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).
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).
The indicated power of the main engine is ________ kW (rounded off to two decimal places).
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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.
Updated On: Apr 25, 2025
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Solution and Explanation
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.
\[
\boxed{2631.58 \, {kW}}
\]
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