 Welcome to lecture 4 incompressible fluid flow related to fluid drive. This is under module 2 fundamentals of fluid flow and properties. Control volume method is applied in analysis of fluid power in general. This is because the movements of fluid particles in a space are not much of our interest as in general fluid mechanics based on Newtonian theory rather we are interested in the movement of volume of fluid through the conduit. Now let us consider conservation of mass. The rate of mass flow into the control volume equals the rate of mass flow out plus the rate at which mass accumulates inside. Now let us consider a control volume. Fluid is flowing inside this volume and then we consider a velocity in normal direction at a point on the volume which is Vn and obviously which is having a straight component V. Now mathematically we can write rho Vn dash is equal to minus dm Cv by dt is equal to minus d by dt integration of rho dv. Now in this case As is the normal surface area rho is the density which we have already discussed. Vn is the normal velocity and this normal velocity it is importantly is positive when directed outwards. That means the direction shown is positive. That must be equal to time rate of change of accumulated mass in control volume. m Cv is the accumulated mass. Now this means that the minus sign is given here. The reason is that if the mass is accumulated inside then Vn will be in the opposite directions. If it is going out then it will be minus. That must be equal to the elemental mass of control volume. We should consider the rate of change of elementary mass of control volume. Now application of this conservation of mass or momentum theory is that let us consider a vessel with a conduit. Now flow in to this control volume through the left side where the area is A1. V1 is the velocity and the area surface area of this vessel is At and height of a datum line is L where the area is A2 and velocity is V2. This means that along this line when the flow is going in velocity is V1, flow is going out velocity is V2. Area of this outlet is A2. Area of this inlet is A1 and At is the surface area of this vessel and L is the height from this datum line. Now for steady state condition we can write rho V1 A1 is equal to rho V2 A2. What does it mean the steady state conditions that there will be no change in this volume where L is not varying. Now for unsteady flow into tank that is this vessel L will vary and the expression becomes rho V1 A1 minus rho V2 A2 is equal to rho A2 DL by DT. This means DL by DT is will be positive if this height is increasing and this is automatically it will be negative L is decreasing. Now conservation of momentum the rate of change of momentum of the system in any direction is expressed as the sum of the rate of change of momentum of the material inside the control volume which we have considered as a VA small VA in the same direction. And net rate of outflow of momentum through the control surface in the same direction it is expressed as Fx is equal to D by DT integration of rho Vx DVA over the volume VA the control volume VA this Vx is the velocity in x directions plus rho integration of rho Vx Vn and Das it will be in the y directions as you look into this equations only we will considering here the expression is same velocity in y direction and here also we consider the velocity in y directions similarly in z directions we consider velocity z here and also here for steady flow first term of right hand side is 0. If it is a steady flow then there will be no change in this whereas there may change in this surface area and accordingly this equations will be reduced. Now steady flow example let us consider another tank it is a closed it is having an inlet and a outlet. Now let us consider at the inlet area is a 1 oil is having density rho 1 and velocity v 1 at outlet it is rho 2 velocity v 2 and area is a 2. Now rho 1 and rho 2 may be equal for incompressible fluid if there is no change in temperature. Now for this type of flow you will find that if we do not apply a load this will move because this velocity obviously will be higher than this and a force will be generated here. Now to balance this for equilibrium we apply a load fx and therefore the relation we can write inflow of x momentum is equal to v 1 rho 1 a 1 v 1 that is equal to rho 1 a 1 v 1 square is it clear this is the volume flow rate and with the velocity. So we get this is the x momentum similarly outflow we get it rho 2 a 2 v 2 square. Now equating we get the external force fx is equal to rho 2 a 2 v 2 square minus rho 1 a 1 v 1 square this means that we have to apply this force for the equilibrium of this body. Let pressure condition be p 1 is equal to p 2 this means that here pressure and here pressure will be same then from continuity consideration incompressible flow rho 1 a 1 v 2 is equal to rho 2 a 2 v 2 and therefore fx will be rho 1 a 1 v 1 v 2 minus v 1. I think I have made a mistake it is not pressure conditions the pressure conditions remain same for which rho 1 and rho 2 remain same here I would like to mention the pressure remains same as well as temperature also remains same. So that we get rho 1 is equal to rho 2. Now unsteady flow example how the pressure difference p 1 minus p 2 is related to rate of change of