LINEAR MOMENTUM APPLICATIONS AND THE MOMENT OF MOMENTUM 1.More linear momentum application...

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


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)



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)



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 ??



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


SOLU: now we

obtain

- (ans)



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)



SOLU: dropping in the known numbers, we get

which verifies our result in the original CV configuration

- (ans)


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



SOLU:

1. Let us apply the momentum eqn to what was given

- (1)



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



SOLU: let us evaluate term AA

- (3)

or

- (4)

and now sub AA back into (2), we get

- (5)



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


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



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)



(6) is valid for each particle in the system, let us use a summation to characterize the system


- (7)

then we pull the differential outside the summation sign to acquire

- (8)

which says



and we well know that when a CV and system are instantaneously coincident


- (9)

such that we can write does this look familiar? (it better !) [RTT!]

which says

- (10)



Therefore, for an inertial and non-deforming CV, we combine (8)-(10)


- (11)

Here lies the moment of momentum eqn !


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?


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)



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


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)



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)



SOLU: therefore

or

- (3)

- (4)

recalling

we have

- (6)

- (5)



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)



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)


LINEAR MOMENTUM APPLICATIONS AND THE MOMENT OF MOMENTUM 1.More linear momentum application...

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