 illnesses you earlier I think because this is the opportunity when we can discuss about the problems on reaction machines. So, therefore, before starting these rotodynamic pumps I like to discuss about the problems that I gave you earlier for this for you to solve this if you go look quickly to this problem I just will have you I will be having a hurryed discussion on this problem I hope that you have already taken it in your note the problem was like that the diameter of the runner of a vertical shaft turbine is 450 millimeter at the inlet the width of the runner at inlet is 50 millimeter the diameter and width at the outlet are 300 millimeter and 75 millimeter respectively. So, quickly we will be going through this problem the blades occupy 8 percent of the circumference the guide van angle is 24 degree the inlet angle of the runner blade is 95 degree and the outlet angle is 30 degree the fluid leaves the runner without any whirl there is no whirling component the pressure head at inlet is 55 millimeter above that at exit from the runner the fluid friction losses account for 18 percent of the pressure head at inlet actually there is a mistake this will be 18 percent of the difference you write of the difference in pressure head at inlet and outlet it will be 18 percent of the difference in pressure head at inlet and outlet calculate the speed of the runner and the output power well. So, this problem if you solve just you see that how you can solve this problem if I write the Bernoulli's equation at the inlet and outlet of the runner then what we get this is the p 1 the suffix 1 stands for inlet v 1 square by 2 g that means the pressure head and the velocity head is equal to p 2 by rho g that means simply an energy balance per unit weight basis v 2 square by 2 g plus the work head that means if I well reproduce at this w w is the work head that means the work per unit weight plus the loss h l we consider the change in the potential head or the elevation between the inlet and outlet of the runner to be negligible. That means this is the inlet pressure head inlet velocity head of the runner this is the outlet pressure head outlet velocity head of the runner work head plus the head loss here we know from the problem that p 1 by rho g minus p 2 by rho g we can write in this fashion is equal to v 2 square by 2 g minus v 1 square by 2 g plus the work what is the value of this work head do you know this value work head no we do not know this we know only that w plus h l this h l we know. So, this is 55 meter 55 millimeter well I am sorry extremely this is not 55 millimeter this is 55 meter this cannot be so small the difference between the pressure head 55 meter that is 55 meter I am sorry it is 55 meter h l according to the problem is 18 percent 0.18 into 55. So, the problem is like this 9.9 now we do not know v 1 v 2 v 1 and w, but w we can express as w is equal to v w 1 u 1 by g where this is the tangential component of velocity at inlet rotor speed and g. Now, how can we calculate this to do this we will have to find we will have to think in terms of the that means this is the rotor the usual velocity triangles well. So, this is your u 1 this is your v 1 this is your v r 1. So, this is let this is the perpendicular so this is v f 1. So, according to the problem this angle is giving that is the van angle 24 degree you see and this angle is given as 95 degree and here also similarly v r 2 well and this will be u 2 and this will be v f 2 which is equal to v 2 the absolute velocity is radial and this angle is probably given beta 2 you see that beta 2 is also given 30 degree the outlet angle is 30 degree. Now, here we see that in this expression v 1 v 2 and w if we express v w 1 u 1 by g if all the velocities can be expressed in single velocity we can find out that velocity from this equation because we know h l we know p 1 by rho g minus v 2 by rho g. So, now it becomes simply certain trigonometric manipulations that if you consider the inlet triangle inlet velocity triangle you see that v 1 is equal to what is v 1 v 1 is equal to v f 1 this is cos 24 degree or v w 1 rather better we write the v w 1 we have to find out the v w 1 in the work head expression. So, everything better should be expressed in terms of v 1. So, v w 1 is v 1 cos of 24 degree let us think that every all velocities should be expressed in terms of v 1 the inlet velocity. So, this comes out to be just I am giving you it is already solved 0.913 v 1. So, what is that then u 1 I have to find out what is u 1 u 1 by d 1 is u 2 by d 2 well. So, u 1 by d 1 is u 2 by d 2 is obvious because the rotational speed is same. So, it is in the ratio of the diameter. So, u 1 becomes equal to u 2 d 1 by d 2 what is d 1 and d 2 d 1 and d 2 are given. So, therefore, I can write that it equal to 450 you see by 300 450 millimeter is the inlet diameter and 300 millimeter is the outlet diameter it becomes 1.5 u 2. Now, again one thing that you see here in the problem d 1 is given 450 millimeter what is b 1 that is the inlet at the width the inlet at the width is 50 millimeter let this is b b 1 is 50 millimeter well similarly d 2 is given 300 millimeter and b 2 is given now here in this context I like to tell you one thing when b and d d is the diameter at the inlet and b is the width that means if the shaft if the turbine is in the horizontal look position then this b 1 is the vertical width that means pi d b is the flow area that means if