Chapter 10, Problem 91P

Repeat the calculation of Prob. 10-90, except for a test section of square rather than round cross section, with a $30\text{cm}\times 30\text{cm}$ cross section and a length of 80 cm. Compare the result to that of Prob. 10-90 and discuss.

To determine

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

Answer to Problem 91P

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

Explanation of Solution

Given information:

The side 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.

  $A_1=b\times b$   ....... (III)

Here, the side of the of the wind tunnel is $b$.

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

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

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

Write the expression for displacement thickness.

  ${\delta }^{\ast }=\frac{0.048x}{{\left(R_{e_x}\right)}^{1/5}}$   ....... (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)

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).

  $\begin{array}{c}{\delta }^{\ast }=\frac{0.048\times 80\text{cm}}{{\left(1.005\times {10}^5\right)}^{1/5}}\\ =\frac{0.048\times 80\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{{\left(1.005\times {10}^5\right)}^{1/5}}\\ =3.83\times {10}^{-3}\text{m}\end{array}$

Substitute $30\text{cm}$ for $b$ in Equation (III).

  $\begin{array}{c}{A}_1=30\text{cm}\times 30\text{cm}\\ =30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\times 30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.09{\text{m}}^{\text{2}}\end{array}$

Substitute $30\text{cm}$ for $b$ and $3.83\times {10}^{-3}\text{m}$ for ${\delta }^{\ast }$ in Equation (IV)

  $\begin{array}{c}{A}_2={\left[30\text{cm}-\left(2\times 3.83\times {10}^{-3}\right)\text{m}\right]}^2\\ ={\left[30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)-\left(7.66\times {10}^{-3}\right)\text{m}\right]}^2\\ ={\left[0.29234\text{m}\right]}^2\\ =0.085{\text{m}}^{\text{2}}\end{array}$

Substitute $0.09{\text{m}}^{\text{2}}$ for $A_1$, $0.085{\text{m}}^{\text{2}}$ for $A_2$ and $2\text{m}/\text{s}$ in Equation (V).

  $\begin{array}{l}0.09{\text{m}}^{\text{2}}\times 2\text{m}/\text{s}=0.085{\text{m}}^{\text{2}}\times V_2\\ 0.085{\text{m}}^{\text{2}}\times V_2=0.18{\text{m}}^{\text{3}}/\text{s}\\ V_2=\frac{0.18{\text{m}}^{\text{3}}/\text{s}}{0.085{\text{m}}^{\text{2}}}\\ V_2=2.12\text{m}/\text{s}\end{array}$

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

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

Substituting $0.12\text{m}/\text{s}$ for $V_{\text{increase}}$ 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 $6\%$.