 Let us now look at the next example. The problem statement reads like this, consider an axial turbine rotor with a blades speed of 315 meter per second and constant axial velocity of 280 meter per second, flow enters the rotor at an angle of 60 degrees and leaves at an angle of minus 30 degrees, find the blade angle at inlet and outlet and the degree of reaction. So the first step as we did in the previous example is to construct the velocity triangle using the given information. Here the magnitude of the axial velocity is not given but the flow angle at the inlet and the axial velocity are given, the flow angle is given to be plus 60 degrees which means that the inlet absolute velocity vector is at an angle of 60 degrees in the counter clockwise direction from the reference direction. So let us go ahead and do this. So here as you can see the absolute velocity vector is at an angle of 60 degrees in the counter clockwise direction from our reference direction which is the axial direction. The axial velocity is given to be 280 meter per second. Once again you should bear in mind that this diagram is not to scale but qualitatively accurate. The blade speed is also given. So we can draw the blade velocity vector in the following manner, blade speed is given to be 250 meter per second, what is that v theta 1 and u are in the same direction and that is the basis for this choice of direction for the blade speed. So the relative velocity vector at inlet now points from here to here and we may go ahead and complete the inlet velocity triangle like this, what is that? This is v theta 1 and vx1 is given and this angle beta 1 as we can see here is also going to be positive as c1 is in the counter clockwise direction from the reference direction. The flow angle at the exit is given to be minus 30 degrees. So we may draw the absolute velocity vector at the outlet like this and that this is at an angle of 30 degrees in the clockwise direction from the reference direction. It is also given that the axial velocity remains constant. So vx2 is equal to vx1 and we have indicated that using this symbol here and this is v theta 2, so from here to here is v theta 2. So the relative velocity vector at the outlet points from here to here and we can go ahead and complete the outlet velocity triangle also like this. So the inlet and outlet velocity triangles are complete and we may proceed with calculating the required quantities using the appropriate normatric and algebraic relations. Let us proceed. So v theta 1, in this case v theta 1 has to be evaluated using the given axial velocity. So we calculate v theta 1 using the given axial velocity vx1, so this is the flow angle. So tan alpha 1 is going to be opposite side v theta 1 divided by vx1. So we may calculate v theta 1 using that information as 485 meter per second and since v theta 1 which is this is greater than u, c theta 1 is going to be v theta 1 minus u. So c theta 1 is equal to v theta 1 minus u and that comes out to be 135 meter per second and we have already mentioned that v theta 1, the blade angle at the inlet is positive because the relative velocity vector is in the clockwise direction from the reference step. In the similar manner we may evaluate v theta 2 at the outlet as vx2 times tan alpha 2 and since c theta 2, c theta 2 is this segment from here to here, since c theta 2 is greater than u, v theta 2 is going to be c theta 2 minus u, v theta 2 is known. So c theta 2 is v theta 2 plus u. So we may evaluate c theta 2 as 511.66 and we also make a note of the fact that beta 2 is less than 0 since c2 is in a clockwise direction from the reference step. So the blade angle at inlet may now be evaluated as arc tangent of c theta 1 over cx1 which gives us 25.74 with a positive sign and the blade angle at outlet is arc tangent of c theta 2 over cx2 from which we get 61.31 degrees and we attach a negative sign to the blade angle at the outlet. The specific power output may be evaluated as follows, w dot over m dot is equal to the blade speed u times v theta 1 minus v theta 2 and we make a note of the fact that v theta 2 is in the opposite direction to v theta 1. So v theta 1 is in this direction and v theta 2 is in this direction. So which means that v theta 1 and v theta 2 are added together here which gives us a specific power output of 226.32 kilojoule per kilogram. In Pythagoras theorem we can calculate c1 from the respective tangential and axial components at inlet and at outlet that will be equal to this. The degree of reaction is defined as h1 minus h2 divided by h01 minus h02. We have already mentioned that this is equal to the specific power output and the quantity in the numerator may be evaluated as c2 square minus c1 square over 2 since h plus c square over 2 is a constant for an axial machine and it is given that this is an axial turbine. So if we substitute the numbers we get the degree of reaction to be 0.538. Once again I encourage students to go through this example several times and make sure that they understand each and every step clearly particularly construction of the velocity triangles and the triangle at the inlet and outlet using the given information and also paying attention to the sign of v theta 1 and v theta 2 while calculating the specific power output.