2. LINEAR MOMENTUM APPLICATIONS AND THE MOMENT OF MOMENTUM 1.More linear momentum application (continuing from last day) 87-351 Fluid Mechanics and Hydraulics  of course we remember momentum = mass * velocity [ physical interpretation: what are we doing today? ] 2.So now, what is “moment of momentum”?  The moment of momentum equation relates torques to the flow of angular momentum for the contents of a control volume 3.Who cares !?  appreciating the nature of the moment of momentum affords us a tool in the analysis and design of kool things that spin like turbomachines, turbines, etc supercharger turbocharger water pump wind turbine lawn sprinkler

3. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 ] GIVEN: REQD: P 1a = 30 psi P 2a = 24 psi Compute the horizontal components of anchor force necessary to hold the elbow in place 87-351 Fluid Mechanics and Hydraulics

4. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 (cont’d) ] 1. We observe that only the weight force acts in the z- dir, therefore does not contribute to the horizontal anchoring force we are after SOLU: 2. Let us write the x component of the momentum eqn - (1)  careful study of the problem geometry tells us that all flow (and thus momentum) enters and leaves the CV in a y direction, thus the x component of momentum flow is 0  so - (2) 87-351 Fluid Mechanics and Hydraulics

5. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 (cont’d) ] SOLU:3. Now write the y component of the momentum eqn - (3)  again, here we assume uniform velocity profile (or 1D flow) and thus we can bypass the integration  employing continuity we yield - (5) - (4)  now plugging in given data, we compute the mass flux - (6) 87-351 Fluid Mechanics and Hydraulics

6. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 (cont’d) ] SOLU:  finally, solving for FA y  The negative FA y tells us our assumed direction of FA y was not correct - (ans)  **NB: again, like other examples, the anchoring force here is independent of the atmospheric pressure (it cancels out), but what of the force the elbow puts on the fluid itself ?? 87-351 Fluid Mechanics and Hydraulics

7. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 (cont’d) ] SOLU:3. Now let us reconstruct the CV (but only encase the fluid within the bend) 4. Now re-apply the momentum eqn to solve for an internal reaction R y - (7)  here, p1 and p2 must be expressed as absolute because the force between fluid and wall is a complete pressure effect 87-351 Fluid Mechanics and Hydraulics

8. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 (cont’d) ] SOLU:  now we obtain - (ans) 87-351 Fluid Mechanics and Hydraulics

9. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 (cont’d) ] SOLU: - (8) 4. Now let us check our solution for the anchoring force FA y by applying a different CV  just consider the pipe bend without the fluid inside, i.e.,  and we have already solved for R y in (7) 87-351 Fluid Mechanics and Hydraulics

10. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 1 (cont’d) ] SOLU:  dropping in the known numbers, we get  which verifies our result in the original CV configuration - (ans) 87-351 Fluid Mechanics and Hydraulics

11. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 2 ] GIVEN:REQD: Utilizing momentum considerations, develop an expression describing the pressure gradient from section 1 to section 2 87-351 Fluid Mechanics and Hydraulics

13. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 2 (cont’d) ] SOLU : 1. Let us apply the momentum eqn to what was given - (1) 87-351 Fluid Mechanics and Hydraulics

14. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 2 (cont’d) ] SOLU: - (2)  we notice that the velocity profile at section 2 is no longer “uniform”  i.e., we are going to have to do some simple integration term AA 87-351 Fluid Mechanics and Hydraulics

15. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 2 (cont’d) ] SOLU:  let us evaluate term AA - (3)  or - (4)  and now sub AA back into (2), we get - (5) 87-351 Fluid Mechanics and Hydraulics

16. LINEAR MOMENTUM APPLICATIONS [ additional examples, ex 2 (cont’d) ] SOLU: - (ans) 3. It is very interesting for us to note the following regarding the parameters in the solution that effect pressure drop in the conduit  pressure drop in a pipe is affected by the momentum flux related to a change in velocity profile (flow development)  an increase in wall shear means an increase in pressure required to push the flow through the pipe  for a vertical flow, you must also be concerned with the gravity’s pull on the water 2. Let us solve for the pressure drop 87-351 Fluid Mechanics and Hydraulics

