 to this session, we will discuss today the energy transfer in fluid machines part two in continuation of our earlier discussion. Now, last in last discussion we have recognized that the energy transfer to the rotor of the machine by the fluid in terms of the energy per unit weight which is known as head can be expressed as just let me write 1 by 2 g v 1 square minus v 2 square plus u 1 square minus u 2 square plus v r 2 square minus v r 1 square. So, in last in last session we have recognized that the energy per unit weight that is the head transfer to the fluid rotor can be split up into three distinct components where the nomenclatures are like this we can have a recapitulation that where v 1 v 2 u 1 u 2 v r 2 v r 1 are like this that v 1 and v 2 are the absolute velocities of the fluid at inlet and outlet of the rotor u 1 and u 2 are the tangential velocities of the rotor at inlet and outlet either the rotor velocities willing velocities of the rotor at inlet and outlet and v r 1 and v r 2 are respectively the relative velocity of the fluid with respect to the rotor at inlet and outlet. So, if v 1 v 2 u 1 u 2 v r 2 v r 1 are defined like this we can express the head that is transferred to the machine by the fluid as it flows through the rotor vanes can be written like this now we see that these three terms have got their different physical implications now let us see first what is the term v 1 minus v 2 square this implies a change in the velocity head of the fluid or a change in the kinetic head kinetic energy per unit weight of the fluid or simply it can be told as dynamic head dynamic head you can write a change in dynamic head changing dynamic head this this one changing dynamic head. So, therefore, due to the change in the dynamic head of the fluid that means, the change in is absolute velocity as it flows past the vanes the work is being transferred or energy is being transferred to the machine. Similarly, this term represents a change in the head due to the change in its position radial position with respect to the axis of rotation when a fluid has got a rotational velocity and it changes its radial position with respect to the axis of rotation there occurs a change in the head or energy in the fluid now this term can be better understood if we see this one. Let us consider a container where the fluid is flowing in this direction and container is given an angular rotation omega like this. So, basic objective is to show that when a fluid element under a rotational velocity changes its position in radial coordinates with respect to the axis of rotation. Let this is the axis of rotation at this point perpendicular to this plane of the figure about which the container is rotating then we can show that the work is either being done on the fluid element or work is being extracted from the fluid element how we can show now let us consider a fluid element at a radius r of thickness d r and area d a. Now, you know that whenever there is a rotational flow field it induces a pressure gradient a pressure variation in the flow in the direction of the flow exists for which the pressure in the positive direction this direction of the r is higher than that at this upstream plane. So, therefore, if we take the force balance of the fluid element we see the net force acting on the fluid element in the radially inward direction is can be written as p plus d p into d a where d a is the cross sectional area of the fluid element minus p d a well which can be written as d p d a. So, what is d p d a is the net force in the radial inward direction let we denoted by f that is equal to d p into d a. Now, this radial inward force balances the centrifugal force due to the rotational motion of the fluid element. So, this radial inward force balances the centrifugal force of the fluid element under rotational velocity. So, what is the centrifugal force what is the centrifugal force let f c for the fluid element it is the elemental mass d m times the linear velocity due to this rotation that is the tangential velocity v square by the radius or the radial location from the axis of rotation this can be written in terms of the angular velocity as d m omega square r this is the usual expression of the centrifugal force which is acted on this fluid element. Now, if we substitute the mass in terms of the area and the other geometrical dimension and the density of the fluid element we can write it row d a d r. So, row d a d r is the mass of the fluid element. So, this is the angular velocity square into r. Now, at equilibrium this two are equal that means the fluid motion is possible in this direction provided there is a balance between the centrifugal force and the inward radial pressure force. So, if we write this we get the expression d p into d a is equal to row d a d r omega square into r. So, d a cancels the angular velocity out well we can write then d p d r is equal to row omega square r this equation is a very well known equation in the fluid flow with rotational velocity and is known as radial equilibrium equation this is known as radial equilibrium equation I can write it that radial radial equilibrium equation this equation simply implies that when there is a rotational velocity in a flow field and fluid flows in the radial direction then a inward radial pressure gradient is imposed on the flow field which provides the necessary pressure forces to be balanced with the centrifugal force. You know that in any rotational motion of in your solid body there is there are two forces are in balance with each other one is the centrifugal force which tends to make it flying away from the path and another the centripetal force which is a force