Question

Approximate the skin friction drag on a 1m long by 60 cm diameter cylinder, located axially in a wind tunnel, when the air speed is 4.5 m/s. The pressure is atmospheric and the temperature is 50°C.

Answer 1

Answer:

The drag force is $1.76\times10^{-2}\ N$

Explanation:

Given that,

Diameter = 60 cm

Length = 1 m

Air speed = 4.5 m/s

Temperature = 50°C

We need to calculate the Reynolds number

Using formula of Reynolds number

$R_{e}=\dfrac{vl}{\mu}$

Put the value into the formula

$R_{e}=\dfrac{4.5\times1}{1.900\times10^{-5}}$

$R_{e}=2.368\times10^{5}$

We need to calculate the drag coefficient

Using formula of drag coefficient

$C_{d}=\dfrac{2\times0.646}{\sqrt{R_{e}}}$

Put the value into the formula

$C_{d}=\dfrac{2\times0.646}{\sqrt{2.368\times10^{5}}}$

$C_{d}=0.002655$

We need to calculate the drag force

Using formula of drag force

$F_{d}=\dfrac{1}{2}\times C_{d}\times\rho\times A\times v^2$

Put the value into the formula

$F_{d}=\dfrac{1}{2}\times0.002655\times1.095\times0.6\times1\times(4.5)^2$

$F_{d}=0.01766\ N$

$F_{d}=1.76\times10^{-2}\ N$

Hence, The drag force is $1.76\times10^{-2}\ N$