 We now turn to discussion of hydroelectric turbines and the first hydroelectric turbine that we look at is the Francis turbine which is a reaction turbine. The Francis turbine as can be seen here is usually oriented because access particularly upwards. Water from the penstock enters the wall root casing which is seen here, the construction view of which is shown here. So the water at high pressure from the penstock enters the wall root casing and it then flows through this runner which is shown in red from larger radius to a smaller radius and the water then leaves axially through this draft tube. A close-up view of the Francis turbine runner is shown here. It can be seen that this is the entry to the runner and this is the exit for the water enters radially, it is a radially inward flow turbine and exits in this direction axially. These are usually massive runners and capable of producing large amounts of power. So they are used under medium head and medium discharge conditions and the runner as we have already said is a radially inward flow type. The pressure in the scroll casing where the water enters from the penstock is the hydrostatic pressure due to the entire head from the reservoir level up to the turbine level and the pressure decreases as the fluid flows through the runner and exits. Now Francis turbine runners are designed usually in such a way as to cause a decrease in pressure due to both the terms in this equation. Notice that dp is 0 not only because dr is negative because it is a radial inward flow turbine but also because dc is positive. So the rotor is designed in such a way that the dc is also positive so that this also contributes in a negative fashion to this equation. So the pressure decreases along the streamline due to a combination of both these terms. The advantage of both these terms contributing to the pressure change is that the rotor becomes very compact and the rotational speed can also be kept as low as possible which actually reduces the centrifugal stress on the rotor blades and other parts of the rotor. Since the pressure changes in the runner from the high value at the scroll casing to discharge pressure which is much lower it must always run full and the casing on the scroll and other parts of the casing has to be sufficiently thick to withstand the hydrostatic stress and as we have already mentioned the Francis turbine is categorized as a reaction turbine on account of the factor the pressure changes through the runner. Here is a view of the blade element of the rotor so you can see the blade element here so the flow enters at a higher radius, flows along the blade and then leaves at a smaller radius so this would be characterized as a design operating condition where the relative velocity at entry is tangential to the blade and the velocity here is also tangential to the blade. Now the power produced by the turbine is usually controlled by changing the flow rate through the turbine since the hydraulic power is rho times q times g times h by regulating q the power produced by the turbine by the regulator but the disadvantage of this strategy is that when q is changed the absolute velocity approaching the runner changes magnitude and hence direction I am sorry it changes magnitude and so the relative velocity vector at entry to the blade will no longer be tangential and there are off design operating conditions so in order to mitigate this effect movable guide vanes are provided upstream of the runner so that when the absolute velocity at entry decreases these guide vanes change the inlet flow angle alpha 1 in such a way there the relative velocity vector is always tangential at entry to the blade so this eliminates shock losses at off design operating condition and this guide vanes ensures that C1 is tangential to the rotor blade at any flow rate. In fact we can actually see the guide vanes here so the guide vanes are shown in green here they are located between the casing and the point drop entry of the runner and they serve and they are movable through this mechanism in green that is shown here and they serve to ensure a tangential entry into the runner at any flow rate right close up view or different view of the guide vanes may be seen here so the yellow blades of the guide vanes and they are connected to this mechanism to the hydraulic piston here and they are they can be rotated depending upon the flow rate to always ensure a tangential entry into the runner at any flow rate. So what we will do next is to record an example involving a francis turbine so the problem statement reads like this a francis turbine operates with a discharge of 4.5 meter cube per second head of 150 meters of water and the rotational speed of 450 rpm the inlet radius r1 is 0.6 meters and the water enters the rotor at an angle of 72 degrees plus 72 degrees the exit radius r2 is equal to 0.48 meter and the water leaves the rotor without slope determine the power generator the blade angle beta 1 and beta 2. So at the inlet the blade speed may be evaluated as 2 pi n times r1 divided by 60 and if we substitute the numbers this comes out to be 28.27 meter per second. So if we neglect any loss in the in the penstock then the velocity absolute velocity v1 at the entry to the guideway or the runner may be evaluated as square root of 2gh which comes out to be 54.25 meter per second. Now at the inlet we know that the tangential component v theta 1 is v1 times sin alpha 1 and if we substitute the numbers with alpha equal to 72 we get this to be 50.69 meter per second and at the outlet v theta 2 is 0 since it is given that the water leaves the rotor without slope. Therefore the power p may be evaluated in the Euler-Turban equation as rho times 2 times v theta 1 times u1 which comes out to be 6448 kilowatts. At the inlet we may write the rate of velocity vr1 as v1 times cosine alpha 1 which gives vr1 as 16.764 meter per second which is the same as cr1. Since v theta 1 which is 50.69 is greater than u1 which is 28.27 we may write c theta 1 as v theta 1 minus u1 which gives c theta 1 as 23.32 meter per second. We have already made the observation that beta 1 is going to be positive so I think in this velocity triangle the relative velocity vector is in a counterclockwise direction from the reference direction so we expect beta 1 to be greater than 0 so with the given values of or with the calculated values of cr1 and c theta 1 we may evaluate beta 1 as arc tangent of c theta 1 divided by cr1 which gives us 54.29 degrees. Furthermore the runner height and inlet may also be evaluated by or from the given flow rate. Notice that at the inlet the flow rate q into the runner may be written as 2 pi r is the 2 pi r1 times the height of the runner which is b times the radial velocity at the inlet. So, q is equal to 2 pi r1 times b times vr1 from which we can calculate b as 0.0712 meters. At the outlet the radial velocity may be evaluated as q divided by 2 pi r2 times b and if we substitute the known values we get that the velocity to be 5.24 meter per second and this is the same as cr2 since v theta 2 equal to 0 c theta 2 is equal to u2 which may be evaluated to be 22.62 meter per second and as we have already noted from the velocity triangle we expect the angle beta 2 to be negative because the relative velocity vector is in the clockwise direction from the reference direction. So known the values of cr2 and c theta 2 you may evaluate beta 2 as r tangent c theta 2 divided by cr2 which gives beta 2 to be minus 77 degrees.