Hydraulic Power Unit - Valve Closure Time 1

[J.Cho]

Hi, I am a 6 months experienced graduate mechanical engineer in the offshore oil and gas industry.

Currently checking the hydraulic power unit system calculation, but have a few questions in relation to valve closure time.

I have the valve actuator swept volume data (2.019 L), and it shall be closed within 8.6 seconds for process safety purposes.
In the existing system design, a 1.35 L/min pump was selected, but I could not understand the logic.
1.35 L/min = 0.0225 L/sec, resulting in the valve closure within 2.019/0.0225 = 90 seconds.

I found an article explaining that the valve swept volume will be supplied by an accumulator and have understood the accumulator sizing.
If the required swept volume to open/close the valve is provided by an accumulator, can I get your advice on how to calculate the valve closure time, please?

Thank you.
Joe

 

[LittleInch]

If you've got a typical HPU then you have an accumulator quite a bit bigger than the swept volume.

You can't calculate the time, you set the flow rate by adjusting a control valve somewhere in the discharge line.

It might be in the HPU, it might be on the actuator.

If you post a hydraulic diagram, we might be able to point it out.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

What type of actuator do you have. If its a linear piston operating a worm gear, you will need the gear ratio and the number of turns, or degrees of rotation required to close the valve.

Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[J.Cho]

Hi, @LittleInch and 1503-44. Thank you for your kind reply.
I cannot upload the diagrams and stuff, directly relating to the ongoing project due to security reasons.

The system is analogous to the article written below:

https://hoanghung.weebly.com/profes...equipment-for-hydraulic-system-for-instrument

The existing HPU system is likely to have a problem with leaking, possibly caused by pump sizing.

 

[1503-44]

The accumulator only provides the volume of fluid needed to close the valves.
The flow rate 0.0225 L/s determines how fast the swept volume is supplied.
In your attached calculation, the swept volume is the volume of one valve's actuator when full. 2.019 L

Therefore the valve's actuator travels from initial position to final position as it is being filled, which takes 90 seconds. When the actuator is full the valve is closed.

It would appear that your pump's flow rate is 1/10 of what you will need just to close one (1) valve to meet your required closing time of 8.6s

I do not know why the example problem says you normally need 40 to 50L to open all valve's at black start and then uses only 8.7 L for one valve to get his answer. That is only one valve.


Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[LittleInch]

Ok, but do you have a further question?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[J.Cho]

I appreciate your help.

@1503-44, for clarification,

In your comment, which states that "The accumulator only provides the volume of fluid needed to close the valves." and "The flow rate 0.0225 L/s determines how fast the swept volume is supplied.", I wonder about the relationship between the accumulator and pump flow rate.
In my understanding, the accumulator could also affect the required flow rate in that accumulators have stored volume in them. To be more specific, if the minimum required flow rate is 2.019 L / 8.6 sec = 0.235 L/s, the pump will partially supply the flow rate of 0.235 L/s, and the accumulator also contributes to filling the flow rate (ex. Pump flow rate = 0.200 L/s + Accumulator flow rate = 0.035 L/s = 0.235 L/s). Is it correct?

If this is correct, I wonder if there are any ways to partition the flow rate into the pump and accumulator flow rates.

After calculating the motor size with the flow rate of Q = 0.235 L/s = 14.1 L/min, P = 500 bar, and pump efficiency of e = 70%, the motor should have a power capacity of 16.7 kW, using the equation: (Q*P)/(600*e) = 16.7 kW. I heard from one of the principal engineers that the HPU typically requires 1 kW of power.

 

[LittleInch]

You are correct.

Generally no way to segregate.

That's why you have accumulators because valve operation is intermittent so you can afford to have a smaller pump working for longer than the valve operation to store the energy in the accumulator.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[J.Cho]

@LittleInch

Thanks for the huge help. I think now I am getting to know about the HPU!

If there are any calculation methods or ways to verify how much a pump can be smaller to utilise an accumulator, should it be tested in laboratories or performance tests?
I also wonder how much smaller pump engineers normally select (Practically).
I've also checked API 674 and 675, but there are no equations to calculate/verify them.

