Photo collection by Th.Schiöler Applied mathematics concerning water-lifting 1: The flow sheet

The flow sheet

To find out the horsepower of a steam engine James Watt constructed an indicator which could make a diagram showing the force F as a function of the displacement x. Such a diagram describes in an excellent way the work W delivered to the device. The work divided by the time is equal to the power. The power is measured in watt not Watt. Nomenclature F is the force supplied to the dhoon………. N x is the displacement measured in………... m W is the supplied work measured in ……… Nm We will look at the indicator diagram as regards the dhoon. The force F is drawn on the axis of abscissas and the displacement on the ordinate. The area, which results from the graph is in direct relation to the work done. The weight of the counterbalance is not only a question about physical laws but also a question about occupational therapy. By changing the weight of the counterbalance, the ratio between the two quantities of work is changed. The area to the right of x2 is that part of the work where the man can make use of his weight without using his bag. The little area to the left of x2 is that part of the work which needs the use of his back muscles. In the diagram, this force is negative. If too small counterbalance is used we should look at the two areas located to each side of x1. This means that the muscles of the back should lift more. If the counterbalance is made too big (x3) the dhoon will rise by itself, this is a drawback to the man as he has to use too much force to have the counterbalance lifted. Instead of changing the weight of the counterbalance, it is much easier to push the fulcrum back and forth until the man has a feeling of having the right ratio between the two quantities of work. If the head is increased the shape of the “amoeba” will changes its shape. If we suppose that the axes are provided with figures, we can find the total work. Let us say that the area is about 20 cm2 and one cm2 is equal to 10 Nm; if so the total work is equal to 200 Nm. If then this work is done during 10 seconds, the mean power will be 20 watt.  Next chapter: Powerfunction of a man