 In this module, we will discuss hydro-turbo-machines. It may be recalled that hydro-turbo-machines or those turbo-machines that handle an incompressible fluid such as water or gasoline or kerosene or any other liquid. In the first part of this module, we will discuss pumps and in particular centrifugal pumps and in the second part of this module, we will discuss hydraulic turbines. Now, as we have already seen earlier, if we apply SFE to the rotor of a turbo-machine, assuming no heat loss, we get this equation. We also neglect elevation changes across the rotor between inlet and outlet. This is usually negligibly small. So, if we do that, we end up with this expression. Now, keeping in mind the definition of specific enthalpy, which is U plus pV, where V is the specific volume, we may write this as U plus p over rho. For an incompressible liquid, rho is practically a constant. So, we may write this term, the enthalpy change, H2 minus H1 as P2 minus P1, if we assume that the temperature of the fluid remains the same as it passes through the turbo-machine. In other words, if you are looking at an incompressible fluid, let's say a liquid, which is being pumped in a pump or maybe even passing through a hydraulic turbine and there is no significant change in the temperature of the fluid and so there is no change in the, significant change in the internal energy of the fluid. So, we may write H2 minus H1 as P2 minus P1 divided by rho. So, if we substitute this here and rearrange, we get the following P2 minus P1 over rho equal to V1 square minus P2 square over 2 minus Wx dot over m dot. Wx dot is the thermodynamic work or rate at which work is done in the thermodynamic sense and this is positive for a turbine and negative for a pump or a work absorbing machine. So, we may rewrite this expression in the following manner. Gather P2 and V2 square over 2 on this side, P1 on this side and then write it like this. So, we may write it like this where the P here is the power of the machine and it is a positive number. So, the sign is accounted for here. So, the power is always a positive number and the negative sign is chosen in case it is a turbine and the positive sign is chosen in case it is a compressor or a pump that is a work absorbing or power absorbing machine. If it is a power producing machine then the negative sign is chosen. We have also chosen to write m dot as the product of the density times the volume flow rate. So, this is rho times Q. Now, since the power itself may be written as rho times G times Q times H, that is the hydraulic power, this term may be written as the head H across the turbo machine. So, in case it is a centrifugal pump, the head increases across the pump, that is P2 is more than P1. So, there is an increase in head. In case it is a hydraulic turbine, the pressure decreases across the rotor or runner. So, the outlet pressure is less than the inlet pressure in that case. So, this actually may be thought of as Bernoulli's equation applied across the rotor of a turbo machine that handles liquids with Z2 minus Z1, Z1 across the rotor being very, very small. So, this is an equation that is used quite extensively in the analysis of both pumps as well as hydraulic turbines. The brake horsepower of a turbo machine is equal to the torque T times omega, where omega is the angular velocity. So, the brake horsepower is the shaft power that is generated by the turbo machine in case it is a power producing machine or it is a shaft power that is supplied to the turbo machine because it is a power absorbing machine. So, we can define the efficiency of the overall efficiency of the turbo machine as the hydraulic power P divided by the brake horsepower. So, this is either rho GQH divided by BHP for a power absorbing machine or it is BHP divided by rho GQH for a power generating machine like a turbine. So, this is the definition of efficiency that we will use throughout. Now, let us look at the particular case of centrifugal pump. First we will develop the theory and develop certain theoretical expressions for the performance parameters, do some calculations and then compare the theoretical expressions and the theoretical characteristics against the actual characteristics of the centrifugal pump. So, the blade element of a centrifugal pump looks like this we have already seen this earlier. So, this is section or taken at a certain axial distance equal to constant in a centrifugal impeller. So, let us say at an x equal to constant section and this is what the blade element looks like this is the inlet, this is the exit. So, the angles are also marked here and the reference direction in this case is the radial direction because it is a radial machine. So, alpha 1 here is the angle that the absolute velocity makes with the reference direction and as shown here this is a negative quantity because it is in a clockwise direction. The blade angle beta 1 as can be seen here it is in the anticlockwise direction. So, this is positive and the same is true here also notice that u 1 is not equal to u 2 obviously because it is a radial machine. So, the basic idea here is given the geometric details of the impeller such as r 1, r 2, beta 1, beta 2 and the width of the rotor v 1 and v 2 and inlet and outlet and the angular speed omega we will derive theoretical expressions for the performance parameters of the pump and the performance parameters of the pump are the power that is absorbed by the pump and the head change across the impeller of the head generated by the impeller as functions of the volume flowing. So, the h versus q and the p versus q are called the characteristic curves of the pump. So, given these geometric details and the angular velocity how do we relate h to q and how do we relate p to q in a theoretical manner is what we will take up next. So, based on the velocity triangle shown notice that at any r equal to constant section if you look at the velocity triangle this is the radial component of velocity C 1 I am sorry C r 1 is equal to v r 1. So, the radial component of the relative velocity and the axial and the absolute velocity are equal and that is equal to this in this velocity triangle and it is equal to this in this velocity triangle. So, the volume flow rate at any radius r is nothing but or may be written as q equal to v r times 2 pi r times b from which it follows that v r which is equal to C r may be written as q over 2 pi r times b. Now it is also clear from here that C r the radial component of velocity is equal to C times cosine beta 2. So, I can replace C r with C cosine beta 2 or C cosine beta at any radius r and write C equal to q over 2 pi r b cosine beta. Now it is also clear from here that v theta at any radial location which is this segment here or this segment here is equal to u minus C theta at any radius. So, v theta equal to u minus C theta which is equal to u minus C sin beta from the velocity triangle. So, if I replace C with this expression I get v theta equal to this. Now from the Euler-Turbo-Missionary