we multiply it with v f 1 it will give you the flow rate q now this must be equal to v f 2 pi d 2 b 2 now where both d 2 b 2 and d 1 b 1 are given then we will check whether v f 1 is equal to v t f 2 or not if they are different the areas that d 2 b 2 and d 1 that product of d 1 b 1 is different from d 2 b 2 then v f 1 v f 2 are not necessarily the same for a particular problem if it is so we can accept this but usually in all designs v f 1 and v f 2 are made same which I like to tell that in any problem if you see that this 4 quantities out of this 4 quantities one is not given then we will make the equality that means this flow rate will be same under steady condition and we will make that v f 1 is equal to v f 2 so that automatically d 1 b 1 becomes d 2 b 2 in this problem the values are given such that this is this equality holds good that means from the continuity if you make v f 1 pi d 1 b 1 is equal to v f 2 pi d 2 b 2 that means the flow rates are equal then at inlet and outlet then we get automatically v f 1 is equal to v f 2 so therefore we can write that v f 1 let us write here v f 1 is equal to v f 2 and that is is equal to v 2 this is v f 2 now from this outlet velocity triangle we see that v v 2 v 2 is what v 2 is in terms of u 2 v 2 is u 2 v 2 is v 1 in terms of v 1 let us see v 2 is v f 1 rather here you see v 1 sin 24 then what is v 2 is v 1 sin 24 is 0.406 so therefore we can find out u 2 from here u 2 is what tan that is v f 2 tan 30 and v f 2 is 0.406 tan 30 tan 30 because oh see your tan cot 30 or tan 60 you are correct perpendicular by base so u 2 is base by perpendicular cot 30 or tan 60 all the same cot 30 let me check with the calculations that well so therefore u 2 comes out to be 0.703 v 1 so now you see u 2 0.703 v 1 if I put here then u 1 becomes equal to u 1 becomes equal to 1.05 v 1 that means this if I put u 2 0.703 v 1 that means v 1 we get in terms v w 1 we get in terms of v 1 u 1 we get in terms of v 1 well then you see that in this expression we can write this v w 1 u 1 by g v w 1 is 0.913 v 1 and u 1 is 1.05 v 1 by g so v 2 what is v 2 v f 2 v 2 we see that v f 2 v 2 is 0.406 v 1 0.406 v 1 v 2 is 0.406 v 1 so therefore this right hand side of this expression h l is known 9.9 that is 18 percent of the difference in pressure rate at inlet and outlet so therefore from this equation we can find out the value of v 1 well when the value of v 1 is found out then what we have to find out in this problem we have to find out the well we have to find out the what we have to find out calculate the speed of the runner and the output power well the speed of the runner is very simple now when v 1 is found out u 1 is found out or u 2 any one of this and speed of the runner is pi d n by 60 d 1 is u 1 so or pi d 2 n by 60 any one you make u 2 so you can find out in the rotational speed in r p now to find out the power developed what you have to do you have to write the power developed what is power developed power developed we will have to find out rho q v w 1 u 1 because here the power developed by the runner is v w 1 u 1 into rho q how to find out q q we have to find out already I have told now v f 1 pi d 1 into b 1 but in the problem it has been given the blades occupy 8 percent of the circumference that means the effective flow area will be 0.9 so you can find out the flow rate well so flow rate value will come if you 11.62 meter per second because all the velocities are known v f 1 also known when v 1 is known all the flow velocities unknown so q is equal to 11.62 meter per second and power p becomes equal to 593.06 kilo watt all right simple problem so you see this problem again in your hall so that you can understand well now I will come to the discussion on I have given you another problem on axial flow turbines please you see it is a very simple problem you solve it if you cannot I will discuss it in the next class if possible so now I will come to the rotodynamic machines rotodynamic sorry rotodynamic pumps well now I will discuss the rotodynamic pump please rotodynamic pump so you know the pump as we have discussed earlier is a machine where the mechanical energy is converted into the stored energy of the fluid and you know the pump or compressors if you recall the classification then you will see that pump are those machines which handle incompressible fluid or liquids that means in a pump the mechanical energy is converted to the stored energy of the liquid well the rotodynamic machines again on the other hand are those machines which work on the principle of fluid dynamics that means there is a continuous motion of the fluid through the machines and because of this motion relative to the machine a part of the machine as you know known as rotor the energy transfer between the fluid and the machine take place so therefore a rotodynamic pump is a machine where the fluid that is the liquid gains in its stored energy while flowing through a moving parts of the machine or while flowing through the rotor of the machine this is precisely a rotodynamic pump as similar to turbines the rotodynamic pump is also classified into different categories depending upon the direction of the fluid one is the axial flow rotodynamic machines where the flow is axial that means in a direction parallel to the axis of rotation understand that means the inlet and outlet is at the same radial location from the axis of rotation similarly the radial flow machines are there for