Engineering, 08.04.2020 04:28 markaylarowland8

Consider a NACA 2412 airfoil in a low-speed flow at zero degrees angle of attack and a Reynolds number of 8.9·106 . Calculate the percentage of drag from pressure drag due to flow separation (form drag). Assume a fully turbulent boundary layer over the airfoil. Assume that the airfoil is thin enough so that the skin-friction drag can be estimated by the flat-plate results discussed earlier. Now consider the same airfoil at an angle of attack of 6 degrees. Again, calculate the percentage of drag from pressure drag due to flow separation. What does this tell you about the rapid increase in drag coefficient as the angle of attack increases?

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Answer from: ben2005

Following are the calculation to the given question:

Given:

$\text{NACA 2412 airfoil}\\\\\to Re = 8.9 \times 10^6$

Calculating the turbulent flow when the drag is coefficient:

$\to \bold{Cdf = \frac{0.074}{ Re^{0.2}}} \\\\$

$\bold{= \frac{0.074}{ (8.9 x 10^6)^{0.2}}}\\\\= \bold{\frac{0.074}{ 24.54}} \\\\ = \bold{0.003}$

From sides of plates:

$\to Cd = 2 \times 0.003 = 0.006\\\\$

For $\bold{ 0^{\circ}\ angle}$ of attack for $\bold{NACA\ 2412}$ :

$\to Cdf = 0.006\\\\$

Also,

$\to Cd = Cdf + Cdp\\\\\to 0.006 = 0.006 + Cdp\\\\\to Cdp=0.006- 0.006 \\\\\to Cdp=0 \\\\$

Therefore, the drag pressure =0

Now, for $\bold{ 0^{\circ}\ angle}$ of attack for $\bold{NACA\ 2412}$ : :

$\to Cd = 0.0075\\\\$

Also,

$\to Cd = Cdf + Cdp\\\\\to 0.0075 = 0.006 + Cdp\\\\\to Cdp = 0.0075 - 0.006 = 0.0015$

Calculating the percentage of drag pressure:

$= (\frac{Cdp}{ Cdf}) \times 100$

$= (\frac{0.0015}{0.0075}) \times 100 \\\\ = (\frac{15}{75}) \times 100 \\\\ = (\frac{1}{5}) \times 100 \\\\ =0.2 \times 100\\\\= 20 \%\\$

Hence, drag pressure is $\bold{20\%}$ of drag pressure due to separation flow.

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Answer from: justiceisbae1

a) pressure drag is zero (0)

b) pressure drag is 20%

Explanation:

Ans) Given,

NACA 2412 airfoil

Re = 8.9 x 10^6

We know, for turbulent flow ,drag coefficient, Cdf = 0.074 / Re^0.2

=> Cdf = 0.074 / (8.9 x 10^6)^0.2

=> Cdf = 0.003

For both side of plate, Cd = 2 x 0.003 = 0.006

For zero degree angle of attack for NACA 2412, Cdf = 0.006

Also, Cd = Cdf + Cdp

=> 0.006 = 0.006 + Cdp

=> Cdp = 0

Hence, pressure drag is zero

Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

=> 0.0075 = 0.006 + Cdp

=> Cdp = 0.0075 - 0.006

=> Cdp = 0.0015

Hence, Pressure drag percentage = (Cdp / Cdf) x 100

=> Pressure drag percent = (0.0015/0.0075) x 100 = 20 %

Hence, pressure drag is 20% of pressure drag due to flow seperation Ans) Given,

NACA 2412 airfoil

Re = 8.9 x 10^6

We know, for turbulent flow ,drag coefficient, Cdf = 0.074 / Re^0.2

=> Cdf = 0.074 / (8.9 x 10^6)^0.2

=> Cdf = 0.003

For both side of plate, Cd = 2 x 0.003 = 0.006

For zero degree angle of attack for NACA 2412, Cdf = 0.006

Also, Cd = Cdf + Cdp

=> 0.006 = 0.006 + Cdp

=> Cdp = 0

Hence, pressure drag is zero

Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

=> 0.0075 = 0.006 + Cdp

=> Cdp = 0.0075 - 0.006

=> Cdp = 0.0015

Hence, Pressure drag percentage = (Cdp / Cdf) x 100

=> Pressure drag percent = (0.0015/0.0075) x 100 = 20 %

Hence, pressure drag is 20% of pressure drag due to flow seperation

Answer from: Quest

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Answer from: Quest

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Consider a NACA 2412 airfoil in a low-speed flow at zero degrees angle of attack and a Reynolds numb...

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