Control Volume Revisited

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1 Civil Engineering Hydraulics Control Volume Revisited Previously, we considered developing a control volume so that we could isolate mass flowing into and out of the control volume Our goal in developing this approach was to simplify complex problems into simpler problems for analysis 2 1

2 Control Volume Revisited We choose boundaries to isolate internal mass flows from external mass flows We then extended this to consider energy moving into and out of the control volume We developed a relationship conserving energy using the Bernoulli equation when appropriate 3 Control Volume Revisited In a manner similar to that undertaken in Statics and Dynamics, we looked at the control volume as being a boundary to isolate internal from external events Now we will add another element to our analysis and continue using the control volume approach to simplify analysis 4 2

3 Control Volume Revisited The development of a usable control volume is critical in the solution of our problems Problems that are exceptionally computationally complex can be reduced considerably by the consideration of a viable control volume Take the time to consider what the control volume provides before setting up your analysis 5 Linear Momentum In Statics, we considered systems that were not under acceleration in any direction so we were able to make the statement a = F = ma = 6 3

4 Linear Momentum In Dynamics, the systems considered no longer were restricted to systems with an acceleration equal to so you moved from the special case where the vector sum of the forces were equal to to the more general case F = ma 7 Linear Momentum The critical element of this expression is the fact that both acceleration and force are vector expressions having both magnitude and direction F = ma 8 4

5 Linear Momentum By definition, acceleration is the time rate of change of velocity so if we consider constant mass we can rewrite the expression as dv F = ma = m dt 9 Linear Momentum And since mass is a scalar quantity, we can move it into the derivative to generalize the expression dv d mv F = ma = m = dt dt 1 5

6 Linear Momentum The last term represents the change of linear momentum with respect to time Linear momentum being defined as the product of a mass times its linear velocity dv d mv F = ma = m = dt dt 11 Linear Momentum This simple statement implies that a change in the force acting on a body will change the momentum of the body and A change in momentum of a body will produce a force in response dv d mv F = ma = m = dt dt 12 6

7 Linear Momentum One is always connected with the other Notice that both force and velocity are vector quantities and therefore they have both magnitude and direction That is a critical point dv d mv F = ma = m = dt dt 13 Angular Momentum If we look at the angular momentum of a body, we consider the moment or torque that produces the rotation The expression is slightly different than for linear momentum M = Iα 14 7

8 Angular Momentum M is the moment or torque causing the rotational acceleration I is the moment of interia about the axis of rotation And α is the angular acceleration about the axis of rotation M = Iα 15 Angular Momentum If we use ω to represent angular velocity then the expression can be rewritten as dϖ M = Iα = I dt 16 8

9 Angular Momentum If we use ω to represent angular velocity then the expression can be rewritten as dϖ M = Iα = I dt 17 Angular Momentum Both expressions are similar in form but they do differ in the units Both expressions do involve vector quantities 18 dϖ M = Iα = I dt dv F = ma = m dt 9

10 Force Balance In Statics, when we looked at the forces acting on a free body diagram we had three types of forces Externally applied loads l Reactive forces from supports l Weight l 19 Force Balance On a control volume, the forces acting on the system are expanded to include Gravitational forces weight l Pressure forces l Viscous forces l And any other force acting on the system l 2 1

11 Force Balance It will be critical to our analysis to be able to evaluate both the magnitude and direction of forces acting on the system And to understand that we are dealing with vector quantities in this type of balance and so we have to use vector operations for our analysis 21 Application of Linear Momentum 22 There are a number of theoretical developments of how the principle of the conservation of momentum is applied and an outline of all the special cases that can be utilized 11

12 23 24 A reducing elbow is used to deflect water flow mass flow at a rate of 14 kg/s in a horizontal pipe upward 3 while accelerating it. 12

13 The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 113 cm2 at the inlet and 7 cm2 at the outlet. 25 The elevation difference between the centers of the outlet and the inlet is 3 cm. The weight of the elbow and the water in it is considered to be negligible

14 27 Determine a the gage pressure at the center of the inlet of the elbow and b the anchoring force needed to hold the elbow in place. They have already selected the control volume for you in this problem

15 The Force to Hold a Deflector Elbow in Place Force directions are positive to the right and upward. 29 What we know kpa 1Pa ρ water := 1 kg 3 m 2 Area1 := 113cm 2 Area2 := 7cm 3 z1 := cm z2 := 3cm kg mdot := 14 sec 15

16 We have three sets of expressions that we can use to analyze this system. 1. We know that the mass flow into the system must be equal to the mass flow out of the system. 2. We know that the energy in must equal to the energy out. 3. And we know that any change in momentum in the system must be reacted to by a force applied to the system We know that the mass flow into the system must be equal to the mass flow out of the system. m 1 = m

