Question:medium
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
Correct Answer: 280
Solution and Explanation
Step 1: Work out how fast the propeller would advance with no slip.
At $n = 120$ rpm $= 2$ rev/s and pitch $P = 4$ m, one revolution would ideally push the propeller forward by the pitch, so the theoretical (pitch) speed is $V_p = nP = 8$ m/s.
Step 2: Apply the real slip to get the actual advance speed.
Real slip $S_R = 25\%$ means the propeller actually advances less than the pitch speed by that fraction: $V_a = V_p(1 - S_R) = 8 \times 0.75 = 6$ m/s.
Step 3: Back out thrust from the efficiency definition.
Since $\eta_B = TV_a/P_D$, thrust is $T = \eta_B P_D / V_a$. With $\eta_B = 0.6$, $P_D = 2800$ kW and $V_a = 6$ m/s: $T = (0.6 \times 2800)/6 = 1680/6 = 280$ kN.
Final Answer:
The propeller produces 280 kN of thrust. \[ \boxed{T = 280 \text{ kN}} \]
At $n = 120$ rpm $= 2$ rev/s and pitch $P = 4$ m, one revolution would ideally push the propeller forward by the pitch, so the theoretical (pitch) speed is $V_p = nP = 8$ m/s.
Step 2: Apply the real slip to get the actual advance speed.
Real slip $S_R = 25\%$ means the propeller actually advances less than the pitch speed by that fraction: $V_a = V_p(1 - S_R) = 8 \times 0.75 = 6$ m/s.
Step 3: Back out thrust from the efficiency definition.
Since $\eta_B = TV_a/P_D$, thrust is $T = \eta_B P_D / V_a$. With $\eta_B = 0.6$, $P_D = 2800$ kW and $V_a = 6$ m/s: $T = (0.6 \times 2800)/6 = 1680/6 = 280$ kN.
Final Answer:
The propeller produces 280 kN of thrust. \[ \boxed{T = 280 \text{ kN}} \]
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