# For the BD-5J, calculate  (a) The maximum value of CL/CD (b) The maximum value of CL^(1/ 2) /CD

For the BD-5J, calculate

(a) The maximum value of CL/CD

(b) The maximum value of CL^(1/ 2) /CD

(c) The velocities at which they occur at sea level

(d) The velocities at which they occur at 10,000 ft

Worldtech Asked on 13th December 2023 in

(a) $\left ( \frac{C_{L}}{C_{D}} \right )_{max} = \frac{1}{2\sqrt{k\times C_{D,0}}} = \frac{1}{\sqrt{0.04\times 0.016}} = 19.7642$

(b)$\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max} = \frac{\frac{C_{D,0}}{3K}}{\frac{4C_{D,0}}{3}}=\frac{\sqrt{\frac{0.016}{3\left ( 0.04 \right )}}}{\frac{4\left ( 0.016 \right )}{3}} = \frac{0.36515}{0.02133}=17.12$

(c)For $$\left ( \frac{C_{L}}{C_{D}} \right )_{max}$$$C_{L} = \sqrt{\frac{C_{D,0}}{K}}=\sqrt{\frac{0.016}{0.04}} = 0.63246$

$S = 4.4\,m^{2};\rho = 1.225\,kg/m^{3},W = 3700\,N$

$V\left ( \frac{C_{L}}{C_{D}} \right )_{max} = \sqrt{\frac{2W}{SC_{L}\rho }} = \sqrt{\frac{2\left ( 3700 \right )}{4.4\left ( 1.225 \right )\left ( 0.63246 \right )}}= 46.59\,m/s$

For $$\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}$$$C_{L} = 0.36515 V\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}=V\left ( \frac{C_{L}}{C_{D}} \right )_{max}\sqrt{\frac{0.63246}{0.36515}}= 61.316\,m/s$

(d) At 10,000 ft, $\frac{{\rho}’}{\rho} = \sigma = 0.73833$

${V}'{\left(\frac{C_{L}}{C_{D}} \right)}_{max}= V\left ( \frac{C_{L}}{C_{D}} \right )_{max}\times \frac{1}{\sigma} = 46.59\times \frac{1}{0.73833}$

$\Rightarrow {V}’\left ( \frac{C_{L}}{C_{D}} \right )_{max} = 54.22\,m/s$

${V}'{\left(\frac{C_{L}^{1/2}}{C_{D}} \right)}_{max}= V\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}\times \frac{1}{\sigma} = 61.316 \times \frac{1}{\sqrt{0.73833}} = 71.36\,m/s$

techAir Answered on 13th December 2023.