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A twin turbo prop is flying with power loading of 12. What will be the maximum velocity?(a) 48.225 mph(b) 48.225 m/s(c) 48.225 km/h(d) 225.48 m/hThe question was asked by my college professor while I was bunking the class.My query is from Thrust Weight Ratio-2 in division Thrust-Weight Ratio and Wing Loading of Aircraft Design

Answer»

Right option is (a) 48.225 mph

Easiest explanation: Given, Power LOADING P.L. = 12.

P.L. is given by,

W/hp =1 / (a*Vmax^c), for twin TURBO prop a=0.012, c=0.5.

Hence, MAXIMUM velocity Vmax is given by,

12 = 1 / (0.012*Vmax^0.5)

0.012*Vmax^0.5 = 1/12.

Vmax^0.5 = 1 / (0.012*12) = 6.944

Takin log at both sides,

0.5*LN (Vmax) = ln (6.944) = 1.937

Now, taking anti-log,

Vmax = e^(1.937/0.5) = e^3.875 = 48.225mph.



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