Q10 (10 Marks) Ship Resistance & Propulsion 🔥 Repeated 2x in exams
SC&S • Written Exam

A ship 150 m in length, 24 m breadth, displaces 25000 tonne when floating at a draft of 9 m in, sea water of density 1025 kg/m3. The ship's propeller has a diameter of 5.8 m, a pitch ratio of 0.9 and a blade area ratio of 0.45. With the propeller operating at 2 revs/sec, the following results were

recorded:

Apparent slip = 0.06

Thrust power = 3800 Kw

Propeller efficiency = 64%

The taylor wake fraction Wt = 0.5Cb-0.05

Calculate each of the following for the above condition:

(a) The ship's speed

(b) The real slip ratio

(c) The thrust per unit are of blade surface

(d) The torque delivered to the propeller

Appeared In: Aug 2019Jun 2019

Verified Model Answer (Text Solution)

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Given:

$$L \space = \space 150m$$

$$B \space = \space 24m$$

$$D \space = \space 9m$$

$$\Delta=25000t$$

$$d \space = \space 5.8m$$

$$p \space = \space 0.9m$$

$$BAR \space = \space 0.45m$$

$$n \space = \space 2 \space rev/sec$$

$$Apparent \space slip \space = \space 0.06m$$

$$T_p \space = \space 3800kw$$

$$η_{prop} \space = \space 64\%$$

$$W_f \space = \space 0.5C_b - 0.05$$

$$p\space=\space{{P}\over d}\space$$

$$0.9 \space = \space {{P} \over 5.8}$$

$$Pitch \space = \space 5.22m$$

$$V_T \space = \space {{P \times N \times 3600} \over 1852}$$

$$V_T P \space = \space 20.29 \space knots$$

$$App. \space slip \space = \space {{V_T - V} \over V_T}$$

$$0.06\space=\space{{20.29-V}\over20.29}$$

$$Ship's \space speed \space (V) \space = \space 19.07 \space knots $$

$$C_{b}=\frac{\Delta}{L\times B\times D\times\rho}$$

$$=\frac{25000}{150\times24\times9\times1.025}$$

$$C_{b}=0.752$$

$$given, \space W_f \space = \space 0.5C_b - 0.05$$

$$W_{f}=0.5\times0.752-0.05$$

$$W_{f}=0.326$$

$$W_F \space = \space {{V - V_a} \over V}$$

$$0.326 \space = \space {{19.07 - V_a} \over 19.07}$$

$$V_a \space = \space 12.85Knots$$

$$Real\space slip\space=\space{{V_T - V_a} \over V_T}\space\space{}$$

$$=\frac{20.29-12.85}{20.29}$$

$$Real \space slip \space = \space 0.367 $$

$$Thrust \space power \space (T_p) \space = \space d_p \times η_{prop}$$

$$3800=d_{p}\times0.64$$

$$d_{p}=5937.5kw$$

$$Thrust \space power \space T_p \space = \space Thrust \times V_a $$

$$3800 \space = \space Thrust \times 12.85 \times {{1852} \over 3600}$$

$$Thrust=574.83kw$$

$$Blade \space area \space = \space {{\pi} \over 4}d^2 \times BAR $$

$$=\space{{\pi}\over4}\times5.8^2\times0.45$$

$$Blade \space area \space = \space 11.88m^2$$

$$Thrust \space per \space area \space = \space {{574.82} \over 11.88} \space = \space 48.34 \space KN/m^2$$

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