# Enclosure: George Fleming’s Description of a Steam Engine, [ca. 30 October 1815]

# Enclosure

# George Fleming’s Description of a Steam Engine

[ca. 30 Oct. 1815]

Mod. operandi.

pour in the top of ye mainpipe water enough to cover the valves, from the position of cock 4 the steam now passes directly into the lower chamber of ye steam cylinder, & from ye position of cock 5. the steam previously collected over pistons is at ye same time passing to ye condenser, under these circumstances the pistons 1 & 2 & the bar 3.8. connecting them must rise, when they have reached the top, the arm at 8. will have raised the stop rod 6.7. the sweep 6.11 will be at 9.11. the arm 7. will be at 10. & the cocks 4 & 5. will be turned in ye other direction, the cock 4 must then direct the steam to the upper chamber of ye steam cylinder & cock 5. will at same time convey the steam from the lower chamber of the steam cylinder to the condenser, the pistons 1 & 2 must then descend with ye bar 3.8 connecting them, & while descending the arm 8. will force down the arm 7 now at 10. to its former place & the wheel & cocks will be turned as at first. A few strokes up & down expels from all the pipes the Atmosc air; on the principle of the common pump the water must then rise to the upper valves in pipes E & D. & in pipe C. it will rise up to piston 2, while piston 2 descends the water it forces down cannot escape thr’o valve. e. but must rise thr’o valve. g. & discharge an equal column from ye top of pipe E & while piston 2 descends1 it is followed by a stream of water from pipe D. by Atmosc pressure wch cooperates with the steam pressing on piston. 1. when piston 2 ascends it forces up the water now above it which cannot escape thr’o valve. f. but must go by valve. h. into pipe E & discharge an equal quantity from the top, the water now follows piston 2 ascending from the same cause & contributes the aid of Atmosc pressure to the piston 2 both ascending & descending.—

Rationale—

The power of an engine thus constructed must then be compounded of two forces viz. Steam & Atmosc pressure, the power of Steam is again in the compound ratio of its temperature & the surface pressed on, while Atmosc pressure is simply as ye surface—

It has been ascertained that Atmosc pressure is nearly equal to 15 lbs on every square inch, & that steam at 212° Fahrenheit presses also 15 lbs on every inch, as long as it is unnecessary to raise the steam much over 212° F. there can be no danger of explosion, for any vessel can confine steam while its elasticity is but equal to the pressure without the boiler.

On these principles we can calculate the power of such an engine, let the boiler be of a moderate size with an escape valve one inch square & a Thermometer so placed as to indicate the temperature let the steam cylinder & piston 1 be 18 inches diameter let the receiver, piston 2. & all the other pipes be 6 inchs diamr & the steam at 212° Fae it is required what weight & what height it will heave a column of water—

We may first count 34 feet gained by Atmosc pressure on the well, upon ye principle of the common pump, this needs no demonstration for if piston 2 worked within 34 feet of the wells surface the water would follow to that height, & if piston 2. worked at the wells surface, it would be pressed up by a force equal to ye weight of 34 feet of water on the same principle.

As every circle is equal to a triangle whose base is equal to ye circumference & perpendicular equal to Radius, we have the area of piston 1. 3.1416 × 18 × 9⁄2 = 254.4696 inches the area of piston 1. which multiplied by 15 the force of Steam at 212° Fa. gives 3817.0440 lbs for the Steam pressure on piston. 1. The pipes being all equal and 6 inchs diamtr we find the number of cubic inchs in one foot 6 × 3.1416 × 3⁄2 × 12 = 339.2928 cubic inches in each foot of this pipe, measuring 1.4688 Galn & weighing 12.4848 lbs in every foot; divide 3817.0440 lbs steam pressure by 12.4848 the weight of each foot of water & you have the number of feet in height to which this engine will heave the water & multiplying the number of feet so obtained by 12.4848 you have the weight.

the Logm for ye steam pressure is | 3.5817271 | |

the Logm for ye weight of 1 foot of water is | 1.0963812 | |

2.4853459 | = 305.73 feet the height to which the |

It is easily seen by calculation how the proportions of the parts composing such an engine may be altered or modified so as to suit the different quantities of water or heights to which it may be necessary to heave it. for example, what height would this same steam power raise water in pipes of 4 inchs instead of 6 circles are to each other as the squares of their diamtr therefore 4 2 : 6 2 : : 305.73 : 687.89 feet in height raised by same Steam pressure in pipes of 4 inchs. Or let it be necessary to raise the water only 20 feet for a Mill or other mechanical purpose requiring much power. Let the Steam cylinder & piston 1, be 18 inchs diamr let the receiver, piston 2 & all the other pipes be 18 inches diamr let the lower end of ye receiver be 6 feet above the well, let piston 2 be at the bottom of the receiver, & let all the pipes be full of water; it is asked what force of steam to the inch is sufficient to raise the water to this height.—

There is then 6 feet of water in the pipe below piston 2. sustained by the Atmosc pressure below. & 14 feet of water above pressing on piston 2 with a like Atmosc pressure on top of pipe E. the difference between what we gain from Atmosc pressure below piston 2 & weight we must encounter of water & Atmosc pressure above piston 2 is the resistance the Steam must counteract.—

each foot of those pipes contains 13.2187 Gns weighing 112.3589 lbs that multiplied by 6 the height of piston 2 above the well gives 674.1534 lbs of water already sustained below piston 2 by Atmoc pressure on the well, & we have 14 feet water above piston 2 equal to 185.0618 Gns weighing 1573.0246 lbs: the atmosc pressure on pipes of 18 inches diamr at 15 lbs the inch is 3817.0440 and it is equal at the top & bottom therefore 3817.0440 + 1573.0246 − 3817.0440 − 674.1534 = 2247.32803 lbs the resistance the steam must counteract; divide the resistance so found by 254.4696 the area in inchs of piston 1, quotient 8.8314 lbs is wt pr square inch the steam must press to counterpoise this resistance, this is a low degree of temperature little more than half the Atmoc pressure without; if ye Steam pressure be raised 2 or 3 lbs to the inch this engine may make 20 double strokes to the minute, & if the receiver is 4 feet long it will discharge 105.7496 × 20 = 2114.9920 Gallns in one minute—

Corollary—

If the diamr of the pipes be lessened (cæteris paribus) the same quantity of water will be discharged in same time, for [when]4 the power is the same the resistance is diminished. When the water is raised any height less than 34 feet the difference between what you gain from Atmoc pressure below piston 2. & the weight of water & Atmosc pressure above piston 2. is the resistance. When the water is to be raised exactly 34 feet the whole resistance is the Atmosc pressure on top of the pipe. When the water is to be raised any height above 34 the Atmosc pressure above & below being equal calculate the weight of water in all pipe F. & you have the resistance.—