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For a Hartnell governor, the loads on the spring at the lowest and highest equilibrium speeds are 1150 N and 85 N, respectively. If the lift of the governor is 1.5 cm, then spring stiffness would be
- a)700 N/cm
- b)710 N/cm
- c)725 N/cm
- d)690 N/cm
Correct answer is option 'B'. Can you explain this answer?
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For a Hartnell governor, the loads on the spring at the lowest and hig...
let S be the stiffness of the spring, h be the lift of governor, F1 and F2 are the force exerted on sleeve at lowest and highest equilibrium speed
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For a Hartnell governor, the loads on the spring at the lowest and hig...
Calculation of Spring Stiffness for Hartnell Governor
Given data:
Load at lowest equilibrium speed = 1150 N
Load at highest equilibrium speed = 85 N
Lift of the governor = 1.5 cm
To calculate spring stiffness, we can use the formula:
k = (W1 - W2) / δ
Where k is the spring stiffness, W1 is the load at the lowest equilibrium speed, W2 is the load at the highest equilibrium speed, and δ is the lift of the governor.
Substituting the given values, we get:
k = (1150 - 85) / 1.5
k = 710 N/cm
Therefore, the spring stiffness for the Hartnell governor is 710 N/cm, which is option B. The spring stiffness is a measure of how much force is required to produce a given amount of deformation or displacement in the spring. Higher spring stiffness means the spring requires more force to be compressed or extended by the same amount, while a lower spring stiffness means less force is required for the same displacement. In the case of the Hartnell governor, the spring stiffness plays an important role in maintaining the equilibrium speed of the governor by providing the necessary centripetal force.
Given data:
Load at lowest equilibrium speed = 1150 N
Load at highest equilibrium speed = 85 N
Lift of the governor = 1.5 cm
To calculate spring stiffness, we can use the formula:
k = (W1 - W2) / δ
Where k is the spring stiffness, W1 is the load at the lowest equilibrium speed, W2 is the load at the highest equilibrium speed, and δ is the lift of the governor.
Substituting the given values, we get:
k = (1150 - 85) / 1.5
k = 710 N/cm
Therefore, the spring stiffness for the Hartnell governor is 710 N/cm, which is option B. The spring stiffness is a measure of how much force is required to produce a given amount of deformation or displacement in the spring. Higher spring stiffness means the spring requires more force to be compressed or extended by the same amount, while a lower spring stiffness means less force is required for the same displacement. In the case of the Hartnell governor, the spring stiffness plays an important role in maintaining the equilibrium speed of the governor by providing the necessary centripetal force.
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For a Hartnell governor, the loads on the spring at the lowest and hig...
1150-85/1.5 =710
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For a Hartnell governor, the loads on the spring at the lowest and highest equilibrium speeds are 1150 N and 85 N, respectively. If the lift of the governor is 1.5 cm, then spring stiffness would bea)700 N/cmb)710 N/cmc)725 N/cmd)690 N/cmCorrect answer is option 'B'. Can you explain this answer?
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