 In this lecture, we will discuss another vertical axis reaction turbine that works with water namely the Kaplan turbine. The Kaplan turbine is very similar to the Francis turbine in that they are both reaction turbines and they are both vertical axis hydraulic turbines. However, the Kaplan turbine is an axial turbine as maybe seen in this figure. The flow through the rotor is axial in the case of the Kaplan turbine. The layout of the Kaplan turbine is quite similar to that of the Francis turbine. Water from the dam reservoir is brought through the penstock to the scroll casing which is shown on both sides here in sectional view. The water then goes through a set of guide vanes, radial guide vanes that are located here and after passing through the radial guide vanes, the flow changes direction and flows axially and then passes through the rotor in the axial direction. It then enters the draft tube like we saw for the case of the Francis turbine. The main difference is that this is an axial turbine in contrast to the Francis turbine which is a radially inward flow turbine. Kaplan turbines are generally used for power generation under low head and high discharge conditions and the runner as we already mentioned is of axial flow type. The pressure in the scroll casing is the hydrostatic pressure due to the entire head because the water from the dam reservoir is brought through the penstock pipe to the scroll. So, the pressure here is the hydrostatic pressure due to the entire head and the pressure decreases as the fluid flows through the runner and leaves. Now, in this design, the decrease in pressure along the streamline is solely due to DC being positive by virtue of the blade chip. Note that DU in this case is 0 because it is an axial flow turbine. So, if you take a streamline that passes through the rotor, there is hardly any change in the radius of the streamline. So, DU is 0 and the change in pressure is entirely due to the increase in relative velocity across the runner. As the pressure changes in the runner, like in the case of the Francis turbine, the turbine must always run full and the casing also has to be sufficiently thick to withstand the hydrostatic stress and the Kaplan turbine is obviously a reaction turbine because the pressure changes through the runner. And as mentioned before, the effective head in the case of the Kaplan turbine can also be increased by the installation of a draft tube which is customary in almost all Kaplan turbine installations. Now, in contrast to the Francis turbine, the purpose of the guide means is slightly different in the case of a Kaplan turbine. So, here the guide means are radial and what they attempt to do is to impart a tangential or circumferential component of velocity to the fluid as it leaves the guideway. So, as the fluid leaves the guideway, swirl component is introduced. So, the flow goes from being a radially swirling flow at the end of the guideway to an axially swirling flow once it reaches the runner. Because the angular momentum of the fluid is, I am sorry, the angular momentum of the fluid is conserved when it goes from the guideway into the runner, the swirl component increases as it increases near the hub, whereas near the tip, it more or less remains the same as the value at the exit of the guideway. So, the swirl introduced by the guideway is uniform across its height. So, in the following passage, the flow changes from being a swirling radial flow to a swirling axial flow. And so, the swirl component of velocity v theta approaching the blade varies inversely from hub to tip as in a free vortex flow going to conservation of angular momentum. So, at the entry to the rotor blade, then r times v theta is a constant. So, certain amount of v theta is induced in the flow by the guidewings here and when the flow approaches the inlet of the rotor blade, r times v theta is a constant. So, which means v theta is very high near the hub of the blade and lesser near the tip. And in addition to this, the blade speed, u also varies from the hub to tip in direct proportion to the radius since u is equal to r omega. So, these two facts together result in the blade being highly twisted from hub to tip. So, look at this blade. You can see that from hub to tip, the blade is highly twisted, which is in contrast to the rotor of a Francis turbine. But we will do an example and calculate the blade angles at the hub, the mid-radius and at the tip to demonstrate how the flow becomes twisted as we go from the hub to the tip. Usually, in the case of Kaplan turbines, the flow leaves the blade without any swirl so that v theta is equal to 0 at exit. So, that is the customary design for a Kaplan turbine. Now, a blade element of a Kaplan turbine blade is shown here. So, here we can see that u1 is equal to u2 because it is an axial machine. And we also see that c2 is greater than c1. So, that dc is positive and so the pressure decreases in the runner. Generally, Kaplan turbine installations or such that add entry to the blade, beta 1 is usually negative, alpha 1 is usually positive. So, you can see that this is the reference direction, which is the axial direction. So, v1 is in counterclockwise direction to the reference direction. So, alpha 1 is positive and c1 is in a clockwise direction to the reference direction. So, beta 1 is negative and here beta 2 is also negative because c2 is in the clockwise direction and the exit velocity triangle shows clearly that v theta 2 equal to 0 since v2 is perpendicular to u2. Let us now try to work out an example involving Kaplan turbine. Small scale Kaplan turbine has a power output of 8 megawatts and available head at the turbine entry of 13.4 meters and the rotational speed of 200 rpm. The inlet guide veins have a height of 1.6 meter. The diameter at the trailing edge surface is 3.1 meter. The runner diameter is 2.9 meter and the up to tip