17 We know that the mass flow into the system must be equal to the mass flow out of the system. m 1 = m 2 water flow at a rate of 14 kg/s ρq1 = ρq2 This is dm/dt ρv1 A1 = ρv 2 A2 cross-sectional area of the elbow is 113 cm2 at the inlet This is A1 33 We can use these along with our expression to solve for the velocity at the inlet of the system m 1 = ρv1 A1 m V1 = 1 = ρa1 14 kg s 1m kg cm 2 1cm m 2 water flow at a rate of 14 kg/s This is dm/dt cross-sectional area of the elbow is 113 cm2 at the inlet This is A

18 We can use these along with our expression to solve for the velocity at the inlet of the system m 1 = ρv1 A1 m V1 = 1 = ρa1 14 kg s 1m kg cm 2 1cm m Velocity1 := 2 water flow at a rate of 14 kg/s This is dm/dt mdot ρ water Area1 cross-sectional area of the elbow is 113 cm2 at the inlet m Velocity1 = s This is A1 35 And we can use the mass balance for the control volume to solve for the velocity at the outlet ρv 2 A2 = ρv1 A1 Velocity2 := Velocity1 Area1 Area2 m Velocity2 = 2 s 36 water flow at a rate of 14 kg/s This is dm/dt cross-sectional area of the elbow is 113 cm2 at the inlet This is A1 18

19 There is nothing else that the conservation of mass expression can tell us for this problem One of the things that the problem asks for is the gage pressure at the inlet, point 1 Velocity2 := Velocity1 Area1 Area2 m Velocity2 = 2 s 37 water flow at a rate of 14 kg/s This is dm/dt cross-sectional area of the elbow is 113 cm2 at the inlet This is A1 We can use the energy balance on the control volume to determine this pressure V1 2 P1 V 22 P2 gz1 + + = gz ρ1 2 ρ2 We know the elevations at point 1 and 2, the velocities at points 1 and 2, and the fact that the pressure at point 2 is atmospheric

20 We can use the energy balance on the control volume to determine this pressure gz1 + v12 p1 v2 p + = gz ρ 2 ρ z1 =, z2 = 3cm p2 = gage m m,v = 2 s 2 s kg ρ = 1 3 m v12 p1 v2 + = gz ρ 2 v1 = 1.24 p1 = 1kPa 39 Since we have a change in both magnitude and direction of velocity across the control volume, we have a change in momentum across the control volume We can look at these in component form using the unit vectors to describe the inlet and outlet velocities. V1 = V1i V2 = V2 cos3 i + sin3 k 4 2

21 The change in momentum across the control volume can then be expressed as V1 = V1i V 2 = V 2 cos3 i + sin 3 k 2 cos3 i + sin 3 k mv 1i mv 41 This change in momentum must be exactly offset by the forces acting on the system F = mv 2 cos3 i + sin 3 k mv1i 42 21

22 The forces acting on the system include the reaction forces needed to support the system as well as the pressure forces acting on the system. The problem statement provided the reasoning behind neglecting the weight of the water and the weight of the value. 2 cos3 i + sin 3 k mv 1i F = mv 43 Since atmospheric pressure is constant at all points in the system, we can only consider the forces developed by gage pressures. Again, we only consider these at the boundaries of the control volume so the only force developed by pressure is that developed normal to the flow into the system at 1. F = mv 2 cos3 i + sin 3 k mv1i 44 22

23 From our diagram you can see that the direction of this force is in the positive i direction. 2 cos3 i + sin3 k mv 1i F = mv Fpressure = p1,gage Area1 i 45 The only other force in the system will be the support force, our unknown. 2 cos3 i + sin3 k mv 1i F = mv Fpressure = p1,gage Area1 i FR = FR i + FR k x 46 z 23

24 Now we can substitute into the conservation of momentum expression to obtain. 2 cos3 i + sin 3 k mv 1i F pressure + FR = mv 2 cos3 i + sin 3 k mv 1i = P1, gage Area1 i + FR i + FR k mv 47 x z Setting the coefficients in the i and k directions equal we have 2 cos3 i + sin 3 k mv 1i = P1, gage Area1 i + FR i + FR k mv 2 cos3 mv 1 = P1, gage Area1 + FR mv x z 2 sin 3 = FR mv x 48 z 24

25 Isolating the unknowns and solving 2 cos3 mv 1 P1, gage Area1 = FR mv 2 sin 3 = FR mv 49 x z Isolating the unknowns and solving & 2 cos 3 mv & 1 P1, gage Area1 = FRx mv & 2 sin 3 = FRz mv FRx := mdot Velocity1 + mdot Velocity2 cos 3deg P1gage Area1 FRx = N FRy := mdot Velocity2 sin 3deg FRy = 14 N 5 25

26 Homework A reversing elbow is placed such that water makes a 18 U-turn before it is discharged. The elevation difference between the centers of the inlet and the exit sections is.3 m. Determine the anchoring force needed to hold the elbow in place. The water flows at a rate of 14 kg/s. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 113 cm2 at the inlet and 7 cm2 at the outlet. The weight of the elbow is negligible. Homework If the flowrate is doubled in the system shown in Problem 1, how are the forces to hold the system in place changed? 26

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