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Chapter 1: Problem 106

Often the assumption is made that the flow of a certain fluid can be considered as incompressible flow if the density of the fluid changes by less than \(2 \%\). If air is flowing through a tube such that the air pressure at one section is 9.0 psi and at a downstream section it is 8,6 psi at the same temperature, do you think that this flow could be considered an incompressible flow? Support your answer with the necessary calculations. Assume standard atmospheric pressire.

Short Answer

Expert verified
The answer is based on the result of the calculation step. If the calculated \(ρ_{\%}\) is less than 2%, then the flow of air through the tube is considered to be incompressible.

Step by step solution

01

Initial setup

We need to convert the pressures given in psi to Pascals (Pa). 1 psi = 6894.76 Pa. So, initial pressure (P1) = 9.0 psi = \(9.0 * 6894.76\) Pa = 62052.84 Pa. Final pressure (P2) = 8.6 psi = \(8.6 * 6894.76\) Pa = 59304.536 Pa.
02

Calculation of initial and final densities

Initial air density (ρ1) under the atmospheric pressure P1 can be calculated by using the Ideal Gas law, that is \(ρ=P/RT\) where R is the specific gas constant for air which is 287 J/(kg*K) and T is the temperature in Kelvin. Here, we will keep the T general because temperature is the same for both the situations. So \(ρ1 = P1/RT\).Initial air density (ρ2) under the atmospheric pressure P2 is calculated similarly. So, \(ρ2 = P2/RT\).
03

Calculation of density change percentage

The percentage change in density can be defined as: \(ρ_{\%} = ((ρ2 - ρ1)/ρ1) * 100\% \). Substituting \(ρ1\) and \(ρ2\) from step 2 we get \(ρ_{\%} = ((P2/RT - P1/RT)/(P1/RT)) * 100\% \). Simplifying this yields \(ρ_{\%} = ((P2-P1)/P1)*100\% \). Substitute the values for P1 and P2 from Step 1.
04

Answering the question

We can consider the flow of air as incompressible if the density change is less than 2%. Hence, if the calculated \(ρ_{\%}\) is less than 2, then the flow can be considered as incompressible.

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