LINEAR MOMENTUM APPLICATIONS AND THE MOMENT OF MOMENTUM 1.More linear momentum application (continuing from last day) 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

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 Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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 ?? Fluid Mechanics and Hydraulics

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 Fluid Mechanics and Hydraulics

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

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) Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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

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

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 Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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 Fluid Mechanics and Hydraulics

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 Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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 Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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 ! Fluid Mechanics and Hydraulics

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? Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics

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

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) Fluid Mechanics and Hydraulics

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) Fluid Mechanics and Hydraulics