Momentm Eqation Interest in the momentm eqation: Qantification of proplsion rates esign strctres for power generation esign of pipeline systems to withstand forces at bends and other places where the flow changes. Revisitation of Newton s Second Law ecept we consider it in its general form: The rate of change of momentm = Resltant force in the direction of measrement. The ontrol Volme A control volme is an imaginary srface that is sed to help s apply the momentm eqation to a particlar problem. It is based on the following concept: The a lg ebraic sm of the eternal forces acting on the flid in the control volme in a given direction, = The rate of change of momentm in the given direction as a reslt of the flid pas sin g throgh the control volme, ( ) ρq 1 Necessary becase body is not made p of a fied assembly of particles Its volme is the same however Imaginary 1
ts across eternal agents Applying the ontrol Volme oncept The general steps necessary to set p a problem prior to analysis are: (1) raw the system, then draw the imaginary control volme which represents the part of the system to be analysed. If it helps, think of the control volme as cling-film enclosing part of the real system. () Use arrows to show the direction of travel of the flid entering and leaving the control volme. Label them 1 and to represent velocities. (3) Label the aes, and y for a two-dimensional problem in the horizontal plane, and z for a problem in a vertical plane. The aes are positive in the initial direction of the flid as it enters the control volme. (4) raw the eternal forces acting on the control volme. This incldes the eternal pressre forces (PA) acting on the ends of the pipe, and the resltant force, R (5) Vector qantities sch as velocity, pressre forces and the nknown resltant force mst be resolved in the direction of the aes before the vales are pt into the momentm eqation. If yo do not know the direction in which RX and RY act initially, jst gess. Having applied the momentm eqation, yor gess is correct if the answer obtained for RY or RX is positive, and wrong if it is negative. (6) All forces acting in the same direction as the positive aes are positive; those actions in the opposite direction are negative. Use these signs when evalating X, Y or Z. Momentm is a vector and has a direction. erivation of the Steady low Momentm Eqation (diagram drawn in class).
Small streamtbe implies niform velocity over AB and. Now in time δt, the streamtbe moves to A B onsider an elemental mass δm. Its momentm at time 0 = Its momentm at time δt = δ m AB δ m A' B' ' ' = AB δ m ABB' A' δm + δm ' ' Net increase = Momentm final - Momentm initial = δ m δm ' ' ABB' A' since the flow is steady. If δt is sfficiently small is constant in ABB A and. = δ m δm Therefore, net momentm increase ( ) ' ( ) ABB' A' Now δ m = m& δt = δm (Mass of flid entering system in nit time delta t) ' ' ABB' A Net increase in momentm = m& ( ) Newton s Second Law 1 = m& ( ) 1 and y = m& ( ) y y1 3
and z = m& ( ) z z1 In steady flow, & ( is perpendiclar to A) m = ρ = 1A1 1 ρ A or a single streamtbe δ = m& ( 1 ) = ρδa ρ1δa1 1 1 Total force in, y and z directions can be given as: = 1 δ = ρδa ρ1δa1 1 y z = y y y1 δ δ = ρδa = ρδa ρ1δa1 1 = z z z1 ρ1δa1 1 Integrals can only be evalated when the distribtion of velocities across the area is known. It is assmed that flow at the inlet and otlet section is straight and parallel. The control volme shold be appropriately chosen. Momentm orrection actor Rate of momentm flow taken as ρ Bt it really is ρ da = βρ or flly developed trblent flow in a circlar pipe for eample, ma = y R 1 7 4
β = 1.0 or laminar flow, β = 4 3 Applications Jet Striking a Srface (diagram drawn in class) X direction taken normal to the srface so that the momentm flow ot of the control volme is zero. Rate at which momentm enters the control volme is: ρ 1 da 1 1 1 = 1 cosθ ρ11 da Therefore, the rate of increase of momentm is Total force in direction on = cosθ ρ11 da1 flid cosθ ρ11 da = 1 Since pressre is atmospheric both at entry and at eit, the hydrodynamic force on the flid is provided only by the solid srface. Therefore, 5
Hydrodynamic force by the flid on the srface = 1 + cosθ ρ11 da That is the force is eqal and opposite. If both the density and velocity are niform, then = 1 1 1 ρ A cosθ = 1 1 1 ρ Q cosθ onsider now the component of the oncoming momentm parallel to the srface, i.e., along the Y direction. It is assmed that the flid is ideal and that the srface is smooth, so no shear forces are possible. Ths the component ρ 1 Q 1 1 sinθ is not changed at all. There will therefore be no force eerted in a direction at right angles to the direction. or an ideal flid, y =0 or a real flid, y 0 low in a Pipe Bend (diagram drawn in class) orce on flid in direction = p θ + 1A1 p A cos 6
where is the force eerted by the pipe bend on the flid in the direction. The resltant force in -direction is ρ Q( cosθ ) Therefore, ( cosθ 1 ) = p1 A1 p A cosθ ρ Q + 1 ( cosθ 1 ) + p A cos p1 A1 = ρ Q θ orce on bend is To calclate y orce on flid in y direction = y sinθ p A where y is the force eerted by the pipe bend on the flid in the y direction. The resltant force in y-direction is ρq ( sinθ ) ( θ ) p sinθ y = ρq sin + A orce on bend is y orce on a Solid Bondary in a lowing lid (drawn in class, if discssed) To show that this force can be determined by measring the velocity and pressre in the wake of the body. onsider the body immersed in a flid in which the velocity far removed from the body is U and the pressre is p 7
The control volme has oter bondaries far enogh from the solid body for pertrbations of velocity and pressre to be negligible. Inner bondary of the control volme is on the solid bondary. onsider a small element of the plane of length δy. mass flow rate throgh the element = ρδy rate of increase of momentm for this flid ( ) = ρ δy U Total rate of increase of momentm for this flid = ρ ( U )dy Total force on flid in the -direction is given by (1) the component, of the force eerted by the body on the flid and () the net pressre force in the -direction given by A B p since AB =. The momentm eqation is: dy pdy = ( p p) dy = ( U ) + ρ ( p p) It shold be noted that beyond points and, = U and p = p, and so the limits might be replaced by and +. So, + { ( ) ( ) } U p p = dy ρ dy 8