Chapter 7, Problem 7.3P

To determine

Calculate the boundary layer thickness at both occasions and compare it?

Answer to Problem 7.3P

If the flow is turbulent throughout the flow, the $\delta =0.02735m$

Or $\delta =0.02134m$

Therefore, fully turbulent flow will have a boundary layer thickness which is almost $28\%$ higher.

Explanation of Solution

Given information:

Laminar to turbulent transition point is given as ${\mathrm{Re}}_{exit}$

$x=1.5m$

Velocity of the flow is equal to $40m/s$

${\mathrm{Re}}_{x.crit}=1.2\times {10}^6$

For laminar flow,

$\frac{\delta }{x}=\frac{5.0}{{\mathrm{Re}}_x^{1/2}}$

The Reynolds’s number is defined as,

${\mathrm{Re}}_x=\frac{\rho Vx}{\mu }$

In above equation,

$V$ - Velocity

$\rho$ - Density

$\mu$ - Dynamic viscosity

$\delta$ - Boundary layer thickness

For turbulent flow,

$\frac{\delta }{x}=\frac{0.16}{{\mathrm{Re}}_x^{1/7}}$

Assume, air at $20°C$ will have,

$\begin{array}{l}\rho =1.2kg/m^3\\ \mu =1.8\times {10}^{-5}kg/ms\end{array}$

Calculation:

Calculate the distance from leading edge at critical point,

$\begin{array}{l}{\mathrm{Re}}_{x.crit}=\frac{\rho Vx}{\mu }\\ 1.2\times {10}^6=\frac{\left(1.2kg/m^3\right)\left(40m/s\right)x_{critical}}{1.8\times {10}^{-5}kg/ms}\\ x_{critical}=0.45\end{array}$

Calculate the relevant boundary layer thickness,

$\begin{array}{l}\frac{{\delta }_{critical}}{x_{critical}}=\frac{5.0}{{\mathrm{Re}}_x^{1/2}}\\ {\delta }_{critical}=\frac{5.0\left(0.45\right)}{{\left(1.2\times {10}^6\right)}^{1/2}}\\ {\delta }_{critical}=0.00205m\end{array}$

Find the relevant leading edge distance for turbulent flow,

$\begin{array}{l}\frac{{\delta }_{critical}}{x_{turbulent}}=\frac{0.16}{{\mathrm{Re}}_x^{1/7}}\\ \frac{0.00205m}{x_{turbulent}}=\frac{0.16}{{\left(\frac{\left(1.2kg/m^3\right)\left(40m/s\right)\left(x_{turbulent}\right)}{1.8\times {10}^{-5}kg/ms}\right)}^{1/7}}\\ x_{turbulent}=0.073\end{array}$

Now, at $x=1.5m$

Find the effective length,

$x_{effective}=1.5m+x_{turbulent}-x_{critical}=1.5+0.073+0.45=1.123m$

Find the effective Reynolds’s number,

$\begin{array}{l}{\mathrm{Re}}_{effective}=\frac{\rho Vx}{\mu }\\ =\frac{\left(1.2kg/m^3\right)\left(40m/s\right)\left(1.123m\right)}{1.8\times {10}^{-5}kg/ms}\\ =2.995\times {10}^6\end{array}$

Calculate the boundary layer thickness,

$\begin{array}{l}\frac{\delta }{x_{effective}}=\frac{0.16}{{\mathrm{Re}}_x^{1/7}}\\ \frac{\delta }{1.123}=\frac{0.16}{{\left(2.995\times {10}^6\right)}^{1/7}}\\ \delta =0.02134m\end{array}$

Now, calculate the boundary layer thickness for all turbulent flow,

$\begin{array}{l}\frac{\delta }{x_{effective}}=\frac{0.16}{{\mathrm{Re}}_x^{1/7}}\\ \frac{\delta }{1.5}=\frac{0.16}{{\left(\frac{\left(1.2kg/m^3\right)\left(40m/s\right)\left(1.5m\right)}{1.8\times {10}^{-5}kg/ms}\right)}^{1/7}}\\ \delta =0.02735m\end{array}$

Conclusion:

If the flow is turbulent throughout the flow, the $\delta =0.02735m$

Or $\delta =0.02134m$

Therefore, fully turbulent flow will have a boundary layer thickness which is almost $28\%$ higher.