Introduction
“…but we weary the unlearned, who
need not know so much, and trouble the
learned, who know it already.”
From the preface, “To the Reader” by the Translators of the King James Bible, 1611
Releasing Stored Fluid Energy
Efficiently
With varying loads and variable
accumulator pressure.
Presented by Dan Helgerson CFPS, CFPAI
Hydraulic Foreman, Energy Project Manager
Cascade Steel Rolling Mills, Inc.
for the
Energy Efficient Hydraulics and Pneumatics Conference
Chicago, IL
November 15-17, 2011
The Potential Energy in a gas accumulator is a factor of the pressure
acting on the available liquid volume. Our challenge is to release that
energy in a controlled yet efficient way.
What is the Potential Energy of the
lifted weight?
25,000 lb
12”
Cylinder
4”
Bore
Stroke 12”
4”
25,000 lb x 12” = 300,000 lb/in
When we multiply the volume of fluid
below the piston by the pressure
required to lift the load, we find:
150.796 in3 x 1,989.44 psi = 300,000 lb/in
The Potential Energy of the lifted load is precisely equal to the pressure
and volume in the cylinder.
If we remove the weight and the rod, we can view this as a gas
accumulator.
An increase in pressure in the accumulator will produce a corresponding
increase in the potential energy.
When Potential Energy is released to become the Kinetic Energy of useful
work, we tend to take advantage of the relatively incompressible liquid at
the expense of the pressure of the stored fluid.
When moving multiple, varying loads from a single
source of stored (Potential) energy, our minimum stored
pressure is based on the maximum load pressure that
may be seen.
Pressure reducing valves and/or flow controls are added
to limit the velocity of the fluid going to the various
actuators.
The volume of fluid directed to the actuators is seen as a
constant and the pressure as the excess that we have to
dissipate.
We are going to build a case for a system where the pressure becomes
the determining factor and the volume of fluid directed to the actuator is
the variable.
Daniel Bernoulli informed us that, given a constant
flow, there would be a discrete pressure drop (ΔP)
through a fixed orifice.
ΔP
Daniel Bernoulli
To maintain a constant flow with a varying pressure,
there will have to be a varying orifice.
The higher the upstream pressure, the greater the ΔP
will be for a given flow.
The upstream pressure will always be higher than the
downstream pressure.
With hydraulics, a restrictive flow control will always
be an energy consumer and will release the energy as
heat.
When we use a restrictive type flow control as a
flow divider we see the same kind of thing.
The upstream pressure will always be higher than
the highest downstream pressure;
by at least 80 psi.
If the source is the stored
energy in an accumulator, it
will sometimes be
substantially higher.
In this system you will notice that
there is no flow control.
Unless the flow is from a limited source like a fixed displacement pump,
there will be additional energy lost as the supplied fluid is squeezed
through a flow control.
Something very different happens when we use a
displacement flow divider.
The flow is divided evenly as before,
But the input pressure is much
lower.
disp1
p1
How can this be???
The input pressure is found by
multiplying the downstream
pressures times the displacements
and then dividing by the total
displacement.
We will assign some values to the
flow divider and let the math show
us the result.
disp2
p2
(disp1 x p1) + (disp2 x p2)
disp1 + disp2
T = Disp. x psi ÷ 2π
The displacement flow divider is essentially two or
psi = T x 2π ÷ Disp.
more positive displacement hydraulic motors that
are connected by a common shaft.
psi = 477.46 x 6.2832 ÷ 4
They could be vane motors, piston motors, gear
motors, or any combination.
The common shaft between the motors has to
develop enough torque to push the exhausting
fluid against the downstream load.
We know that torque is a factor of pressure and
T
displacement so we can determine the torque load
on the common shaft by adding the torque on one
side (T1) to the torque on the other side (T2).
The torque developed by the input pressure acting
on the total displacement has to match the
combined torque of the two outputs.
We will assign a displacement of 2 cubic inches to
each motor.
We will assign a resistive pressure of 1000 psi to
the upper motor and 500 psi to the lower motor.
psi = 750
1000 psi
2 cu
in
T1
500 psi
2 cu
in
T2
TT11 == 2318.31
x 1000
÷ 6.2832
lb/in
T22 = 2159.15
x 500lb/in
÷ 6.2832
T1 + T2 = 477.46 lb/in
Now, instead of a high pressure drop turning into
wasted energy, all the pressurized fluid is being
used for useful work.
psi = 750 psi = 1000
2 cu
in
There is no magic here.
We did not create some new source
of power.
