Governed speed with reduced torque
A fictional shaft-power calculation states that a propeller governor holds rotational speed at 2,000 revolutions per minute throughout its specified operating range. Torque initially equals 500 newton metres and later falls to 400 newton metres, with the scenario explicitly keeping both points inside that governing range. Use angular speed equal to 2πtimes revolutions per minute divided by 60, and power equal to torque times angular speed. One learner argues that lower torque must make the propeller slow down, despite the stated governing condition. The exercise asks for the two powers and an explanation of the distinction. It supplies no real engine limit, power-lever setting, propeller oil-flow direction or abnormal procedure. The numbers are hypothetical and should not be transferred to a real aircraft’s operating schedule.
1. Calculate the two shaft powers.
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Angular speed is 2π×2000/60≈209.44 radians per second. Power is approximately 104.72 kilowatts at 500 newton metres and 83.78 kilowatts at 400 newton metres.
2. Must speed decrease under the stated assumptions?
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No. The scenario explicitly states that the governor holds 2,000 RPM within its range. Torque and power decrease while governed rotational speed remains constant; behavior outside the governing range is not supplied.