equation the power is equal to rho times q times u times v theta at outlet minus u times v theta at inlet. What is that? v theta 1 here is in the same direction as u 1 and v theta 2 is in the same direction as u 2 here. So, there is no change in direction of I am sorry v theta is in the same direction as u 1 here and v theta is in the same direction as u 2 here. So, there is no change in direction. So, there is no change in the sign of this term it remains negative sign. So, we may substitute for v theta from here. So, v theta 2 is u2 minus q times tan beta 2 divided by 2 pi r2 times v2 and similarly for v theta 1. So, this is the expression for power. If we divide this by rho gq then this is nothing but the head that is generated by the impeller. So, h is equal to p over rho gq which looks like this. Now normally the impeller is designed so that at the design point which means design flow rate and the given value of omega the flow enters the impeller radially which means alpha 1 is equal to 0. Alpha 1 is equal to 0 and it also follows that v theta 1 is equal to 0. So, we may write the flow rate at design condition like this. So, v theta 1 is equal to 0, then q may be written in terms of u and the other quantities at the inlet using the fact that v theta 1 equal to 0. So, q star equal to 2 pi u1 r1 v1 times cotangent of beta 1 and the h, since this term goes to 0, h is equal to u2 over g times this term and if I pull the u2 outside and use the fact that q is given by this at design condition I end up with an expression like this for h star and p star is nothing but rho times g times q star times h star. So, we set out to derive theoretical expressions connecting the head produced by the machine with q and the power for the machine with q. We have expressions like this at the design condition. We need to have expressions for all flow rates. Now if we neglect v theta 1 at off design conditions, notice that v theta 1 will not be equal to 0 at off design condition blade angles beta 1 and beta 2 and omega r1 r2 are such that at a design point or design operating condition v theta 1 equal to 0 or alpha 1 equal to 0. To all other condition it will not be 0, but for the sake of simplicity if we simply neglect v theta 1 at off design conditions then we may write h equal to this. So, in this expression we simply neglect this and we can write h in terms of q like this and p also in terms of q of course, p is nothing but rho g times q times h. So, now it is clear that this is the characteristic that we are looking for h as a function of q and it is clear from here that h is linear in q and the sign depending upon the exit angle beta 2 whether it is a falling characteristic or a rising characteristic or a constant characteristic depends upon the value for beta 2. So, if you have beta 2 equal to 0 which means that it is a radial blade at exit. So, beta 2 becomes equal to 0 then the tangent to the blade at exit lines up with the radial direction. So, it is a radial blade that means if beta 2 is 0 h is a constant for all values of q equal to q 2 square over g. Now, if beta 2 is less than 0 then we have a forward curved blade and in this case this is negative. So, this becomes positive. So, this is a rising characteristic it increases linearly with q and if beta 2 is greater than 0 then we have a backward curved blade and this is positive. So, h actually decreases with q. So, this is a falling characteristic. So, let us see how we get this connotation from here. So, we can see here that if it is a radial blade then the blade actually must look like this because beta 2 is radial at the exit 0 it has to be radial at inlet also that is the way it is going to be for the radial blade and if the blade is radial then the h is a constant and the power varies increases linearly with q flow rate q that is what we are seeing here. Now, in the case of a forward curving blade remember forward curving refers to the fact that the blade curves in the direction of rotation of the impeller. So, since the impeller rotates like this the blade also curves in this direction. So, in this case the blade angle beta 2 is actually in the clockwise direction. So, it is negative which is why we say it is a forward curving blade. So, if it is forward curving beta 2 is negative and head increases it is a rising characteristic h increases with q. Now, in case it is a backward curving blade like this it is against the direction of rotation beta 2 is in the counter clockwise direction and is positive. So, this is a falling characteristic here and at the pressure increases I am sorry the power increases reaches a maximum and then starts to decrease that is the theoretical characteristics. These are the theoretical characteristics for the centrifugal pump people. Where we have made many simplifications most notably we have assumed that v theta 1 is 0 at all conditions which is which is not a very good assumption to make, but this gives us some insights on the general trends to be expected from various types of impellers. Now, before we proceed to see the actual characteristic of a centrifugal pump let us work out an example. So, we have taken this example from Fox and McDonald, Fox, McDonald and Pritchard. So, the geometric details of the blade are given radius at inlet and outlet, blade width at inlet and outlet, blade angle at inlet and outlet, omega is also given. We are asked to calculate the theoretical head, theoretical power and flow rate and the design operating condition and the same quantities at the different flow rate of 3600 meter cube per hour. So, from the given data we may evaluate the blade speed at the inlet and outlet u1 as 2 pi n r1 over 60 which comes out to be 9.817 meter per second and u2 as u1 times r2 over r1 and it comes out to be 32.73 meter per second. So, at the design point q star is given as 2 pi u1 r1 v1 cotangent beta 1. So, if we substitute the known values into this we get q star to be 0.3045 meter cube per second and the design h is also known. So, h star is given by this expression and if we substitute the numbers this comes out to be 101.32 meters and the power required at the design operating condition is simply rho times g times q star times h star and that comes out to be 302.66 kilowatts. We are also asked to calculate the shutoff head. The shutoff head is simply nothing but the head that is generated by the machine at zero flow rate. So, that is this quantity here. So, if you look at this expression we said q equal to 0. So, the shutoff head is nothing but u2 square over g and this comes out to be 109.17 meter. Next, we are asked to calculate head and power input for flow rate of 3600 meter cube per hour which corresponds to 1 meter cube per second. So, we substitute this into the general expression for h what is that we have neglected b theta 1. So, if you substitute this into the expression for h the given flow rate is 1 meter cube per second and we get the head to be 83.4 meters and the power to be 818.154 kilowatts corresponding to this case.