Chapter 6: Momentum Analysis of Flow Systems

Chapter 6 MOMENTUM ANALYSIS OF FLOW SYSTEMS 6–1 NEWTON’S LAWS Newton’s laws: Relations between motions of bodies and the forces acting on them. Newton’s first law: A body at rest remains at rest, and a body in motion remains in motion at the same velocity in a straight path when the net force acting on it is zero. Therefore, a body tends to preserve its state of inertia. Newton’s second law: The acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass. Newton’s third law: When a body exerts a force on a second body, the second body exerts an equal and opposite force on the first. Therefore, the direction of an exposed reaction force depends on the body taken as the system.  Newton's second law:   F  ma  m  d V d (mV )  dt dt Linear momentum: The product of the mass and the velocity of a body. Newton’s second law is usually referred to as the linear momentum equation. Conservation of momentum principle: The momentum of a system remains constant only when the net force acting on it is zero. Linear momentum is the product of mass Newton’s second law is also expressed as the and velocity, and its direction is the rate of change of the momentum of a body is direction of velocity. equal to the net force acting on it. The counterpart of Newton’s second law for rotating rigid bodies is expressed as 𝑀 = 𝐼 𝛼 where 𝑀 is the net moment or torque applied on the body, I is the moment of inertia of the body about the axis of rotation, and 𝛼 is the angular acceleration. It can also be expressed in terms of the rate of change of angular momentum 𝑑𝐻 𝑑𝑡 as Angular momentum equation:    d  d (I  ) d H M  I  I   dt dt dt  Angular momentum about x-axis:  M x  Ix (6 - 2) d x dH x  dt dt The conservation of angular momentum Principle: The total angular momentum of a rotating body remains constant when the net torque acting on it is zero, and thus the angular momentum of such systems is conserved. The rate of change of the angular momentum of a body is equal to the net torque acting on it. 6–2 CHOOSING A CONTROL VOLUME • A control volume can be selected as any arbitrary region in space through which fluid flows, and its bounding control surface can be fixed, moving, and even deforming during flow. • Many flow systems involve stationary hardware firmly fixed to a stationary surface, and such systems are best analyzed using fixed control volumes. • When analyzing flow systems that are moving or deforming, it is usually more convenient to allow the control volume to move or deform. • In deforming control volume, part of the control surface moves relative to other parts. Examples of (a) fixed, (b) moving,and (c) deforming control volumes. 6–3 FORCES ACTING ON A CONTROL VOLUME The forces acting on a control volume consist of Body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and Surface forces that act on the control surface (such as pressure and viscous forces and reaction forces at The total force acting on a points of contact). control volume is composed Only external forces are considered in the analysis. of body forces and surface forces; body force is shown    Total force acting on control volume:  F   Fbody   Fsurface on a differential volume element, and surface force is shown on a differential surface element. The most common body force is that of gravity, which exerts a downward force on every differential element of the control volume.  Gravitational force acting on a fluid element:  d Fgravity  p g dV   g  g k Gravitational vector in cartesian coordinates:  Total body force acting on control volume:  Fbody   CV   p g dV  mCV g Surface forces are not as simple to analyze since they consist of both normal and tangential components. Normal stresses are composed of pressure (which always acts inwardly normal) and viscous stresses. Shear stresses are composed entirely of viscous stresses. The gravitational force acting on a differential volume element of fluid is equal to its weight; the axes have been rotated so that the gravity vector acts downward in the negative z-direction. When coordinate axes are rotated (a) to (b), the components of the surface force change, even though the force itself remains the same; only two dimensions are shown here.   