rho \cdot A \cdot v
\]

Where:
- \(\dot{m}\) is the mass flow rate
- \(\rho\) is the density of the gasoline
- \(A\) is the cross-sectional area of the fuel line
- \(v\) is the velocity of gasoline

First, calculate the cross-sectional area \(A\) of the fuel line:

\[
A = \pi r^2
\]

Substitute the given radius:

\[
A = \pi \times (2.56 \times 10^{-3})^2 \approx 2.06 \times 10^{-5} \, \text{m}^2
\]

Next, rearrange the original formula to solve for velocity \(v\):

\[
v = \frac{\dot{m}}{\rho \cdot A}
\]

Substitute the known values:

\[
v = \frac{7.53 \times 10^{-2}}{735 \times 2.06 \times 10^{-5}} \approx 5.00 \, \text{m/s}
\]

#### Final Answer:

**Enter the calculated speed of the gasoline through the fuel line:**
- **Number:** [Input Box]
- **Units:** [Units Drop-down]

Please verify the results by substituting the values and
Transcribed Image Text:### Fuel Flow Rate Calculation **Current Attempt in Progress** A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of \(7.53 \times 10^{-2}\) kg/s. The density of the gasoline is 735 kg/m\(^3\), and the radius of the fuel line is \(2.56 \times 10^{-3}\) m. What is the speed at which gasoline moves through the fuel line? #### Calculation Input - **Mass flow rate (\(\dot{m}\)):** \(7.53 \times 10^{-2}\) kg/s - **Density of gasoline (\(\rho\)):** 735 kg/m\(^3\) - **Radius of the fuel line (r):** \(2.56 \times 10^{-3}\) m #### Solution To find the speed at which the gasoline moves through the fuel line, we can use the relationship between mass flow rate, density, and velocity. The formula to use is: \[ \dot{m} = \rho \cdot A \cdot v \] Where: - \(\dot{m}\) is the mass flow rate - \(\rho\) is the density of the gasoline - \(A\) is the cross-sectional area of the fuel line - \(v\) is the velocity of gasoline First, calculate the cross-sectional area \(A\) of the fuel line: \[ A = \pi r^2 \] Substitute the given radius: \[ A = \pi \times (2.56 \times 10^{-3})^2 \approx 2.06 \times 10^{-5} \, \text{m}^2 \] Next, rearrange the original formula to solve for velocity \(v\): \[ v = \frac{\dot{m}}{\rho \cdot A} \] Substitute the known values: \[ v = \frac{7.53 \times 10^{-2}}{735 \times 2.06 \times 10^{-5}} \approx 5.00 \, \text{m/s} \] #### Final Answer: **Enter the calculated speed of the gasoline through the fuel line:** - **Number:** [Input Box] - **Units:** [Units Drop-down] Please verify the results by substituting the values and
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