pump that is radial flow pump and as you know if it is a pump where fluid gains the energy the radial flow machines will be radially outward flow this is because the fluid will gain in centrifugal head at the centrifugal energy per unit while in case of turbine it will be radially inward where the fluid loses its centrifugal head which is got in terms of work of time so it is outward radial flow turbine I sorry pump so this outward radial flow pump is referred to as commonly as centrifugal pump the what centrifugal comes from the concept that the centrifugal head is impressed on the fluid or it is gained by the fluid so therefore a radial flow rotodynamic pump is termed as centrifugal pump so basically we see that centrifugal pump is the converse of francis turbine because francis turbine is a radially inward flow machine where energy of the fluid is given up to the rotor and we get mechanical power similarly in a centrifugal pump it is a radially outward flow machine where the situation is such the fluid gains its energy from the mechanical energy imparted to the moving rotor so it is just the converse of francis turbine so before coming to the discussion of the shape of the rotor the blades the velocity triangles and the work transfer we first concentrate on a general system a general pumping system what you mean by pumping or a general pumping system and the net head develop by a pump there are certain terminologies static head, net head develop this we will study through the study of a general pumping system so now let me tell you this thing clear what pumping refer to hydraulic machines implies conveying of water from a lower head to a higher head from a lower reservoir to a upper reservoir you see that very common and popular meaning of pumping when you talk that when you tell that water is to be pumped or in a liquid is to be pumped we simply means that it has to be taken from a lower reservoir to higher reservoir that means from a place from a location to a location where the height is more this is the common implication of the what pumping as referred to hydraulic machines let us understand the general pumping system with this common implication of the what pumping so let us come to this diagram which will give you a clear understanding of the general pumping system let us consider just you see that this is a pump this is a centrifugal pump this may not be this may not necessarily be a centrifugal pump but the figure look is drawn that is a centrifugal pump pump is here now you see this is the lower reservoir where the liquid is there from where the liquid has to be taken to a upper reservoir here so this is the upper reservoir this is the lower reservoir there is a terminology that is known as sump s-u-m-p lower reservoir is known as sump now this is a pipe which takes the water from the lower reservoir and the water or liquid flows through this pipe and goes to the pump so this pipe is referred to as suction pipe suction pipe why this is because usually this water is kept at atmospheric pressure this is open to atmosphere so therefore for the liquid to flow through this pipe the pressure in this pipe has to be lower than the atmosphere so that the fluid at rest at atmospheric pressure can flow through it so that is the reason for which this pipe is known as suction pipe where the liquid is sucked from the atmospheric pressure now after gaining energy liquid comes out from the pump and flows through a pipe which is known as the delivery pipe because this is the pipe through which the fluid is being delivered by the pump to the upper reservoir well usually the suction pipe diameter is little more than the delivery pipe in most cases it may be equal but if there is a variation it has to be more than this that I will discuss afterwards because of the cavitation problem now you see that at any point in the flow of the liquid through suction pipe pump and delivery pipe the total energy comprises the pressure energy kinetic energy and the potential energy now we take a reference datum here from where the elevations are or the potential heads are defined now let us see just a minute there is a mistake let us well this will be up to this now we see that what is the total energy of the fluid at the upper reservoir surface this is because the fluid is at rest and at atmospheric pressure so total energy of the fluid at the upper reservoir is h s plus h a these are the terminologies at present without any name you just understand that h s is this height and h s is the height of this level from the reference datum so h s plus h a is the total energy at this point well what is the energy at this point it is simply h a the energy at this point is h a why because at this point the pressure is atmospheric pressure so we consider the pressure energy to be zero when the pressure is atmospheric pressure that means we calculate the pressure energy as p by rho g where p is the gauge pressure so therefore when the pressure energy is zero kinetic energy is zero so total energy is h so difference in total energy of the liquid at upper reservoir and the liquid at lower reservoir is h s this h s this is known as static head of the this h s is known as static head h s is known as static head static head that means this is precisely the difference in the total energy of the liquid at the lower reservoir upper reservoir and lower reservoir and this is simply the difference in elevation head between the