which makes it possible to have the rotational motion which is inward towards the center of rotation. So, this centripetal force is provided by the pressure gradient through this pressure forces this is the well known radial equilibrium equation. Now, if I write this in a little different form the same equation can be written as d p by row is equal to d p by row is equal to omega square r d r now if I integrate this equation d p by row integrate this equation omega square r d r between two points one and two between two points one and two which physically indicates the two points let one is at the inlet and two it is at the outlet it may be any two points in the flow field one is an upstream point and two is an downstream point which may be at the inlet and outlet in a flow passage well then we can write this one by two d p by d row is equal to half omega square r two square minus omega square minus omega square which is nothing but half the linear velocity of the tangential velocity due to the rotation at the point two the section two minus u one square what is the meaning of this now what is this d p by row from one to two integral this is the flow work flow work now I come to the concept of flow work now if you recollect thermodynamic general energy question you know what is flow work let us recapitulate little bit of thermodynamic concept you know when you have a closed system when you have a closed system and it interacts with the surrounding in terms of work either work is being developed by the system to the surrounding or is absorbed from the surrounding to the system mechanical work if you consider the most usual form is by the displacement of the system boundary by the displacement of the system boundary for a closed system because the mass within the system is fixed and under reversible condition this work transfer is written as p d v where d v is the we cut d v is the change in the volume so the integral is made between the two state points one and two well but what happens when the system is a is an open system that means in thermodynamics we know there are two types of system and the closed system at the mass is fixed with the same identity that is known as control mass system usually we tell a system another system is there where the mass is not fixed with the identity there is a continuous flow of mass in and flow of mass out when the volume is control volume is fixed known as control volume system and usually we tell open system or a control volume so in case of an open system or a control volume that means an open system open system or a control volume there is a continuous influx of mass and energy continuous mass coming and mass going out similarly in this control volume if it interacts with surrounding in the form of work that means if it develops work or it absorbs work which comes in our case of fluid machine a fluid machine is an open system continuously the fluid comes into the machine at one part and it goes out of the machine by virtue of with the machine develops work to the surrounding in the form of shaft work in some machines it is being developed to the surrounding the shaft work is being obtained by us and in some cases the machine absorbs the work from the surrounding that means the work is being put to the shaft in the form of the shaft work these are case of compressors and pumps while the work is obtained in case of turbine so how to find out this work in this case in this cases we write the steady flow energy equation this is the little recapitulation of your thermodynamic concept so that you can recognize or appreciate the term dp by d rho as the flow work so what is that if we write the thermodynamic equation general energy equation of thermodynamics at section one and two then we can write that the internal energy at one u one associated with the mass flux plus the pressure energy which is written in thermodynamics in terms of the specific volume rather than the density plus we write the kinetic energy v one square by two per unit mass basis if we write so if we consider the potential energies at this and this sections are given like this if we denote z one and z two at the elevations from reference datum so this quantity represents the amount of energy influx per unit mass with the mass flow coming into the control volume similarly the amount of energy going out from the control volume associated with the mass flux out of the control volume per unit mass will be the same energy quantities with their values at the outlet section denoted by the suffix two. I am sorry per unit mass means this will be g z one so this will be g z two and if we consider the work is out coming out of the open system or control volume plus the work done per unit mass here w is the work done per unit mass at the time being we neglect the heat flow if you consider the heat flow you can take in this heat flow either on the right hand side or on the left hand side depending upon whether you consider the heat is coming into the system or going out of the system. Now therefore we see that if we replace this as the enthalpy you know from the definition of enthalpy it is the internal energy plus the product of pressure specific volume. So we can write h one plus v one square by two plus g z one is equal to h two plus v two square by two plus g z two plus w. Now under all usual operating conditions of engineering system we have found that the change in kinetic energy and the potential energies are much smaller as compared to the changes in the enthalpy in such systems. So therefore we can neglect the changes in kinetic and potential energies as compared to the change in enthalpy and we can write under such