 We now turn to the next example, which is given here. The pressure at the entrance of a transistor bind runner is 238.6 kPa. The shaft turns at 210 rpm. At the exit, the flow leaves without swirl. The inlet radius is R1 equal to 910 millimeters and the exit radius is R2 equal to 760 millimeters. The relative velocity entering the runner is C1 equal to 10.2 meter per second and the flow angle of the relative velocity leaving the runner is beta2 equal to minus 72 degrees. The blade height is constant at B equal to 0.6 meters. Find the power delivered by the turbine and the pressure at the exit. At the exit, the blade speed may be calculated from the rotational speed and the outlet radius as U2 equal to 2 pi n R2 divided by 60 equal to 16.7 meter per second. Since V theta2 equal to 0, we may evaluate the radial velocity at exit PR2 equal to V2 equal to U2 divided by tangent of beta2 as 5.43 meter per second. From mass conservation across the exit and inlet of the rotor, we may evaluate VR1 like this which then gives VR1 at 4.54 meter per second and this is the same as CR1. Now, at the inlet, the blade speed may be calculated as 20 meter per second. From the inlet velocity triangle, C1 cosine beta1 is equal to CR1. So, if you look at the inlet velocity triangle, C1 cosine beta1 is equal to CR1. So, this is CR1. So, we can evaluate beta1 as arc cosine CR1 divided by C1 which gives us two possible answers plus minus 63.5 degrees. So, now we have to decide which one of these values we can use or which one of these values actually is physically realistic. Let us go ahead with both the values and we will make this decision at a later point. So, we can obtain, depending upon the angle that we use, V theta1 as U1 plus or minus C1 sine beta1 which after substituting the known values gives 29.13 meter per second or 10.87 meter per second for V theta1. Consequently, if we evaluate alpha1 as arc tangent of V theta1 over VR1, we get alpha1 to be 81.14 degrees or 67.33 degrees. Now, 81.14 as can be seen is too extreme. So, let us take a look at the velocity triangle one more time. So, an angle of alpha1, an angle of 81.14 for alpha1, as you can see we will make this absolute velocity vector almost parallel to the U1 or blade speed direction. And so, it would appear as if the absolute velocity vector is parallel to the tangent to the circle at this point. And so, the flow seems to be entering the rotor almost tangentially. So, contrary to the previous example, here when we say tangentially, we mean the absolute velocity entering the rotor tangential to the circle. Whereas in the previous example, and we said the guide vanes allow the water to enter the rotor tangentially. What we meant was the guide vanes adjust C1 so that C1 is tangential to the blade tangent at entry which is how it should be for ideal entry without any shock losses. Whereas in this case, an alpha of 81.14 makes V1 almost parallel to U1 and that is somewhat unrealistic. So, we will discard this choice for alpha and we will use the negative value for beta. So, this is obtained from the positive value for beta. So, we will use the negative value for beta which is minus 63.5. So, notice that the inlet velocity triangle in this case appears to be different from the way we have drawn here. So, alpha 1 is positive whereas beta 1 is negative. So, that means alpha 1 is such that V1 is in the counter clockwise direction to the radial direction. And beta 1 is negative which means C1 is going to be in the clockwise direction to the reference direction. So, we need to keep that in mind when we do the calculations from here onwards. So, we take beta 1 equal to minus 63.5 degrees. So, the absolute velocity at inlet may be evaluated using Pritagal's theorem like this and it comes out to be 11.78 meter per second. The volume flow rate at entry may also be calculated as we did earlier. It is 2 pi over 1 times the height of the rotor times the radial velocity at inlet which gives us 15.575 meter cube per second. Hence, the power generated that is the hydraulic power assuming no losses is equal to rho times Q times V theta 1 times U1 which gives us 3386 kilowatts. Now, if you apply SFE across the rotor and take into account the fact that water is an incompressible fluid and we also assume that the flow is isothermal that there is no change in internal energy. Then we end up with an expression that looks like this. And if you substitute the known values on the right hand side, we get P2 minus P1 over rho as minus 162.76 from which using the given value for P1, we may evaluate P2 as 75.84 kilopascals. So, we can see that the pressure does indeed decrease from inlet to the exit of the rotor. So, this completes this for the example. The layout of Francis Stelbine is shown in this illustration which is reproduced from the fluid mechanics textbook by Fox McDonnell and Pritchard. In fact, any reaction machine whether it is Francis Stelbine or the Kaplan turbine that we are going to discuss next uses an almost identical layout. The turbine is a vertical axis turbine in these cases. Water from the dam or reservoir at a higher elevation is brought through a penstock to the turbine runner. The fluid passes through the scroll casing and then the runner and then exits into a reservoir downstream which is usually called a tail brace. Now, the draft tube which is shown here is a diverging passage that is used to increase the effective head in such a reaction turbine installations. Let us see how this is accomplished. Now, when the draft tube is absent, notice that the free surface of the tail brace would be in line with the exit of the runner. So, the effective head on the turbine would then be the difference in elevation between the free surface of the water in the reservoir and the free surface of the water in the tail brace. So, that would be the effective head. In other words, this would be the effective head in the absence of a draft tube. The presence of the draft tube allows the effective head to be increased in the following manner. Since it is a diverging passage, the velocity or kinetic energy of the fluid that leaves the runner can be converted into a pressure head here. And the tail brace in this case has to be lowered from the level that it was at before. Now, notice that since the free surface of the tail brace is at atmospheric pressure, my Pascal's law point here would also be at atmospheric pressure, which then implies that the pressure at the exit of the runner is less than atmospheric. So, the effective head acting on the runner would then be this height plus an additional height which is the length of the draft tube. So, the effective head is increased by pretty much the length of the draft tube in this case. So, this increases the power since the power is proportional to the effective head that acts on the runner. But the disadvantage or a downside to having the draft tube is that since the pressure at the exit of the runner is more negative, there is a likelihood that cavitation may occur at the exit of the runner. So, this has to be taken into account when designing the draft tube. Plus, since the draft tube is a diverging passage, there is also a danger of flow separation in this diverging passage. So, this also has to be taken into account while designing the draft tube. This completes our discussion of Francis Turbine.