Surface force acting on a differential surface element: d Fsurface   ij  n dA   Fsurface    ij  ndA Total surface force acting on control surface:      CS  F   Fbody   Fsurface    g dV    ij  n dA  CV Total force:   CS     F   Fgravity   Fpressure   Fviscous   Fother totalforce body force surface forces Components of the stress tensor in Cartesian coordinates on the right, top, and front faces. A common simplification in the application of Newton’s laws of motion is to subtract the atmospheric pressure and work with gage pressures. This is because atmospheric pressure acts in all directions, and its effect cancels out in every direction. This means we can also ignore the pressure forces at outlet sections where the fluid is discharged to the atmosphere since the discharge pressure in such cases is very near atmospheric pressure at subsonic velocities. Atmospheric pressure acts in all directions, and thus it can be ignored when performing force balances since its effect cancels out in every direction. Cross section through a faucet assembly, illustrating the importance of choosing a control volume wisely; CV B is much easier to work with than CV A. 6–4 THE LINEAR MOMENTUM EQUATION Newton’s second law for a system of mass m subjected to net force   dV d  expressed as F  ma  m  (mV ) (6 -13) dt 𝐹 is dt where 𝑚𝑉 is the linear momentum of the system. Noting that both the density and velocity may change from point to point within the system, Newton’s second law can be expressed more generally as   d  F   pVdV dt sys (6 -14) Newton’s second law can be stated as The sum of all external forces acting on a system is equal to the time rate of change of linear momentum of the system. This statement is valid for a coordinate system that is at rest or moves with a constant velocity, called an inertial coordinate system or inertial reference frame.  d (mV )sys dt     d   pV dV   pV (Vr  n ) dA CS dt CV  General:     d  F   pV dV   pV (Vr  n ) dA CS dt CV The net flow rate of  Thesum of all   The time rate of change     external forces   of the linear momentum    linear momentum out of the         acting on a CV  of the contents of the CV  controlsurface by mass flow     Vr  V  VCS  Fixed CV :     d  F   pV dV   pV (V  n ) dA CS dt CV The linear momentum equation is obtained by replacing B in the Reynolds transport theorem by the momentum 𝑚𝑉, and b by the momentum per unit mass 𝑉. The momentum equation is commonly used to calculate the forces (usually on support systems or connectors) induced by the flow.      d  F   pV dV   pV (V  n ) dA CS dt CV In most flow systems, the sum of forces 𝐹 consists of weights, pressure forces, and reaction forces. Gage pressures are used here since atmospheric pressure cancels out on all sides of the control surface. Special Cases     r  F    V (V  n ) dA Steady flow: CS Mass flow rate across an inlet or outlet:   m   (V  n ) dAc  Vavg Ac Ac Momentum flow rate across a uniform inlet or outlet:     Ac    V (V  n ) dAc  Vavg Ac V avg  mV avg In a typical engineering problem, the control volume may contain many inlets and outlets; at each inlet or outlet we define the mass flow rate and the average velocity. Examples of inlets or outlets in which the uniform flow approximation is reasonable: (a) the well-rounded entrance to a pipe, (b) the entrance to a wind tunnel test section, and (c) a slice through a free water jet in air. Momentum-Flux Correction Factor,  The velocity across most inlets and outlets is not uniform. The control surface integral of Eq. 6–17 may be converted into algebraic form using a dimensionless correction factor , called the momentum-flux correction factor.      d  F    V dV    V (V  n ) dA CS dt CV (6 - 17)  Momentum flux across an inlet or outlet:    d  F    V dV   mV avg    mV avg out in dt CV    Ac   V (V  n ) dAc   mV avg      Ac V (V  n ) dAc mVavg     Ac V (V  n ) dAc Vavg AcVavg β is always greater than or equal to 1. Β is close to 1 for turbulent flow and not very close to 1 for fully developed laminar 2 flow.   1 V Momentum-flux correction factor:  Ac  Ac  V  dAc  avg  EXAMPLE 6-1: Momentum-Flux Correction Factor for Laminar Pipe Flow  r  V  2Vavg 1  2   R  2 Laminar flow: 2 1  V  4  dA  c Ac Ac  Vavg   R2 1 0 2  r  1   R 2  2 r dr 2 0 y  4   3  1 3   4 y 2 dy  4  0  R 3 Velocity profile over a cross section of a pipe in which the flow is fully developed and laminar. For turbulent flow β may have an insignificant effect at inlets and outlets, but for laminar flow β may be important and should not be neglected. It is wise to include β in all momentum control volume problems. Steady Flow  Steady linear momentum equation:    F    m V    mV out in The net force acting on the control volume during steady flow is equal to the difference between the rates of outgoing and incoming momentum flows. The net force acting on the control volume during steady flow is equal to the difference between the outgoing and the incoming momentum fluxes. Steady Flow with One Inlet and One Outlet    One inlet and one outlet:  F  m( 2 V2  1 V1 ) Along x-coordinate:  Fx  m( 2V2, x  1V1, x ) A control volume with only one inlet and one outlet. The determination by vector addition of the reaction force on the support caused by a change of direction of water. Flow with No External Forces    d (mV )CV 0    mV    mV out in dt No external forces: In the absence of external forces, the rate of change of the momentum of a control volume is equal to the difference between the rates of incoming and outgoing momentum flow rates.  