A compound yacht engine expands steam in a high-pressure cylinder and then a larger low-pressure cylinder. A receiver connection carries the HP exhaust to the LP stage. This page illustrates a vertical two-cylinder arrangement and compares cylinder areas and static net piston forces. It does not predict running pressure or horsepower.
Compound Yacht Engine Interactive Calculator
Adjust the HP and LP bores and supply gauge pressure. The animation shows a vertical two-cylinder compound arrangement. Results compare areas and equal static net piston forces, including receiver back pressure on the HP piston.
Equation Used
- Supply and comparison receiver pressures are gauge pressures relative to atmospheric exhaust.
- Equal force is an imposed comparison condition, not an observed property of every compound engine.
- Piston-rod area, friction, inertia, valve events, condensation and condenser vacuum are omitted.
- Both bores are drawn at one scale; common stroke, structure and 90° crank separation are illustrative.
- If LP bore is no larger than HP bore, a message identifies that departure from the usual compound arrangement.
Compare cylinder areas and static net piston forces. The receiver value balances those forces under an atmospheric-exhaust assumption; it is not an operating pressure prediction.
Two cylinders driving one crankshaft
Hiscox Figure 234 identifies a Herreshoff-model compound yacht engine and shows end and longitudinal elevations with a direct receiver pipe. The animation follows that arrangement: fixed vertical cylinder bodies, sliding pistons and crossheads, connecting rods, two crank throws and a common shaft carried on a bedplate.
The two piston rods remain straight and guided. Each connecting rod keeps a constant length as its crank turns. Bore widths follow the selected HP and LP diameters. The illustrated common stroke and quartered cranks are chosen to explain motion; the drawing is not a dimensioned reconstruction of an individual engine.
Steam passes from the HP stage through the receiver to the LP stage. The pipe identifies that connection; it does not display simulated valve timing or instantaneous flow. The force cards are a separate static comparison.
What the calculator can compare
Use the controls to examine how bore ratio changes piston area and an equal-net-force pressure comparison. Increasing LP bore gives a larger area ratio and lowers the receiver pressure required by this particular comparison.
Engine power, steam consumption, condenser performance and suitable boiler pressure require additional data and a cycle model. None is established by these three inputs. The displayed pressure range is a calculator range, not an operating recommendation.
Include receiver back pressure
Piston area is A = πD²/4. Let R = ALP/AHP, and express supply and receiver pressures as gauge pressures relative to an assumed atmospheric exhaust.
The static net HP force is FHP = (Psupply − Preceiver)AHP. The LP force is FLP = PreceiverALP. Equating these two forces gives Preceiver = Psupply/(1 + R).
These expressions neglect piston-rod area and mechanical losses. Real cylinder pressures vary through the stroke. Calculating indicated horsepower would require mean effective pressures, stroke, speed and working strokes per revolution; this tool does not provide those quantities.
Default comparison
A 6.5 in HP bore and 13 in LP bore give an area ratio of 4:1. With 150 psig supply and atmospheric LP exhaust, the equal-force comparison receiver pressure is 150/(1 + 4) = 30 psig.
The HP pressure difference is therefore 120 psi. Multiplying by its 33.183 in² area gives about 3,982 lbf. The LP area is four times larger, so 30 psi gives the same net force. This equality is imposed by the comparison equation; it does not prove that a running engine maintains equal forces.
Change either bore or supply pressure to see the new comparison. Pause or scrub the animation to inspect the crossheads and common crankshaft independently of the pressure calculation.
Limits of a bore comparison
A larger LP cylinder provides more piston area for the lower-pressure stage. Its appropriate size depends on the complete steam cycle, not a universal bore ratio. The receiver pressure also depends on operating conditions and valve events.
The animation does not model steam distribution, heat transfer, condensation, bearing loads or vibration. A condenser would change the exhaust pressure assumed here. Neither the drawing nor these static force results establish equipment ratings or operating settings.
Questions about the compound yacht engine
Is the receiver value the pressure a real engine will run at?
No. It is the pressure that equalizes two static net piston forces under the stated atmospheric-exhaust assumption.
Why subtract receiver pressure from HP pressure?
The downstream pressure opposes the HP piston. Using supply pressure alone would overstate its net pressure-area force in this comparison.
Does equal force mean equal power?
No. Power depends on force through travel and speed. Real cylinder pressures and valve events also vary over the cycle.
Why are the cylinders different sizes?
The larger LP piston provides more area for lower-pressure steam. The selected bore ratio controls the drawn widths and calculated area ratio.
Can the LP bore be smaller?
The controls allow that comparison and the animation remains visible, but the page flags that the LP bore is not larger than HP. It is not the usual expanding compound arrangement.
Drawing reference
Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, printed page 70, Figure 234 (PDF page 80): compound yacht engine, Herreshoff model, direct receiver pipe, end and longitudinal elevations.
Animation dimensions illustrate the arrangement. The equal-static-force comparison does not establish operating pressures or manufacturer performance ratings.
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