luence machines. ,A last class we discussed the basic principle of operation for a reciprocating pump. And we have also discussed that how the acceleration heads during suction and delivery strokes are developed due to non-uniform motion of the piston that during the suction stroke acceleration and acceleration takes place for the piston similarly the acceleration and deceleration takes place in delivery strokes, because of which the piston or the pump as a whole has to develop additional heads over and above that of the theoretical head you can say determined by the static lift of the pump. So, today we will discuss the rate of delivery. Now, if you see a pump reciprocating pump in its simplest form you see that during the suction stroke when the piston or the plunger moves from idc to odc in this direction the pressure falls below that in the sum. So, the liquid comes through this inlet or suction pipe to the cylinder. So, therefore, during this stroke when the crank moves 180 degree there is no flow in the delivery line that means the delivery is 0 that means the flow in the delivery is 0 when the piston starts the delivery stroke or starts this motion from odc to idc in that direction. So, it pushes the liquid the pressure is immediately sensed at this point considering the liquid to be incompressible that we have recognized earlier and then the delivery valve opens and the delivery takes place. So, this type of pump is known as a single acting single cylinder obviously is a one cylinder single cylinder single acting pump. So, therefore, if we plot the rate of delivery versus the crank angle that means this is the delivery axis the rate of delivery we will see that first 180 degree revolution that means if we plot it with respect to crank angle on abscissa. So, 0 degree is the angle that crank measured from the inner dead center position. So, 0 to 180 this represent the suction stroke there is 0 delivery. So, delivery takes place only in the delivery stroke from 180 to 360 and as you have noticed the velocity in the delivery pipe varies periodically because of the acceleration and deceleration head where it is 0 and then it goes on increasing attains the maximum the mid of the stroke considering a simple harmonic motion of the piston and simultaneously of the liquid column in the delivery pipe. So, therefore, the velocity follows this sinusoidal distribution and similarly the delivery that is the flow rate that means here we can write the flow rate q dot. So, therefore, we see the delivery occurs only from 180 degree to 360 degree angle of rotation for the crank and that is also a non uniform delivery. Now, this can be eliminated that means to make a continuous delivery this is not a continuous delivery. So, it is intermittent delivery that means again another 360 to 540 there will be a suction and again there will be a delivery. So, this delivery will be intermittent. So, to make a continuous delivery a type is used known as double acting single cylinder pump that means what happens here this cylinder is single, but the delivery and suction takes place in both the sides the connection is like that which means that if we consider this as the i d c i d c and let this as the o d c that means when the piston moves from i d c to o d c this side the suction takes place whereas, this side the liquid for the liquid in this side of the piston the delivery takes place. That means when the piston moves from i d c to o d c the suction valve here or the inlet valve opens at the same time the delivery valve opens. That means this is suction for the liquid in this side of the piston and delivery for the liquid in this side that means the right side of the piston this face. So, therefore delivery takes place in both the strokes while this is a suction stroke for the liquid in this side the delivery takes place through this line. Similarly, when the piston moves from o d c to i d c the delivery valves here opens and the liquid here flows through the delivery. So, it is simultaneous the executing both the delivery and the suction strokes in both 0 to 180 degree rotations of the crank. So, therefore you see in a double acting machine we can just get a diagram of this type. That means from 0 to 180 there is a delivery with a sinusoidal variation because of non-uniformity in the flow because of the acceleration and deceleration of the piston and similarly this is there from 180 to 360. Therefore, we can improve the delivery from an intermittent one to a continuous one, but still the delivery is non-uniform. Now, the non-uniformity in the delivery can be eliminated if we use multi-cylinder pumps, but multi-cylinder pumps if we use two cylinder pumps two cylinder here where the reciprocating motion of the piston is executed by 180 degree phase lag here 180 degree phase lag a two cylinder with 180 degree phase lag unfortunately this does not remove the non-uniformity why this gives the same diagram why this is because when suction for one cylinder will be the delivery for other cylinder and vice versa. So, a superimposition of this two gives the similar thing, but if we use three cylinder if we use three cylinder pump let us consider equally spaced that means 120 degree within the crank angle that means 120 degree apart from one another three cylinder such we can get relatively a uniform discharge if you follow this let this cylinder one two three. Now, if you draw the diagram with respect to the angle made by the piston of the first cylinder with respect to its inner dead center position then you see from zero to 180 is the suction. So, therefore this is a superimposition so red one is the first cylinder. So, red one gives the discharge for the first cylinder from 180 to 360 well the second cylinder they it starts its suction at 120 degree with respect to the first cylinder. So, from 120 to 300 that is 180 degree of crank rotation there will be suction for the second cylinder while from 300 to 120 this side that means it is 