 

[1503-44]

OK. I forgot about why the accumulators are there. To reduce the pump size.
But then why determine the closing time from the pump's flow rate? The flow rate from the accumulator should be used to determine valve closing time. It also appears that the pump is sized based on the flow rate needed for one valve. Is that how we read it? Strange that the example link does not specifically state that. Now, why not start the the required accumulator flow rate calculation with the known value of required valve closing time?

Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[J.Cho]

@1503-44,
I am determining the valve closure time using the pump flow rate because I've got no idea how to calculate the flow rate from the accumulator to the valve actuator.

Would you mind having a look at the calculation below to see if the valve can be closed within 30 sec and if there is anything wrong?


SCSSV valve swept volume Vs = 0.25 L
Tubing volume = 0.52 L
Pressure = 500 bar (Pmin = 450 bar, Pmax = 500 bar)

Accumulator size = 12 * 2 = 24 L
Pump flow rate = 0.98 L/m
Minimum pump required power = 0.942 kW
Motor rating = 1 kW 110 VDC 1500 RPM

It would also be of significant development to me if you could suggest any equations or methodologies to calculate the partial flow rate from the accumulator to the valve actuator.

Regards,
Joe

 

[georgeverghese]

If the valve is a fail close valve with single acting actuator, the accumulator has little to do with closing time. Closing time would then be field set by instrumentation comm engineers at the local closing speed controller on the vent side tubing of the valve actuator, which returns motive oil back to the HPU storage tank. Closing time would also be some what related to the stored energy in the closing spring in the actuator; the more energy stored in the spring, the faster the valve will close.
In a single acting actuated FC valve, the accumulator fills the valve actuator with oil only during the opening stroke, which has nothing to do with process safety time in this case. The gas filled accumulator merely acts as a cushion to limit the pressure decay on the HPU pump side while the valve actuator is being filled.
Faster closing times require a double acting fail close (usually) hydraulically operated actuator, and in this setup, the accumulator plays a role in supplementing the action of the closing spring during closing. These actuators deliver a bigger punch for the same size as single acting actuators, are more compact than single acting ones. Check what type of actuator you have and talk to your instrumentation engineer if you need fast closing speed.
Typical thumb rule for closing speed on a single acting air operated actuated valve is 1.5-2inch (of valve bore size) /second. So a 10inch valve would close within 5-7seconds.
Looks like what you have is a requirement for a fast acting fail close shutdown valve, which has resulted in a local HPU and most likely double acting fail close actuator. So closing speed is influenced by the actuator closing spring AND accumulator stored oil volume.

 

[LittleInch]

Joe,

You're not listening to me. You cannot "calculate" this time as there are no moving parts here.

You have a store of energy in your accumulator which you are using to do work (move the valve actuator).

How fast that is is up to you ( and I don't know where anyone gets a ridiculously exact 8.6 seconds from). Did someone multiply the OD of the pipe (8.65 incheas by 1 second an inch?.

Anyway, somewhere in your piping from the accumulator to the valve actuator, and probbably within the valve actuator itself, will be a variable control valve / needle valve which is adjusted in the factory or on site to allow the required flowrate that you need to open or close your valve in your specified time.

This is like saying how long does it take me to fill up my cup from the tap in the kitchen. answer - It depends on how much you open the tap.

Is that more clear to you now?

In designing HPU's you often look at how many valves are being actuated at the same time and how many times per day to determine the optimum size of the accumulator or pump. Unless you have a constant demand or a demand for many hours a day, many valve HPU's have a small pump which pumps up an accumulator over a long period and then discharges / flows out over a short period to reduce costs.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

Noting that the accumulator delivery capacity must be able to, at a minimum, supply the maximum flow rate needed at the valve required to meet the valve closing time, it would be wise to design the accumulator system to be able to at least do that rate. That flow rate remains = swept volume/valve closing time. I would ignore the pump, considering it not to be pumping at the time the actuator travel is initiated.

Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[hydtools]

Calculate the net force to close the valve. Find the mass of the moving parts. Acceleration = F/m = a
Actuator travel distance = 1/2 × a × t^2
Assumes constant acceleration.
Solve for t.

Ted

 

[1503-44]

That's really best done by getting the actual torque and closing times from the valve manufacturer, unless you dre designing the valve yourself. They tend to vary a lot.

Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[LittleInch]

Break torque and make torque in valves are usually the key figures. Valves don't close fast enough to make acceleration an issue

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.