Chapter 19, Problem 19.1P

The wing on a Piper Cherokee general aviation aircraft is rectangular, with a span of 9.75 m and a chord of 1.6 m. The aircraft is flying at cruising speed $\left(\text{141 mi/h}\right)$ at sea level. Assume that the skin-friction drag on the wing can be approximated by the drag on a flat plate of the same dimensions. Calculate the skin-friction drag:

a. If the flow were completely laminar (which is not the case in real life)

b. If the flow were completely turbulent (which is more realistic) Compare the two results.

To determine

The value of skin friction drags for laminar flow.

Answer to Problem 19.1P

The value of skin friction drag for laminar flow is $38.38\text{N}$ .

Explanation of Solution

Given:

The length of the wing is $l=9.75\text{m}$ .

The chord of the wing is $C=1.6\text{m}$ .

The cruising speed of the aircraft is $V=141\text{mi}/\text{h}$ .

Formula used:

The expression for the area of wing is given as,

  $A=l\times C$

The expression for the Reynolds number is given as,

  $\mathrm{Re}=\frac{{\rho }_{0}VC}{{\mu }_0}$

Here, $\mathrm{Re}$ is the Reynolds number, ${\mu }_0$ is the dynamic viscosity of the air, ${\rho }_0$ is the density of air.

The expression for the coefficient of drag is given as,

  $C_d=\frac{1.328}{\sqrt{\mathrm{Re}}}$

The expression for the drag is given as,

  $D=C_d\cdot \frac{{\rho }_0{V}^{2}A}{2}$

Calculation:

The area of the wing can be calculated as,

  $\begin{array}{l}A=l\times C\\ A=9.75\text{m}\times \text{1}\text{.6}\text{m}\\ A=15.6{\text{m}}^{\text{2}}\end{array}$

It is known that the density of the air is $1.225\text{kg}/{\text{m}}^{\text{3}}$ and the dynamic viscosity of the air is $1.78\times {10}^{-5}\text{Pa}\cdot \text{s}$ .

The Reynolds number can be calculated as,

  $\begin{array}{l}\mathrm{Re}=\frac{{\rho }_{0}VC}{{\mu }_0}\\ \mathrm{Re}=\frac{1.225\text{kg}/{\text{m}}^{\text{3}}\times \left(141\text{mi}/\text{h}\times \frac{0.4470\text{m}/\text{s}}{1\text{mi}/\text{h}}\right)\times 1.6\text{m}}{1.78\times {10}^{-5}\text{Pa}\cdot \text{s}}\\ \mathrm{Re}=6.90\times {10}^6\end{array}$

The coefficient of the drag can be calculated as,

  $\begin{array}{l}{C}_d=\frac{1.328}{\sqrt{\mathrm{Re}}}\\ C_d=\frac{1.328}{\sqrt{6.90\times {10}^6}}\\ C_d=5.05\times {10}^{-4}\end{array}$

The drag can be calculated as,

  $\begin{array}{l}D=C_d\cdot \frac{{\rho }_0{V}^{2}A}{2}\\ D=5.05\times {10}^{-4}\times \frac{1.225\text{kg}/{\text{m}}^{\text{3}}\times {\left(141\text{mi}/\text{h}\times \frac{0.4470\text{m}/\text{s}}{1\text{mi}/\text{h}}\right)}^2\times \text{15}\text{.6}{\text{m}}^{\text{2}}}{2}\\ D=38.38\text{N}\end{array}$

Conclusion:

Therefore, the value of skin friction drag for laminar flow is $38.38\text{N}$ .

To determine

The value of skin friction drags for turbulent flow.

Answer to Problem 19.1P

The value of skin friction drag for turbulent flow is $240.93\text{N}$ .

Explanation of Solution

The length of the wing is $l=9.75\text{m}$ .

The chord of the wing is $C=1.6\text{m}$ .

The cruising speed of the aircraft is $V=141\text{mi}/\text{h}$ .

Formula used:

The expression for the area of wing is given as,

  $A=l\times C$

The expression for the Reynolds number is given as,

  $\mathrm{Re}=\frac{{\rho }_{0}VC}{{\mu }_0}$

Here, $\mathrm{Re}$ is the Reynolds number, ${\mu }_0$ is the dynamic viscosity of the air, ${\rho }_0$ is the density of air,

The expression for the coefficient of drag is given as,

  $C_d=\frac{0.074}{{\left(\mathrm{Re}\right)}^{\frac{1}{5}}}$

The expression for the drag is given as,

  $D=C_d\cdot \frac{{\rho }_0{V}^{2}A}{2}$

Calculation:

The area of the wing can be calculated as,

  $\begin{array}{l}A=l\times C\\ A=9.75\text{m}\times \text{1}\text{.6}\text{m}\\ A=15.6{\text{m}}^{\text{2}}\end{array}$

It is known that the density of the air is $1.225\text{kg}/{\text{m}}^{\text{3}}$ and the dynamic viscosity of the air is $1.78\times {10}^{-5}\text{Pa}\cdot \text{s}$ .

The Reynolds number can be calculated as,

  $\begin{array}{l}\mathrm{Re}=\frac{{\rho }_{0}VC}{{\mu }_0}\\ \mathrm{Re}=\frac{1.225\text{kg}/{\text{m}}^{\text{3}}\times \left(141\text{mi}/\text{h}\times \frac{0.4470\text{m}/\text{s}}{1\text{mi}/\text{h}}\right)\times 1.6\text{m}}{1.78\times {10}^{-5}\text{Pa}\cdot \text{s}}\\ \mathrm{Re}=6.90\times {10}^6\end{array}$

The coefficient of the drag can be calculated as,

  $\begin{array}{l}{C}_d=\frac{0.074}{{\left(\mathrm{Re}\right)}^{\frac{1}{5}}}\\ C_d=\frac{0.074}{{\left(6.90\times {10}^6\right)}^{\frac{1}{5}}}\\ C_d=3.17\times {10}^{-3}\end{array}$

The drag can be calculated as,

  $\begin{array}{l}D=C_d\cdot \frac{{\rho }_0{V}^{2}A}{2}\\ D=3.17\times {10}^{-3}\times \frac{1.225\text{kg}/{\text{m}}^{\text{3}}\times {\left(141\text{mi}/\text{h}\times \frac{0.4470\text{m}/\text{s}}{1\text{mi}/\text{h}}\right)}^2\times \text{15}\text{.6}{\text{m}}^{\text{2}}}{2}\\ D=240.93\text{N}\end{array}$

The value of skin friction drag for turbulent flow is $240.93\text{N}$ and The value of skin friction drag for laminar flow is $38.38\text{N}$ .

By comparing these two results, it is showing that the skin drag force in turbulent flow is 6.27 times greater than the skin drag force in laminar flow.

Conclusion:

Therefore, the value of skin friction drag for turbulent flow is $240.93\text{N}$ .