300 plus 180. So, it will be the discharge that means first 60 degree 300 to 360 this will be the part of the discharge curve 60 degree then for another 120 degree this is the discharge curve. So, this will be maximum here which is 30 degree. So, this two curve it represents this part and this part is the discharge curve for second cylinder and similarly 120 to 300 is the suction for the second cylinder very simple for the third cylinder which is 120 degree apart phase lag from the second one. So, that will be 0 120 plus 120 240. So, this suction starts from 240 and it will go up to this point 60 degree here the 240 plus 180 well. So, 120 this side and 60 again it comes back 60. So, from 60 it starts its delivery stroke. So, this one is the delivery of the third cylinder. So, if you know superimpose this delivery curve this one this one and this one and we show it from 0 to 360 degree of crank revolution with respect to the first cylinder then we get a continuous curve like that a more or less continuous curve that means discharge becomes more or less uniform if we use more number of cylinder. If we use large number of cylinders this non uniformity will go down almost we will get a straight curve. So, therefore, we conclude that the rate of delivery may be made uniform in spite of the deceleration and acceleration of the pistons provided we use number of cylinders in line with some phase lag of their reciprocating motion execution of their reciprocating motion. Next is very important phenomena that air vessel application of air vessel what is an air vessel application of air vessel in a reciprocating pump. Now, we have seen that until and unless we use a multi cylinder in the pumps we have for a single cylinder what we have we have an intermittent discharge and discharge is non uniform. So, to make the discharge uniform for a single cylinder air vessels are used in delivery and suction side what is this let me draw the diagram first for air vessel these are vessel is very simple. Now, what is done what is an air vessel now to make the basic purpose is to make the discharge uniform in the delivery line air vessels are used of course, it has multi functions that I will come ultimately we will arrive at that conclusion. So, air vessels are large vessels large container close that one end with compressed air inside it with air inside it which is attached to both delivery and suction lines very close to the cylinder. So, it should not be attached somewhere here as close as possible to the cylinder. So, that this length of the delivery and suction pipes from the cylinder to the air vessels are as small as possible. Now, what happens it is very simple these air vessels work as a fly wheel similar to that of a fly wheel in a machine. So, it takes the extra air from the mean discharge a sorry extra water from the mean discharge when it is more than the mean discharge and it gives water when the mean when the water flowing to the delivery or suction pipe is less than the mean discharge. Let us understand the now first of all we have to find out we have to under we have to understand that the delivery in the first delivery side we consider the delivery side the flow is not uniform it is sometimes more sometimes less if we find out the uniform flow rate uniform flow rate can be found out by finding out the volume swept. Now, let the stroke length is l which is is equal to two times the crank radius and if the cross sectional area of the piston is a. So, this is the volume swept per unit per one cycle. So, if I know the omega the crank revolutionary speed we can tell the rate of volume flow is l a into omega by two pi this is the volume swept by the piston this is an average volume flow rate. So, if this volume flow rate divide we divided it by the cross sectional area of the delivery pipe let a d then we get the average velocity of flow, but this velocity of flow varies in the delivery because of the acceleration and deceleration. So, that the flow rate through the delivery pipe and simultaneously through the suction pipe varies what happens what is the principle of operation for air vessel. Now, let us consider the delivery side when the flow is very high that means flow is more than this mean discharge or mean flow then the pressure here is more. So, what happens when the pressure here is more the liquid rushes to the air vessel. So, the pressure of the air is so adjusted that it must equate to a pressure at here which corresponds to the mean discharge. So, therefore, whenever the discharge is more than the mean discharge when it happens at the beginning of the delivery stroke that when the piston starts from its odc. So, pressure here is more so that the discharge is more than the mean discharge the discharge rises there is an acceleration that means the flow rate increases. So, for during that period the extra water flows out flows into the air vessel. So, pressure is going down which means that from this part onwards the flow is reduced and ultimately brought down to this value. The reverse happens that during the subsequent deceleration part that is the latter part of the delivery stroke when the flow rate is less than the mean discharge that mean deceleration takes place the flow rate decreases. So, therefore, the pressure here is reduced the air pressure is so adjusted that when the pressure here is reduced the accumulated water in the air vessel rushes and joints here. So, that the flow becomes more than the reduced flow and ultimately flows through the delivery pipe. So, here a constant pressure is maintained to supply a constant flow through the delivery pipe from the air vessel onwards to the overhead tank. So, here we get a constant q mean that is this value which is constant that means this way the air vessel either supplies water or absorbs water to maintain a uniform flow rate through the delivery pipe from the air vessel connections onwards. Similar thing happens for the air vessel in the suction side