Farm Power · Farm Power · 2 marks · Numerical answer
An engine has torque at maximum power $T_{maxP}=125$ N·m, speed at maximum power $N_{maxP}=2400$ rpm, high-idle speed $N_{HI}=2600$ rpm, maximum torque 160 N·m and speed at maximum torque 1450 rpm. Find governor regulation in percent, to two decimals.
Use the governor-regulation convention based on mean no-load and full-load speeds: $$R=\frac{N_{HI}-N_{maxP}}{(N_{HI}+N_{maxP})/2}\,100=\frac{2600-2400}{2500}\,100=\mathbf{8.00\%}.$$ Torque-rise data are not required. A full-load-speed denominator would instead give 8.33%,so the denominator convention is material to matching the official key.
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