 Hello friends, today we are going to study numerical on inward flow reaction turbine. Myself, SP Mankani, Assistant Professor, Department of Mechanical Engineering, Valchand Institute of Technology, Solapur. At the end of this session, students will be able to analyze velocity diagram, students will be able to apply theoretical knowledge to solve numerical on reaction turbine. So now we are going to study the numerical on inward flow reaction turbine and inward flow reaction turbine has external and internal diameters as 0.9 meter and 0.45 meter respectively. Turbine is running at 200 rpm and width of turbine at inlet is 200 mm, the velocity of flow through the inner, the velocity of flow through the runner is constant and is equal to 1.8 meter, the guide blades makes an angle of 10 degree to the tangent of the wheel and the discharge that the outlet of the turbine is radial, draw the inlet and outlet velocity triangles and determine the absolute velocity of water at inlet of runner, the velocity of wheel at inlet, the relative velocity at inlet, the runner blade angles, the width runner at outlet, mass of water flowing through the runner per second, headed the inlet of the turbine, power developed and hydraulic efficiency of the turbine. So here he has given the conditions as diameters d1 and diameter d2 and the running at a speed of 200 rpm is the value of n he has given and the width of the turbine the b1 value is given at inlet, this value is b1, the velocity of flow through the runner is constant and is equal to 1.8 meter per second. So this is the condition as vf1 is equal to vf2, guide blades makes an angle of 10 degree, this value is given as a value of alpha to the tangent of the wheel and the discharge at the outlet of the run, outlet of the turbine is radial. So this indicates radial as beta is equal to 90 degree, beta equal to 90 degree, that is the discharge at outlet is 90 degree, beta is equal to 90 degree angle, similarly vw2 wheel velocity at outlet is 0. So as far as the given condition as external diameter, internal diameter, speed, the width b1, vf1 equal to vf2, 1.8 meter per second, alpha equal to 10 degree, beta is 90 degree and vw2 is equal to 0. So now in the next, so this is the velocity triangle we are going to be representing for the reaction turbine, so this is the inlet portion and this is for the outlet condition. So here he has given as this values as the alpha angle is given and similarly the value of d1 is given, d2 is given, the n speed is given, b1 is given and alpha angle is given as a 10 degree, based on this one, velocity triangle from the inlet to the outlet triangle. We are going to solve the values one by one, so with the given condition as we are going to calculate the value of u1, that is the tangential velocity u1 at inlet condition, pi d1n by 60, that is 9.424, as the value of n is given as 200 and the d1 diameter is given as 0.9, by substituting these values we are going to calculate the value of u1. Similarly we go for the outlet tangential velocity u2, so as pi d2n divided by 60, d2 is given the value as 0.45 and the n is 200, so by substituting those values we are going to get the value as 4.712 meter per second. The absolute velocity of water at inlet v1, v1 in the velocity triangle, so this value we are going to be identifying as a v1 value. So from inlet velocity triangle v1 sin alpha, so this is equal to vf1, v1 sin alpha, v1 sin of this angle alpha if we are going to be taken we are going to get the value of vf1, so that is vf1 this value. So as if this velocity of this one we are going to take as a one as a vertical component, another as a horizontal component in the velocity triangle. So v1 is equal to vf1 upon sin alpha, that is v1 sin alpha is equal to vf1, so that value v1 is equal to vf1 upon sin alpha. By substituting this vf1 value which is given as 1.8 meter, 1.8 meter per second and sin alpha, alpha is given as a 10 degree, alpha is given as 10 degree angle. By substituting this value we are going to get as a 10.365 meter per second as a velocity v1. So the next velocity of virl, virl velocity at the inlet vw1, this is vw1, virl velocity at inlet for calculating this vw1, v1 horizontal component of this v1 we are going to take it v1 cos of alpha, so that we are going to get as a vw1 value. So here virl velocity at inlet vw1 equal to vw1 is equal to v1 cos alpha, so alpha is 10 degree, v1 just now you have calculated 10.365, by substituting these two values we are going to get this as a 10.207 meter per second. So