The Steamboating Forum

Hey guys I was wondering if anyone has or knows where to find the tables or formulas , for determining the design hp for a compound marine engine.
As I have said before my experience lies in the area of single acting steam engines, and nothing in my current library deals with double acting engines.
I know it will have something to do with bore,stroke and steam pressure. but I cant seem to find the appropriate formula to determine IHP . Just trying to find a ballpark number to help find correct bore and stroke to get about between 24-30 ihp in a compound . My failure may be due to my own laziness and or Im
not looking in the right book.
As always I welcome all input,thanking you in advance L.E.E.

Some terminology issues, single or double acting refers to steam pushing on one or both sides of the piston, this has nothing to do with if the engine is a compound engine.

A compound engine expands the steam from high pressure to exhaust pressure using a series of cylinders, which is more efficient than a "single expansion" or more commonly named, "simple engine". The compound engines can be "double expansion" (common for steam launches), "tripple expansion" (typical of steamships 100 years ago, and powering all liberty ships), or even "quadruple expansion" found on some large marine plants about 100 years ago. A compound engine can be either single acting or double acting, although most steam launches that have compound engines use double expansion, double acting configurations.

The horsepower produced by a compound engine is very roughly equal to PLAN/33,000, where P is the mean effective pressure acting on a cylinder (pounds per square inch), L is the length of stroke (feet), A is the cylinder area (square inches), and N is the number of power strokes per minute.

For example, a double acting compound engine having a HP cylinder of 5 inch bore (19.6 square inch), with inlet at 150 psi, exhaust to the LP cylinder at 50 PSI. The LP cylinder is 8 inch bore (50 square inch), with inlet at 50 psi, exhaust to atmospheric pressure, both cylinders 6 inch stroke, running at 400 RPM.

The HP cylinder gives PLAN/33000 = 100 x 0.5 x 19.6 x 800 / 33000 = 23.7 HP
The LP cylinder gives PLAN/33,000 = 50 x 0.5 x 50 x 800 / 33000 = 30.5 HP


The HP cylinder will only give about 80% of the value calculated above, or 19 HP, and the LP cylinder will only give about 60% of the value calculated above, or 18 Hp, or a total of 37 horsepower for this example.

I have been asked in the past to estimate the true power (indicated) of compound engines (including triple expansion etc.)
The following formula is taken from the foremost authority of the 1880-1910 period, the results are probably lower than most would expect and are therefore quite accurate as the actual "effective" M.E.P. in the cylinder is usually well below what people hope for.
For triples/quads the I.P. cylinder can be ignored in the formula.

estimated (indicated) power = D squared x WP x S x R/ (r+2) x 140000

where
D is L.P. cylinder dia. (inch)
WP is working boiler pressure (P.S.I. gauge)
S is stroke (inch)
R is R.P.M.
r is ratio of L.P. to H.P. cylinder VOLUME (AREA if stroke is same for all cylinders)
Regards Jack

estimated (indicated) power = D squared x WP x S x R/ (r+2) x 140000

where
D is L.P. cylinder dia. (inch)
WP is working boiler pressure (P.S.I. gauge)
S is stroke (inch)
R is R.P.M.
r is ratio of L.P. to H.P. cylinder VOLUME

Trying a comparison to the above listed 5 x 8 x 6 compound engine example:

D = 8 inches
WP = 150 PSIG
S = 6 inches
R = 400 RPM
R = 2.55, (r+2) = 4.55


According to this formula:

D squared = 8 x 8 = 64


Power = D squared x WP x S x R/ (r+2) x 140000

Power = 64 x 150 x 6 x 400/4.55 x 140000

Power = 64 x 150 x 6 x 87.9 x 140000

Power = A very VERY Big Number, much bigger than the horsepower of any engine ever built

Assuming the 140,000 should have been in the denominator:

Power = D squared x WP x S x R / [ (r+2) x 140000 ]

Power = 64 x 150 x 6 x 400 / [4.55 x 140000]

Power = 64 x 150 x 6 x 400 / 637000

Power = 36.17

This number is almost exactly what is calculated above, so both are in agreement.

It should be noted that some boundary conditions apply to the formula:
estimated (indicated) power = D squared x WP x S x R/ [ (r+2) x 140000 ]

This formula is applicable only to double acting engines, with atmospheric pressure exhaust. Running with a single acting compound, (such as the Westinghouse type engine) only half the power would be available. Running with a Vacuum Exhaust, more power would be available.

As it say in the book it is an ESTIMATED power.
Regards Jack

Many thanks gentlemen for your fine answers and explanations. I would like to apologize for the lack of clarity in my question. I've been working 14 hour days for the last 2 years and I'm a little fuzzy just now. I took an extra engine building contract and its a bit much for me. But I said I could do it ,( I dont advertise strictly word of mouth ),so If I take a contract I've gotta finish it. thankfully I'll have finished the extra work soon and I will be able to go back to 10 hour days