 description of rotodynamic machines as I have told earlier the rotodynamic machines are those machines where there is a continuous motion of fluid and a part of the machine known as rotor and because of this continuous relative motions between the fluid and the rotor of the machine it is possible for an energy transfer to take place between the fluid and the rotor. So, therefore, the basic principle of this machine is based on the fluid dynamic principle, fluid dynamic principle which is basically the utilization of useful work due to the force exerted by a fluid striking on a series of curved vane which is mounted on the periphery of a disk that is rotating the periphery of a disk that is attached to a rotating shaft. So, therefore, to understand the basic principle of a rotodynamic machines we should understand clearly the force interaction and the energy transfer that takes place while a stream of fluid passes through a curved vane. So, this is a little recapitulation of what you have already studied at your basic fluid mechanics course that we study here the interaction of force and energy in the flow of fluid along a curved vane. Now, see here this is a curved vane which is moving with a velocity u and the jet of fluid is striking the vane with a velocity v 1 is the velocity absolute velocity with which the fluid strikes the vane and the fluid after flowing through this vane comes out with a velocity v 2 this is the velocity v 2. Since the vane is moving with a velocity u, so the jet appears to strike the vane that means with respect to the vane the jet strikes it with a velocity v r 1 which is the velocity of the jet relative to the vane. Similarly, it is going out with a velocity v r 2 that is the relative velocity of the fluid with respect to the vane. Now, this relative velocity is at inlet and outlet are determined just by vectorial subtraction from v 1 the velocity u of the vane and from v 2 the velocity u of the vane. So, this vectorial subtraction is shown in terms of the velocity triangles as we have already read at the inlet and outlet. Now, let the suffix 1 refer to inlet condition and suffix 2 refers to outlet condition. Now, we see in this triangle this is v 1 that inlet velocity of the fluid this is the u the vane velocity and this is the v r 1 that is the relative velocity of fluid with respect to vane at inlet. This component this perpendicular component to the motion of the vane is denoted as v f 1 and is usually known as flow velocity. Similarly, the component of the fluid velocity absolute fluid velocity in the direction of the vane motion is conventionally symbolized as v w the suffix 1 is at the inlet. So, this v w I tell you this is a conventional symbol w stands for whirl whirling component this is because in actual case this velocity of the vane is in the tangential direction because the motion of the vane mounted on the periphery is a rotating motion. So, therefore, the linear velocity of the vane is in the tangential direction and that is why this component is known as tangential component or whirling component for which a conventional symbol w is given as this suffix. Similar is the case in case of an outlet velocity triangle this is the vane velocity this is the relative velocity of the fluid with respect to vane and this component is the whirling component or the component of the flow velocity in the direction of the vane velocity and this is the flow velocity that is the direction of the that is the sorry the component of the fluid velocity in the direction perpendicular to the vane velocity. Now, our basic purpose in this case is to analyze what is the force exerted by the fluid on the vane or vice versa vane on the fluid and by virtue of the vane motion which is the what is the amount of energy that is being transferred or developed due to this force due to this action of the fluid on the vane. So, to analyze this as you know we apply the momentum theorem now to apply the momentum theorem we have to take a control volume of the fluid like this which is just adjacent to the vane. Now, you see that this type of analysis can be done on the basis of both system approach and the control volume approach in a system approach what is done the new terms law is applied in a sense that you consider a particular mass of fluid and consider it is change of momentum as it flows along the vane find out the change of momentum in a specific direction and in control volume the same thing is done, but the version is different we find the momentum efflux net momentum efflux in a particular direction and equates this with the force in that particular direction. So, if you if we apply this theorem for a steady state situation the situation is steady then we find that if f x is the force acting on the control volume in the direction x then it will be the net rate of momentum efflux from the control volume in that direction x, because we are interested in the direction x that is the direction of the vane velocity the force in that direction. So, the expression in the right hand side is either the net rate of momentum efflux x momentum efflux from the control volume or from a system approach it is the change of momentum change of momentum in the x direction of a fluid mass taken as a system in either way you can see it and that becomes equal to the force is equal to the change of momentum times the mass flow rate. Now, you see the velocity at the outlet is v r 2 now here we have to consider the relative velocities, because in this case the control volume is moving with a velocity use since the vane is moving with the velocity use this is an inertial control volume. So, the coordinate axis will be fixed to this control volume. So, therefore, the velocities which we have to take at the relative velocity. So, you see the component of the velocity in the direction of vane velocity here the if beta 2 is the angle made by v r 2 with the direction of vane velocity it will be minus v r 2 cos beta 2, because this direction is opposite to that of the vane velocity or to that of the positive direction of the specified axis x. This is the momentum efflux minus the momentum efflux that means v r 1 cos v r beta 1 beta 1 is the