flow rate of a frictionless incompressible fluid in a uniform tube of length l. We have considered a tube circular tube and we have taken a segment of length l this is of uniform cross sectional area a and the tube is also rigid. Now flow in is q 1 and flow out is q 2 and we also for generality we consider that density here was rho 1 and density this side is rho 2 pressure is here p 1 and p 2. We consider that pressure is changing maintaining continuity we can write rho q 1 is equal to rho q 2 again in this case we have consider rho 1 is equal to rho 2 that means density is not changing in most of the fluid power analysis for small conduit if length is not very large and if there is not too much change in pressure and the temperature then we may consider rho 1 is equal to rho 2 this simplicity is required to avoid the complication in the calculations and in most of the cases the error will be negligibly small. Therefore, we can write q 1 minus q 2 is equal to 0 this is for incompressible fluid. Now the momentum equation gives p 1 a minus p 2 a is equal to rho q v minus rho q v minus t by dt rho a l v a into l is the volume v is the velocity and rho is the density that is the mass. So, rate of change of mass minus the mass in now this gives clearly p 1 minus p 2 into a that is the force is equal to rho a l d by dt q by a if you equate this we will arrive here and this gives clearly rho l d q by dt as a is constant this a will cancel out this means that the force net force available due to this flow is mass density into length into time rate of change of flow this q is flow rate. So, this is time rate of change of flow rate. So, final equations is coming p 1 minus p 2 rho l a d q by dt conservation of angular momentum in a steady flow centrifugal pump let v t 1 is the tangential velocity at radius r 1 at entry. If you think of the centrifugal pump then you will find that entry at this towards the centre and outlet is towards the outer periphery. Now in fact it works on hydro kinetic energy in that case when the impeller inside is rotating with the volume of fluid initially there is air then this fluid mass gets momentum and it is discharged outside in turns it creates a suction pressure at the inlet. Anyway after let us consider it is pumping an incompressible fluid in that case let v t 1 is the tangential velocity at radius r at entry this means that if I consider a circle at the entry then tangential velocity of the mass of the fluid. Again let us consider a control volume now same thing when it is being discharged the tangential velocity is v t 2 at radius r 2 then we can write the ideal input torque is equal to t is equal to r 2 v t 2 minus r 1 v t 1 into rho again density into the flow rate. Now let us consider one dimensional frictionless incompressible streamline flow with area change. Now in this case again we have considered a conduit which is having inlet area is a 1 outlet area is a 2 in this closed vessel or closed tank. Let us consider at the inlet the u 1 is the intrinsic internal energy per unit mass of fluid. Now this is a little difficult to have the concept because inside the fluid the internal intrinsic internal energy may change while it is flowing from one side to the other due to addition of some external energy or going out the energy from the inside of the conduit or the control volume. Now this one again having a density rho 1 velocity v 1 pressure p 1 z 1 is the height of this fluid from a datum line and w is the weight flow rate of fluid. In outlet side also we consider u 2 rho 2 v 2 a 2 p 2 z 2 and w clearly this in this case what we have considered in the system that z 1 is equal to z 2 because this is a horizontally placed we have neglect that part rho 1 is also is equal to rho 2 because there is the change in pressure not changing the density or whatever change is there that is negligibly small and also u 1 and u 2 that means there is no change in intrinsic internal energy. Now then simply we can develop these equations and we can neglect also z 1 and z 2 as they are equal of these equations. Now these equations is very well known equations Bernoulli's equations and with this equation we will see next that if we now add some energy then what will be the changes in these equations. Now here as I have told this is a frictionless that means there is no friction between the fluid and the conduit we have neglected that part that that is there always there in a conduit there will be friction but we have neglected that one and again this is we are what we are developing this equation that is for the incompressible streamline flow no turbulent flow also then equation will be different only thing there will be change in area. Now what we have done we have added some external work so that is symbolized by this say let us consider that we have a fan or impeller is being driven and it is a energy being added there now also we are adding some heat and there is also this is the heat flow inside now why we are considering such things the in fluid power usually this external work will be there on the fluid and this heat is automatically generated due to the temperature rise due to the change in pressure due to change in internal energy that we can consider in this form as well in some cases heat also added from the