Question

A ship moving at a steady forward speed of \(10 \, {m/s}\) experiences a total resistance of \(140 \, {kN}\). The Quasi Propulsive Coefficient (QPC) is \(0.70\); the propeller shaft losses are \(5\%\) and the mechanical efficiency of the main engine is \(80\%\).
The indicated power of the main engine is ________ kW (rounded off to two decimal places).

Hint:

Quasi Propulsive Coefficient (QPC) relates effective power and delivered power: \[ {QPC} = \frac{{EHP}}{{DHP}}, \quad {EHP} = R \cdot V, \quad {Adjust for shaft/mechanical losses}. \] Always move step-by-step from resistance to indicated power through efficiency stages.

Answer 1

Step 1: Calculate Effective Power (EHP). \[ {EHP} = \frac{R \cdot V}{1000} = \frac{140 \times 10^3 \times 10}{1000} = 1400 \, {kW} \] Step 2: Use QPC to find Delivered Power (DHP). \[ {QPC} = \frac{{EHP}}{{DHP}} \Rightarrow {DHP} = \frac{{EHP}}{{QPC}} = \frac{1400}{0.70} = 2000 \, {kW} \] Step 3: Account for 5% shaft loss to find Brake Power (BHP). \[ {BHP} = \frac{{DHP}}{1 - {Shaft Loss}} = \frac{2000}{0.95} \approx 2105.26 \, {kW} \] Step 4: Account for 80% mechanical efficiency to get Indicated Power (IHP). \[ {IHP} = \frac{{BHP}}{{Mechanical Efficiency}} = \frac{2105.26}{0.80} = 2631.58 \, {kW} \] Step 5: Round off to two decimal places. \[ \boxed{2631.58 \, {kW}} \]