Chapter 10, Problem 92P

A laminar flow wind tunnel has a test is 30cm in diameter and 80 cm length. The air is at $20°C$ . At a uniform air speed of 2.0m/s at test section inlet, by how much will the centerline air speed air speed accelerate by the end of the test section?

Answer: Approx 6%

To determine

The center line air speed acceleration at the end of the test section.

Answer to Problem 92P

The center line air speed acceleration at the end of the test section is $6\%$.

Explanation of Solution

Given information:

The diameter of the wind tunnel is $30\text{cm}$ and the length of the wind tunnel is $80\text{cm}$ the temperature is $20°\text{C}$ the uniform air speed is $2\text{m}/\text{s}$.

Write the expression for the Reynolds number at the end of the test section.

  $R_{e_x}=\frac{Vx}{v}$   ....... (I)

Here, the velocity of the air is $V$ the length of the test section is $x$ the kinematic viscosity is $v$.

Write the expression for the increase the velocity by equation of continuity.

  $A_1{V}_1=A_2{V}_2$   ....... (II)

Here, the area at the beginning of the test section is $A_1$ the velocity of the air at the beginning of the test tube is $V_1$, the area at the end of the test section $A_2$ and the velocity of the air at the end of the test section is $A_2$

Write the expression for the area at beginning.

  $A_1=\frac{\pi }{4}{D}^2$   ....... (III)

Here, the diameter of the wind tunnel is $D$.

Write the expression for the area at the end of the test section.

  $A_2=\pi {\left(R-{\delta }^{\ast }\right)}^2$   ....... (IV)

Here, the displacement thickness is ${\delta }^{\ast }$ and the radius of the wind tunnel is $R$.

Write the expression for displacement thickness.

  ${\delta }^{\ast }=\frac{1.72x}{\sqrt{R_{e_x}}}$   ....... (V)

Write the expression for the velocity increment.

  $V_{\text{increase}}=V_2-V_1$   ....... (VI)

Write the expression for the percentage of velocity increase at the end if the test section.

  $\%U_{\text{increase}}=\frac{V_{increase}}{V_1}\times 100$   ....... (VII)

Write the expression for the radius of the wind tunnel.

  $R=\frac{D}{2}$   ....... (VIII)

Here, the diameter of the wind tunnel is $D$ and the radius of the wind tunnel is $R$

Calculation:

Refer to the Table A-9 "properties of air" to obtain the value of kinematic viscosity $\left(v\right)$ as $1.516\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$ at $20°\text{C}$ air temperature.

Substitute $1.516\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$ for $v$

  $80\text{cm}$ for $x$, $2\text{m}/\text{s}$ for $V$ in Equation (I).

  $\begin{array}{c}{R}_{e_x}=\frac{\left(2\text{m}/\text{s}\right)\left(80\text{cm}\right)}{1.516\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{\left(2\text{m}/\text{s}\right)\left(80\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{1.516\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{0.16{\text{m}}^{\text{2}}/\text{s}}{1.516\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =1.055\times {10}^5\end{array}$

The value of Reynolds number is less than $3\times {10}^5$ hence the flow remains laminar throughout the length.

Substitute $80\text{cm}$ for $x$ and $1.055\times {10}^5$ for $R_{e_x}$ in Equation (V).

  $\begin{array}{c}{\delta }^{\ast }=\frac{1.72\times 80\text{cm}}{\sqrt{1.005\times {10}^5}}\\ =\frac{1.72\times 80\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{\sqrt{1.005\times {10}^5}}\\ =\frac{1.376\text{m}}{317.017}\\ =4.236\times {10}^{-3}\text{m}\end{array}$

Substitute $30\text{cm}$ for $D$ in Equation (VIII).

  $\begin{array}{c}R=\frac{30\text{cm}}{2}\\ =15\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.15\text{m}\end{array}$

Substitute $0.15\text{m}$ for $R$ in Equation (III).

  $\begin{array}{c}{A}_1=\pi \times {\left(0.15\text{m}\right)}^2\\ =0.0225\pi {\text{m}}^{\text{2}}\\ =0.071{\text{m}}^{\text{2}}\end{array}$

Substitute $0.15\text{m}$ for $R$ and $4.236\times {10}^{-3}\text{m}$ for ${\delta }^{\ast }$ in Equation (IV).

  $\begin{array}{c}{A}_2=\pi \times {\left(0.15\text{m}-4.236\times {10}^{-3}\text{m}\right)}^2\\ =\pi \times {\left(0.145764\text{m}\right)}^2\\ =0.0212\pi {\text{m}}^{\text{2}}\\ =0.067{\text{m}}^{\text{2}}\end{array}$

Substitute $0.071{\text{m}}^{\text{2}}$ for $A_1$, $0.067{\text{m}}^{\text{2}}$ for $A_2$ and $2\text{m}/\text{s}$ in Equation (V).

  $\begin{array}{l}0.071{\text{m}}^{\text{2}}\times 2\text{m}/\text{s}=0.067{\text{m}}^{\text{2}}\times V_2\\ 0.067{\text{m}}^{\text{2}}\times V_2=0.142{\text{m}}^{\text{3}}/\text{s}\\ V_2=\frac{0.142{\text{m}}^{\text{3}}/\text{s}}{0.067{\text{m}}^{\text{2}}}\\ V_2=2.12\text{m}/\text{s}\end{array}$

Substitute $2.12\text{m}/\text{s}$ for $V_2$ and $2\text{m}/\text{s}$ for $V_1$ in Equation (VI).

  $\begin{array}{c}{V}_{\text{increasing}}=2.12\text{m}/\text{s}-2\text{m}/\text{s}\\ =0.12\text{m}/\text{s}\end{array}$

Substitute $0.0142\text{m}/\text{s}$ for $V_{\text{increasing}}$ and $2\text{m}/\text{s}$ for $V_1$ in Equation (VII).

  $\begin{array}{c}\%U_{\text{increase}}=\frac{0.12\text{m}/\text{s}}{2\text{m}/\text{s}}\times 100\%\\ =0.06\times 100\%\\ =6\%\end{array}$

Conclusion:

The center line air speed acceleration at the end of the test section is $6\%$.