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Question 4(10+10=20 Point): A symmetric airfoil operates at a small angle of attack (alpha =2^circ ) in subsonic flow. The lift coefficient is given by C_(L)=2pi alpha (with a in radians). (a) Calculate the lift coefficient (C_(L)) (b) If the chord length is 1.5 m, wing span is 2 m and the airspeed is 30m/s,rho _(infty )=1.225kg/m^3 determine the lift force.

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Question 4(10+10=20 Point):
A symmetric airfoil operates at a small angle of attack (alpha =2^circ ) in subsonic flow. The lift coefficient
is given by C_(L)=2pi alpha  (with a in radians).
(a) Calculate the lift coefficient (C_(L))
(b) If the chord length is 1.5 m, wing span is 2 m and the airspeed is 30m/s,rho _(infty )=1.225kg/m^3
determine the lift force.

Question 4(10+10=20 Point): A symmetric airfoil operates at a small angle of attack (alpha =2^circ ) in subsonic flow. The lift coefficient is given by C_(L)=2pi alpha (with a in radians). (a) Calculate the lift coefficient (C_(L)) (b) If the chord length is 1.5 m, wing span is 2 m and the airspeed is 30m/s,rho _(infty )=1.225kg/m^3 determine the lift force.

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(a) To calculate the lift coefficient $(C_{L})$, we need to convert the angle of attack from degrees to radians. <br /><br />Given that the angle of attack is $2^{\circ}$, we can convert it to radians using the formula:<br /><br />Radians = Degrees * (π / 180)<br /><br />So, the angle of attack in radians is:<br /><br />Radians = 2 * (π / 180) = 0.0349 radians<br /><br />Now, we can calculate the lift coefficient using the given formula:<br /><br />$C_{L} = 2\pi \alpha$<br /><br />Substituting the value of α in radians, we get:<br /><br />$C_{L} = 2\pi * 0.0349 = 0.2196$<br /><br />Therefore, the lift coefficient $(C_{L})$ is 0.2196.<br /><br />(b) To determine the lift force, we can use the lift equation:<br /><br />$L = \frac{1}{2} \rho_{\infty} v^2 S C_{L}$<br /><br />Where:<br />- $L$ is the lift force<br />- $\rho_{\infty}$ is the air density<br />- $v$ is the airspeed<br />- $S$ is the wing area<br />- $C_{L}$ is the lift coefficient<br /><br />Given values:<br />- Chord length = 1.5 m<br />- Wing span = 2 m<br />- Airspeed = 30 m/s<br />- Air density = 1.225 kg/m³<br /><br />First, we need to calculate the wing area $(S)$ using the formula:<br /><br />$S = \frac{1}{2} c b$<br /><br />Where:<br />- $c$ is the chord length<br />- $b$ is the wing span<br /><br />Substituting the given values, we get:<br /><br />$S = \frac{1}{2} * 1.5 * 2 = 1.5 m^2$<br /><br />Now, we can calculate the lift force using the lift equation:<br /><br />$L = \frac{1}{2} * 1.225 * (30)^2 * 1.5 * 0.2196$<br /><br />$L = 0.5 * 1.225 * 900 * 1.5 * 0.2196$<br /><br />$L = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.225 * 900 * 1.5 * 0.2196 = 0.5 * 1.
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