I am praticisning question 2 a,b,c however the answers seems like I have negative climbrate  and i would like some detailed  feedback

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It can be shown that the rate of climb for a propeller driven aircraft is: Pornp PV'SCD. 2kW cos y W 2W
pVS a) Assuming that 7p is independent of the airspeed V, find an expression for the airspeed at which
the climb rate is maximized, assuming that cosy ~ 1. Using the following data, W = 11800N k... 

My though process is to take the derivative of the climb velocity equation then I would equal to zero to get  the max velocity function. Then I would substitute the equation back into the derivative to gain max velocity hence climb rate.

Here is the matlab calculations I Have done

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syms P_br nth W rho V S C_D k g cosg=1 cosg = 1 V_v=((P_br*nth)/W) - ((rho*v^3*s*c_D) /(2*W)+(2*k*W)
/(rho *v*5) ) VV= max_v=diff(V_v,V) V=( (4*5*W^2*k) / (3*C_D*S^2*rho*2) )^(1/4) Por nth 2 WK CosV3p
W=11800; %weight in newtons W SV p 2 W rho=1.10; % density of air in kg/m*3 max_V = S=... 

 

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dVv = 0 dV dVv 2 * S * W *k 3 * CDO * S * VP_o dV S * V-p 2 * W 3 * CDo * S * V2 * p 2 *S *W *k 2 * W S
* V-p 3 * CDo * $2 * V4 * p2 = 4 * S * W2 * k 4 * S * W2 * k V4 = 3 * CDO * $2 * p2 Max V 4 * S * W2 * k V
= 3 * CDO * 52 * p2)-