23. THE MOMENT OF MOMENTUM  We consider the motion of a single fluid particle and apply Newton’s second law [ derivation ] - (1)  Here  F particle represents the resultant external force acting on the particle  Now, take the moment of both sides of (1) - (2)  Here r is the position vector from the particle to the origin of the inertial coord sys we are operating in 87-351 Fluid Mechanics and Hydraulics

24. THE MOMENT OF MOMENTUM  Here we apply the product rule to the LHS of (2) [ derivation (cont’d) ] - (3)  of course, we can write - (4)  and we know - (5)  Combining (1) through (5) - (6) 87-351 Fluid Mechanics and Hydraulics

25. THE MOMENT OF MOMENTUM  (6) is valid for each particle in the system, let us use a summation to characterize the system [ derivation (cont’d) ] - (7)  then we pull the differential outside the summation sign to acquire - (8)  which says 87-351 Fluid Mechanics and Hydraulics

26. THE MOMENT OF MOMENTUM  and we well know that when a CV and system are instantaneously coincident [ derivation (cont’d) ] - (9)  such that we can write  does this look familiar? (it better !)  [RTT!]  which says - (10) 87-351 Fluid Mechanics and Hydraulics

27. THE MOMENT OF MOMENTUM  Therefore, for an inertial and non-deforming CV, we combine (8)-(10) [ derivation (cont’d) ] - (11)  Here lies the moment of momentum eqn ! 87-351 Fluid Mechanics and Hydraulics

28. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 ] GIVEN: REQD: a. What resisting torque is necessary to fix the sprinkler head? b. What is the resisting torque associated with the sprinkler rotating at a constant 500 rev/min? c. How fast will the sprinkler spin if no resisting torque is applied? 87-351 Fluid Mechanics and Hydraulics

29. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 (cont’d) ] SOLU: 1. (a) For a stationary sprinkler head, the velocities entering and leaving the CV will appear as  presently we are concerned with steady flows or “steady in the mean” flows, thus the unsteady term in the moment of momentum eqn goes 0 - (1) 87-351 Fluid Mechanics and Hydraulics

30. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 (cont’d) ] SOLU : 2. (a) Now, consider just the flux term  most times we are only concerned with the axial component of the moment of momentum, for this application term BB becomes 0 - (1) term BB - (2)  similarly, the RHS of (1) reduces to the axial torque of shaft - (3) 87-351 Fluid Mechanics and Hydraulics

31. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 (cont’d) ] SOLU: - (3)  so then, we have  the CV is fixed and non def. and the nozzle flow is tangential, thus V  2 = V 2, so we can write - (4)  then, numerically - (5) - (ans) 87-351 Fluid Mechanics and Hydraulics

32. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 (cont’d) ] SOLU: 1. (b) When the sprinkler is rotating at 500 rpm the flow field in the CV is steady in the mean, so let us consider the absolute velocity leaving each nozzle  thus  so (and we know W 2 = 16.7 m/s) - (1) - (2) 87-351 Fluid Mechanics and Hydraulics

33. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 (cont’d) ] SOLU:  therefore  or - (3) - (4)  recalling  we have - (6) - (5) 87-351 Fluid Mechanics and Hydraulics

34. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 (cont’d) ] SOLU:  note here that the torque resisting rotation at a rotation rate of 500 rev/min is significantly less than that required to hold the head fixed - (ans) 87-351 Fluid Mechanics and Hydraulics

35. THE MOMENT OF MOMENTUM: APPLICATION [ example 1 (cont’d) ] SOLU: 1. (c) If no resistance torque is offered to the shaft, a terminal rotational velocity will be reached  we recall from earlier in the example  combining these, we can write the shaft torque expression as - (1) - (2)  but T shaft goes to 0 for no resistance, thus  so - (ans) 87-351 Fluid Mechanics and Hydraulics