MOMENTUM PRINCIPLE. Review: Last time, we derived the Reynolds Transport Theorem: Chapter 6. where B is any extensive property (proportional to mass),


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1 Chapter 6 MOMENTUM PRINCIPLE Review: Last time, we derived the Reynolds Transport Theorem: where B is any extensive property (proportional to mass), and b is the corresponding intensive property (B / m ). Fluid Mechanics, Spring Term 2011 a) b) c) Recall: The system (e.g., B sys ) is a fluid volume that moves along with the fluid particles. The control volume (cv) may move relative to the fluid; for example, it may be fixed in space. The control surface (cs) is a closed surface that contains the control volume. Meaning: Term a) describes the physical change that takes place for a set of fluid particles. Term b) describes the change observed in a given control volume. Term c) describes the difference: For example, if the property B has not changed for the fluid particles, but B has changed inside the control volume, then this must be because new fluid particles have entered the cv which carried a different amount of B with them.
2 Still review We then applied the Reynolds Transport Theorem to the property of mass (M ). Different ways of writing the results: with Today, we use the Momentum Principle: We will do 2 things: 1) Use the law of conservation of momentum (Newtonʼs 2nd Law). 2) Set B = mv, (where B is the extensive property) and substitute this into the Reynolds Transport Theorem. Notice the similarities to last lecture: 1) Use conservation of mass, dm sys / dt = 0 2) Set B = m Newtonʼs 2nd Law: or Definition of (linear) momentum: The momentum conserved is that of the system: Particles move about in a box. No particles leave or enter, hence the box is a system. Conservation of momentum: Particles can exchange momentum with each other, but that does not affect the total momentum of the system. The only way to change the total momentum is to apply a force to the box.
3 Therefore: Notice the similarities and differences between conservation of mass and momentum: (conservation of mass) (conservation of momentum) 1) Both equations are applied to the system. 2) Mass is conserved absolutely (never changes in classical physics); Momentum is conserved unless a force is applied. 3) Mass conservation is a scalar equation; Momentum conservation is a vector equation (3 equations). Reynolds Transport Theorem with B = mv becomes Notice the little v and the big V in the last term. The big V comes from the Reynolds Transport Theorem. It is the fluid velocity relative to the control surface. and using Newtonʼs 2nd Law we get The little v comes from the momentum definition. It is the fluid velocity relative to the reference frame. v and V are the same only if the control volume is at rest relative to the reference frame.
4 V wall Forces acting on a control volume (Fig. 6.1) V fluid V wall V cv The integral over cv is the change in momentum in the control volume. The last term is again a convective correction in going from the Lagrangian to the Eulerian frame: V fluid In the channel flow illustrated, nothing changed between 1 and 2; only the control volume moved. But the momentum in cv increases because new particles with greater momentum have flowed into cv. cv consists of fluid only. Cv includes the section of the pipe. Choice 1: control volume is entirely within the fluid. Choice 2: The control volume includes the entire section of the pipe. where! is a viscous shear stress acting between the fluid and the pipe wall (its direction depends on whether the fluid is moving up or down). where W p is the weight of the pipe.
5 Momentum Accumulation (Fig 6.2) Momentum equation (general): Steady flow through a nozzle. The momentum of each fluid particle passing through the nozzle changes with time. If v is constant across inlet and outlet: Recall (Chapter 5) the mass flow rate For steady flow, the integrated momentum inside the cv does not change with time. There is no accumulation of momentum in the cv. (Momentum equation for inlet and outlet ports) Momentum Diagram: The change in momentum in the cv may be visualized with a momentum diagram. (Notice that weʼre drawing changes in momentum, i.e., accelerations!) Momentum Equation in Cartesian Coordinates (inlet and outlet ports with constant velocities) (inflow) (outflow) Net outflow of momentum: (a fairly obvious, special form of the momentum equation )
6 Systematic Approach to Solving Problems: 1) Problem Setup Select appropriate control volume. Select inertial reference frame. 2) Force analysis and diagram Sketch body forces on force diagram (gravity). Sketch surface forces: pressures, shear stresses, supports and structures 3) Momentum analysis and diagram Evaluate momentum accumulation term. If the flow is steady, this term is zero. Otherwise evaluate volume integral and add to momentum diagram. Sketch momentum flow vectors on momentum diagram. For uniform velocity, each vector is Example 6.1: Rocket on test stand. Exhaust jet has: Diameter d = 1 cm Speed v = 450 m/s Density " = 0.5 kg/m 3 Assume p in jet is atmospheric p. Neglect momentum changes inside rocket motor. What is the force F b acting on support beam? Example 6.1: Solution Only involves vertical momentum. 1) Forces: Example 6.1 (continued) Substitute forces and momentum into the momentum equation: 2) Momentum change: There is no momentum accumulation because the structure is stationary and because we neglect momentum changes in the rocket motor. or
7 Example 6.2: Concrete flowing onto cart on a scale Example 6.2: Solution Stream of concrete: Given: Density = " Area = A Speed = v Cart + concrete: Weight = W Determine tension in cable and weight recorded by scale. Forces and momenta involve x and z directions. Forces in x: Forces in z: Example 6.2 (continued) Example 6.2 (continued) Since flow is steady and cart does not move, there is no accumulation of momentum inside the cv. The momentum equations in x and z give Momentum changes in x and z: (with ) thus providing T and N as the answer. (Notice in this problem the cv is stationary; hence, the velocities v and V are the same).
8 Example 6.4: Water jet deflected by a vane Example 6.4: Solution Given: Speed of incoming jet = v 1 Speed of outgoing jet = v 2 Diameter of jet = D 1) Forces: Note that mass flow rates have to be equal: 2) Momenta: Net momentum change: Example 6.4 (continued) Example 6.7: Water flow through a 180 o reducing bend Again, substitute into momentum equation: Given: Discharge = Q Where is given by Pressure at center of inlet = p 1 Volume of bend = V Weight of bend = W What force is required to hold the bend in place?
9 Example 6.7: Solution Example 6.7 (continued) Additional difficulties: 1) We donʼt know the velocities. 2) We donʼt know the outlet pressure. 1) Get velocities from continuity equation (incompressible!): 2) Get p 2 from Bernoulli equation: Note: z1 & z2 are at center of pipe. Why? Notice that there is no momentum change in the vertical. In the vertical we only have the force balance Forces in xdirection: Momentum in xdirection: Momentum balance in x: Using continuity and Bernoulli equations (previous slide), all variables are known except R x.
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