try to understand what happens in the suction side. At the initial stage when this piston moves from I D C then what happens due to the acceleration of the piston the suction pressure falls below that of the ideal or theoretical suction pressure determined by the height of the pump from the sum level this height. So, therefore, what happens more liquid the pressure here falls. So, liquid from the sum comes at a rate which is more than the mean discharge, but what happens when the pressure here falls to a value lower than that of the theoretical suction pressure then the water from this air vessel also comes because the flow takes place in this direction. So, which increases the pressure at this point so that from the sum it takes the flow rate which is q mean that means a reduced value not a high value. Similarly, at the subsequent part of the or the later part of the suction stroke when the pressure here increases then the flow rate decreases to the suction pipe, but what happens when there is an air vessel the pressure rises here. So, therefore, water again goes to the air vessel so that the same amount of flow is taken from the sum by the suction pipe. So, therefore, this part of through this part of the suction pipe always a uniform flow comes from the sum. Similarly, this part of the delivery pipe always a uniform flow comes through the delivery pipe. So, therefore, if you look as a hole to the pump you see that the incorporation of these two air vessels makes the flow through this part of the delivery pipe and through this part of the suction pipe almost constant and this is equal to the mean discharge. The discharge or flow rate varies only in this part of the delivery and suction pipe that is from the air vessel to the cylinder. Since, this part is very small because the air vessels are connected very close to the cylinder the fluctuations take place only in the small part of the tube that also does not create any serious danger from the stress consideration. So, therefore, we get almost a uniform discharge to the delivery pipe and a uniform flow rate to the larger part of the suction pipe. So, what advantages do we get from here first advantage you see let us first write that advantages from suction side suction side what advantage we get advantages. So, distinct advantages suction side we see that now suction side as we have seen earlier in our diagram for head discharge diagram that this pressure e falls below that of a this is h s rho g this is the ideal or theoretical suction pressure which is determined by this height of the pump above the sump level. Now, because of the acceleration if there was no acceleration the pressure minimum pressure in the pump was this. So, therefore, it is only the height of the pump which should be the criteria to determine the cavitation you understand this any, but because of this acceleration at the beginning of the suction stroke that we explained in the last class I explained the pressure falls here e. So, therefore, the speed of the pump I told in the last class is restricted because of the cavitation because if the speed is more the acceleration will be more at the beginning of the suction stroke. So, therefore, pressure e that point e will go still below that means the minimum pressure in the pump will go down that should not go below the vapour pressure of the working fluid at the existing temperature. So, therefore, here we see when the acceleration and deceleration is coming into picture then the minimum pressure point goes down. So, that the cavitation there is likelihood for the cavitation to occur. So, here we see when we put the air vessel very close to the cylinder. So, this pressure this point goes very close to that. So, this minimum point pressure is boosted high which is which corresponds to the mean discharge pressure. So, therefore, the likelihood of cavitation is avoided. So, the first advantage is the cavitation may be avoided cavitation may be avoided. So, cavitation may be avoided is related to what cavitation may be avoided means we can go for a higher speed of the pump higher speed of the pump. That means even at the higher speed these are interrelated the cavitation may not be there. If we use an air vessel in the suction side similarly higher length higher length higher length of the suction pipe we can use these are the important things from the practical point of view. So, we can use a higher length of the suction pipe we can go for a higher speed of the pump. So, that the head developed will be more without cavitation without the likelihood of cavitation to occur. Similarly, in the delivery side what we get advantage delivery side the first advantage is what first advantage uniform discharge uniform or constant discharge uniform or constant discharge almost uniform or constant discharge because whatever is the variation of discharge in this part of the pipe it is smoothen out by the flow of water in either direction from air vessel to delivery pipe or delivery pipe to air vessel to make the flow in the latter part to be almost constant. And another very important thing is that less power because a large amount of power is eliminated which was consumed to accelerate the liquid mass here the acceleration of liquid take place for this amount of liquid mass. So, larger length of the delivery pipe the liquid is not accelerated. So, a large amount of power is saved which was earlier consumed to provide the acceleration head to a large amount of liquid mass that means it is the liquid mass in the entire delivery tube. So, therefore, less power requirement less power requirement. So, these are the advantages for incorporation of or for incorporating the air vessels in a reciprocating pump. We mainly get the uniform discharge we can increase the speed of the pump with avoiding cavitation we can use the higher length of the suction pipe which means that we can place the pump at a higher height from the sum. So, we can location of the pump may