Answer:
(a) P = 121kW (b) P = 823kW
Explanation:
Given
U = 5m/s, D = 1.5m, L = 22m, ρ = 998kg/m³
(a) Parallel, L/D = 15,
Re = 998×5×22/0.001 = 1.1×10⁸,
Area = π ×D²/4 = π × 1.5²/4 = 1.77m²
From Table estimate Cd,frontal = 1.1
Fd = 1/2 × Cd,frontal × ρ × U² × Area
Fd = 1/2 × 1.1 × 998 × 5² × 1.77 = 24289N
P = F×U = 24289 × 5 = 121445W ≈ 121kW
(b) Normal, Re = 998×5×1.5/0.001 = 7.5×10⁶,
Area = DL = 1.5 × 22 = 33m²
From Table estimate Cd,frontal = 0.4
Fd = 1/2 × Cd,frontal × ρ × U² × Area
Fd = 1/2 × 0.4 × 998 × 5² × 33 = 164670N
P = F×U = 164670 × 5 = 121445W ≈ 823kW
Fd is the drag force exerted by the fluid (water) on the submerged body. This force tends to oppose the motion of the submerged cylinder through it. This force increases with area and also with velocity. So for large area the force of resistance of the water would be large and for a small area it will also be small. The same is true for velocity.
In order for the ship to tow this cylinder successfully, it must provide enough power to overcome the resisting force of the water (the drag force as the name implies.
This force is given by the relation,
Fd = 1/2 × Cd,frontal × ρ × U² × Area
Where Fd = Drag force
Cd , frontal = coefficient of drag
ρ = density of fluid (water,)
U = velocity other body in the fluid
A = surface area of the object made available to the fluid in the direction of travel.
Power = F × U
