102 STEAM 



SOLUTION. Each elbow has a resistance the same as that 

 of 60 diameters of pipe, or 60X3 = 180 in. = 15 ft., and four 

 elbows have the resistance of 4X15 = 60 ft. of pipe. Each 

 stop-valve has a resistance that is equivalent to adding 40X3 

 = 120 in. = 10 ft. of pipe, and, as there are six valves, their com- 

 bined resistance is that of 6 X 10 = 60 ft. of pipe. The equiva- 

 lent length of pipe is, therefore, 300 + 60+60 = 420 ft. 



Steam Pipes for Engines. In practice, the velocity of flow 

 of steam in the supply pipes of engines and pumps is usually 

 not greater than 6,000 ft. per min., although it is increased to 

 as much as 8,000 ft. per min. in some cases. For exhaust 

 pipes, a common value is 4,000 ft. per min. The assumptions 

 made are that the cylinder is rilled with steam at boiler pressure 

 at each stroke and that a volume of steam equal to the volume 

 of the cylinder is released at each stroke, so that the flow is 

 practically continuous. The areas of the steam and exhaust 

 pipes may then be calculated by the formula 



A S 



a , 

 s 



in which a =area of steam or exhaust pipe, in square inches; 

 A =area of cylinder, in square inches; 

 5 = piston speed, in feet per minute; 

 s = velocity of steam in pipe, in feet per minute. 

 EXAMPLE. Find the areas of the steam and exhaust pipes 

 for an engine whose cylinder is*20 in. in diameter and whose 

 piston speed is 450 ft. per min. 



SOLUTION. By the formula, the area of the steam pipe, 

 assuming that 5 = 6,000 ft. per min., is 

 .7854X202X450 



600 



Similarly, for the exhaust pipe, assuming that 5 = 4,000, the 

 area is 



.7854X202X450 



4,000 



= 35.3 sq. in. 



