What is Boiler Horsepower?

In practice a number of other factors have been related to boiler horsepower. In such boilers as the ones used on traction engines or in small plants where horizontal return tubular boilers are used it is often said that there needs to be 10 square

feet of heating surface per horsepower. Some other factors are 13 square inches of flue cross sectional area to minimize flue gas pressure drop. And similarly, it is said to be good practice to proportion the flue diameter to be about ^l/₃₀th of its length. Also, with coal burning installations it is the practice to provide ? square foot of grate area per boiler horsepower.

But, so far we have not come up with a term that really relates directly to just how much steam a certain engine requires to run it at capacity. So now let us take a look at that part of the problem and see if one 'engine' horsepower equals one 'boiler' horsepower.

Without getting too deeply involved in mathematics, let us recall that once engine horsepower is equal to: 2PLAN 33,000

That is, horsepower in a double acting single cylinder steam engine is equal to the pressure in the cylinder (psig) times the length of stroke In Feet times the area of the piston In Square Inches times the RPM divided by 33,000. Everything about that formula is perfectly straightforward except for the pressure and that takes a bit of thinking.

A look at the Idealized Indicator Diagram will help to understand how the 'mean effective pressure' (m.e.p.) is determined. Here we have plotted a diagram which indicates what the pressure is in the cylinder at any part of the stroke. At point '1' the valve opens and steam pressure from the boiler is admitted and it pushes the piston along its stroke. Since steam can enter from the boiler the pressure remains constant until at, say 30% stroke, cut-off occurs and the steam is allowed to expand. This it does along line 2-3-4 and at point 4 the exhaust valve opens and the pressure drops to atmospheric in our case or point 5. At this point the piston begins its return travel and pushes the steam out the exhaust until the stroke reaches point 6 where the valve closes again and the energy stored in the flywheel starts to compress the remaining steam up to point 7 where the inlet opens and the pressure jumps up to point 1 and the cycle is complete. What we want to know is what is the average pressure represented by line 1-2-3-4-5.

We get that by determining the area under that curve and dividing by the length of the base or stroke. Some people have expensive plani meters for getting this area but let us just count squares. I got 16.3 but than there is the area under 6-7 which we must subtract. That left me with 15.6 squares each of which is 20 psig by 20%. So 15.6 times 20 times 20 and divided by the base (100) equals 62.5 psig m.e.p.

Let us take a practical case of an 8? x 10 double acting steam engine that runs at 200 rpm. That is really an 8? inch x 0.833 foot engine as far as our formula is concerned. When I put in all of those numbers and remembered that Pi is equal to 3.14161 got 35.8 horsepower (indicated). Now we can figure out how much steam this engine will require. But we must think of our 8?' x 10' engine as being an 0.708 foot x 0.833 foot engine so we can figure out the number of cubic feet the engine requires each revolution of the crankshaft. But don't forget that steam flows out of the boiler for only 30% of the stroke in our example. And, it will be useful to know that a pound of 100 psig steam takes up 3.878 cubic feet.