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?

Answers: 3

Show answers

Answers

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.

Learn more:

link

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

correct me if i’m wrong but i believe it’s false

Answer from: Quest

answer:

ask me the quest

explanation:

Another question on Engineering

Engineering, 04.07.2019 18:10

Steel is coated with a thin layer of ceramic to protect against corrosion. what do you expect to happen to the coating when the temperature of the steel is increased significantly? explain.

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Abrake has a normal braking torque of 2.8 kip in and heat-dissipating cast-iron surfaces whose mass is 40 lbm. suppose a load is brought to rest in 8.0 s from an initial angular speed of 1600 rev/min using the normal braking torque; estimate the temperature rise of the heat dissipating surfaces.

Answers: 3

Answer

Engineering, 04.07.2019 18:10

At 12 noon, the count in a bacteria culture was 400; at 4: 00 pm the count was 1200 let p(t) denote the bacteria cou population growth law. find: (a) an expression for the bacteria count at any time t (b) the bacteria count at 10 am. (c) the time required for the bacteria count to reach 1800.

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Compute the pressure drop of 30°c air flowing with a mean velocity of 8 m/s in a circular sheet-metal duct 300 mm in diameter and 15 m long. use a friction factor, f 0.02, and pair = 1.1644 kg/m a. 37.26 pa b. 25.27 pa n c. 29.34 pa d. 30.52 pa

Answers: 1

Answer

You know the right answer?

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

Questions