upper and lower reservoir this is the terminality static head now we see the follow this diagram here we see that how the energy changes now let us consider a point a where the fluid is at total energy h a now while it flows through this pump this is a typical strainer through which the liquid enters the suction pipe of the pump where the mechanical impurities are eliminated and comes to a point b which is almost at the same elevation here so the head or the total energy drops to a little which accounts to h in in is a suffix that means this is the loss at the inlet to the suction pipe then while it flows through the suction pipe now this is the point c that is at the inlet to the pump there is a pressure gauge connection and d is the point at the outlet to the pump there is another pressure gauge connection which measures the pressure at the delivery and pressures at the suction so at the inlet to the pump c is a point where you see there is an appreciable loss of head which is shown by h f 1 that is the head loss at in suction pipe which counts for the fictional head loss while flowing through the suction pipe along with the head losses due to pipe bends so all major and minor losses are taken into account these are the typical flame joints to make this pipe bends which takes place in course of flow of the fluid from the inlet point b to the inlet to the pump c then what happens while the fluid comes to the pump it gains energy this is the basic principle of the fluid machines where the energy is imparted to the fluid while it flows through the rotor and stator of the pump and when it comes out of the pump d it gains total energy or head is developed by the fluid the basic principle of the pump so there is an straight increase of the head to a point d now this line is h a plus h s that means this line implies the total energy of the fluid at this point now the point d the energy of the fluid is more than this point this energy is required to overcome the friction in the delivery pipe that means when it comes to the exit point of the delivery pipe b the line slows downward that means to e where this d to e this loss this loss is h f 2 is denoted by h f 2 that means this is the frictional losses in the delivery pipe that is the frictional loss and other losses of course the losses due to bend in the delivery pipe then what happens at the exit plane e the fluid is delivered to the upper reservoir which has got a flow velocity or the kinetic energy this entire kinetic energy is loss that is the exit loss you know the exit loss is v square by 2 where v is the exit velocity that means the entire kinetic energy this physically implies that entire kinetic energy is converted into intermolecular energy so that fluid which was having a velocity v at this exit plane is now coming to raise that means this energy is lost so therefore this is known as the exit energy this is converted in terms of intermolecular energy in the upper reservoir so there is a loss e to f so that means f is the point where the fluid is at rest in the upper reservoir and having the energy h a plus h s so this is the flow diagram for the energy in course of flow of the fluid through the suction pipe pump and the delivery pipe i think you have understood this well ok then i tell you certain terminologies you know that h s is the static head now if we write the Bernoulli's equation between the point a and the point c just to see first here between the point a and the point c then what we get ok before that let me explain few other terminologies total head this is most important after that we will write the Bernoulli's equation total head at inlet to the pump total head at inlet to pump and total head at outlet to total head at total head at outlet sorry pump at outlet from the pump what is total head at inlet to the pump if we consider p is the pressure suffix one is the inlet to the pump the corresponding velocity that is the velocity of the fluid at the inlet to the pump that means the velocity of the fluid in the suction pipe plus z 1 let z 1 is the as i have shown z 1 is the elevation from a reference datum at the inlet similarly z 2 is that at the outlet and p 1 by rho g and p 2 by rho g at the pressure rates these are shown by the pressure gauges so therefore total head at the outlet of the pump is p 2 by rho g well plus v 2 square by 2 g plus z 2 so this is the total head that means the total energy per unit weight this is the total energy per unit weight at the outlet so difference of this let this is denoted by h 1 this is denoted by h 2 so this h 2 minus h 1 is the head gained by the fluid that is p 1 by rho g plus v 1 square by 2 g plus z 1 minus p 2 by rho g well plus v 2 square by 2 g plus z so this is oh it is opposite yes p 2 this is v 2 this is z 2 this is p 1 v 1 z 1 and this is known as head developed by the pump that means this is the head developed by the pump this is the head developed by the pump and this is the head developed by the pump which is gained by the fluid this is the change in the head from its outlet to inlet this is known as conventionally manometric head manometric head of the pump please ask me the question if any question you want to ask manometric head not manometric head the word manometric head comes from the concept that I will tell you now so this is the head developed by the pump that is the outlet head of the pump minus the inlet head that is head at the outlet minus head at