Question:
A propeller of 4 m pitch is rotating at 120 rpm. It has a behind-hull propeller efficiency of 60% and a real slip of 25%.
If the power delivered to the propeller is 2800 kW, then the thrust produced by it is ______ kN (answer in integer).
A propeller of 4 m pitch is rotating at 120 rpm. It has a behind-hull propeller efficiency of 60% and a real slip of 25%.
If the power delivered to the propeller is 2800 kW, then the thrust produced by it is ______ kN (answer in integer).
If the power delivered to the propeller is 2800 kW, then the thrust produced by it is ______ kN (answer in integer).
Show Hint
Get the advance speed from pitch speed and slip, then use behind-hull efficiency to find thrust.
Updated On: Jul 28, 2026
Show Solution
Verified By Collegedunia
Correct Answer: 280
Solution and Explanation
Step 1: Find the pitch speed and then the speed of advance.
The rotational speed is \(n = 120\) rpm \(= 2\) rev/s, so the pitch speed is \(V_p = nP = 2 \times 4 = 8\) m/s. The real slip relates pitch speed to the actual speed of advance by \(S_R = \dfrac{V_p - V_a}{V_p}\), so \(V_a = V_p(1 - S_R) = 8 \times (1 - 0.25) = 6\) m/s.
Step 2: Use the behind-hull propeller efficiency definition.
Behind-hull propeller efficiency is the useful thrust power over the delivered power: \(\eta_B = \dfrac{T V_a}{P_D}\), where \(T\) is thrust, \(V_a\) is speed of advance and \(P_D\) is power delivered to the propeller.
Step 3: Rearrange and substitute.
\(T = \dfrac{\eta_B P_D}{V_a} = \dfrac{0.6 \times 2800}{6} = \dfrac{1680}{6} = 280\) kW/(m/s), which in proper units is \(280\) kN since power divided by velocity gives force.
Final Answer:
The thrust produced is 280 kN, matching the official key exactly. \[ \boxed{T = 280 \text{ kN}} \]
The rotational speed is \(n = 120\) rpm \(= 2\) rev/s, so the pitch speed is \(V_p = nP = 2 \times 4 = 8\) m/s. The real slip relates pitch speed to the actual speed of advance by \(S_R = \dfrac{V_p - V_a}{V_p}\), so \(V_a = V_p(1 - S_R) = 8 \times (1 - 0.25) = 6\) m/s.
Step 2: Use the behind-hull propeller efficiency definition.
Behind-hull propeller efficiency is the useful thrust power over the delivered power: \(\eta_B = \dfrac{T V_a}{P_D}\), where \(T\) is thrust, \(V_a\) is speed of advance and \(P_D\) is power delivered to the propeller.
Step 3: Rearrange and substitute.
\(T = \dfrac{\eta_B P_D}{V_a} = \dfrac{0.6 \times 2800}{6} = \dfrac{1680}{6} = 280\) kW/(m/s), which in proper units is \(280\) kN since power divided by velocity gives force.
Final Answer:
The thrust produced is 280 kN, matching the official key exactly. \[ \boxed{T = 280 \text{ kN}} \]
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