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:

Propulsion power chain to remember: \[ P_E \xrightarrow[\text{QPC}]{P_D = P_E/\text{QPC}} P_D \xrightarrow[\eta_s]{P_B = P_D/\eta_s} P_B \xrightarrow[\eta_m]{P_I = P_B/\eta_m} P_I \] Always convert $R$ to N and power to kW at the end.

Answer 1

Correct Answer: 2631.58

Step 1: Calculate the effective power required by the vessel. 
Effective power represents the power needed to overcome the total resistance while moving at a given speed. It is calculated using:

\[ P_E = R \times V \]

Here, the resistance is $R = 140$ kN (or $140000$ N) and the ship speed is $V = 10$ m/s. Substituting these values:

\[ P_E = 140000 \times 10 = 1{,}400{,}000 \text{ W} = 1400 \text{ kW} \]

Step 2: Find the power delivered to the propeller.
The Quasi Propulsive Coefficient (QPC) relates effective power to the power delivered at the propeller:

\[ \text{QPC} = \frac{P_E}{P_D} \]

Rearranging to find delivered power:

\[ P_D = \frac{P_E}{\text{QPC}} = \frac{1400}{0.70} = 2000 \text{ kW} \]

Step 3: Include shaft losses to determine brake power.
A shaft loss of 5% corresponds to a shaft efficiency of $\eta_s = 0.95$. The relationship between delivered power and brake power is:

\[ P_D = \eta_s \, P_B \]

Solving for brake power:

\[ P_B = \frac{P_D}{\eta_s} = \frac{2000}{0.95} = 2105.26 \text{ kW} \]

Step 4: Convert brake power to indicated power.
Mechanical efficiency links brake power and indicated power:

\[ \eta_m = \frac{P_B}{P_I} \]

Given $\eta_m = 0.80$, the indicated power is:

\[ P_I = \frac{P_B}{\eta_m} = \frac{2105.26}{0.80} = 2631.58 \text{ kW} \]

Step 5: Final answer.
 

\[ \boxed{2631.58 \text{ kW}} \]