Hydraulic jacking system power loss

[Mechanic878]

Hi everyone,

I am new to this forum and would appreciate your assistance on following design problem.

I am designing a heavy duty hydraulic jacking system for liftboat. When the cylinders retract, the hull goes up. When the cylinders extend, the hull goes down. It is a high pressure system operating at 315bar working pressure. There are counterbalance valves connected to both rod and bore side of the cylinders and the system is supposed to descend by gravity. The cylinders are controlled by 4/3 proportional directional valves.

I have calculated the power loss and consequently the waste heat that has to be dissipated during HULL UP operation, but now I have to calculate the power loss during HULL DOWN. My question is, how to calculate the power loss across the counterbalance valves? And also, what is the reasonable amount of total system power loss in % of the input power I should expect as a result?

Regards,
Mechanic878

 

[3DDave]

All the potential energy stored in the mass of the elevated hull has to go somewhere; some will heat the fluid at the CB valves, the rest in dynamic losses in the lines and any other restrictions in the return path. It really depends on the CB valve as to what the cracking pressure is; that will set how much additional energy is pumped into driving fluid into the cylinder to lower the hull.

 

[HPost]

Multiply the pressure drop across the CB valve by the flow rate across the CB.

LPM x BAR = kW

You can measure the pressure drop or get the data from the supplier. The flow can be calculated from the cylinder velocity, if you know the cylinder volume(s).

Whatever power you have will turn directly to heat and the temperature rise depends on the specific heat capacity of the oil.

For a conventional hydraulic system, that is one without load sensing or pump power control, a 15% loss is about average. That includes pump losses and system losses.

HPost

 

[HPost]

Correction...

Pressure in Pascals x flow in cubic metres per seconds is power in watts

(Pressure in Bar x flow in litres per minute)/600 is power in kW

 

[Mechanic878]

Thank you very much for your replies.

Basically this is the procedure for calculating the power loss I followed so far for the rest of the hydraulic system components. What is my doubt about the counterbalance valve is following.

From the CB valve characteristic curve as per manufacturer's catalog I see that at 115 LPM, which is my calculated flow rate during HULL DOWN operation, it yields a pressure drop of only 25 bar. After the CB valve the oil passes through the proportional control valve and then goes to tank. Under these circumstances my understanding is that the CB acts as a relief valve and I thought the pressure drop should be higher (since working pressure is 315 bar). Am I wrong?

 

[EngineerTex]

A couple of things, Mechanic878:

First, let's make sure that I understand the original problem. Can you confirm that when the weight of the boat is held by the cylinders, the pressure is on the rod end of the cylinder and not on the blind end of the cylinder? If this is correct, then the following apply:

1) The 25 bar at 115LPM is the amount of pressure loss across the valve due to the flow of the fluid. This is in addition to the amount of weight-induced pressure on the pressure side of the cylinder and it has nothing to do with the load hanging on the cylinder. The 25 bar is added to the pressure drop across the valve when you're running the cylinder at the full 115LPM whether the boat is hanging from the cylinder or the cylinder is sitting on your bench in the shop. If you are running it more slowly, then the pressure drop will be lower, as I'm sure you'll notice in the charts.

2) You add the induced pressure based on the force on the blind side of the cylinder, taking into account the type of counterbalance. Is it a 1.5:1? 3:1? 5:1? 10:1? By the way, I wouldn't suggest a 10:1 for heavy loads. At 10:1 and above, it's really not a counterbalance valve any more - it's more like a pilot-operated check valve and you WILL have problems with fine control of the load.

For example:

If you have a 3:1 counterbalance valve and an annulus to bore ratio of 2:1 and a load-induced pressure of 2000 psi on the rod side of the cylinder, the pressure drop across the counterbalance valve at 115LPM will be:

2000 psi (from the load)
+
386 psi (25 bar due to the high flow rate)
+
[(2000 psi)/3]*2 <<-- this comes from the cracking pressure of the counterbalance valve from the blind side of the cylinder times two because of the cylinder bore to rod area ratio of 2:1
=
3719 psi

Then take the pressure times the flow rate and convert units, which gives you your heat generation across the counterbalance valve.

(115Lpm)*(3719psi)*(144sqin/sqft)*(1cuft/28.13liters)*(1min/60sec)*(746Watts/horsepower)*(1horsepower/(550ft*lbs/sec)) = 49kW [someone please check my work]

This is minimum heat creation, since this doesn't include the pressure drop across the proportional directional valve, which can be as much or more as the flow-induced pressure drop across the counterbalance valve. Your actual heat creation will be higher than this.

Engineering is not the science behind building. It is the science behind not building.

 

[HPost]

First of all, Tex has mixed the units up, so it's too difficult to check the working.

Having read the original question, I am not sure this is right. If it has CB valve on, it can't decend under gravity.

The total pressure loss is from the CB valve and the proportional valve in series. The delta P is cumulative across the whole system.

The pressure loss is not a function of the original pressure.

Can you please clarify exactly what it is you are trying to achieve?

