Aerodynamics Question From Foundations of Aerodynamics by Kuethe and Chow, Probl
ID: 1996673 • Letter: A
Question
Aerodynamics Question
From Foundations of Aerodynamics by Kuethe and Chow, Problem 6.5.1
A wing with an elliptical planform is flying through sea-level air at a speed of 45 m/s. The wing loading WIS = 1000 N/m^2. The wing is untwisted and has the same section from root to tip. The lift curve slope of the section m_0 is 5.7. The span of the wing is 10 m, and the aspect ratio is 5. Find the sectional-lift and induced-drag coefficients. Find also the effective, induced, and absolute angles of attack. What is the power that is required to overcome the induced drag of the wing?Explanation / Answer
Given data:
Free stream velocity (v0) = 45m/s
Wing loading (W/S) = 1000N/m2
Slope (m0) = 5.7
Wing span(b) = 10 m
Aspect ratio(AR) = 5
To find:
Rho = 1.225kg/m3 ( density of air at sea level)
Since the wing is an elliptical platform the sectional lift coefficient is equal to wing lift coefficient.
= (0.8062)2/(3.14*5)
CDi = 0.041
= 0.8062/5.7
= - 0.041/0.8062
= - 0.05 rad
= 0.1414 + 0.05
= 0.1922 rad
Di = CDi * 0.5*rho*v2*S
= [(CL)2 * 0.5*rho*v2*S]/ (pi*AR)
Di = [{W/(0.5*rho*v2)*S}2* 0.5*rho*v2*S]/ (pi*AR)
AR = b2/S
5 = 102/S,
S = 20 m2
W/S = 1000, W = 1000*20 = 20000N
Equating we get
Di = [1/(pi*0.5*rho*v2)]*[W/b]2
= [1/(3.14*0.5*1.225*452)]*[20000/10]2
= 1026.54 N
Power Required PR = Di * v0
= 1026 * 45
= 46194 N
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.