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:
| 5 + 4 + 3 + 2 + 1 | |
| ——————— | = 3, and 3 × 10 = 30 square feet, as before. |
| 5 |
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:
| 5 + 0 | |
| ——— | × 10 = 25 square feet, |
| 2 |
or 25 foot-pounds of work done.
The object of [Figs. 3], [4] and [5] is to show how foot-pounds of work may be represented graphically by the areas of diagrams, and also to make it clear that this remains true whatever the form of the diagram. It is also evident that knowing the area, the average height or pressure may be found by dividing by the length, and vice versa.
Fig. 6. The Ideal Work Diagram of a Steam Engine
[Fig. 6] shows the form of work diagram which would be produced by the action of the steam in an engine cylinder, if no heat were lost by conduction and radiation. Starting with the piston in the position shown in [Fig. 2], steam is admitted at a pressure represented by the height of the line OY. As the piston moves forward, sufficient steam is admitted to maintain the same pressure. At the point B the valve closes and steam is cut off. The work done up to this time is shown by the rectangle YBbO. From the point B to the end of the stroke C, the piston is moved forward by the expansion of the steam, the pressure falling in proportion to the distance moved through, until at the end of the stroke it is represented by the vertical line CX. At the point C the exhaust valve opens and the pressure drops to 0 (atmospheric pressure in this case).
As it is always desirable to find the work done by a complete stroke of the engine, it is necessary to find the average or mean pressure acting throughout the stroke. This can only be done by determining the area of the diagram and dividing by the length of the stroke. This gives what is called the mean ordinate, which multiplied by the scale of the drawing, will give the mean or average pressure. For example, if the area of the diagram is found to be 6 square inches, and its length is 3 inches, the mean ordinate will be 6 ÷ 3 = 2 inches. If the diagram is drawn to such a scale that 1 inch on OY represents 10 pounds, then the average or mean pressure will be 2 × 10 = 20 pounds, and this multiplied by the actual length of the piston stroke will give the work done in foot-pounds. The practical application of the above, together with the method of obtaining steam engine indicator diagrams and measuring the areas of the same, will be taken up in detail under the heading of Steam Engine Testing.