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Control valve stroking time estimation

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Instr13

Electrical
Jun 3, 2002
12
US
1- How can I estimation a ball valve full stroking time with spring return piston actuator. Any formula or guide line.
The valve is in ON/OFF application for depressurizing a reactor.
2- Does it stroking time differ if two actuator relieve through the same solenoid?
 
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1. A general rule of thumb used is 1 second per inch of ball size, i.e., 8" ball valve = 8 seconds.
2. Yes the stroking time will be slower if two actuators exhaust through one solenoid valve. However, this depends on the size of the solenoid valve.

If you require fast closing then you need to get the air out of the actuator quickly. TO do this you would need to fit Quick Exhaust Valves at the actuator inlet ports.
 
Stroking times are dependant on the mass of air in the actuator and the flowrate that can be achieved by the instrumentation package to vent the mass. The driving force in this venting is usually a spring package. If the Cv values of the venting instrumentation is known, the valve vendor copuld help in determining the expected stroke time.

As atad said if two actuators are connected to the same solenoid, the stroking time will increase by approx. 2x if the actuator volumes are identicle.
 
Thanks for your information. The actuator is spring return piston 25 liters(6.61 gallon)volume.
The solenoid has Cv=1.7 and the pressure on the piston side is 80 Psig(before venting).
1- Any idea about the formula and stroking time calculation for single actuator or two actuator venting through the same solenoid?
2- Does it require to include spring force into the calculation for venting/stroking time calculation?
 
I may be talking out of turn here, but don't you need to establish the speed of operation for the valve service and then taylor the actuator to achieve the duty?

If however you have the control arrangement you mention already in service then the best way to check the speeds of operation is to physically open and close the valve.

1. Formula to try is as follows;
V = Displacement of actuator (in³)
C = Constant factor (80psig = 0.480)
t = Required stroking time (seconds)
Cv = Flow coefficient

t = V x C / Cv x 29

OR

V = t x Cv x 29 / C

2. You don't need to consider the spring as one the pressure is below 80 psig the spring is assisting closure. Equally you should note that any excess pressure above 80psig is detrimental to your speed of closure, i.e., until the pressure is down around the 70psig mark, the actuator will remain open so although you have initiated closure the valve has not necessarily moved for several seconds. As I said in my earlier thread this can be overcome by fitting Quick Exhaust Valves at the actuator inlet ports to get the air out quickly in order to speed up the closure.
 
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