RATING AND GENERAL PROPORTIONS OF STEAM ENGINES
The capacity or power of a steam engine is rated in horsepower, one horsepower (H. P.) being the equivalent of 33,000 foot-pounds of work done per minute. The horsepower of a given engine may be computed by the formula:
| APLN | |
| H. P. = | ——— |
| 33,000 |
in which
| A | = | area of piston, in square inches, |
| P | = | mean effective pressure per square inch, |
| L | = | length of stroke, in feet, |
| N | = | number of strokes per minute = number of revolutions × 2. |
The derivation of the above formula is easily explained, as follows: The area of the piston, in square inches, multiplied by the mean effective pressure, in pounds per square inch, gives the total force acting on the piston, in pounds. The length of stroke, in feet, times the number of strokes per minute gives the distance the piston moves through, in feet per minute. It has already been shown that the pressure in pounds multiplied by the distance moved through in feet, gives the foot-pounds of work done. Hence, A × P × L × N gives the foot-pounds of work done per minute by a steam engine. If one horsepower is represented by 33,000 foot-pounds per minute, the power or rating of the engine will be obtained by dividing the total foot-pounds of work done per minute by 33,000. For ease in remembering the formula given, it is commonly written
| PLAN | |
| H. P. = | ——— |
| 33,000 |
in which the symbols in the numerator of the second member spell the word “Plan.”
Example:—Find the horsepower of the following engine, working under the conditions stated below:
- Diameter of cylinder, 12 inches.
- Length of stroke, 18 inches.
- Revolutions per minute, 300.
- Mean effective pressure (M. E. P.), 40 pounds.
In this problem, then, A = 113 square inches; P = 40 pounds; L = 1.5 feet; and N = 600 strokes.
Substituting in the formula,
| 40 × 1.5 × 113 × 600 | ||
| H. P. = | ————————— | = 123. |
| 33,000 |
The mean effective pressure may be found, approximately, for different conditions by means of the factors in the following [table of ratios], covering ordinary practice. The rule used is as follows: Multiply the absolute initial pressure by the factor corresponding to the clearance and cut-off as found from [Table II], and subtract the absolute back pressure from the result, assuming this to be 17 pounds for non-condensing engines, and 3 pounds for condensing.
Example 1:—A non-condensing engine having 3 per cent clearance, cuts off at 1⁄3 stroke; the initial pressure is 90 pounds gage. What is the M. E. P.?
The absolute initial pressure is 90 + 15 = 105 pounds. The factor for 3 per cent clearance and 1⁄3 cut-off, from [Table II], is 0.71. Applying the rule we have: (105 × 0.71) - 17 = 57.5 pounds per square inch.
Example 2:—A condensing engine has a clearance of 5 per cent. It is supplied with steam at 140 pounds gage pressure, and has a ratio of expansion of 6. What is the M. E. P.?
The absolute initial pressure is 140 + 15 = 155. The factor for a ratio of expansion of 6 (1⁄6 cut-off) and 5 per cent clearance is 0.5, which gives (155 × 0.5) - 3 = 74.5 pounds per square inch.
The power of an engine computed by the method just explained is called the indicated horsepower (I. H. P.), and gives the total power developed, including that required to overcome the friction of the engine itself. The delivered or brake horsepower (B. H. P.) is that delivered by the engine after deducting from the indicated horsepower the power required to operate the moving parts. The brake horsepower commonly varies from 80 to 90 per cent of the indicated horsepower at full load, depending upon the type and size of engine.
In proportioning an engine cylinder for any given horsepower, the designer usually has the following data, either given or assumed, for the special type of engine under consideration: Initial pressure, back pressure, clearance, cut-off, and piston speed.
These quantities vary in different types of engines, but in the absence of more specific data the values in [Table III] will be found useful. The back pressure may be taken as 17 pounds per square inch, absolute, for non-condensing engines, and as 3 pounds for condensing engines as previously stated.
The first step in proportioning the cylinder is to compute the approximate mean effective pressure from the assumed initial pressure, clearance, and cut-off, by the method already explained. Next assume the piston speed for the type of engine to be designed, and determine the piston area by the following formula:
| 33,000 H. P. | |
| A = | ——————————. |
| M. E. P. × piston speed |
This formula usually gives the diameter of the piston in inches and fractions of an inch, while it is desirable to make this dimension an even number of inches. This may be done by taking as the diameter the nearest whole number, and changing the piston speed to correspond. This is done by the use of the following equation.
| First piston speed × first piston area | |
| —————————————— | = new piston speed. |
| new piston area |
In calculating the effective piston area, the area of the piston rod upon one side must be allowed for. The effective or average piston area will then be (2A - a)⁄2, in which A = area of piston, a = area of piston rod. This latter area must be assumed. After assuming a new piston diameter of even inches, its effective or average area must be used in determining the new piston speed. The length of stroke is commonly proportioned to the diameter of cylinder, and the piston speed divided by this will give the number of strokes per minute.
Example:—Find the diameter of cylinder, length of stroke, and revolutions per minute for a simple high-speed non-condensing engine of 200 I. H. P., with the following assumptions: Initial pressure, 90 pounds gage; clearance, 7 per cent; cut-off, 1⁄4; piston speed, 700 feet per minute; length of stroke, 1.5 times cylinder diameter.
By using the rules and formulas in the foregoing, we have:
M. E. P. = (90 + 15) × 0.63 - 17 = 49 pounds.
| 33,000 × 200 | ||
| A = | —————— | = 192.4 square inches. |
| 49 × 700 |
The nearest piston diameter of even inches is 16, which corresponds to an area of 201 square inches. Assume a piston rod diameter of 21⁄2 inches, corresponding to an area of 4.9 square inches, from which the average or effective piston area is found to be (2 × 201) - 4.9⁄2 = 198.5 square inches.
Determining now the new piston speed, we have:
| 700 × 192.4 | |
| ————— | = 678.5 feet per minute. |
| 198.5 |
Assuming the length of stroke to be 1.5 times the diameter of the cylinder, it will be 24 inches, or 2 feet.
This will call for 678.5 ÷ 2 = 340 strokes per minute, approximately, or 340 ÷ 2 = 170 revolutions per minute.