Chapter 11, Problem 48P

The local atmospheric pressure in Denver, Colorado (elevation 1610 m) is 83.4 kPa. Air at this pressure and at 25’C flows with a velocity of 9 m/s over a 2.5-rn 5-rn flat plate. Determine the drag force acting on the top surface of the plate if the air flows parallel to the (a) 5-rn-long side and (b) the 2.5-m-long side.

To determine

(a)

The drag force acting on the top surface of the plate when the air flows parallel to the 5 m long side.

Answer to Problem 48P

$F_D=1.58\text{N}$

Explanation of Solution

Given information:

$\text{Size of plate}=2.5\text{m}\times 5\text{m}$

$\text{Atmospheric pressure}=83.4\text{kPa}$

$\text{Temperature}={25}^{\circ }\text{C}=298\text{K}$

$\text{Velocity}=9\frac{\text{m}}{\text{s}}$

Concept used:

$\rho =\frac{P}{RT}$

$\mathrm{Re}=\frac{\rho VL}{\mu }$

$C_f=\frac{0.074}{{\mathrm{Re}}_L^{1/5}}-\frac{1742}{{\mathrm{Re}}_L}$

$F_D=C_{f}A\frac{\rho V^2}{2}$

$\begin{array}{c}\mathrm{Re}\to \text{Reynolds number}\\ F_D\to \text{Drag force}\\ C_f\to \text{Average friction coefficient}\\ \rho \to \text{Density}\\ \mu \to \text{Dynamic viscosity}\\ A\to \text{Area}\\ V\to \text{Velocity}\end{array}$

Calculation:

The gas constant is, $R=0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$.

$\begin{array}{c}\rho =\frac{P}{RT}\\ =\frac{83.4}{0.287\times 298}\\ =0.975\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

For the given temperature, $\mu =1.849\times {10}^{-5}\frac{\text{kg}}{\text{m-s}}$.

$\begin{array}{c}{\mathrm{Re}}_L=\frac{\rho VL}{\mu }\\ =\frac{0.975\times 9\times 5}{1.849\times {10}^{-5}}\\ =2.373\times {10}^6>{\mathrm{Re}}_{\text{critical}}\left(5\times {10}^5\right)\end{array}$

$\begin{array}{c}{C}_f=\frac{0.074}{{\mathrm{Re}}_L^{1/5}}-\frac{1742}{{\mathrm{Re}}_L}\\ =\frac{0.074}{{\left(2.373\times {10}^6\right)}^{1/5}}-\frac{1742}{2.373\times {10}^6}\\ =0.0032\end{array}$

$\begin{array}{c}{F}_D=C_{f}A\frac{\rho V^2}{2}\\ =0.0032\times \left(5\times 2.5\right)\times \frac{0.975\times 9^2}{2}\times \left(\frac{\text{1}\text{N}}{1\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =1.58\text{N}\end{array}$

Conclusion:

The drag force acting on the top surface of the plate when the air flows parallel to the 5 m long side is $1.58\text{N}$.

To determine

(b)

The drag force acting on the top surface of the plate when the air flows parallel to the 2.5 m long side.

Answer to Problem 48P

$F_D=1.48\text{N}$

Explanation of Solution

Given information:

$\text{Size of plate}=2.5\text{m}\times 5\text{m}$

$\text{Atmospheric pressure}=83.4\text{kPa}$

$\text{Temperature}={25}^{\circ }\text{C}=298\text{K}$

$\text{Velocity}=9\frac{\text{m}}{\text{s}}$

Concept used:

$\rho =\frac{P}{RT}$

$\mathrm{Re}=\frac{\rho VL}{\mu }$

$C_f=\frac{0.074}{{\mathrm{Re}}_L^{1/5}}-\frac{1742}{{\mathrm{Re}}_L}$

$F_D=C_{f}A\frac{\rho V^2}{2}$

$\begin{array}{c}\mathrm{Re}\to \text{Reynolds number}\\ F_D\to \text{Drag force}\\ C_f\to \text{Average friction coefficient}\\ \rho \to \text{Density}\\ \mu \to \text{Dynamic viscosity}\\ A\to \text{Area}\\ V\to \text{Velocity}\end{array}$

Calculation:

The gas constant is, $R=0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$.

$\begin{array}{c}\rho =\frac{P}{RT}\\ =\frac{83.4}{0.287\times 298}\\ =0.975\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

For the given temperature, $\mu =1.849\times {10}^{-5}\frac{\text{kg}}{\text{m-s}}$.

$\begin{array}{c}{\mathrm{Re}}_L=\frac{\rho VL}{\mu }\\ =\frac{0.975\times 9\times 2.5}{1.849\times {10}^{-5}}\\ =1.186\times {10}^6>{\mathrm{Re}}_{\text{critical}}\left(5\times {10}^5\right)\end{array}$

$\begin{array}{c}{C}_f=\frac{0.074}{{\mathrm{Re}}_L^{1/5}}-\frac{1742}{{\mathrm{Re}}_L}\\ =\frac{0.074}{{\left(1.186\times {10}^6\right)}^{1/5}}-\frac{1742}{1.186\times {10}^6}\\ =0.003\end{array}$

$\begin{array}{c}{F}_D=C_{f}A\frac{\rho V^2}{2}\\ =0.003\times \left(5\times 2.5\right)\times \frac{0.975\times 9^2}{2}\times \left(\frac{\text{1}\text{N}}{1\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =1.48\text{N}\end{array}$

Conclusion:

The drag force acting on the top surface of the plate when the air flows parallel to the 2.5 m long side is $1.48\text{N}$.