2 cu
in
psi = 500
We just demonstrated that the simple displacement type flow control may
be more efficient in converting an energy source into useful work than is
a restrictive flow control.
This is from an actual application on a piece of mobile equipment.
The customer came to our counter and asked for an 11 gallon/minute pump.
However,
We sold
So,
it turned
him
his 11
a pump
gpm
out his
pump
that
power
produced
became
source
a11
22
operated
gpm
gpmatpump.
1200
at 2400
rpm.
rpm.
1500 psi
11
gpm
11
gpm
11
22
gpm
1200 rpm
2400
So, what is happening here?
22 gpm was being charged with 1500 psi but only 11 gpm
was doing work.
The remaining 11 gpm had to release its energy as heat.
If we had added a 50/50 displacement flow divider, we would have had a
very different result.
1500
750 psi
11
gpm
11
gpm
22
gpm
2400 rpm
All the pressurized fluid would be used for useful
work; either to drive the motor or else to assist in
driving the motor.
Flow Dividers
The pressure upstream of a restrictive
flow divider will always be higher than
the greatest pressure demand.
The pressure upstream of a 50/50
displacement flow divider will always
be lower than the greatest pressure
demand.
Once potential energy is stored in a liquid as pressure,
it must be either used for useful work or it will be lost
as heat.
Use it or Lose it!
If the liquid is used with reduced pressure, energy is
lost.
If the pressure is used with a reduced flow, energy is
conserved.
Using Fluid Stored at a Higher Pressure Than
is Required by the Load
This application has a torque load that requires 800 psi
and 20 gpm to rotate at the desired speed.
An accumulator stores the fluid it
receives from some source at 3000 psi.
3000
880 psi
3000 psi
Even if we introduce a
pressure reducing
valve into the circuit,
we only relocate and
distribute the pressure
loss.
800 psi
20 gpm
880 psi
This requires a pressure drop of
2200 psi across the flow control.
kW
x gpm
1714) xx .7457
.7457
kW == (psi
(2200
x 20 // 1714)
19 kW is being lost as Btu’s.
Now we will introduce a simple 50/50 displacement device.
When we shift the directional valve something very different happens.
One side of the displacement device is used as a motor driving the other
side as a pump.
10 gpm is supplied from
The power input is 10
the accumulator,
gpm at 3000 psi.
1500 psi
1600
3000 psi
800 psi
The power output is 20
gpm at 1500 psi.
and 10 gpm is drawn
from the reservoir.
20 gpm
This requires a
pressure drop of 700
psi across the flow
control.
kW = (psi
(700xxgpm
20 / 1714)
/ 1714)x x.7457
.7457
6 kW is being lost
as Btu’s.
We just made a Flow Augmenter / Pressure Transformer
By adding the displacement device we reduced the volume of fluid
required from the accumulator by 50%.
We reduced the energy loss by 13 kW.
We reduced the heat load by about 25,000
Btu’s per hour.
3000 psi
What would be the
minimum pressure for
the accumulator?
1500 psi
20 gpm
800 psi
By
making
sidetoofthe
the
displacement
flowcauses
divider
The
fixed
pump
needs
to rotate
at a specific
RPMwe
in add
order
If we
add displacement
a one
control
variable
motor
that
itvariable,
to maintain
a
additional
flexibility.
to
produce
the 20
required.
constant
RPM,
wegpm
will be
able to use whatever pressure is available and
adjust the motor displacement to match the need.
Disp. x P x RPM
Disp.
x
P
2π
Power = Torque x RPM
Torque =
Power
=
2π
Constant
Constant
For any given load pressure and RPM there
We
no longer
need the restrictive
will
bewill
a discrete
displacement
that willflow
match
control.
the supply pressure
to the load and to that
specific RPM.
3000 psi
2000
5000
800 psi
Dispv
1.33
.73
.38
cu
in
cu in
Dispv = LDispf /(S-L)
Where:
Disp
v = 800 x 2 ÷ (5000
(3000 – 800)
(2000
Dispv = Variable Displacement
L
= Load Pressure
S
= Supply Pressure
Dispf = Fixed Displacement
20 gpm
Dispf
2 cu
in
This will also hold true when we replace an
electric motor with a hydraulic motor.
3000 psi
2000
5000
Power =
Torque x RPM
Constant
Torque =
Disp. x P
2π
Disp. x P x RPM
2π
Power =
Constant
For any given combination of
power and RPM, there will be a
discrete product of displacement
and pressure.
Using RPM as the controlling
factor, the displacement will
always reflect the available input
pressure and the required output
torque.