angle made by the relative velocity with the vane direction. So, this component is in the direction of the vane velocity or in the positive direction of x. So, minus sign is that because it is the efflux minus efflux. So, both the terms are with a minus sign. So, it comes out. So, minus m dot now this v r 1 cos beta 1 or v r 2 cos beta 2 if you see from this triangle. So, this comes out to be v w 1 and v w 2. So, therefore, we say that force on the vane is equal to minus efflux that means this is the force that is being acted on what that is being acted on the control volume. So, the force acting on the vane is in the opposite direction that means if this is the efflux it is the in opposite direction of efflux minus efflux. Now, power developed due to the motion of the vane is then force into the velocity that is m dot v w 1 plus v w 2 into u u is the vane velocity. So, this way the we can develop an expression for the power developed due to the action of the fluid passing over a curved vane. I think you have understood it all right. So, from this two triangles you can get from here triangular relationship geometry that it is v w 1 plus v w 2 and this this minus sign is because the force which is acting on the fluid element or the control volume is in the opposite direction to the specified axis that means in a direction opposite to the vane motion. So, therefore, the force on the vane is in the direction of the vane motion. However, the expression for power developed is written as the multiplication of efflux and u they are in the same direction. So, they are absolute values are taken well. Now, yes please v r 2 cos beta 2 is not v w 2 plus u and v w 2 plus v w 2 plus v and here it is v w 1 minus u that cancels out actually. So, ultimately you get v w 1 plus v w 2 yes correct v r 2 cos beta 2 is not v w 2 it is v w 2 plus u on the other end v r 1 cos beta 1 is also not v w 1 it is v w 1 minus u. If you substitute that automatically it cancels out and becomes because I felt that you have already done it at your basic fluid mechanics course. So, this thing you know well very good I am happy that you are asking questions. Now, we come to the basic equation of energy transferring rotodynamic machines. Now, in a rotodynamic machines what happens is that the rotor of the machine is a rotating wheel on which the vents are mounted and the wheel is mounted on a shaft where the rotation is imparted. So, in this case the same principle is applied here and we analyze this in the similar fashion with the help of a diagram here which is the general representation of a rotor or the representation of a rotor of a generalized fluid machines. Now components of flow velocity in a generalized fluid machines here the most general sense we consider the rotor where the fluid enters at a velocity v 1 at any point whose radius of rotation from the axis is r 1. Now, before that I like to mentioning that there are few assumptions for this analysis one assumption is that the flow is steady. So, there is no mass accumulation no mass depletion anywhere in the system and number two assumption is that flow is uniform over any cross section normal to the flow velocity which is very important and that means that the velocity vector at a point is the representative of the flow over a finite area. That means we analyze with respect to a velocity vector at a point and we assume that this is uniform over the entire flow area that is an area normal to the flow velocity. So, that this is the representative of the entire flow through the fluid machines well. So, with these assumptions now we consider that at any point the velocity vector is v 1 that is the inlet point a very general case whose radius of rotation from the axis of rotation is r 1. Similarly, the fluid goes out or discharges at a point from the rotor whose radius of rotation from the whose radius sorry whose radius from the axis of rotation is r 2. Now, the velocities v 1 and v 2 can be resolved into three components there may be an arbitrary angle at which the velocity flow velocity strikes the rotor which can be resolved into three reference directions. One is in the direction of tangential one in in the direction of tangent a tangential direction which is the tangent to the rotor at that point another is the direction which is the axial direction that means it is parallel to the axis of the shaft and another is the radial direction which is perpendicular to the axial direction. So, these three perpendicular mutually perpendicular directions the velocities are resolved one is the tangential direction another is the axial direction another is the radial direction. So, these three perpendicular directions and accordingly symbolized as v w one is the suffix at inlet that is the tangential component whirling component that is why the suffix w is used the suffix a v a is the axial component that means component parallel to the axis of the rotation. And as I have told earlier the symbol f is used v f 1 for the inlet that is the flow velocity that is in the radial direction. Similarly the velocities are resolved in tangential direction as v w 2 at the outlet axial direction v a 2 and the flow direction v a 2. Now, the rotor is moving with an angular velocity omega it is a constant angular velocity this is a steady state problem well now let us apply the momentum theory or the new terms laws of motions either with respect to a system or a control volume here. Now, here the momentum which will be considered is the angular momentum this is because here the work transfer takes place due to the rotation of the shaft. So, we will be considering the angular momentum or moment of the momentum it is very simple if we consider a system approach our version will be that considering a fluid mass as it passes from the inlet to outlet what is its change in angular momentum if we consider a control volume of a fluid then what will be the net rate of a flux of the angular momentum from the control volume. Now, here one thing is very important we are not bothered about the path in the rotor it is only the inlet and outlet that decides the change because if the inlet and outlet conditions are fixed kinematic conditions are