Question:

An engine’s torque-speed characteristics is given below:

\( T_{{\text{maxP}}} = 125 \, \text{N.m}, \, N_{{\text{maxP}}} = 2400 \, \text{rpm}, \, N_{{\text{HI}}} = 2600 \, \text{rpm}, \, T_{{\text{max}}} = 160 \, \text{N.m}, \, N_{{\text{maxT}}} = 1450 \, \text{rpm} \)

Where:

\( T_{{\text{maxP}}} = \text{torque at maximum power}, \quad T_{{\text{max}}} = \text{maximum torque}, \quad N_{{\text{maxP}}} = \text{speed at maximum power}, \quad N_{{\text{maxT}}} = \text{speed at maximum torque}, \quad N_{{\text{HI}}} = \text{speed at high idle}. \)

The Governor’s regulation is _________% (Rounded off to 2 decimal places).

Show Hint

To calculate the Governor's regulation, use the formula \( {Regulation} = \frac{N_{{HI}} - N_{{maxP}}}{N_{{maxP}}} \times 100 \), where \(N_{{HI}}\) is the high idle speed and \(N_{{maxP}}\) is the speed at maximum power.
Updated On: Apr 14, 2025
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Solution and Explanation

The Governor's regulation is defined as the relative change in speed from no load to full load speed, expressed as a percentage. It is calculated using the following formula: \[ {Regulation} = \frac{N_{{HI}} - N_{{maxP}}}{N_{{maxP}}} \times 100 \] Where:
\( N_{{HI}} \) is the high idle speed (2600 rpm),
\( N_{{maxP}} \) is the speed at maximum power (2400 rpm).
Substitute the values into the formula: \[ {Regulation} = \frac{2600 - 2400}{2400} \times 100 = \frac{200}{2400} \times 100 = 8.33% \] Thus, the Governor’s regulation is 8.33%.
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