d (mV )CV d VCV    mCV  (m a )CV  mCV a dt dt     Fthrust  mbody a    mV    mV in out The thrust needed to lift the space shuttle is generated by the rocket engines as a result of momentum change of the fuel as it is accelerated from about zero to an exit speed of about 2000 m/s after combustion. NASA  6–24 A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ=45°θ=45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03 at both the inlet and outlet. 6–32 A 5-cm-diameter horizontal jet of water with a velocity of 30 m/s relative to the ground strikes a flat plate that is moving in the same direction as the jet at a velocity of 20 m/s. The water splatters in all directions in the plane of the plate. How much force does the water stream exert on the plate? 6–5 REVIEW OF ROTATIONAL MOTION AND ANGULAR MOMENTUM Rotational motion: A motion during which all points in the body move in circles about the axis of rotation. Rotational motion is described with angular quantities such as the angular distance , angular velocity , and angular acceleration . Angular velocity: The angular distance traveled per unit time. Angular acceleration: The rate of change of angular velocity. d d (l r ) 1 dl V  dt  dt  r dt  r d d 2 1 dV at   2   dt dt r dt r V  r and at  r The relations between angular distance , angular velocity  , and linear velocity V. Newton’s second law requires that there must be a force acting in the tangential direction to cause angular acceleration. The strength of the rotating effect, called the moment or torque, is proportional to the magnitude of the force and its distance from the axis of rotation. The perpendicular distance from the axis of rotation to the line of action of the force is called the moment arm, and the torque M acting on a point mass m at a normal distance r from the axis of rotation is expressed as M  rFt  rmat  mr 2 Magnitude of torque: r 2  m    r 2  m    I  mass  mass  M  I is the moment of inertia of the body about the axis of rotation, which is a measure of the inertia of a body against rotation. Unlike mass, the rotational inertia of a body also depends on the distribution of the mass of the body with respect to the axis of rotation. Analogy between corresponding linear and angular quantities. r 2  m    r 2  m    I  mass  mass  Magnitude of angular momentum: H     H  I Angular momentum equation:   M  I  I    d  d (I  ) d H   dt dt dt Angular momentum of point mass m rotating at angular velocity  at distance r from the axis of rotation. Angular velocity versus rpm: 2 n   2 n (rad/min)  (rad/s) 60 Wshaft  FV  Fr  M  shaft power: Wshaft   M  2 nM Rotational kinetic energy: 1 2 KE r  I  2 The relations between angular velocity, rpm, and the power transmitted through a shaft. During rotational motion, the direction of velocity changes even when its magnitude remains constant. Velocity is a vector quantity, and thus a change in direction constitutes a change in velocity with time, and thus acceleration. This is called centripetal acceleration. V2 ar  r  r 2 Centripetal acceleration is directed toward the axis of rotation (opposite direction of radial acceleration), and thus the radial acceleration is negative. Centripetal acceleration is the result of a force acting on an element of the body toward the axis of rotation, known as the centripetal force, whose magnitude is Fr  mV 2 /r Tangential and radial accelerations are perpendicular to each other, and the total linear acceleration is determined by their vector sum:    a  at  ar 6–6 THE ANGULAR MOMENTUM EQUATION Many engineering problems involve the moment of the linear momentum of flow streams, and the rotational effects caused by them. Such problems are best analyzed by the angular momentum equation, also called the moment of momentum equation. An important class of fluid devices, called turbomachines, which include centrifugal pumps, turbines, and fans, is analyzed by the angular momentum equation. A force whose line of action passes through point O produces zero moment about point O.  