the inlet now usually it is found that the difference in delivery and suction pipe diameters are so small that the difference between the velocity head that v 2 square by 2 g and v 1 square by 2 g is neglected similarly you know very well it is very very much obvious that difference in elevations at the inlet to outlet is much small compared to change in other quantity so that h 2 minus h 1 that head develop can be simply written as p 2 by rho g minus p 1 by rho g so therefore you see here that this differences in pressure head simply that means which is registered by any pressure measuring device is a well representation of the head developed by the pump or the manometric now the word manometric head comes that if you neglect the kinetic energies a change in kinetic energies then the head developed is simply given by the difference in piezometric head that means p 2 by rho g plus z 2 minus p 1 by rho g plus z 1 even if we do not neglect the elevation head now this difference in piezometric head across the pump that means difference in piezometric head between point d and c can be registered if there is a manometer attached to this point because you know a manometer reads the difference in piezometric head the manometer deflection straightforward gives the difference in piezometric head not the pressure head only difference in piezometric head or piezometric pressure that means the pressure plus the change in the elevation so therefore this name is given that it is manometric head that means the head which is being registered by a manometer manometric head so if we neglect the change in the elevations also this is simply given by p 2 by rho g minus p 1 by rho g it is not usual to connect a manometer because the difference in head is so high if manometric it will require a very high height is the long height or very higher value of this height of the manometric tube so that the pressure gauges are attached to inlet and outlet ends and simply the differences of pressures are rate and the difference in pressure energy is well representative of the head developed by the pump so we now see that h s is known as this static head and this h 2 minus h let this is h simply h h is the head developed by the pump now if we write the Bernoulli's equation at the two points that means at inlet a and at inlet s at the inlet to the pump c that means at the sum that the liquid point liquid at the point a that means the inlet to the suction pipe that is liquid at some and the inlet to the pump c then what we get we can write that this is sorry this at a pressure is 0 velocity is 0 so only is h a you see that h a is its elevation when it comes to the inlet of the pump with our nomenclature suffix 1 is the refers to inlet p 1 by rho g v 1 square by 2 g plus you see here z 1 that is the elevation at the inlet from the same reference datum so therefore this is z 1 plus the losses h in as already we have denoted by this two term these two terms constitutes the losses in the intake pipe at the pipe and the total losses in the flow through the pipe so h in plus h f 1 is the total losses that takes place while the fluid flows from a point at the sum to the inlet to the turbine now if we inlet to the pump sorry sorry if we write the Bernoulli's equation between point d and point f if we now write the Bernoulli's equation from point d to point f what we will get point d that means p 1 by rho g sorry p 2 by rho g plus v 2 square by rho g plus z 2 if you recall the figure that z 2 is the elevation of the outlet end of the pump is what is equal to there is no pressure at the free surface or the fluid at rest in the upper atmosphere in the upper reservoir no velocity plus h a plus h s plus the loss in the delivery pipe which takes care of the frictional losses in the delivery pipe plus the losses in the pipe bends plus the exit loss that means this is the total loss that takes place while the fluid flows flows from the outlet end of the pump to the upper reservoir similarly this is the total loss that takes place when the fluid flows from the lower reservoir that is summed to the inlet to the pump any query please ask me any query industry very simple Bernoulli's equation alright now if ok now if we substitute h a from this equation to this equation then we simply get this head developed if you just substitute this equation and take this p 1 by rho g v 1 square by 2 g z 1 this side so it is a very simple expression we get h in plus h f 1 plus h f 2 plus h please tell me that whether you have got any difficulty up to this or not I think there should not be any difficulty as far as algebraic steps are concerned but from the concept actual point of view do you have any difficulty this is the energy equation that is the Bernoulli's equation between the two points this left hand side represents the liquid at the sum the lower reservoir so total energy per unit weight or head is the h a that is the elevation head this is the pressure head at inlet to the pump velocity head at inlet to the pump this is the elevation head this is the total loss alright similarly this equation represents the Bernoulli's equation between the two point that one is the delivery from the pump that means these are the two points d and f that means the point this corresponds to the point at the delivery end of the pump that means at the inlet to delivery pipe where this is the pressure energy per unit weight that is pressure head that is the