ratio is 0.4. Swimming the efficiency to be 92 percent, determine the radial and tangential components of velocity at exit from the guide veins and the flow angle, component of axial velocity at the runner, absolute and relative flow angles upstream and downstream of the runner at the hub mid radius and tip. So, the efficiency is given to be 0.92 and the output power is given to be 8 megawatts. So, the hydraulic power is equal to rho times g times q times h. So, the output power is 0.92 times this from which we can get the flow rate to be 66.15 meter cube per second. So, clearly looking at the head and the flow rate, we can see that this installation is ideal for a Kaplan turbine because Kaplan turbine works with low head and high flow rate at the exit of the guide veins. Now, the flow rate is known. So, at the exit of the guide vein, the flow area is the perimeter pi times dg times height of the guide veins multiplied by the radial component of velocity. So, from which we get the radial component of velocity to be 4.245 meter per second. Now, the blade element at the tip or a blade element at any other location is expected to produce the same power. So, if I apply Euler-Turbo-Missioner equation to the blade element at the tip, we get P equal to rho times q times v theta 1 times u1 since v theta 2 equal to 0. u1 may be evaluated as pi times n times d1 over 60 which is equal to 30.37 meter per second. So, it follows that at the entrance to the rotor at the blade tip, the swirl velocity comes out to be 3.982 meter per second. Note that the basis for this expression is the fact that every blade element or any blade element in the rotor is expected to generate the same power. So, what we have calculated here is v theta 1 the swirl velocity at entrance to the blade at the tip section. Since the runner is a free vortex design, the tangential velocity at the exit of the guide vein is nothing but so r times v theta remains constant. So, the tangential velocity at the exit of the guide vein is v theta 1 times the tip radius divided by the radius at the exit of the guide vein. So, basically what we are doing here is the following. So, we know v theta at the tip of the blade and we know the diameter at the tip here and the diameter at the exit of the guide vein. So, by using the fact that v theta times r is a constant in a free vortex design, we can evaluate v theta at the exit of the guide vein that is what we are doing here. So, the subscript g here denotes guide vein exit of guide vein. So, v theta g is equal to 3.725. We already know the radial velocity at the exit of the guide vein. So, the absolute flow angle at the exit of the guide vein may be evaluated as r tangent of v theta g divided by v r g which is positive 41.27 degrees. In the runner, the flow area is nothing but pi times d tip square minus d hub square divided by 4. And when we multiply this by the axial velocity, we get the flow rate in the runner and we substitute the numbers, we get the axial velocity in the runner to be equal to 11.922 meter per second. And from the velocity triangle, we know that v x1, which is this, is also equal to c x1. So, c x1 is also known now. Now at the tip, alpha 1 is arc tangent of v theta 1 divided by v x1 and this comes out to be 18.47 degrees. Notice that from the velocity triangle at the tip of the blade, alpha 1 is arc tangent of v theta 1 which is this segment divided by v x1. So, we get alpha 1 to be positive 18.47 degrees. Notice that u1 which is the blade speed is 30.37 and that is greater than v theta 1 which is 3.982 which means that c theta 1 is u1 minus v theta 1 equal to 26.388 and the blade angle at inlet is negative. And this blade angle may be evaluated as arc tangent of c theta 1 divided by c x1, which is 65.69 degrees and we attach a negative sign to this value. Continuing with the tip, so we have completed the calculation at the inlet section, at the inlet of the runner at the tip section. Now we can go to the exit. At the exit beta 2 is arc tangent of c theta 2 divided by c x2. And since the exit v theta 2 equal to 0, c theta 2 becomes equal to u2 and c x2 is also equal to v x2. So, we can see from here that c theta 2 equal to u2. So, this is c theta 2 that is equal to u2 and this is c x2 and that is equal to v2 or that is also equal to v x2. So, we can rewrite this expression as beta 2 equal to arc tangent of u2 divided by v x2. Now u2 equal to u1 because it is an axial machine. So, u2 is equal to 30.37 meter per second and v x2 equal to v x1. Again, the flow area remains the same. So, we take v x2 to be equal to v1 since the volume flow rate remains the same. So, that value is also known. So, beta 2 comes out to be 68.57 and we attach a negative sign to this. So, as we have calculated all the quantities that are asked for in the problem, namely the angle of the guide vanes, flow angle at the exit of the guide vanes, flow angle at the inlet of the rotor. Notice that these two values are different because the flow changes from being radial at the exit of the guide vanes to axial at the inlet to the runner, which is in contrast to the Francis turbine. We have also calculated the blade angle at the inlet to the runner and the blade angle at exit to the runner. So, it is straightforward to repeat this exercise for the mid radius section and the hub section and students are encouraged to do this exercise for these two sections and make sure that they are able to get these values. Now, looking at the values, we can see that the blade angle at the hub is minus 10.43 degrees and it increases all the way to minus almost 66 degrees at the tip. And again, at the outlet, the angle changes from minus 45 to about minus 68. So, this sort of variation at the inlet shows clearly why the blade appears to be so highly twisted in the case of a in the case of a Kaplan turbine. This completes our discussion of the Kaplan turbine.