Moment of a force:   M  rF Magnitude of the moment of a force: M  Fr sin  The moment of a force 𝐹 about a point O is the vector product of the position vector 𝑟 and 𝐹. The determination of the direction of the moment by the right-hand rule.  Moment of momentum:   H  r  mV   H sys   (r  V )  dV Moment of momentum ( system):  sys   d H sys d  Rate of change of moment of momentum:   (r  V )  dV dt dt sys    d H sys M  dt    M   (r  F )      d H sys d     (r  V ) dV   (r  V )(Vr  n ) dA CS dt dt CV  General:     d    M   (r  V ) dV   (r  V )(Vr  n ) dA CS dt CV  The sum of all   external moments    acting on a CV   The net flow rate of   The time rate of change    angular momentum    of the angular momentum      out of the control   of the contents of the CV    surface by mass flow   General:     d    M   (r  V ) dV   (r  V )(Vr  n ) dA CS dt CV  The sum of all   external moments    acting on a CV     Vr  V  VCS   The net flow rate of   The time rate of change    angular momentum    of the angular momentum      out of the control   of the contents of the CV    surface by mass flow   Fixed CV :     d    M   (r  V ) dV   (r  V )(V  n ) dA CS dt CV  Vr  V The forces acting on the control volume consist of body forces that act throughout the entire body of the control volume such as gravity, and surface forces that act on the control surface such as the pressure and reaction forces at points of contact. The net torque consists of the moments of these forces as well as the torques applied on the control volume. The angular momentum equation is obtained by replacing B in the Reynolds transport theorem by the angular momentum 𝐻, and b by the angular momentum per unit mass 𝑟 × 𝑉. Special Cases During steady flow, the amount of angular momentum within the control volume remains constant, and thus the time rate of change of angular momentum of the contents of the control volume is zero.      Steady flow:  M   (r  V )(Vr  n ) dA CS An approximate form of the angular momentum equation in terms of average properties at inlets and outlets:     d     M   (r  V ) dV   (r  mV )   (r  mV ) out in dt CV      Steady flow:  M   (r  mV )   (r  m V ) out in The net torque acting on the control volume during steady flow is equal to the difference between the outgoing and incoming angular momentum flow rates.  M   rmV   rmV out in scalar form of angular momentum equation A rotating lawn sprinkler is a good example of application of the angular momentum equation. © John A. Rizzo/Getty Images RF The net torque acting on a control volume during steady flow is equal to the difference between the outgoing and incoming angular momentum flow rates. Flow with No External Moments    d H CV   0   (r  mV )   (r  mV ) out in dt No external moments : In the absence of external moments, the rate of change of the angular momentum of a control volume is equal to the difference between the incoming and outgoing angular momentum fluxes. When the moment of inertia I of the control volume remains constant, the first term on the right side of the above equation becomes simply moment of inertia times angular acceleration. Therefore, the control volume in this case can be treated as a solid body, with a net torque of       M body  I body    (r  mV )   (r  mV ) in out This approach can be used to determine the angular acceleration of space vehicles and aircraft when a rocket is fired in a direction different than the direction of motion. Radial-Flow Devices Radial-flow devices: Many rotary-flow devices such as centrifugal pumps and fans involve flow in the radial direction normal to the axis of rotation. Axial-flow devices are easily analyzed using the linear momentum equation. Radial-flow devices involve large changes in angular momentum of the fluid and are best analyzed with the help of the angular momentum equation. Side and frontal views of a typical centrifugal pump. The conservation of mass equation for steady incompressible flow V1  V2  V V V1,n  2 rb 1 1  (2 rb 1 1 )V1, n  (2 r2b2 )V2, n and V2,n Euler's turbine equation: angular momentum equation V  2 r2b2  M   rmV   rmV out in Tshaft  m(r2V2,t  rV 1 1,t ) Tshaft  m(r2V2 sin  2  rV 1 1 sin 1 ) When V1,t   r1 and V2,t   r2 Tshaft,ideal  m (r22  r12 ) Wshaft   Tshaft  2 nTshaft   2 n An annular control volume that encloses the impeller section of a centrifugal pump. Lawn sprinklers often have rotating heads to spread the water over a large area. © Andy Sotiriou/Getty Images RF