TITLE:          How A Gear Pump Works
AUTHOR:    Martin L. Culpepper
COURSE:     2
YEAR:          G

MAIN FUNCTIONAL REQUIREMENT:     Convert mechanical power into fluid power.

DESIGN PARAMETER:     Pump (a gear pump is one type of pump which can satisfy this functional requirement)

GEOMETRY/STRUCTURE AND PARTS:

EXPLANATION OF HOW IT WORKS:

1. One shaft is driven by a motor or some other means
2. The gear mounted to this shaft (driving gear) engages the other gear (driven gear)
3. Fluid on the inlet side flows into and is trapped between the rotating gear teeth and the housing
4. The fluid is carried around the outside of the gears to the outlet side of the pump
5. As the fluid can not seep back along the path it came, nor between the engaged   gear teeth (they create a seal,) it must exit the outlet port.

In some gear pumps, there are side plates usually made of brass which can be replaced or re-ground when the gap between the face of the gear and the end housing becomes too large

DOMINANT PHYSICS:

 Variable Description Metric Units English Units Pin Power input to shaft Watts Horsepower Pout Power output to fluid system Watts Horsepower Ploss Power loss (i.e. to coloumb friction and viscous dissipation) Watts Horsepower w Shaft rotational speed rad/s RPM Dp Pressure increase between inlet and outlet Pascals psi Q Flow rate through the pump liters3/s in3/s hm Mechanical efficiency --- ---

The pump takes power from a rotating shaft:

Pin = T x w

A portion of this power is dissipated in the pump through coloumb friction and viscous dissipation.  This is not easily quantified theoretically and is often determined experimentally.  This power will be denoted at Ploss.

Ploss = f(friction, viscous effects......)

Some fluid will seep through the gap between the sides of the gears and the endplates (see figure below.)  This gap must be small in order to maintain the pressure increase across the pump.  Increasing the gap diminishes the pumps ability to hold a pressure difference between the inlet and outlet.  The gap is typically around 0.0005 inches.

The power which can then be derived from the fluid which comes out of the pump is:

Pout = (Dp x Q) = Pin -  Ploss = T x w - Ploss

This can also be expressed using the efficiency:

Pout = hm x  Pin

LIMITING PHYSICS

The performance/use of the pump is limited by its:

Efficiency

hm=of a pump is Pout/Pin.   This is a function of the fluid viscosity, clearance between internal components, friction between mating components, and other variables.

Typically, gear pumps have efficiencies around 85%.

Bearings

Many external gear pumps use journal bearings to support the rotating shafts.  In order for these bearings to work, a minimum speed is required (depends upon pressure of the pump.)  In addition to imposing limits on the operational speed, in many cases, the bearings determine the maximum pressure the pump can operate at.  Should the pressure drop across the pump be too large, the journal bearings will not be able to support the loads on the shafts (which come mainly from the pressure difference.)

IF NEEDED, PLOTS/GRAPHS/TABLES:

None to include here

WHERE YOU CAN FIND GEAR PUMPS:

These pumps have few moving parts, making them inexpensive.  These pumps are typically used where low to medium pressure (about 2500 - 4000 psi) is needed and mechanical efficiency is not extremely important (typical efficiency is about 85%.)

You can find gear pumps on the following machines:

1. Usually your car's oil pump
2. Hydraulically driven lawn care equipment
3. Some hydraulically driven log splitters
4. Hydraulic power units on trucks and construction equipment
5. Metering applications (gear pumps are good at controlling volume flow rate)