Homework Answers
[volume] = [mass] / [density] so the volume flow rate of the gasoline is just [mass flow rate] / density
= [5.79 * 10^-2 kg/s] / [735 kg/m^3]
=7.877 * 10^-5 m^3 /s
The cross sectional area (A) of the fuel line of radius r is
given by
A = πr² = π * (2.95 * 10^-3)²
A = 2.73 * 10^-5 m²
Linear flow rate = [volume flow rate] / [cross sectional
area]
= [7.877 * 10^-5 m^3/s] / [2.73 * 10^-5 m^2]
= 2.88 m/s
Calculation may be wrong but concept is right
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