 Let us look out an example that requires us to construct a velocity triangle and calculate a few quantities of interest for an axial turbine. So the problem statement reads as shown here. Steam enters the rotor of an axial turbine at 500 meter per second. The blade speed is 200 meter per second. The nozzle angle is minus 65 degrees and the blade angle at exit is 70 degrees. Note that nozzle angle is the angle at which the nozzle before the inlet to the rotor is oriented and steam issues from this nozzle and hits the rotor. So essentially this is the inlet flow angle. If the mass flow rate is 1.8 kilogram per second, the relative velocity remains constant, determine the blade angle at inlet, absolute velocity and flow angle at exit, the power produced and the axial thrust. So we start by constructing the velocity triangle at the inlet and exit of this rotor. And we start to do this, we start with the absolute velocity vector, which has a magnitude of 500 meter per second and is oriented in a clockwise direction from the reference direction at an angle of 65 degrees clockwise because the sign is given to be negative. So here is the absolute velocity vector at inlet magnitude of 500 meter per second and it is located at an angle of minus 65 degrees from the reference direction or 65 degrees in the clockwise direction. Note that the velocity triangle that we will draw would not be to escape but will be accurate qualitatively. So this is the axial direction because it is an axial machine that is our reference direction. So this is the axial direction. So this is the axial component of the absolute velocity. This is the tangential component of the absolute velocity at the inlet. Now the blade speed is given to be 200 meter per second. So it is actually possible to draw the blade velocity vector in two orientation, one which is shown in red color like this, the other one which is shown in black like this. Now we need to make a decision on which one of this is actually the one that is appropriate for the problem under consideration. Notice that if we go with this choice, both the choices would actually need to legitimate velocity triangles. For instance, if we take this choice, then the relative velocity vector in this case would point from here to here. So then C1 plus U would be equal to the absolute velocity vector at inlet which is a legitimate velocity triangle. Now for this choice, the relative velocity vector would point from here to here. So C1 plus U is equal to V1 which is also a legitimate velocity triangle. So the decision on which one of this is appropriate for us is made as follows. In this case, if you choose this as the blade velocity direction, then notice that V theta1, the tangential component of the incoming fluid is opposite in direction to the blade velocity vector which means that the sting that comes out of the nozzle will actually impinge on the rotor blade and serve as a braking jet. In other words, its effect would be to break the rotor which is not what the application that we have in mind. So we discard this choice for that reason and use this choice. Notice that in this case, V theta1 and U are oriented in the same direction. So we discard this and proceed with this. We can now proceed to complete the inlet velocity triangle which then looks like this. So C1 plus U equal to V1, the inlet velocity triangle. And it can also be seen that the inlet blade angle beta1 is going to be a negative number because C1 is located in the clockwise direction from the reference direction. And this is C theta1. This segment is C theta1. Notice that Cx1 is equal to Vx1. On the outlet side of the velocity triangle, it is given that the relative velocity makes an angle of 70 degrees or the blade angle at the outlet makes an angle of 70 degrees. Since the relative velocity vector is tangential to the blade profile, this implies that the relative velocity vector at exit makes an angle of 70 degrees. And since it is positive, it is in the counterclockwise direction from the reference direction. So we may actually translate this information into a relative velocity vector that looks like this. So the blade angle of the angle of the relative velocity vector is 70 degrees in the counterclockwise direction from our reference direction. And it is further given that C2 is equal to C1. Note that as I mentioned earlier, the diagram that you are trying to draw here is qualitatively accurate but not to escape. This is why some of this information is explicitly mentioned in the velocity triangle which it may not be clearly discernible from the triangle itself and that is not the intent also. So now we can go ahead and complete the exit velocity triangle. So V2 is equal to C2 plus U. So the exit absolute velocity vector looks like this and this is the flow angle at exit. Notice that the flow angle at the exit is also positive because the velocity vector V2 is in the counterclockwise direction from our reference direction. And this segment here is V theta2 and this is equal to Vx2. So now we may go ahead and actually use trigonometric relations to carry out and the calculations and compute the quantities that we require. So let us proceed keeping this velocity triangle in mind. Let us now proceed. So starting with the inlet V theta1 is nothing but V sin alpha1 which comes out to be 450 meter per second. Vx1, the axial component of the inlet absolute velocity is V1 cosine alpha1 and we make note of the factor this is also equal to Cx1. Now since V theta1 is greater than U that is at the inlet the tangential component V theta1 is greater than the blade speed C theta1 is equal to V theta1 minus U. So C theta1 is equal to V theta1 minus U and we already have mentioned the factor beta1 is going to be negative. So we make a note of that here as well. Beta1 may now be evaluated we have Cx1, we have C theta1. So beta1 is arc tangent of C theta1 divided by Cx1. So that comes out to be 50.15 degrees and the negative sign is attached to this angle. C1 may also be evaluated now as a square root of using Pythagoras theorem as a square root of C theta1 square plus Cx1 square and it comes out to be 329 meter per second and it is given in the problem that C2 is equal to C1. So we take this to be equal to C2 and once C2 is known from this velocity triangle C theta2 and Cx2 may be evaluated since the blade angle beta is known. So C theta2 is C2 sine beta2 which is like this and Cx2 is C2 cosine beta2 which is 112.78 which is also equal to Vx2. Now since C theta2 is actually greater than the blade velocity here V theta2 is equal to C theta, I am sorry since C theta2 which is this is greater than the blade velocity V theta2 is equal to C theta2 minus U. So V theta2 is equal to C theta2 minus U and this is 109.87 meter per second and we have already noted the fact that alpha2 is positive since V2 is in the counterclockwise direction from the reference direction. So we make a note of that here and alpha2 may now be evaluated as arc tangent of V theta2 over Vx2 which comes out to be 44.25 degrees and absolute velocity may be evaluated using Pythagoras theorem in this fashion. So now we can evaluate the power that is produced by the rotor which we have from Euler turbine equation as m dot u times V theta1 minus V theta2. You substitute the numbers notice that V theta2 in this case is in the opposite direction to V theta1 so that means both V theta1 and V theta2 contribute to the torque on the rotor in the same sense. So this actually becomes V theta1 plus V theta2 so the power produced by the rotor is 202.7 kilowatt. You may recall that this was a fact that we had mentioned earlier during our initial discussion of Euler, this is the initial development of Euler-Taboum-Missionary equation that the negative sign should be used with care or should be used only when V theta1 contributes to the torque in an opposite sense to V theta2 otherwise they have to be added together. Now axial thrust is the force that is exerted on the rotor in the axial direction. Notice that this quantity here is the change in is the negative or the change in axial momentum of the fluid. The change in axial momentum of the fluid would be m dot times V x2 minus V x1 by so that is the axial force that is exerted on the fluid by Newton's third law an equal and opposite force is exerted on the rotor which is why we have written this as m dot times V x1 minus V x2 so this comes out to be 177.354 Newton's in the positive x direction. Notice that V x2 minus V x1 since V x2 is less than V x1 this implies that force has been exerted on the fluid in the negative x direction since its axial velocity decreases which means that the force exerted on the rotor is equal and opposite so this is in the positive x direction. Since the relative velocity is constant the degree of reaction is zero and the rotor is an impulse rotor. So I urge the student to carefully go through this example several times and make sure that he or she understands the fundamental concepts how the concepts that we discussed earlier are now actually utilized for solving a practical situation and for calculating quantities of practical interest.