be made at a higher height from the sum and at the delivery side uniform discharge and less power requirement. Now, let us find out a very simple analytical expressions for the flow of water from air vessels. Now, you see what is the mean discharge? Mean discharge if we write I have seen that the length that the stroke length stroke length I earlier also that means the same thing that is this is the mean discharge stroke length into area of the piston area of cylinder or same thing area of cylinder. This is the swept volume volume swept by the piston in one revolution. So, in terms of time that means the flow rate per unit time we multiply with omega by 2. Now, stroke length in terms of the crank radius is 2 r obvious because it two radius movement of the crank this executes one stroke. That means this stroke is the distance between i d c to o d c this is 2 r that means half revolution of the crank. So, this is the crank radius well so therefore, 2 r into a let a is the area of the cylinder into omega by 2 pi. So, this is the mean discharge based on the swept volume of the cylinder, but what is the instantaneous discharge if you remember the instantaneous velocity in cylinder or in delivery pi what was the instantaneous velocity in the delivery pi d by a by a into r what is that r omega cos cos or sin sin omega t r omega sin omega t all right. So, this is the instantaneous discharge. So, therefore, the difference between the instantaneous discharge and this is the instantaneous velocity not the discharge then it will be multiplied with the area this is the instantaneous velocity. So, the instantaneous discharge is a r omega sin. So, therefore, v minus q dot will be a r omega sin omega t minus a sorry minus a r omega by pi clear. This was the velocity of the piston r omega sin omega t this we develop just to recapitulate the series that there is any mistake x is equal to r minus r cos theta this is the displacement of the piston from the inner dead center position after a time t from when the piston moves by an angle theta from his inner dead center position. So, d x d t is the displacement of the piston r omega sin omega t and we have assumed that this is the velocity with which the liquid in the piston also moves. So, therefore, the velocity in the delivery pipe will be from the continuity principle a by a again you multiply with a we get the flow rate or simply we can find out the rate of flow as this which will be same for both in the cylinder and the piston. So, this value only we have written here. So, this can be written as a r omega let us put theta as the omega t which is the instantaneous crank angle now you see an interesting thing. So, the sin of this quantity will depend upon the value of theta it may be greater than 0 it may be less than 0 when it is greater than 0 means the mean velocity or the instantaneous oh sorry this is not the instantaneous velocity then it is the instantaneous flow rate. So, I have written v and again I have made it q there is some problem. So, it is instantaneous q. So, this part is the instantaneous. So, when it is greater than 0 means the instantaneous flow is more than the mean flow that means the liquid goes to the air vessel when it is less than 0 means the mean flow is less than the actual flow sorry when it is greater than 0 means the actual flow is that means q is greater than q dot. That means the actual flow is more than the mean flow that means the water flows to the air vessel when it is less than 0 the mean flow is less than the the instantaneous flow is less than the mean flow q dot mean rather you write this is mean otherwise difficult. So, when it is greater than 0 then q dot is greater than q dot mean that means the water flows to the air vessel when it is less than 0 q dot is less than q dot mean that means the air water comes from the air vessel the most interesting part is that when sin theta is 1 by pi when sin theta is equal to 1 by pi then this this gives 2 angle theta is some 16 degree no theta is some 18 degree or some minute 34 minute or like that and another is one 61 degree 26 minute like that I will verify it is approximately this angle this gives that means sin theta is 1 by pi gives 2 angles of theta between 0 to 360 degree and those 2 crank angles the instantaneous value coincides with the mean value there is no flow of water either to the air vessel or from the air vessel well there is no flow. So, corresponding to these equations sin theta is 1 by pi it is almost equal to 18 degree another is equal to 161 degree that you can see from the trigonometric table is approximate values like that. So, this is the mathematical part for the air vessel. So, you can very well understand what air vessel does it just supplies the surplus air when it is needed that means the instantaneous flow rate is less than the mean flow or it absorbs or takes the excess surplus air when the instantaneous flow is more than the mean flow well any question please any question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . the air vessel is a large area so that initially you fill it with some compressed air predetermined pressure so expansion little expansion and contraction of the pressure by the movement of the liquid column that the height of the liquid it increases or decreases does not change the pressure much that is one of the very important characteristics feature for the design of air vessel it will be large like a search tank in a compressor as you see in the search tank in a compressor you have seen is very big air compressor also does the same thing so the discharge becomes uniform when it comes from a search tank never air is taken directly from the delivery line of a compressor it is taken from the search tank the search tank volume is much more than the volume of the cylinder of the reciprocating compressors same principle okay next I just like to tell you is a straightforward application no doubt straightforward application of these principles