fixed and the mass flow rate is steady. So, the change in momentum or the moment of the momentum whatever you say depends upon the inlet and outlet conditions well now if we write the momentum moment of the momentum at the inlet for a unit mass at inlet what will be its value at inlet at inlet at inlet moment of the momentum moment of the momentum moment of the momentum is equal to that is the moment of the tangential momentum that means v w 1 times the r 1 radius from the axis of rotation it is per unit mass per unit similarly the same thing at outlet at outlet the same thing at outlet at outlet at outlet is equal to v w 2 into r 2. So, therefore, per unit mass the change in the moment of the momentum of a fluid mass or the net rate of a flux of the moment of the momentum per unit mass from a control volume will be v w 2 and that multiplied by the mass flow rates that means this will be the net rate of angular momentum a flux or the rate of change of angular momentum rate of angular momentum net rate of angular momentum a flux when we refer it to a control volume that is a control volume approach control volume of the fluid or it is the net rate of change of angular momentum for a single system as it passes from inlet to outlet. So, in both the cases that equals to the torque that is the angular momentum theorem that is the angular momentum theorem applied to a system or to a control volume that torque is equal to the rate of change of angular momentum of a system or torque is equal to the net rate of angular momentum a flux from a control volume at steady state. So, that is equal to the torque that is being imparted on the fluid by the rotating disk. Now, the energy rate of energy that is being imparted to the fluid will be nothing but this torque into the angular velocity omega and that if we multiply the angular velocity and recognize that omega r 1 is the velocity of the linear velocity that tangential velocity of the rotor and inlet and omega r 2 is the linear or tangential velocity of the rotor at outlet and denoting them by the symbol u we can write v w 2 u 2 minus v w 1 u 1. So, therefore, we see that energy transfer per unit time the rate of energy transfer in the fluid as it passes from inlet to outlet becomes equal to the mass flow rate m dot flowing times v w 2 u 2 minus v w 1 u 1 where u 2 and u 1 are the tangential velocity that is the linear velocity of the rotor at the outlet point and u 1 is that at the inlet point because in a generalized case we have to consider that inlet and outlet are not in the same radius from the axis of rotation there is not at the same radial plane. So, this is the general equation and here in this equation we see that this is the energy rate of energy fluid now if we write the rate of energy as e dot then we can write this is just the negative of that m dot into v w 1 u 1 minus v w where we tell that here e dot is the rate of energy that is being supplied to the rotor by the fluid conventionally we take the positive sign of this energy transfer when it is being transferred by the fluid to the rotor or the machine. So, therefore, we always express in this fashion this can be expressed as per unit mass basis that means the energy transfer per unit mass e dot by n dot is equal to v w 1 u 1 minus v w 2 u 2 as you know in fluid flow we define a term head what is head what is the definition of a sometimes we call that is a pressure head is a velocity head potential head total head what is the definition of a height yes energy can be expressed in terms of the height height of the fluid column because head is expressed in terms of the linear dimension but, what is the definition it implies the concept of energy what is that the height of a fluid implies the concept of pressure sometimes it concept of energy implies the concept of energy also it is the energy per unit weight for a fluid flow it is the energy per unit weight if you see the dimension we will see the dimension of energy per the height for example, the height of a liquid column exerts a pressure at the base and at the same time it has due to that pressure has a pressure energy and pressure energy per unit weight is that height. So, therefore, it is the energy per unit weight is the height. So, therefore, we express it as the energy per unit weight that means, we take g and the expression comes like this g v w 1 u 1 minus. So, therefore, these three equations they are dependent equations any one of them is called as Euler's equation these are known as Euler's equations according to the name of the scientist this is known as Euler's equation this should not be confused with the Euler's equation of motion. This is the Euler's equation for fluid machines this is comes from the equation of motion, but another Euler's equation you have already read in your fluid mechanics class that is the equation of motion for an inviscid fluid is a generalized equation of motion for flow of an inviscid fluid, but these Euler's equation is the equation for the energy transfer between the fluid and the photodynamic machines. This is the form where the rate of energy transfer is given that is the energy transfer per unit mass and this expression is the energy transfer per unit weight or head sometimes we call it as head here you can write head h that is the head transfer. Now, it is very simple you can understand that when when v w 1 u 1 is greater than v w 2 u 2 this quantity is more than this this e dot is positive h is positive that means energy is being transferred to the rotor we get mechanical energy from the fluid from the stored energy in the fluid while v w when it is the reverse that is v w 2 is greater than v w 2 u 2 I am sorry v w 1 u 1 that means this is greater than this this is greater than this then this is negative means the energy is transferred from the rotor to the fluid as it passes from its inlet to outlet. So, if for turbines as we know that classifications depending upon the direction of energy transfer this is always greater than this the v w 1 u 1 is always greater than v w 2 u 2 and in case of compressors where the mechanical energy is being embedded to the fluid by the rotor of the machine v w 2 u 2 is greater than v w 1 