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: Compute Effective Power (EHP).
Effective power is: \[ P_E = R \times V \] Given $R = 140$ kN $= 140000$ N and $V = 10$ m/s: \[ P_E = 140000 \times 10 = 1{,}400{,}000 \text{ W} = 1400 \text{ kW} \] Step 2: Use QPC to find Delivered Power to propeller (DHP).
Quasi Propulsive Coefficient: \[ \text{QPC} = \frac{P_E}{P_D} \Rightarrow P_D = \frac{P_E}{\text{QPC}} \] \[ P_D = \frac{1400}{0.70} = 2000 \text{ kW} \] Step 3: Account for propeller shaft losses to get Brake Power (BP).
Shaft losses $= 5% \Rightarrow$ shaft efficiency $\eta_s = 0.95$.
\[ P_D = \eta_s \, P_B \Rightarrow P_B = \frac{P_D}{\eta_s} \] \[ P_B = \frac{2000}{0.95} = 2105.2632 \text{ kW} \] Step 4: Use mechanical efficiency to get Indicated Power (IP).
Mechanical efficiency $\eta_m = \frac{P_B}{P_I} = 0.80$: \[ P_I = \frac{P_B}{\eta_m} = \frac{2105.2632}{0.80} = 2631.5790 \text{ kW} \] Step 5: Conclusion (rounded to two decimals).
\[ \boxed{P_I = 2631.58 \text{ kW}} \]