crank past the dead centers at the ends of the stroke, and gives a uniform motion to the shaft. The various parts of the engine are carried on a rigid bed K, usually of cast iron, which in turn is bolted to a foundation of brick or concrete. The power developed is taken off by means of a belted pulley attached to the main shaft, or, in certain cases, in the form of electrical energy from a direct-connected dynamo.

When in action, a certain amount of steam (1⁄4 to 1⁄3 of the total cylinder volume in simple engines) is admitted to one end of the cylinder, while the other is open to the atmosphere. The steam forces the piston forward a certain distance by its direct action at the boiler pressure. After the supply is shut off, the forward movement of the piston is continued to the end of the stroke by the expansion of the steam. Steam is now admitted to the other end of the cylinder, and the operation repeated on the backward or return stroke.

Fig. 2. Section of Cylinder, showing Slide Valve

An enlarged section of the cylinder showing the action of the valve for admitting and exhausting the steam is shown in Fig. 2. In this case the piston is shown in its extreme backward position, ready for the forward stroke. The steam chest L is filled with steam at boiler pressure, which is being admitted to the narrow space back of the piston through the valve N, as indicated by the arrows. The exhaust port M is in communication with the other end of the cylinder and allows the piston to move forward without resistance, except that due to the piston-rod, which transfers the work done by the expanding steam to the crank-pin. The valve N is operated automatically by a crank or eccentric attached to the main shaft, and opens and closes the supply and exhaust ports at the proper time to secure the results described.

Work Diagram

Having discussed briefly the general principle upon which an engine operates, the next step is to study more carefully the transformation of heat into work within the cylinder, and to become familiar with the graphical methods of representing it. Work has already been defined as the result of force acting through space, and the unit of work as the foot-pound, which is the work done in raising 1 pound 1 foot in height. For example, it requires 1 × 1 = 1 foot-pound to raise 1 pound 1 foot, or 1 × 10 = 10 foot-pounds to raise 1 pound 10 feet, or 10 × 1 = 10 foot-pounds to raise 10 pounds 1 foot, or 10 × 10 = 100 foot-pounds to raise 10 pounds 10 feet, etc. That is, the product of weight or force acting, times the distance moved through, represents work; and if the force is taken in pounds and the distance in feet, the result will be in foot-pounds. This result may be shown graphically by a figure called a work diagram.

Fig. 3. A Simple Work Diagram

In Fig. 3, let distances on the line OY represent the force acting, and distances on OX represent the space moved through. Suppose the figure to be drawn to such a scale that OY is 5 feet in height, and OX 10 feet long. Let each division on OY represent 1 pound pressure, and each division on OX 1 foot of space moved through. If a pressure of 5 pounds acts through a distance of 10 feet, then an amount of 5 × 10 = 50 foot-pounds of work has been done. Referring to Fig. 3, it is evident that the height OY (the pressure acting), multiplied by the length OX (the distance moved through), gives 5 × 10 = 50 square feet, which is the area of the rectangle YCXO; that is, the area of a rectangle may represent work done, if the height represents a force acting, and the length the distance moved through. If the diagram were drawn to a smaller scale so that the divisions were 1 inch in length instead of 1 foot, the area YCXO would still represent the work done, except each square inch would equal 1 foot-pound instead of each square foot, as in the present illustration.

Fig. 4. Another Form of Work Diagram

In Fig. 4 the diagram, instead of being rectangular in form, takes a different shape on account of different forces acting at different periods over the distance moved through. In the first case (Fig. 3), a uniform force of 5 pounds acts through a distance of 10 feet, and produces 5 × 10 = 50 foot-pounds of work. In the second case (Fig. 4), forces of 5 pounds, 4 pounds, 3 pounds, 2 pounds, and 1 pound, act through distances of 2 feet each, and produce (5 × 2) + (4 × 2) + (3 × 2) + (2 × 2) + (1 × 2) = 30 foot-pounds. This is also the area, in square feet, of the figure Y54321XO, which is made up of the areas of the five small rectangles shown by the dotted lines. Another way of finding the total area of the figure shown in Fig. 4, and determining the work done, is to multiply the length by the average of the heights of the small rectangles. The average height is found by adding the several heights and dividing the sum by their number, as follows:

This, then, means that the average force acting throughout the stroke is 3 pounds, and the total work done is 3 × 10 = 30 foot-pounds.

Fig. 5. Work Diagram when Pressure drops Uniformly

In Fig. 5 the pressure drops uniformly from 5 pounds at the beginning to 0 at the end of the stroke. In this case also the area and work done are found by multiplying the length of the diagram by the average height, as follows: