Answered

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?

Answer :

adebayodeborah8

Answer:

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  

Following are the calculation to the given question:

Given:

[tex]\text{NACA 2412 airfoil}\\\\\to Re = 8.9 \times 10^6[/tex]

Calculating the turbulent flow when the drag is coefficient:

[tex]\to \bold{Cdf = \frac{0.074}{ Re^{0.2}}} \\\\[/tex]

            [tex]\bold{= \frac{0.074}{ (8.9 x 10^6)^{0.2}}}\\\\= \bold{\frac{0.074}{ 24.54}} \\\\ = \bold{0.003}[/tex]

From sides of plates:

[tex]\to Cd = 2 \times 0.003 = 0.006\\\\[/tex]

For [tex]\bold{ 0^{\circ}\ angle}[/tex] of attack for [tex]\bold{NACA\ 2412}[/tex]:

[tex]\to Cdf = 0.006\\\\[/tex]

Also,

[tex]\to Cd = Cdf + Cdp\\\\\to 0.006 = 0.006 + Cdp\\\\\to Cdp=0.006- 0.006 \\\\\to Cdp=0 \\\\[/tex]

Therefore, the drag pressure =0

Now, for  [tex]\bold{ 0^{\circ}\ angle}[/tex] of attack for [tex]\bold{NACA\ 2412}[/tex]:  :

[tex]\to Cd = 0.0075\\\\[/tex]

Also,

[tex]\to Cd = Cdf + Cdp\\\\\to 0.0075 = 0.006 + Cdp\\\\\to Cdp = 0.0075 - 0.006 = 0.0015[/tex]

Calculating the percentage of drag pressure:

[tex]= (\frac{Cdp}{ Cdf}) \times 100[/tex]

[tex]= (\frac{0.0015}{0.0075}) \times 100 \\\\ = (\frac{15}{75}) \times 100 \\\\ = (\frac{1}{5}) \times 100 \\\\ =0.2 \times 100\\\\= 20 \%\\[/tex]

Hence, drag pressure is [tex]\bold{20\%}[/tex] of drag pressure due to separation flow.  

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