Cheers

Adrian
Hydraulic Systems Specialist
Caterpillar UK

 

[3DDave]

What am I missing - if there is a 2000 psi tank with a valve and a 20 psi tank with a valve, won't the first tank's valve see 2000 psi drop if opened to tank and the second tank's valve see only 20 psi when similarly opened?

In this case if it takes 2000 psi to keep the hull from descending, won't there be 2000 psi across the valves to tank, plus the pressure * area ratio to open the CB valve?

 

[Mechanic878]

EngineerTex,

I refer to your question about the liftboat descent. Actually, during descent there will be pressure on both sides of the cylinder because the pump is delivering hydraulic fluid to the bore side, whereas the load of the hull induces pressure on the rod side and makes it extend.

HPost

Yes, the boat doesn't really descend by gravity. More precisely said it is gravity assisted, because the pump works at lower pressure and higher flowrate to fill the blind side of the cylinder.

 

[HPost]

Right...so your cylinders are upside down, they retract to pull the hull up and extend to lower the hull.

In this case, the motion control valves you have fitted are more like over centre valves in that they are piloted open. A counterbalance valve is just a relief valve that opens itself under increasing upstream pressure.

To lower the hull, you pump oil into the full bore of the cylinders, causing them to extend. The pressure generated in the full bore is sent to the motion control valves attached to the ports on the rod end of the cylinders. The pilot signal, together with the pressure in the rod end of the cylinder causes the motion control valve to open and the hull lowers...with the assistance of gravity.

The total pressure loss you will be seeing is from the pump outlet port, thorough the the line to proportional direction control valve, through more line (hose or rigid tube), across the cylinder. Then, in addition, through the motion control valve, through more line, then back through the directional control valve and then whatever you have between the directional control valve and the reservoir.

Total losses = Line pressure losses + CB valve losses + Prop DCV losses + any other losses through.

You have 25 BAR delta P across the motion control valve, most likely more across the directional control valve. On top of that, you will have the losses through the other valves and pipework. Any and all resistance to flow creates a pressure drop.

Your available energy/power must be at least equal to your (max pressure x max flow rate)/ 510. This power will cover your losses, which will be around 15% in total.

In effect, your system needs at least 71.03kW. That is to provide power for 315 BAR @ 115 LPM

If your total power required to lift your load and overcome the pressure drops is less than this, you will run into trouble at some point.

Your power loss through the motion control valve (CB valve) will be 4.8kW. That is (115 LPM x 25 BAR)/600.

Do you know the pressure drop values for the proportional valve?

If your power is used up in lifting the hull, you wont have enough power left to be able to push the oil through the motion control valve. Depending on what pump you have, the system will either go slow or the motor will be over power.

Can I ask why you have chosen to pull the hull up with the cylinders, rather then push?

It doesn't make any difference really, the power requirement remains the same, I am just curious.

I think all of your questions are now answered. You now know to calculate the power and you know what losses you need to accommodate in a typical hydraulic system. On your hull down operation, the total pressure drop is from the cylinder port and all the way back to the reservoir.

Cheers

Adrian Wright CEng MIMechE
Engineering Specialist
Hydraulic Systems Team
Caterpillar (UK) Ltd

 

[EngineerTex]

Mechanic878,

Ok - then what I wrote was correct.

Am I to understand that the reason you posted this was to determine the capacity of a heat exchanger required to prevent the oil from overheating? If so, then the heat rejection I put up there as 49kW should be correct or close. The power from the pumping unit would be:

(115LPM/[cylinder bore to annulus ratio])*([induced annular pressure due to weight of boat]/[counterbalance valve ratio])

But that power will not be factored into the heat rejection calculation for lowering, since that fluid is not going back to the tank and, presumably, through your heat exchanger.

Am I correct in my understanding that this was to size a heat exchanger?

Engineering is not the science behind building. It is the science behind not building.

 

[Mechanic878]

Thank you everyone for your prompt replies. They are really helpful to me.

HPost,

I attach a sketch to illustrate the principle of operation of the jacking system. It is designed to pull the hull up with the cylinders, because of specific requirements of the application, mostly due to the requirement of a bigger force to pull the liftboat legs off the seabed.

From what you are suggesting, the power loss through the overcenter valve during HULL DOWN will be(115 LPM x 25 BAR)/600, which is actually the same I determined based on the valve characteristic during HULL UP. Can we say in general that the power losses of the entire system during HULL UP and HULL DOWN shall be the same? I have to mention that the required speed is same, the load of the hull is also same.

Referring to EngineerTex's question. Yes, I posted this thread in order to determine accurately the size of the heat exchanger for the system. The system is a new development, hasn't been built yet, and has to be submitted to class for approval.

Regards,
Mechanic878

 

[HPost]

Not exactly, if your cylinder velocity is the same, the flow retract the cylinder, to lift the hull...will be lower. I imagine that is controlled by your proportional valve.

Anyway, whatever the flow is, if you multiply that by the delta P for the across the over centre valve, that will tell you how much heat will be going into the oil.