condition w is simply equal to h one minus h two that means the change in the enthalpy from the inlet to outlet is given as the work transfer provided the heat transfer between the open system and the surrounding is neglected that means the system is properly insulated. So that the heat interactions with the surrounding is prevented. So nowadays we know that for any open system it is the enthalpy difference which gives the work done. Now therefore the enthalpy difference we get as the work transfer. Now if we consider the process to be reversible reversible means without friction and any other dissipative effect and no heat transfer is there then we can write from thermodynamic equation first of all you know the general thermodynamic equation T d s is d h minus d d t where this s is the specific entropy h is the specific enthalpy and v is the specific volume this is a generalized thermodynamic relations under reversible adiabatic condition which is isentropic condition we can write d s zero. So therefore it is clear that d h is nothing but v d p or integral of d h is equal to integral of v d p for between any two sections one and two which can be written as h two minus h one is integral of v d p. So therefore we see this integral of v d p refers to this enthalpy differences which equals to the work transfer which equals to the work transfer as we have discussed in an open system this v is the specific volume of the system. So therefore if we come to our earlier slides then we see that this left hand side is d p by rho which is nothing but the flow work because one by rho is v. So flow work means that is the work transfer between a open system here the turbo machines or the fluid machines with the surrounding and this flow work becomes equal to half u two square minus u one square. So therefore we see some work is being done on the fluid system if it changes its radial location from one to two where at two the radial position or the value of the radial location is more than that at one that means if it goes further from the centre of rotation then this is positive some work is being done on the fluid element and vice versa takes place that when the fluid changes its radial position in a way that it comes nearer to the axis of rotation then this is negative then work is being released by the fluid. So therefore if we look to this diagram well then we can realize therefore from this equation that whenever a fluid moves from this position to this position that means from an upstream position to this downstream position away from the axis of rotation work is being imparted on the fluid and this work is stored in the fluid as energy and this energy is mainly in the form of pressure energy why because it is very simple because of this radial pressure gradient the fluid when moves from this place to a place where at two where the radius from the axis of rotation is more then the fluid attains a higher pressure that means fluid possess a higher pressure and therefore the work done on the fluid is stored in the fluid as pressure energy. So this way we can tell that movement of a fluid element with a rotational motion from one radial location to other radial location physically implies that head is either given to the fluid and it stores it or head is being extracted from the fluid from its stored head. So this is the physical implication of this term that in a fluid machine as the fluid flows in a rotational flow field from one radial location to other radial location. So either the head is gained by the fluid or head is developed by the fluid. So this is the contribution and that head gained or developed is in the form of the pressure it. So that is why this gain in head or loss in head is known as static head here the way it is written this head is the head developed by the fluid. So therefore you see the positive value of it indicates that fluid comes from a higher radius to a lower radius from the center of rotation. So u 1 is more than u t that means it releases some of its static head or mainly the pressure head contributes this thing to the turbo machines to be developed. So the contribution of the second term is like that what is the contribution of the third term which is very important and also very interesting physical significance just it is obvious from mathematical expression it is the change of relative velocity. And one interesting thing is that where everything is that the first one is the inlet and second one is the outlet here it is just the opposite that means it is the change of velocity head relative based on the relative velocity from the outlet to inlet what does it mean. Now what is the concept of relative velocity let us think in this fashion that relative velocity is the velocity with respect to the moving vane that means if the vane could have been fixed the inlet velocity could have been v r 1 and outlet velocity could have been v r 2. Now you consider a fixed vane where the inlet velocity is v r 1 and outlet velocity is v r 2. Then what are the what are the possibilities are under which there can be a change in the velocity number one is the friction on the vane that due to the friction on the vane vane is at rest. So the velocity may change where the outlet velocity will be lower than the inlet velocity will be reduced because of the friction another opportunity is there what is that if it is not a single vane open to atmosphere if there are number of vane in a closed casing. And if the flow takes place through the passage of two vane gliding over one vane then even if the vane is at rest consider series of vane at rest then the whether the fluid velocity will change or not depend upon the flow cross