Question

A laminar flow wind tunnel has a test section that is 26 cm in diameter and 80 cm in length. The air is at 20°C. At a uniform air speed of 2.0 m/s at the test section inlet, by how much will the centerline air speed accelerate by the end of the test section? The kinematic viscosity of air at 20oC is ν = 1.516 × 10-5 m2/s.

Answer 1

The centerline air speed in the wind tunnel will accelerate by approximately 9.972 m/s by the end of the test section.

Step-by-step explanation:

The centerline air speed will accelerate by the end of the test section in the laminar flow wind tunnel. To calculate this, we can use the equation for flow rate, Q = A * V, where Q is the flow rate, A is the cross-sectional area, and V is the velocity.

Given that the diameter of the test section is 26 cm, the radius is 13 cm (0.13 m). The cross-sectional area can be calculated using the formula for the area of a circle, A = π * r^2, where π is approximately 3.14.

So, A = 3.14 * (0.13 m)^2 = 0.16706 m^2.

We know the velocity at the test section inlet is 2.0 m/s, and we want to find the change in velocity, DV, at the end of the test section.

To find DV, we can rearrange the equation Q = A * V to solve for V:

V = Q / A

Substituting the given values, V = (2.0 m/s) / (0.16706 m^2) = 11.972 m/s.

Therefore, the centerline air speed will accelerate by approximately 9.972 m/s by the end of the test section.

Answer 2

Increase of 6%

Step-by-step 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,

`Re_x = (Vx)/(\upsilon)`

Where,

X is the lenght of the test section

`\upsilon= 1.516*10^(-5)m^2/s` at 20°c (kinematic viscosity at tables)

V= Velocity of air.

Replacing,

`Re_x = ((2)(0.8))/(1.516*10^(-5))`

`Re_x = 1.055*10^(5)`

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,

`\delta^*=(1.72x)/(√(Re_x))`

`\delta^* = (1.72*0.8)/(√(1.055*10^5))`

`\delta^* = 4.23532*10^(-3)m`

That is size of the boundary layer at the end of the test.

We know for conservation of mass that

`A_1V_1 = A_2V_2`

Where,

`A_1 =`Area at the beginning

`A_2 =` Area at the end

`V_1 =`Velocity at the beginning

`V_2 =`Velocity at the end

Area at the beginning is

`A_1 = \pi R^2`

`A_1 = \pi (0.15)^2`

`A_1 = 0.071m^2`

Area at the end is

`A_2 = (R-\delta^*)^2`

`A_2 = \pi (0.15-4.236*10^(-3))^2`

`A_2 = 0.067m^2`

We can now reach the velocity at the end, given by

`A_1V_1 = A_2V_2\\V_2 = (A_1V_1)/(A_2)\\V_2 = ((0.071)*2)/(0.067)\\V_2 = 2.12m/s`

The increase at the end was 0.12m/s

That is 6% from the initial velocity, around to 6% the air speed accelerated