linear momentum applications and the moment of momentum 1.more linear momentum application...
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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 HydraulicsTRANSCRIPT
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
superchargerturbocharger water pump
wind turbine
lawn sprinkler
LINEAR MOMENTUM APPLICATIONS[ additional examples, ex 1 ]
GIVEN:
REQD:
P1a = 30 psi
P2a = 24 psi
Compute the horizontal components of anchor force necessary to hold the elbow in place
87-351 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)
87-351 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)
87-351 Fluid Mechanics and Hydraulics
LINEAR MOMENTUM APPLICATIONS[ additional examples, ex 1 (cont’d) ]
SOLU: finally, solving for FAy
The negative FAy tells us our assumed direction of FAy 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
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 Ry
- (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
LINEAR MOMENTUM APPLICATIONS[ additional examples, ex 1 (cont’d) ]
SOLU: now we
obtain
- (ans)
87-351 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 FAy by applying a different CV
just consider the pipe bend without the fluid inside, i.e.,
and we have already solved for Ry in (7)
87-351 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)
87-351 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 2
87-351 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)
87-351 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
87-351 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)
87-351 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
87-351 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 dFparticle 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
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
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
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
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
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
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
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
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 Vq2 = V2, so we can write
- (4)
then, numerically
- (5)
- (ans)
87-351 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 W2 = 16.7 m/s)
- (1)
- (2)
87-351 Fluid Mechanics and Hydraulics
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
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
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 Tshaft goes to 0 for no resistance, thus
so - (ans)
87-351 Fluid Mechanics and Hydraulics