outside. So what will be the equation now we introduce the energy equation and using the control volume concept then what we find that the same equation which we are consider earlier we consider this u 1 and u 2 we consider that there will be change in this in changing energy as we are working on that we are also adding the heat and we have we have also kept in this height term here and then the energy added is written in this form this is the work added to this minus decrease by dt divided by w by g. Now w s is the shaft or shear work done and q h the heat flow to control volume the here the question is that if we why it is negative the heat flow in actually reduces the this energy. Now if we do not add any external work to the system then dx by dt is equal to 0. Therefore the total equation will be reduced in this form. Obviously we have considered z 1 is equal to z 2 and we can write dq by dt is equal to w by g u 2 minus u 1 is equal to w by g c h t 2 minus t 1 whereas c h is the specific heat t 1 is equal to temperature before heating and t 2 is equal to temperature after heating. Now we consider frictionless flow through nozzles and orifices. Now here we have considered again vessels or large reservoir and where we are having a small hole which is an orifice. Now this is known theory to you you will find that oil will flow out and you will find that the stream will have a little less area than the orifice area at a certain distance and then again this area will increase. Now the minimum area as you know it is called vena contractor. Now for this one we can write the equation in this form. Here the velocity at the entry we may consider 0. We are considering this area a little away in the left side from the orifice and then we are considering the velocity v 2 at the vena contractor. Then this equation can be written in this form that means velocity here we can write 2 by rho p 1 minus p 2. Therefore the mass flow rate can be written in the form of rho into q is equal to rho a 2 into v 2 because a 2 is the area here is equal to rho a 2 square root of 2 by rho p 1 minus p 2. Now the a 2 is the area at the vena contractor which is not known it is not it is neither we can measure the area there. Now to take care of that what we do that rho q into c d into a 0 square root of twice rho p 1 minus p 2 where c d is the coefficient of discharge and a 0 is the area of the orifice. Now again you know that coefficient of discharge is equal to the area contraction coefficient and another coefficient of velocity. So combining this c d we use for fluid flow. Now in case of fluid power with incompressible flow for a suitable orifice normally this c d value remain more or less constant. Of course that depends on the orifice area in most of the calculation in fluid power particularly in oil hydraulics the c d value may be taken as 0.62. Later when we will come to valve flow I will show you that how c d can be taken more or less constant for fluid power analysis. Also a care is taken to make the orifice there will there are various type of orifice in fluid power components starting from a circular hole to the variable area. But the care is taken so that c d more or less given constant. The final equation here 2.4.24 is q is equal to c d a 0 square root of 2 by rho p 1 minus p 2 this we should remember because very often we need to use this equation. Other equations which I have shown normally when we analyze the inside flow in a valve then it will be required. But normal cases this equation will be more useful. Now viscous flow through the capillary passage. Now here one important factor we consider which is Reynolds number. Reynolds as you know is a scientist he developed these equations where R e is equal to U d by rho. Now this is the this is kinematic viscosity already we know that this is dynamic viscosity by density. Here of course we have used in other form normally V is rho. Now U is equal to the velocity of fluid in conduit. Now normally this Reynolds number becomes very high where d we consider d is equal to hydraulic diameter that is equal to 4 into flow section area whatever may be the area we consider 4 into flow section area divided by flow section perimeter. Let us consider it is a circular one then what we will do 4 into pi R square divided by 2 pi R 4 into pi R square divided by 2 pi R. So this becomes 2 pi R that is equal to pi d. So this is hydraulic diameter. Now in case of fluid power we consider a reduced Reynolds number which is expressed as R star is equal to U L by sorry this is not rho nu into h by L whole square. Normally you will find this is much much less than 1 and where the L is the length of capillary passage and h is equal to height of the gap or capillary passage. This is for an example even in a pipe we can consider L is the length and h simply the inside diameter of the conduit. But this is more useful where the flow between two parallel plates or some capillary passage. Now with this I finish this lecture and this note is prepared based on this two books the control of fluid power by Martin and McLeod and Blackburn, Rathoff and Shearer by fluid power control. Also I have followed the another book hydraulic control system which is by merit. But most of the equations that I have followed from these two books. Thank you.