(G) Main Dimensions of Internal Combus tion Engines.Among the many methods of computing the main dimensions of the internal combustion engine, not one is wholly reliable or exact. The one giving the best practical re sults is that of a German authority, Hugo Gueldner. It is based on the practical amount of air, Lh, required in one hour, if the fuel consumption per hour is Ch, both Lh and Ch being in cubic feet for gaseous, and in pounds for liquid fuel. Using the former equation, nw = ni X nm, we have Ne 33,000 X 60 =--- Q 778 Ne 2545 77w H (a) where H is the lower heat value in Britiih Thermal Units of the fuel consumed. Con sidering that the actual charge volume drawn in during one suction stroke consists of the Ch fuel, Ca = 30 X n and a corresponding amount of air, La Ch XL , the sum of the two, Vat, 30 X np is equal to the piston displacement,I12I 4 X S, times the volumetric efficiency, nr, which is the ratio of the actual charge volume to the piston dis placement volume. The equation for the de termination of the main dimensions will then be X S + Lat = 4 97 ?iv From equation (a) : Ne X 2545 Ch H X md from this we get Lst = X 2545 X L HXnwX 30 X n 84.8 Ne (I +L) .3
Hence: (b) . . . . D = Are (1 + L) = the S n nv nw tive diameter of the cylinder for a four stroke cycle engine, measured in feet, and 108 (1 + L) S = stroke in feet; D2 n H qv nw 108 Ne (1 + L) n = number of revolutions S H nn nw per minute. For two cycle engines, it is obvious that the factor 54 would he used instead of 108. In the above equations L is the air in cubic feet actually aced to horn one cubic foot of the gaseous, or one pound or the liquid fuel, while N. is the effective horse power for one cylinder end only. In view of the amount of experimental data available, a selection of L, and nw can easily be made. For this purpose the tables 1 and 2 are inserted.
Example.A four-cycle, single-acting, one cylinder anthracite producer gas engine is to develop 170 horse power, at a piston speed of 800 feet per minute and a stroke-diameter ratio = 1.35. What are its main dimensions? Solution. From Table No. 1 use se =.90; From Table No. 2 use sw=.26; El == 140 B. T. U.
L = 1.5 cube feet.
800 And since n = 2S = 2.7D Revolutions per minutes.
These values substituted in equations (b) and solving for D = si 108 X N. (1 + L) SXnXHXouXqi, 108 X 170(1+1.5) X 2.7