sectional area. That means you simply consider a flow through a fixed duct simply flow through a fixed duct when a flow takes place through a fixed duct the flow velocity changes from upstream to downstream section under two conditions one is the friction with the wall. And another is that if the flow area changes where the pressure changes and velocity changes this is the consequence of continuity and Bernoulli's theory there due to the change in the flow area the velocity changes because of the continuity for example if the flow area is converging the velocity increases and due to this change in the momentum there is a change in the pressure that is even for an ideal fluid according to Euler's equation or Bernoulli's equation. So because of the change in velocity due to his flow through a varying area passage even a static duct is there either it is formed by two vane or it is formed or it is made by walls just like a duct. So velocity of fluid can change here also now you see that relative velocity change can take place in the similar fashion that means if the area flow area between the vane changes then there can be a change in the relative velocity otherwise the change in the relative velocity is not there only there will be a little change due to friction. And if the vane surface is very smooth this change due to friction is not much change due to friction always makes the outlet velocity slightly lower than the inlet velocity. But the relative magnitude of the outlet relative velocity with respect to the inlet relative velocity due to the change in the cross sectional area in the vane passage will depend upon the fact whether the area gradually decreases or converges in the direction of flow or diverges in the direction of flow. If we allow a converging area in the direction of flow then vr 2 will be more than vr 1 you understand. So in that case what will happen if vr 2 is more than vr 1 consider a fixed vane vr 2 is more than vr 1 means pressure will be less at outlet than the inlet. That means fluid releases its pressure or contributes or gives away its pressure. If the vane is at rest means what practically we will have to give some support to keep him at rest. So if it is if moves in developing work the release in the pressure of the fluid gives some energy that is fluid releases the pressure energy. In the in opposite case when we allow a diverging passage then what happens the vr 2 decreases or according to Bernoulli's theory pressure increases. That means when vr 2 is less than vr 1 because of a diverging passage then pressure is increased that means fluid gains the pressure energy. So therefore you see that the change in the relative velocity means either fluid gives the pressure energy to the turbo machines the rotor of the turbo machines or it gains pressure energy from the rotor of the turbo machine which is manifested in terms of the increase or decrease in pressure. So therefore here you come in this expression vr 2 minus vr 1 square says that if vr 2 is more than vr 1 which is possible only if the flow through the vane passage flow area of the vane passage is converging in the direction of flow. Then vr 2 is more than vr 1 then this quantity implies a change in the pressure energy. Because if we write the Bernoulli's equation then we will see that p 1 by rho plus v r 1 square by rho that means considering the vane to be in a static condition then p 2 by rho plus v r 2 square by rho. So therefore you see that v r 2 square minus v r 1 square now you see v r 2 square minus v r 1 square by 2 g is nothing but p 1 minus p 2 by rho g. So this is simply the change in the pressure energy which is equal to that due to the change in the dynamic head based on the relative velocity. Since this is manifested in terms of the pressure head this we will not tell as the dynamic head transfer. Now we can tell that both of these terms two terms contribute in the energy transfer of the fluid machines in terms of the static energy that means the pressure energy of the fluid. So therefore they are told or they are they are called as change in static head change in static head. So therefore today we conclude that the general expression for the head transfer to the turbo machines by the fluid is given by 1 by g v w 1 u 1 minus v w 2 u 2 this is known as Euler's equation where v w 1 and v w 2 are the whirling or tangential component of velocities of the fluids at inlet and outlet and u 1 and u 2 are the tangential velocities of the rotor at inlet and outlet and with the help of velocity triangles at inlet and outlet we can express this into three components which are very interesting from their physical point of view where we can get an idea that this total head which is being transmitted to the turbo machines by the fluid is contributed by the dynamic head of the fluid and static head of the fluid. So this part is contributed by the dynamic head of the fluid where the absolute velocity of the fluid is changed if it is lowered then it gives the head otherwise it gains the head similarly the change in the rotor velocity as the fluid passes from one radial location to other radial location and along with the change in the relative velocities of the fluid from its inlet to outlet provided a varying area of passage is given for the flow of fluid contributes to the change in the static head of the fluid that means either the fluid loses its static head what is the static head that is the pressure head loses its pressure energy and it is given to the turbo machines or it extracts from the turbo machine the pressure energy or the static energy. So this is the three components for the head. So today I will end here. Thank you.