 The next example reads like this, the shaft of a small turbine turns at 20,000 rpm and the blade speed is 250 meter per second. The axial velocity leaving the stator is vx2 equal to 175 meter per second. The angle at which the absolute velocity leaves the stator is alpha2 equal to 67 degrees and the relative velocity leaves the rotor at beta3 equal to minus 60 degrees and the absolute velocity leaves the rotor at alpha3 equal to minus 20 degrees. Find the mean radius of the blade beta2 vx3 v2 specific power delivered by the stage and a degree of reaction. So, we have to be very careful in transferring this information given into a sketch. So, the sketch is actually shown here. What is that in this sketch, Professor Coppola has used w for relative velocity as opposed to the c that we are using. Please bear this in mind when you look at the diagram. So, let us see. So, this is the stator and it is given that the absolute velocity leaving the rotor is minus 20. It should be borne in mind that alpha1 is equal to alpha3, so alpha1 is equal to minus 20 degrees and since the blade velocity direction u is actually like this or reference direction or axial direction is perpendicular to the blade velocity direction, so that is like this. So, since alpha1 is minus 20 that is in the clockwise direction by 20 degrees from the reference direction. So, this is v1 entering the stator. Now at the exit of the stator alpha is plus 67, so that is the absolute velocity leaving the stator. So, the velocity diagram at exit to the stator would look like this and this is c2 and this is u. So, c2 plus u equal to v2, notice that this is the blade angle of the stator at exit beta2. Now here we have the rotor, notice that this beta2 here is actually the blade angle of the rotor at entry, so the relative velocity diagram is shown like this, notice that the blade angle of the stator at exit is alpha2 and beta2 here is the blade angle at entry to the rotor and it is also given that vx2 is 175 meter per second. So, vx2 would be this value, this segment if you extend this then this would be vx2, vx2 is given as 175 meter per second. Now at the exit of the rotor the blade angle is given to be minus 60 degrees and the flow angle is given to be minus 20 degrees. So, once again this is our reference direction, so the blade angle of minus 60 degrees c3 looks like this and the blade angle of minus 20 degrees looks like this and the velocity diagram at exit is v3 equal to c3 plus u, it may also be noted that u1 is equal to u2 equal to u3. So, with the help of this velocity triangles and using appropriate trigonometric relations we can now proceed with the calculation. The students must actually review this part of this work example a few times to convince themselves and to understand properly how the given information in terms of the rotor stator blade angles is transformed or is transformed into velocity triangle information like this. So, the mean rotor radius is nothing but the blade speed divided by the angular speed of the rotor, so that is 250 meter per second divided by 2 pi n over 16 which gives the mean radius to be 11.94 centimeters. At the rotor inlet v theta2 is nothing but vx2 times tan alpha2, so v theta2 may be calculated as 412.4 meter per second, so v theta2 which is this segment here is nothing but vx2 times tan alpha2, so that is 412.4 and c theta2 which is this segment here is nothing but v theta2 minus u, so c theta, so the blade angle of the rotor at inlet theta2 is arc tangent of c theta2 over cx2 and that is 42.8 degrees and as we know from here this is in the counter clockwise direction, so this is positive. At the rotor outlet which is here v theta3, notice that this is v3 to c theta3, so c theta3 would be this entire segment, so c theta3 minus u would be equal to v theta3, so v theta3 equal to c theta3 minus u and v theta3 is nothing but vx3 times tan alpha3, c theta3 is cx3 times tan beta3. Eliminate or rewrite this expression, notice that cx3 is equal to vx3, so if I rewrite this then I get vx3 to be u over tan beta3 minus tan alpha3 and this gives me vx3 to be equal to 100, so once I know vx3 I can now go back and calculate v theta3 as vx3 times tan alpha3 which gives me 66.5 meter per second for this and using Pythagoras theorem I can calculate the absolute velocity at exit v3 equal to 194.4 meter per second. The specific work may be evaluated from oil turbine equation like this two times v theta, in this case v theta2 which is this is pointing in this direction and v theta3 which is this is pointing in this direction or the opposite directions, opposite direction that means there is going to be a change in sign here, so the specific work then comes out to be 119.7 kilo joule per kilogram. Using the values for the relative velocity that we have calculated above I can calculate c2 and c3 using Pythagoras theorem like this, so these two values may be easily evaluated.