Question

(a) The residuary resistance of a 1/25 scale model of a ship is 7.68N when tested at 1.646 m/s in fresh water of density 1000 Kg/m3. The frictional resistance of the ship at 12 knots in sea water density 1025 kg/m3 is 148KN. Frictional resistance can be assumed to vary wild speed to the power 1.825

Calculate the effectie power (naked) for the ship at the speed currespon ding to the model test

(b) The following additional data apply to the ship operating in service aat the curresponding speed calculated in (a) with a propeller having a pitch of 4.8m.

Appendage and weather alowance = 24%

Quasi-propulsive coefficient (QPC) = 0.71

Propeller speed = 1.85 rev/sec

Taylor wake fraction = 0.3

Propeller thrust = 650kN

Calculate EACH of the following

(i) The torque delivered to the propeller

(ii) The propeller efficiency

(iii) The real slip ratio

Answer 1

Part (a)

Effective power (naked) for the ship at the speed corresponding to the model test.

Model scale = 1/25, so the linear scale ratio lambda = 25.

Model residuary resistance Rr(m) = 7.68 N at model speed 1.646 m/s in fresh water (density 1000 kg/m3).

Froude's law of comparison: the residuary resistance of the ship is related to that of the model by:

Rr(ship) = Rr(model) x (density_ship/density_model) x lambda^3.

Rr(ship) = 7.68 x (1025/1000) x 25^3 = 7.68 x 1.025 x 15625 = 7.68 x 16015.6 = 123000 N = 123 kN.

Ship speed corresponding to model speed (Froude scaling): Vs = Vm x sqrt(lambda) = 1.646 x sqrt(25) = 1.646 x 5 = 8.23 m/s.

Ship speed in knots = 8.23 x 3600/1852 = 29628/1852 = 16.0 knots.

Frictional resistance of the ship at 12 knots = 148 kN. Frictional resistance varies as speed^1.825.

Frictional resistance at 16 knots = 148 x (16/12)^1.825 = 148 x (1.3333)^1.825.

(1.3333)^1.825 = e^(1.825 x ln 1.3333) = e^(1.825 x 0.2877) = e^0.5251 = 1.6905.

Frictional resistance at 16 knots = 148 x 1.6905 = 250.2 kN.

Total naked resistance at 16 knots = residuary + frictional = 123 + 250.2 = 373.2 kN.

Effective power (naked) = resistance x speed = 373200 x 8.23 = 3071436 W = 3071 kW.

Answer: Effective power (naked) = 3071 kW.

Part (b)

Service data at the corresponding speed (16 knots = 8.23 m/s).

Appendage and weather allowance = 24%, QPC = 0.71, propeller pitch = 4.8 m, propeller speed = 1.85 rev/s, Taylor wake fraction = 0.3, propeller thrust = 650 kN.

(i) Torque delivered to the propeller.

Total resistance in service = naked resistance x (1 + allowance) = 373.2 x 1.24 = 462.8 kN.

Thrust required = total resistance/(1 - t), where t is the thrust deduction factor. However, t is not given; the propeller thrust is given as 650 kN, so we use this.

Delivered power PD = Effective power (service)/QPC = (462.8 x 8.23)/0.71 = 3808.8/0.71 = 5364 kW.

Torque Q = PD/(2 x pi x rev/s) = 5364000/(2 x pi x 1.85) = 5364000/11.6239 = 461464 N-m = 461.5 kN-m.

Answer: Torque delivered to the propeller = 461.5 kN-m.

(ii) Propeller efficiency.

Speed of advance Va = V x (1 - w) = 8.23 x (1 - 0.3) = 8.23 x 0.7 = 5.761 m/s.

Thrust power = T x Va = 650000 x 5.761 = 3744650 W = 3744.7 kW.

Propeller efficiency (open water) = Thrust power/Delivered power = 3744.7/5364 = 0.698.

Answer: Propeller efficiency = 0.698 (69.8%).

(iii) Real slip ratio.

Pitch speed = P x rev/s = 4.8 x 1.85 = 8.88 m/s.

Real slip = (Pitch speed - Va)/Pitch speed = (8.88 - 5.761)/8.88 = 3.119/8.88 = 0.3512.

Answer: Real slip ratio = 0.351 (35.1%).

✓ Verified Model Answer (Text Solution)

Effective power (naked) for the ship at the speed corresponding to the model test.

Service data at the corresponding speed (16 knots = 8.23 m/s).