The historical Cochrane rotary pump illustrated by Hiscox uses one eccentric cylindrical piston and a rigid wing that slides and rocks in a pocket in the casing. A rotating eccentric carries the piston center around the fixed shaft axis while the piston remains tangent to the casing. This is a different arrangement from a twin-lobe pump.
Cochrane Rotary Pump Interactive Calculator
Explore the historical eccentric-piston Cochrane pump. Enter displacement per revolution, shaft speed, volumetric efficiency and pressure rise to estimate mean delivered flow and hydraulic output power. The animation uses fixed illustrative proportions; displacement is an entered pump characteristic.
Equation Used
- Known displacement in L/rev and shaft speed in rpm.
- Entered volumetric efficiency is held constant for each calculation.
- Pressure rise is outlet minus inlet, in psi.
- No shaft-power, cavitation or instantaneous-flow prediction.
Single eccentric piston after Hiscox figure 542 and Asakawa figure 711. The wing bears against an idealized pocket-mouth edge, not a pinned hinge. Fixed illustrative dimensions. Mean-flow calculation uses independently entered volumetric efficiency; pressure does not automatically change that efficiency. Hydraulic output excludes mechanical losses and is not required motor power.
One eccentric piston and a wing pocket
Hiscox figure 542 shows the piston wing entering a pocket in the outer shell. Asakawa figure 711 gives a second view and describes pressure keeping the wing against one side of a V-shaped groove. These historical drawings are the basis of this end-section animation.
The crank axis O is fixed. The piston center C orbits O on the eccentric. Piston radius plus eccentricity equals casing radius, so the tangent point travels around the casing. The wing is rigid to the piston and rocks as it slides into the pocket.
The animation idealizes the loaded contact as an edge at the pocket mouth. The wing face remains against that edge throughout rotation; it is not a pinned joint. Pocket depth and clearances are illustrative, and finite seals, end plates and bearing details are omitted. The arrows identify nominal inlet and outlet, rather than calculated fluid trajectories.
What this historical model represents
The animation shows the eccentric piston and its sliding wing contact. The calculator estimates mean flow from entered displacement, speed and volumetric efficiency; it does not establish pressure rating, self-priming performance or suitability for a particular installation.
For a working pump, use its measured or specified displacement and volumetric efficiency. Suitability for a liquid, suction condition or pressure must come from the actual pump specification and tests.
Mean flow and hydraulic power
For displacement Vd in litres per revolution and speed n in revolutions per minute, theoretical flow is Vd × n. Delivered mean flow is Q = Vd × n × ηv, with volumetric efficiency entered as a fraction. Divide Q by 3.785411784 for US gallons per minute.
The equivalent slip shortfall is Vd × n − Q. It collects the delivery shortfall represented by the entered efficiency; the calculator does not identify a leak path or predict how efficiency changes with pressure.
Hydraulic output power equals pressure rise times delivered flow. With Q in L/min and pressure rise in psi, divide Q × Δp × 6894.757293 by 60,000,000 for kW. Required shaft power also depends on mechanical losses, which are not specified here.
Worked mean-flow example
Enter 1.2 L/rev, 200 rpm, 90% volumetric efficiency and a 60 psi pressure rise. Theoretical flow is 240 L/min. Delivered flow is 216 L/min, or approximately 57.1 US gpm. Equivalent slip shortfall is 24 L/min.
The hydraulic output is approximately 1.49 kW. This is energy delivered to the liquid, not a motor sizing result. At zero pressure rise, the entered flow estimate remains 216 L/min while hydraulic output becomes zero.
How to interpret the controls
Displacement and speed multiply theoretical delivery. Volumetric efficiency determines what fraction is delivered. Pressure changes hydraulic output; it does not change flow unless you also update the entered volumetric efficiency. This independence is an explicit calculation assumption, not a claim that real pump leakage is pressure-independent.
Changing displacement does not stretch the cross-section: the historical drawing does not provide a unique set of pump dimensions for a specified displacement. Speed changes both mean flow and playback speed. The mechanism is shown thirty times slower than the entered shaft speed.
Cochrane pump questions
Does this pump have two lobed rotors?
No. The historical arrangement pictured here has one eccentric piston and a wing entering a casing pocket.
Why is the wing not pinned to the casing?
The historical description specifies a sliding pocket or V-shaped groove. An idealized loaded edge contact in the animation lets the wing slide and rock.
Why does changing pressure leave flow unchanged?
Volumetric efficiency is an independent input. Update it from measurements or a pump specification when operating pressure changes.
Is hydraulic power the motor power required?
No. Hydraulic power is pressure rise times delivered flow. Shaft power must also supply mechanical losses.
Does the animation calculate a leak or fluid trajectory?
No. It shows constrained mechanical motion. The flow and equivalent slip shortfall are mean values from the entered operating assumptions.
Historical references
- Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, printed page 147, figure 542: Cochrane rotary pump.
- Gonpachi Asakawa, Kikai no Moto, figure 711, historical text and diagram reproduced by XSim from the 1940 tenth edition. The description identifies the V-shaped groove and pressure-loaded wing contact.
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