Question

The engine of the The power (W) required to overcome drag is f 650 KW and has a total mass of m-650 Kg car can produce a maximum power o P = FoV where V (m/s) is the vehicle speed. The total drag force FD (N) acting on the car is the sum of the drag on the body and the drag on the wings, The individual drag force components can be calculated using the known drag coefficients. For example the drag on the body, where A (m2) is the reference area, ρ-1.2 (kg/m3) is the fluid density, V (m/s) is the velocity of the car and Co is a drag coefficient. Similarly the down-force FL is given by and the component lift forces determined from FL-CLtpV2. The body of the car has a frontal area of Ab 1.8 m2 and lift/drag coefficients of CLb0.7 and CD,b0.2. The front and rear wings on the vehicle each have a planform area of A.52. The lift and drag coefficient of the wings vary with angle of attack θ and are shown in the curve below: 0.9 2.5 0.6 1.5 0.3 0.5 10 15 20 Angle of Attack The total lateral force that the vehicle tires can resist is given by Fe = 1.8(mg + FL) where Fi is the total down force acting on the vehicle and gravitational acceleration is g lateral force generated going through a corner of radius R is 9.81 m/s2. The

* This is all the information given

QUESTION 1 (iii,iv) PLEASE!!!:

Examine the car performance when the angle of both the front and rear wings is set to θ = 10◦. In all the analysis that follows ignore frictional losses other than drag. Determine:

1. (iii) What is the maximum speed the vehicle can safely corner a bend with R = 30 m (ans. 108Km/hr).

   What is the power required at this speed? (ans. 27.8KW)

2. (iv) What is the maximum speed the vehicle can corner a bend of R = 80 m? (ans. 309Km/hr)

Notes:

* The maximum speed the vehicle will attain on a long straight track is 309Km/hr.

* The required down-force (N) for the vehicle to safely corner a bend with a radius of R = 30 m at 100 Km/hr is 2900N. This force attained with θ = 10◦ is 3850N.

Answer 1

Fdb = cdb 1/2Ab + v^2 cbd = 0.2, Ab = 1.8, t = 1.2 f alpha b = 0.2 times 1/2 times 1.8 times v^2 = 0.216v^2 therefore FDW = CDW1/2 AW + W^2 CDW from the graph is between 0.3 and 0.6 = 0.45 AW = 1.5 FDW = 0.45 times 1/2 times 1.5 times 1.2 times v^2 = 0.405v^2 effective FDW = 2 times 0.405v^2 = 0.81v^2 because there are 2 wings 1 for front and for FD = FDb + FDw = 0.216v^2 + 0.81v^2 + 0.81v^2 = 1.026v^2 we know p = FDV 650 times 10^3 = 1.026v^2 times v v^3 = 6335282651 v = 85.88m/s v = 85.88 times 18/5 km/hr = 309.18 km/hr (iii) v = 100km/hr = 100 times 5/18 = 27.778 m/s FR = mv^2/R = 650 times 27.78^2/30 = 16718.3744N this lateral force is subjected to the types F_F = 1.8(650 times 9.81 times F_L) = 16718.37 FL = 291N