one problem is example how to solve this problem a single acting you write example a single acting reciprocating pump a single acting reciprocating pump having cylinder diameter of 150 millimeter is used to raise the water through a height of 20 meter so diameter of cylinder is 150 millimeter and is used to raise the water through a height of 20 millimeter so 20 millimeter is the total height that is the static lift that is a hs plus hd from the sum to the overhead reservoir this is the total height these are the important dimension cylinder diameter its crank rotates at 60 rpm that is the revolutionary speed is 60 rpm find the theoretical discharge and theoretical power required to run the pump first is the theoretical discharge and theoretical power theoretical discharge and theoretical power is related to the power and discharge which neglects the acceleration and decelerations of the piston next but pyre is it a function of its the relationship between the cylinder and the pipe. Thecking results this is the final Castile which is maximum which is maximum at the beginning of the stroke that means the power developed has to be much more than that of the 20 meter for pumping if you take care of the acceleration here next but it tells if large air vessel the word large is very important that means the pressure in of the air in the air vessel remains constant substantially air vessel is fitted very close if it is not large in this context title this is an approximation that is why there will be a little fluctuations in the discharge if it is theoretically constant that means it is always maintained a constant pressure at that point where the air vessel was connected to the delivery pipe so it does not maintain so because of the fluctuation in the air pressure that is why there is the little fluctuation in the discharge in the delivery pipe from the air vessel connections onwards ok if large air vessel is fitted very close to the cylinder on delivery site very simple find the theoretical velocity in delivery pipe this theoretical velocity is no more connected with this large air vessel this is already known from the swept volume but more than most important is the and the pressure aid in cylinder necessary to overcome friction in delivery pipe take f is equal to 0.0 and this is typical Darcy's friction coefficient find the rate of flow next part from or to air vessel when crank makes angles of 30 90 120 that is the straight forward application of the equations that is we have described from inner dead center position I think all of you can solve this problem there is no need of discussing this problem theoretical discharge is simply the swept volume per revolution you convert it in terms of time ok all right so I think you can solve the problem well find the rate of flow from or to air vessel when crank makes angles of 30 degree 90 degree and 120 degree from inner dead center position all right ok now ok now theoretical discharge q dot theoretical what will be is what is the area of the cylinder pi d square pi into 150 millimeter meter square d square by 4 into the stroke length what is the stroke length stroke length is given what is the stroke length stroke length is not given what r is given what is that pump having cylinder diameter of 150 is used to raise the water through a height of 20 meter its crank rotates at 60 rpm find the theoretical discharge and theoretical power required you just see it to run the pump if delivery pipe is 100 millimeter in diameter and 15 meter long find the so stroke length is not given so stroke length has to be given I am sorry the stroke length is omitted so stroke length has to be given otherwise you cannot find it what is no no length of the cylinder cannot be the stroke length there will be a clearance volume inner dead outer dead length of the cylinder where is the length of the cylinder where is length of cylinder not given length of cylinder is not the stroke length I am sorry should not tell like that so stroke length has to be given this is missing in the problem this data well just you see the procedure then you can find out your 60 rpm that means 60 by 60 that means one revolution per second so this will be the q dot theoretical meter cube per second power theoretical is what rho q dot theoretical g into h what is this h this is 20 h is equal to 20 I am sorry that stroke length is missing is a data however if stroke length is given you can find out so delivery pipe the acceleration head as you know what is the expression of acceleration head what is the expression of acceleration head please l d by g into h is equal to next is a d by capital A and then r omega square cos omega t ok capital A in the numerator and small a in the denominator ok so at the beginning means delivery stroke so this will be 1 180 cos theta cos 0 1 so this will be l d by g you see that a by a d omega c square so you can find out well if large air vessel is fitted very close to the cylinder it is the concept that theoretical velocity will be the theoretical discharge divided by the what is theoretical velocity in delivery pipe is theoretical discharge divided by the cross sectional area of the delivery pipe can it be found delivery pipe diameter is given 100 millimeter that means 0.1 meter into 0.1 square into 4 so if I know q theoretical this is the v theoretical the theoretical velocity and find out the pressure head in cylinder necessary to overcome friction in delivery pipe when this air vessel is incorporated then the theoretical velocity is this and that will be the velocity in the delivery pipe uniform velocity so therefore h f is f l d by d d the diameter of the delivery pipe into v theoretical this v theoretical square by 2 g so this will be the frictional head head required to overcome the friction because the velocity is uniform ok but in other cases you will have to integrate it taking care of the varying heads I am telling this is a very simple problem ok thank you time is up ok