developed by the resistance per minute. Hence, R being the whole resistance, α E N.A.I.R+N.A.l.log. ( 1 + c ) + a + and, N. 4. 1 (R + 2) = . { log. ( 1 + c ) + . E β a + c Let v be the velocity of the piston per minute, then + v4 (R+) = E a+ A a+ B. R a The quantity R in this equation, is the whole resistance acting on an unit of surface of the piston, which comprehends the resistance arising from the motion of the useful load, which call g, and from the friction which may be represented by f+de, where f is the friction of the engine unloaded, and the quantity that that friction is augmented for each unit of useful load g; also, let h be the resistance from imperfect condensation, then R = ç (1 + d) + ƒ + h. Substituting this in equation (h), we have, for, log. (a+c log.(+€) becomes log. (+€)= log. 1=0. Equation (i) is one of Pambour's fundamental equations he determines by means of this equation, the load for a given velocity by finding g from it; g being the load or resistance for an unit of surface of the piston, the load on the whole piston is Ag. and to find the evaporation of an engine to give motion to a load g at the velocity v, we have from equation (¿) ያ This gives the quantity of water which the boiler should be capable of evaporating per minute. The useful effect which an engine can produce in an unit of time, at the velocity v, is evidently Agv. To find the useful effect in terms of the load, multiply To find the horse-power of the engine, we only need divide by 33,000: we have before observed, that a horse is considered to be capable of raising that number of pounds one foot high in one minute. Pambour takes h=4lbs. per square inch, and ƒ=5lbs. and h +ƒ=4·5lbs. per square inch, or, 4.5 × 144648 lbs. per square foot, c=057, 1 + ♪ = 1·14 lbs. From equation (e) and multiplying by l, we have Pambour puts for p' the pressure P in the boiler, and calls it the velocity for the maximum of useful effect. but this can only be approximate, for the pressure of the steam in the cylinder can never equal that in the boiler. By differentiating the equation for a, he finds at what part of the stroke the steam should be cut off so as to render this useful effect a maximum*, which gives Pole's formula, which we have given at page 108, will be Log.t(log. 149.991+log.90) — 39.644 log.90) - 39-644 =(2·1672908 +0.3908485) 39.644 =(2·5581393) - 39.644 t = 361-426-39.644321°-782 Fahrenheit. Log. t = (log. 146-991 + log. 135) — 39.644 log.135) = (2·1672908 +0.4260667) — 39.644 =(2·5933575) - 39.644 t= 392-06439.644 352°-420 Fahrenheit. 1 log. 180) — 39-644 Log.t (log. 146.991+ log. 180) — 39.644 =(2·1672908+0.4510545) — 39.644 t = 415.284 — 39·644 = 375°.640 Fahrenheit. Log.t=(log. 146-991 + log. 240) - 39.644 =(2·1672908+0.4760422) — 39.644 =(2·6433330) - 39.644 t = 439-879 - 39.644 = 400°235 Fahrenheit. To find the volume of the steam compared to the volume of the water that has produced it, when the pressure is 55 lbs. Then by Tredgold's formula, the temperature will be 288°-4 Fahrenheit. |