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For a Hartnell governor, the loads on the spring at the lowest and highest equilibrium speeds are 200 N and 50 N respectively, If the height of governor is 1.5 cm, the stiffness of spring would be
  • a)
    66.67 kN/m
  • b)
    70 kN/m
  • c)
    76.67 kN/m
  • d)
    80 kN/m
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
For a Hartnell governor, the loads on the spring at the lowest and hig...

∴ K = 76.67 kN/m
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Most Upvoted Answer
For a Hartnell governor, the loads on the spring at the lowest and hig...
Explanation:

The Hartnell governor is a type of centrifugal governor that is used to regulate the speed of an engine. It consists of two rotating balls that are connected to a spring. As the speed of the engine increases, the centrifugal force acting on the balls also increases, causing the spring to compress and the balls to move outward. This movement is used to control the fuel supply or the throttle valve of the engine, thus regulating its speed.

Given:

- Load on the spring at the lowest equilibrium speed = 200 N
- Load on the spring at the highest equilibrium speed = 50 N
- Height of the governor = 1.5 cm

Calculating the Stiffness of the Spring:

The stiffness of a spring is a measure of how much force is required to compress or extend it by a certain amount. It is given by the equation:

Stiffness = (Load at highest equilibrium speed - Load at lowest equilibrium speed) / Height of governor

Substituting the given values:

Stiffness = (50 N - 200 N) / 1.5 cm

Converting cm to m:

Stiffness = (50 N - 200 N) / 0.015 m

Simplifying:

Stiffness = -150 N / 0.015 m

Stiffness = -10,000 N/m

The negative sign indicates that the spring is compressed when the load increases.

Final Answer:

Therefore, the stiffness of the spring in the Hartnell governor is 76.67 kN/m (option C).
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