# One way to reduce the drag of a blunt object is to install vanes to suppress the amount of separation. Such a procedure was used on model trucks in a wind tunnel study. For tests on a van-type truck without vanes the CD was 0.78. However, when vanes were installed around the top and side leading edges of the truck body (see the figure), a 25% reduction in CD was achieved. For a truck with a projected area of 8.36 m2, what reduction in drag force will be effected by installation of the vanes when the truck travels at 100 km/h? Assume standard atmospheric pressure and a temperature of 20°C.

Question-AnswerCategory: Fluid MechanicsOne way to reduce the drag of a blunt object is to install vanes to suppress the amount of separation. Such a procedure was used on model trucks in a wind tunnel study. For tests on a van-type truck without vanes the CD was 0.78. However, when vanes were installed around the top and side leading edges of the truck body (see the figure), a 25% reduction in CD was achieved. For a truck with a projected area of 8.36 m2, what reduction in drag force will be effected by installation of the vanes when the truck travels at 100 km/h? Assume standard atmospheric pressure and a temperature of 20°C.
vivek asked 9 months ago

One way to reduce the drag of a blunt object is to install vanes to suppress the amount of separation. Such a procedure was used on model trucks in a wind tunnel study. For tests on a van-type truck without vanes the CD was 0.78. However, when vanes were installed around the top and side leading edges of the truck body (see the figure), a 25% reduction in CD was achieved. For a truck with a projected area of 8.36 m2, what reduction in drag force will be effected by installation of the vanes when the truck travels at 100 km/h? Assume standard atmospheric pressure and a temperature of 20°C.

ਸਾਜਿਦ ਅਹਿਮਦ answered 9 months ago

Drag co-efficient,
Projected area,
Reduced drag co-efficient,  = 0.195
Speed of truck, v = 100km/h = 27.778m/s
Density of air at 20°C , ρ=1.2kg/m³
From drag force equation:
Drag force,
= 0.195×8.36×(1.2×27.778²/2)
= 754.73N
Hence, the Reduced drag force is 754.73N