now, relative velocity at inlet, so this is a relative velocity at inlet vr1, we are going to use this as a Pythagoras theorem by that one. So this is vr1 is equal to under root of vf1 square plus this value, this value is nothing but vw1 minus u1. So now under root vr1 is equal to under root of vf1 square plus vw1 minus u1 bracket square. So now we are going to be substituting this vf1 value and vw1 and vw1, vw1 just now we have calculated and vf1 is equal to vf2 is equal to given as a 1.8 by substituting these values we are going to get this as a 1.963 meter per second. This is the value you have taken it as this value, vr1 is equal to you are going to be getting it as a value as a 1.963 meter per second. That is a component of this one you are going to get the value as vr1 is equal to 1.963 meter per second. Next the runner blade angles means theta angle and alpha angle. Theta angle and alpha angle we are going to show it in the velocity triangle, this is theta angle, this is theta angle and another angle is a phi angle, this is phi angle. We are going to calculate this one, tan of this angle tan of theta, tan of theta is equal to vf1 divided by this value vw1 minus u1, opposite side upon adjacent side, opposite upon adjacent side. So vf1 already know the value as a 1.8 and vw1 just now you have calculated here as a 10.207 and u1 value already calculated as a 9.24. By substituting these values we are going to get theta is equal to 66 degree 29 minutes. So this is a value related with the theta. Similarly we are going for the outlet triangle, outlet angle for the value of phi. So this angle, outlet angle as a phi we are going to get it tan of this angle tan of phi is equal to vf2 opposite side vf2 divided by u2 divided by u2. This opposite side and adjacent side, this one is a high part dose. So vf2 divided by u2 is equal to you are going to get this as a 20.9, vf2 already vf1 is equal to vf2 and u2 just you have calculated as a 4.712. By substituting these two values you are going to get this as a 20.9 as a value of tan phi, but phi is equal to you are going to get as a 20 degree 54 minutes and 4 seconds. Width of the runner at outlet. So for calculating the width of the runner at outlet you are going to take it as a q1 is equal to q2 is equal to q value. So this q1 is pi d1 v1 vf1 is equal to pi d2 v2 vf2. So as pi and pi is going to be cancelled this values the remaining as a d1 v1 and vf1 is equal to vf2 is given in the problem. So based on that one d1 v1 is equal to d2 v2. So based on that one you are going to get it as the value of v1 as a 400 millimeter as the value is width of a runner at outlet. This is the value as v2 value not v1 it is a v2 value v1 is given in the problem. v1 is given in the problem as the value as a 200 millimeter v1 is equal to 200 millimeter is given. Mass of water flowing through the runner per second. So q is equal to pi d1 v1 vf1. So based on this one value we know the value of d1 is given in the problem v1 is given vf1 is given pi value we know it by substituting these values we will get a q value. By using this as a q we are going to get it as a mass is equal to rho into q rho is density. So now rho is kg per meter cube and this q is into meter cube per second. So meter cube, meter cube cancels we are going to get it as a kg per second as a mass that is the value as 1017.8 kg per second by substituting these two values. So next value is expected in the problem as head at the inlet of the turbine that is the value of h head at the inlet of the turbine. So we are going to write this as the equation as h minus v2 square divided by 2g. So based on this one we are going to take as a 1 upon g into bracket vw1 u1 plus or minus vw2 u2. So this value we are going to be simplifying as a 1 upon g vw1 u1 as this vw2 is 0. So now this value is going to be cancelled and only remaining as a 1 upon g vw1 u1. So vw1 so value we are going to be substituting here and u1 we are going to substitute it and we are going to get a value as h is equal to 9.97 meter. So vw2 is 0 and v2 is equal to vf2. So that is shown in the diagram also here v2 is equal to vf2. So last value as power developed that is p is equal to work done per second on runner divided by 1000. We are going to take it as a just now we have calculated this rho q value mass into vw1 u1. So that value we are going to get as a power developed as 97.9 kilo watts. So next is hydraulic efficiency. Hydraulic efficiency is equal to nu h is equal to vw1 u1 upon gh that value we are going to get as a 98.34. Thank you.