Calculate force to lift piston from a given closing pressure

[CraigH]

I am trying to calculate the force required to close the valve on an annular bop. A rubber ring (called a packing unit) closes around the pipe when a force is applied from underneath, as shown here. The actual valve has the hydraulic fluid coming into a chamber that goes the around the outside of the main body, which then lifts a piston which applies the force to the packing unit, as shown here. In the documentation for the bop it says that the hydraulic pressure required to close or open the valve is 3000 PSI.

From this, how do I calculate the force to close the valve?

I assume you just multiply 3000 by the area the pressure is acting on:

$3000*6894.7*( π*R_o^2 - π*R_i^2)$

Where $R_i$ and $R_o$ are the inner and outer radius of the ring, shown in the picture in the second link. Units of these are meters. $6894.7$ is to convert from PSI to $N/m^2$.

But this gives a HUGE force. Am I doing something wrong in the maths? Or am I interpreting the documentation wrong. Does a 1500PSI closing pressure not mean that you have to apply 1500PSI of pressure to the hyrdaulic fluid acing on the outer ring of the piston? Maybe it means that 1500PSI is needed to squish the packing unit so it closes? In this case what information do I need and how do I calculate the force required to close the valve?

Thanks!

 

[Bystander]

But this gives a HUGE force.

You're trying to crush a pipe closed. It's not like kinking a garden hose in your fist.

 

[CraigH]

You're trying to crush a pipe closed. It's not like kinking a garden hose in your fist.

That's true, but for the valve I am looking at the forces calculate at values between 2.5 and 4.0 Mega Newtons. This is ridiculous isn't it? I've never really studied hydraulics, so I can only compare it to an electric motor. For a motor to provide 4.0 Mega Newtons it would have to be geared massively. Is it normal for hydraulics to provide this amount of force? It seems strange without something equivalent to gearing.

Could you please explain it in electric motor terms? It makes sense that a small motor can provide a big force because of conservation of energy
$E_m = E_e$
$E_m=Torque*ω$
$E_e = I*V*time$ and
$Torque=Force*Distance$

In a small motor you can gear it so for the same produced torque you can move less distance and instead use more force. So its creating more force but moving less. The energy is being used more for force and less for distance.

How does a hydraulic pump create so much force? What are the similarities / analogy with motor gearing?

 

[Bystander]

similarities / analogy with motor gearing?

Pump: large force on small piston gives high pressure in hydraulic line/system; actuator in system applies that high pressure to piston with large area to generate a very large force. Pinion:pump::bull gear:piston.

 

[CraigH]

Pump: large force on small piston gives high pressure in hydraulic line/system; actuator in system applies that high pressure to piston with large area to generate a very large force. Pinion:pump::bull gear:piston.

Am I also correct in saying that because the fluid is incompressible, and the volume out of the smaller piston equals the volume into the bigger piston; then the smaller piston will move more than the bigger piston. Is this just the basic principle to why hydraulics can provide such huge force? because you can apply the pressure to a much larger area?

 

[Bystander]

Yes. That's your gearing/leverage/mechanical advantage.

 

[CraigH]

Yes. That's your gearing/leverage/mechanical advantage.

Is there a name for this principle? e.g "hydraulic leverage"

 

[Bystander]

None of which I'm aware --- doesn't mean there isn't.