Spring Design for Emergency Valves of Steam Turbine

[SurajS]

Hello,
I am trying to design a emergency trip lever which is hinged at one end and closes the steam valve under the action of spring force.Valve shuts off by impacting on valve seat. Please Refer the figure. I want to calculate the spring stiffness for the motion of 12 degrees within 1 second. Length of lever= 525mm, width =38 mm. Spring force acting at 301 mm from hinge. Steam Force acting at 140 mm from hinge. Mass of lever considered acting at 245 mm from hinge. Mass of lever = 2.5 kg , Moment of inertia considered =0.23 kg-m^2
I have considered lever as rigid and from newtons law. I am getting differential equation as
y''+0.4ky=0.08k-477, where , k= Stiffness in N/m

My doubts
1. Considering even k= 17000 N/m, differential equation shows cycle time of 0.02 sec which is quite fast since inertia is less. How to asses the safety of the operation?
2. How to calculate energy dissipated when lever closes /impact force on valve seat. here it needed that valve should remain closed as soon as valve hits on valve seat.
3. Does the system need damper or by designing strong cross section, this system can be made practicable?
I am new in this field, Kindly help.
Thanks.
Regards,
Suraj

 

[RVAmeche]

Is this a homework/school problem?

 

[SurajS]

Hello,
The lever is pivoted and not hinged as written in the figure.
y in differential equation is angle rotated by lever in counterclockwise direction from open position.
Thanks.

 

[SurajS]

@ RVAmech: No, I have just started job and got this project.

 

[gerhardl]

Hi SurajS,

You are quite right in that this is a practical problem in addition to the theoretical calculation.

The solution suggested is a mechanical spring and lever solution operated with a set of forces (weight, steam pressure, spring, links) and unsure behavior (tolerances, shatter, calibration, untried construction etc.)

The problem wanted solved is creating a mechanical force by a steam pressure reaching a level of xx bar, at a tolerance of +/-yy bar, force used to shut a valve (valve not defined) within 1 second, valve to remain shut, then reopened by (operation not described in your post).

With this definition you will have more room to find a practical solution by already existing and perhaps commercial available constructions.

Your original construction is somewhat similar to older types weighted down lever arm steam safety valves. Also general safety valve construction and solutions and steam components have been used to solve similar problems.

I will anyway advice you to repost your question at the safety valve forum and perhaps redefine your question asking for suggested solutions. Checking solutions from some of the major steam valve/steamtrap companies might help.

Practical engineering is often the ability to see wider than a singel part.

Good luck!

 

[SurajS]

Thank you for you response gerhardl.

Steam pressure is 4.2 MPa. valve dia =80 mm. So constant force of 846.46 N acts against the lever in vertically upward direction.
And as a project, I am trying to develop this system so commercially available system could be the reference. But I could not get the calculation or any paper related to design of such linkage.
My main doubt : Is the differential equation coming from Newton law, capturing physical behaviour of system. Since system is not going to go through complete cycle of oscillation.
2. Since response time is 35 ms (approx), is it practicable given the linkages?
Please clarify.
Thanks and regards,
Suraj