Answer:
Increase of 6%
Explanation:
According to the description of the problem, the air will have an increase in acceleration as a result of the conservation of the mass.
We proceed to calculate the Reynolds number,
[tex]Re_x = \frac{Vx}{\upsilon}[/tex]
Where,
X is the lenght of the test section
[tex]\upsilon= 1.516*10^{-5}m^2/s[/tex] at 20°c (kinematic viscosity at tables)
V= Velocity of air.
Replacing,
[tex]Re_x = \frac{(2)(0.8)}{1.516*10^{-5}}[/tex]
[tex]Re_x = 1.055*10^{5}[/tex]
Turbulent fluid start from 3*10^6 in Reynolds number. Our fluid remain laminar throughout the length
We can calculate the displacement thickness at the end,
[tex]\delta^*=\frac{1.72x}{\sqrt{Re_x}}[/tex]
[tex]\delta^* = \frac{1.72*0.8}{\sqrt{1.055*10^5}}[/tex]
[tex]\delta^* = 4.23532*10^{-3}m[/tex]
That is size of the boundary layer at the end of the test.
We know for conservation of mass that
[tex]A_1V_1 = A_2V_2[/tex]
Where,
[tex]A_1 =[/tex]Area at the beginning
[tex]A_2 =[/tex] Area at the end
[tex]V_1 =[/tex]Velocity at the beginning
[tex]V_2 =[/tex]Velocity at the end
Area at the beginning is
[tex]A_1 = \pi R^2[/tex]
[tex]A_1 = \pi (0.15)^2[/tex]
[tex]A_1 = 0.071m^2[/tex]
Area at the end is
[tex]A_2 = (R-\delta^*)^2[/tex]
[tex]A_2 = \pi (0.15-4.236*10^{-3})^2[/tex]
[tex]A_2 = 0.067m^2[/tex]
We can now reach the velocity at the end, given by
[tex]A_1V_1 = A_2V_2\\V_2 = \frac{A_1V_1}{A_2}\\V_2 = \frac{(0.071)*2}{0.067}\\V_2 = 2.12m/s[/tex]
The increase at the end was 0.12m/s
That is 6% from the initial velocity, around to 6% the air speed accelerated
