Question

A ship tows a submerged cylinder, which is 1.5 m in diameter and 22 m long, at 5 m/s in fresh water at 208C. Estimate the towing power, in kW, required if the cylinder is (a) parallel and (b) normal to the tow direction.

Answer 1

Answer: a) P = 120kw

b) P = 800kw

Explanation: Please find the attached file for the solution

Answer 2

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