How to build a hydraulic system for 20 ton log splitter

[chandra007]

Hi everybody,

I am building hydraulic system for 20 ton log splitter.

I am interesting to know,how to calculate system requirements.

like engine power,hydraulic pump specifications and hydraulic cylinder specifications.

I am doing some calculations using Force/Area=Pressure

20,000*2.2*4/3.14* d*d.

I am unable to size hydraulic pump.how to get the displacement and flow rate generated by the pump.

Hydraulic tank capacity.some body mentioned spool valve as direction control valve.

how the learned one will approach a hydraulic system building procedure.??

please give some idea to better understand the hydraulics using in any system.

i am thinking like "preparing input for desired out put.


Thanks in advance.

chandra

 

[bcp]

Look on the left side of this page for "Calculators."

Surplus Center - Tech Help

Bruce

 

[chandra007]

Hi bruce,

Thank for that link.

Pumping = 1 Hp = 1 GPM x 1500 Psi
(linear relationship i.e. 2 GPM @ 1500 Psi = 2 Hp

what is the meaning of this?

Engine speed will be same as hydraulic pump speed,if use coupling in between them?

oil flow rate will be same through out the circuit.?

restriction of flow generates pressure,for doing various operations various pressures are required like pushing cylinder and lifting cylinder.

how to adjust pressure ,(based on maximum pressure required for doing one operation?)

please give some idea.

chandra

 

[bcp]

More:

Technical Frequently Asked Questions

Understanding Hydraulic Pumps

http://www.derijcke.com/dl/manual.pdf

Bruce

 

[chandra007]

Thank u bruce lee

chandra

 

[triptester]

Cylinder bore size x pressure = tonnage.
Pump gpm x at rated rpm = speed.
Gpm x pressure = hp required.

Most advertized tonnage ratings of splitters are BS.
4"BORE @3000 PSI =18.8 TONS
4.5"BORE@3000 PSI =23.8 TONS
5"BORE@3000 PSI =29.4 TONS

Cycle times for 24" stroke at rated rpm.
11gpm ,4"bore =12.3 seconds
11gpm ,4.5"bore =16.2 seconds
11gpm ,5"bore =20.1 seconds

16gpm ,4"bore =8.6 seconds
16gpm ,4.5"bore =11.2 seconds
16gpm ,5"bore =14.2 seconds

Hp required
11gpm 2-stage pump @3000psi -5.5 hp min.
16gpm 2-stage pump @3000psi - 8 hp min.
11gpm 2-stage pump

 

[chandra007]

Log splitter calculations from triptester

Hi trip,

4" cylinder with 24" stroke will have volume of 3.14*2*2*24=301 cu.inch

301/231 gives 1.3 gallons.so it occupies 1.3 gallons of fluid/oil in the cylinder.

if pump supplies a flow of 11 GPM,it can occupy 1.3 gallons*1 min/11gpm=7.09 seconds

So it takes 7.09 seconds to completely move the cylinder to its one end.7.09+7.09=14.18 to complete one cycle.(am i correct?)

if i use 22 gpm pump the cycle time will reduce to half.

for 1 HP engine can produce 1 GPM of pump flow at 1500 psi.

,if i use formula (PSI*GPM)/1714=HP,

1500 PSI*1 GPM/1714=0.875HP(but u mentioned 5.5 HP,i am interesting to know the correction here).

my doubt is about pressure.each pump will have different pressure ratings?or pressure will develop for doing operation.?on factor pressure depends in a system?

"if i use 5"*24"stroke cylinder,22gpm pump,3000 psi pump,13HP engine will be suitable to work with 30 ton log splitter."

If i want to run a conveyor drum using engine ,Hydraulic pump,hydraulic motor,how to size different components.how to find the torque of the conveyor drum?"how can feel that these components? will be sufficient to work with"?

Learned members,please give some inputs.

chandra

Cylinder bore size x pressure = tonnage.
Pump gpm x at rated rpm = speed.
Gpm x pressure = hp required.

Most advertized tonnage ratings of splitters are BS.
4"BORE @3000 PSI =18.8 TONS
4.5"BORE@3000 PSI =23.8 TONS
5"BORE@3000 PSI =29.4 TONS

Cycle times for 24" stroke at rated rpm.
11gpm ,4"bore =12.3 seconds
11gpm ,4.5"bore =16.2 seconds
11gpm ,5"bore =20.1 seconds

16gpm ,4"bore =8.6 seconds
16gpm ,4.5"bore =11.2 seconds
16gpm ,5"bore =14.2 seconds

Hp required
11gpm 2-stage pump @3000psi -5.5 hp min.
16gpm 2-stage pump @3000psi - 8 hp min.
11gpm 2-stage pump

 

[LD1]

Re: Log splitter calculations from triptester

Hi trip,

4" cylinder with 24" stroke will have volume of 3.14*2*2*24=301 cu.inch

301/231 gives 1.3 gallons.so it occupies 1.3 gallons of fluid/oil in the cylinder.

if pump supplies a flow of 11 GPM,it can occupy 1.3 gallons*1 min/11gpm=7.09 seconds

So it takes 7.09 seconds to completely move the cylinder to its one end.7.09+7.09=14.18 to complete one cycle.(am i correct?)

NO. It takes less fluid to retract the cylinder. Because the cylinder rod occupies volume inside the cylinder, less oil needs pumped. So you have to figure rod volume and take that OUT of the equation for retract.

On the same tolken, retract will also have less power, but will move faster than the extend stroke.

 

[chandra007]

HI LD,

you are right,cylinder volume will be less ,when i consider rod volume.So less fluid is required here in retraction stroke.

chandra

 

[triptester]

When using formulas to determine horsepower requirements with 2-stage splitter pumps the fact that the pumps have 2 sections must be considered. The small or low volumne section pumps continuously but the large or high volumne section bypasses at between 600-900 psi.
Many people have found they only needed 500-600 psi to split most of their blocks with a 5" cylinder.

Here is a site for hydraulic caclulators, Baum Hydraulics Corp :: Spec Calculator