velocity head and that is the potential head similarly for the reservoir at the point f in your earlier diagram the pressure energy is 0 the kinetic energy is 0 so the potential energy per unit weight datum head that is h a plus h s which is the total height of the upper reservoir from the same reference datum plus the total loss this we have already designated as this two losses h f 2 which takes care of the fictional losses in the delivery pipe along with the band losses that means this takes care of all the losses that is incurred in the delivery pipe plus the exit loss that means the exit kinetic energy which has to be lost because the fluid has to come to rest at the upper reservoir all right so if you see from these two equations that if you substitute h a from the above equation in terms of the pressure velocity and z 1 the datum at the inlet to the pump then we get a very important relation that this is the sum of all the losses that means it is very clear and from the very simple concept one can tell that the head developed by a pump is equal to the static head plus the sum of all the losses and it is very simple because if one sees this figure he sees that okay if a fluid has to be pumped from this point that means from this lower reservoir sum to the upper reservoir the fluid has to gain this much amount of potential energy that means this much head and at the same time a fluid has to be delivered at certain flow rate so to deliver the fluid at certain flow rate it has to come across through various fluid resistances the resistances in the flow path so that resistance in the flow path can be considered in terms of an equivalent head that is the head loss which has to be overcome to come for the fluid to come at the upper reservoir so therefore this potential head or the static head that is the change of potential head between a and m plus the head lost that is the resistance that is the resistance which has to be overcome by the fluid while flowing from this point a to a has to be added and the sum of these two is acting as an resistance or the head that has to be developed by the pump so the total head developed by the pump total head we can realize is a very important formula total head developed total head developed by the pump total head developed by the pump is therefore equal to the static head so you have to immediately recall the terminology static head is the change of vertical height between the lower reservoir to the upper reservoir plus sum of all losses you have to be very careful all losses you will have to take into account which take care which comes into the picture starting from the loss at the inlet to the suction pipe down to the loss at the exit of the delivery pipe so this is the total head that a fluid has that the pump has to develop in pumping the fluid that means in conveying the fluid from the lower reservoir or the pump to the upper reservoir ok all right please any difficulty any difficulty please yes from b to c from b to c there is an increase in head due to elevation but this from b to c even if there is an increase in head due to elevation the loss of head is much more than this increase in elevation so therefore there is a loss in it you are correct when you write this equation you see that from b to c if i write this equation this is from a and b mostly here you see that z 1 is there so you are very correct so in the equation it is taken care of so physically definitely when the fluid comes here fluid gains at the potential head but at the same time there is there are losses due to fluid friction and the pipe bend and more over the pressure here is much less than the pressure here otherwise the fluid cannot flow through this because the fluid flows from a point at the sum because of pressure difference only because there is no velocity you understand so fluid is at the lowest potential energy level at the sum all right so fluid pressure is atmosphere so therefore here the fluid can flow or can enter into the suction pipe provided there is a pressure which is below than that of the atmospheric pressure so therefore entirely in the suction pipe the pressure head is lower than the atmospheric pressure you are correct the total energy here is sum of the pressure energy which is lower than that point a or point b plus the kinetic energy plus the potential energy potential energy and kinetic energy these are the additive part here as corresponding as compared to the point but the losses are much more so that the energy is less more over you can argue from a very simple physical concept that the energy here has to be lower than this otherwise the flow cannot take place so in the direction of the flow energy has to decrease this is known as commonly we tell as phenomenological law that means the flux takes place with the negative gradient or negative potential gradient causes the flux so it has to be so only here it is more this is because the energy is added from outside so when there is no energy added from outside the flux takes place with the negative gradient of the potential so I tell you all these things mathematically but from simple common sense you can tell that the flow of energy takes place with the negative so energy has to decrease in the direction of the flow if there is no energy interaction from external source when the energy is added from outside that is why the point c increases well time is up ok thank you today books are well