Bootstrapping to Save Power - Part 2, Tension Stressing Continued
FLUID POWER - Design Data Sheet 50
The first article in this series on
bootstrapping appeared in
Design Data Sheet 36. The reader should review that
issue for an explanation of the purpose of bootstrapping and the
basic operating principle.
Diagrams in this issue and in
Design Data Sheet 36 illustrate tension stressing as
in a steel cable being spooled between two drums. Essentially the
same circuits and the same components can be used for compression
stressing of a rigid material moving between two rollers. The only
change would be to re-connect the two hydraulic motors to rotate in
the correct direction to produce compression instead of tension
stress. In a later issue we will discuss torsion stressing with
Figure 1. The left side of this diagram shows
hydraulic components for tensioning the cable. The two motors, A
and B, have identical displacement, they are exposed to the same
fluid pressure, and are connected to drums of the same diameter, so
there is no cable movement produced in this part of the circuit.
Cable movement, at any speed desired, is produced by a completely
separate external drive which may be electrical, hydraulic, or
This is a form of closed loop
circuitry. When the system is put into motion, oil discharged from
Motor A, after passing through the filter and heat exchanger, is
available to enter the inlet of Motor B.
For best performance and smoothest
operation at low speeds, piston motors should be used. A small
amount of oil is lost from the loop by internal slippage, and
passes out the external drains. The only flow required from the
variable pump is enough to make up slippage in the two motors.
Complete circuit for cable tensioning with a bootstrap circuit.
Tension is maintained with the fluid
power system. A separate external drive keeps the cable
traveling at the desired rate of speed.
Description of Components,
Pump. The pump should be a variable displacement
piston type unit with a pressure compensator control. It is
required to work on one side of center, only.
Pump volume must be sufficient to make
up internal slippage flow of itself and the two motors. Remember,
that as the pump and motors wear, their slippage flow will
increase, so the pump must be oversized to take care of increased
slippage at some future time. If the pump volume should become
insufficient to supply all slippage losses, the motors could not
develop full tension on the moving cable.
Cable tension depends on the setting
of the pump compensator, and the compensator should be adjusted for
the tension required.
Relief Valve. This
valve is expected only to absorb momentary pressure peaks developed
by mechanical shocks on the system, During the short interval
required by the pump compensator to swing the pump cam plate to a
reduced angle. It should always be set at least 10% or 500 PSI
higher than the pump compensator to prevent accidental continuous
discharge across it which would overheat the system.
Hydraulic Motors A and
B. Fixed displacement, piston-type motors, of
hi-rotational construction, are recommended, although in some
applications gear or vane-type motors may prove satisfactory. Motor
B must be capable of reversible action. It must work as a pump if
its shaft is forced backward against the direction it would run as
Both motors, of course, must have
identical displacement, and should be connected to cable drums of
the same diameter. However, their displacements may be unequal if
cable drum diameters are selected to produce equal torques in
opposite directions. Torques must be in balance.
If drum RPM must be low, say less than
500 RPM, the motors must be of a type which will operate smoothly
at the selected speed. Slow speed motors with built-in speed
reduction, such as Char-Lynn, may not work well when the shaft is
driven backward to work like a pump.
Note: Each motor, by
itself, must be sized to produce the entire torque needed for cable
stress. Torques of both motors may not be added for total torque
when they are pulling against each other.
Heat Exchanger. Most
bootstrap systems, except those operating at high power, will not
need a heat exchanger. Except for mechanical bearing friction, the
only significant power waste entering the hydraulic oil as heat, is
caused by slippage inside the pump and two motors.
If a heat exchanger should be needed,
the return side of Motor A is the best location for its
installation. Pressure is low and oil flow is high. The usual
procedure is to select a model which, with the widest baffle
spacing on the oil (shell) side will handle its flow at a velocity
between 2 and 4 feet per second. Then, specify the number of passes
on the water side (1, 2, or 4) according to volume of cooling water
available. See main index for other information on heat exchangers
and oil cooling.
Filter. In addition
to the pump suction strainer, a good quality micronic filter should
be included, and the ideal installation point is in the return line
from Motor A either ahead of or following the heat exchanger. A
relatively inexpensive model may be used since the pressure is
External Drive. This
can be any kind of fixed or variable speed drive. If hydraulic, it
should be entirely separate, and in no way connected with the
hydraulic tensioning system, except that a common reservoir is
permissible if the two sections are on the same elevation. A common
pump should never be used for both tensioning and external drive
because of possible interaction between the two systems.
Analysis of Oil Flow in the
Figure 2. This is a stripped down flow diagram
for the complete circuit of Figure 1, and
illustrates the oil flow to be expected in various parts of the
circuit, assuming the torques of the motors are balanced so as to
produce no movement of the cable. Flows shown are those which
appear after a typical system has been set in motion by the
external drive to a speed where circulating oil in the loop exceeds
the slippage flow of the motors.
Figure 2. This
is a stripped down version of Figure 1, to illustrate flow in
various parts of the circuit.
Motors A and B are piston-type units being driven at a speed which
will deliver 40 GPM from the outlet of Motor A. Assume also that
slippage flow of each motor is 2 GPM. Find the minimum oil flow
required from the pump under these conditions.
Both motors are forced to rotate at the - same speed because of
being tied together by the moving cable. Motor A requires an
additional 2 GPM for internal slippage or a total of 42 GPM on its
inlet. Motor A delivers 40 GPM to the inlet of Motor B, and 2 GPM
is lost in internal slippage. Motor B delivers only 38 GPM back
into the pump line, and the difference: 42 - 38 = 4 GPM must be
supplied by the pump.
The bottom of the closed loop is
vented to the reservoir. Any slight difference in motor
characteristics would cause a small flow in this line either to or
from the reservoir.
Download a PDF of
Fluid Power Design Data Sheet 50 - Bootstrapping to Save
Power - Part 2, Tension Stressing Continued.
© 1990 by Womack Machine Supply Co. This
company assumes no liability for errors in data nor in safe and/or
satisfactory operation of equipment designed from this