u 1. Now, I will come to the different components of energy transfer different components of energy transfer means that how we can express the basic Euler's equations in a different fashion where we can recognize that the energy transfer between the fluid and the rotor consist of different components which are which have their very interesting physical implications to analyze that let us consider the velocity triangles for a generalized rotor. Now, this is a rotor a generalized rotor where we show the inlet and outlet part of the vane rotating vane with the radius of rotation at r 1 and r 2 that is nothing different from the earlier picture, but with the velocity triangles drawn in inlet and outlet here we see the inlet velocity is v 1 and the relative velocity of the fluid is v r 1 this is the whirling component of the inlet velocity this is the velocity of the rotor at the inlet linear velocity or tangential velocity. Now, in this context I like to tell you that probably you know that the relative velocity should be such that it should approach the vane without shock that for a shock and smooth approach of the fluid to the vane so that fluid can glide along the vane the relative velocity should make an angle with the direction of the vane motion which is same as the angle of the vane at its inlet. So, therefore, the vane angle at the inlet is same as the angle of the relative velocity beta 1 so therefore, we will consider hence forth the angle of the relative velocity with any direction for example, here in the direction of the vane velocity is equal to the vane angle at that point. Similarly, here you see the outlet velocity triangle is shown where v 2 is the velocity of discharge of the fluid and v r 2 is the relative velocity of the fluid at discharge whose angle is beta 2 with the tangential direction that is the same angle the vane makes at the outlet with the tangential direction that is the outlet vane angle. So, this beta 1 is the inlet vane angle now this u 2 is the tangential velocity or the linear velocity that is in the tangential direction of the vane at the outlet. Now, alpha 1 is the angle with angle which the velocity v 1 at the inlet makes similarly alpha 2 is the angle which the velocity v 2 at the outlet makes with the tangential direction. Now, this velocity triangles gives the picture of the relative velocities and the absolute velocities at the inlet and outlet of a generalized rotor the v f 2 and v f 1 at the flow velocities at the outlet and inlet. Now, if we apply certain trigonometric formula for the triangles we see if at the inlet triangle we can write I think you can see that v r 1 square is equal to v 1 square plus u 1 square minus twice u 1 v 1 cos alpha 1 or equal to we can write v 1 square plus u 1 square now v 1 cos alpha 1 is the whirling component or tangential component of the velocity at the inlet u 1 v w 1 and follow similarly from the outlet triangle we can develop the same relationship v r 2 square is equal to v 2 square that means v 2 square plus u 2 square plus u 2 square sorry this 2 in the big minus twice u 2 cos alpha 1 cos alpha 1 cos alpha v 2 cos alpha 2 or we can write v 2 square plus u 2 square minus twice u 2 v w 2 well from the 2 velocity triangles all right. Now, from here therefore, we can write that not from here sorry this is from this step not from this step from this step we can write that v w 1 u 1 is equal to half v 1 square plus u 1 square minus v r 1 and from this step we can write v w 2 u 2 is equal to half v 2 square plus u 2 square minus this thing v r 2 square. Now, if you see this expression with this expression and if you recollect the Euler's equation Euler's equation was that energy transfer or head let us consider the head that is being given to the machine by the fluid if you recollect it was 1 by g 1 by g v w 1 u 1 minus v w 2 u 2 that is the amount of energy per unit weight that is being transferred from the fluid to the rotor. So, just by mere substitution a very simple mathematics I am telling in fluid machines mathematics and not as complicated as you have in fluid dynamics no partial differential equations ordinary differential equation this very simple because in uniform flow we only deal with some algebraic steps. So, if we just substitute this we will get 1 by 2 g 3 distinct terms we will get one is v 1 square minus v 2 square another is u 1 square minus v 2 square minus v 2 square and another is plus v r 2 square minus v r 1 square. So, this is another form of the energy equation it is not an independent equation just by exploiting the relationship between the velocities from the velocity triangles at inlet and outlet we have just express this expression into a different form showing distinctly the three components as the energy transfer components in case of fluid machines and next part will be the discussion or the physical implication of this three components. Thank you today I will stop here that I will discuss in the next class good morning welcome you 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 v 2 square 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 at outlet these are the rotor velocities welding 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 transfer to the machine by the fluid as it flows through the rotor vanes can be written like this. Now, you 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 change in change in dynamic head this this one change in 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, this is the angular velocity of the r. So, d a cancels 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 r two 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. 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 equation 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 as 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 where the volume is control volume is fixed known as control volume system and usually we to tell as 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 these 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 d p 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 are the elevations from reference dot m. 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.