The lower diagram of Figure 07.05.12 shows a cross-sectional view through a cone-shaped turbine T, which has it's intake extended downwards by an additional section TE. Between the turbine and the conical wall KW (shown in grey), water flows from the intake at the bottom E and exits at the upper outlet A. This flow has two components. The first, which is shown in dark blue, flows freely along the conical wall. The second, which is shown in light blue, flows in the grooves or indentations formed by the saw tooth-like turbine "blades".
The upper diagram in the Figure shows a schematic cross-sectional representation of the plan view of this turbine. The ring-shaped water outlet A is shown in light-blue. This outlet is formed between the inside of the conical housing, which has a 24 cm radius at this level, and the cone which has a 22 cm radius. These are marked as R24 and R22 respectively, and between them a 2 cm wide outlet is formed, with a cross-sectional surface area of about 290 cm2 (F250). Also shown in light blue, is the ring-shaped inlet E, formed between a radius of 16 cm and one of 12 cm (R16 and R12), and so is 4 cm wide, with a cross-sectional area of about 350 cm2 (F350).
On the right hand side of the Figure is shown the previous curve D (shown in dark blue), which represents the track of the water flowing in the grooves. Water enters the turbine along its lower edge, at an angle of about 30 degrees and exits from the top of the turbine at an angle of about 60 degrees. Free-flowing water also enters the underside of the turbine at a very low angle and flows upwards until near the outlet it is directed into the grooves where it also exits the turbine at that same steep angle.
In the example above, it was assumed that the inlet water speed was about 7 m/s (V7), i.e. entering at an angle of 30 degrees while moving in the horizontal direction at about 6 m/s (V6), the same speed that the turbine is moving at that level. The inlet, water has a vertical rate of movement of about 3.5 m/s (V3.5). If we were to assume that the water speed at the outlet is also 7 m/s, due to it's steep exit angle of 60 degrees, it's horizontal velocity will be only 3.5 m/s. However, it actually exits at a vertical speed of 6 m/s (see the vector-graphs).
Within pipes, the linear speed of flow is inversely proportional to the cross-sectional area of the pipe. In our particular case, due to the rotational component of motion, the flow also depends on the 'gradient' of the flows, and not just the speed of movement in the axial direction. If water exits at the top at 6 m/s through an opening with a cross-sectional area of 250 cm2, then if the inlet flow has a vertical speed of only 3.5 m/s, then it would require an inlet cross-sectional area of about 22 430 cm , so our cross-sectional area of only 390 cm is a little too small.
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It seems like the efforts to find the best alternative energy sources are seriously being looked into by lots of countries including most US cities. One proof is the signing of the Kyoto Treaty. The main aim of the concerned group and individuals is to lessen the greenhouse gases and pollutants.