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## PowerPoint Slideshow about ' Momentum Conservation' - minowa

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Momentum Conservation

- Newton’s Second Law
- Force = Mass * Acceleration
- Alternate method: From Reynold’s theorem

- Fluid Flow
- Force = Momentum flux + Momentum Accumulation rate

Flux

Accumulation Rate

Example

- Pipe with U turn

P1,V1

F

P2,V2

Use Gage Pressure!

In case of gas, use absolute pressure to calculate density

Example

N

- “L” bend

P1

E

F

Assume the force by the pipe on the fluid is in the positive direction

P2

What will the force be, if the flow is reversed (a) in a straight pipe? (b) in a L bend?

Y

ExamplePressure difference?

Pbm. 5.19

V1=-3 m/s

V2=0 m/s

Vw=Velocity of Sound in water

V2

r1

r2

Vw

P2

P1

V1

Stationary CV

~ 41 atm

Momentum Conservation

- Angular Momentum
- In a moving system
- Torque = Angular Momentum flux + Angular Momentum Accumulation rate

Flux

Accumulation Rate

Example

- Example 4 in book

- Find the torque on the shaft
- In a moving system
- Torque = Angular Momentum flux + Angular Momentum Accumulation rate

Example

- Approach-1. Find effective Force in X direction
- Find the moment of Force
- Assume: No frictional loss, ignore gravity, steady state, atmospheric pressure everywhere

Example

- Approach-2. Using conservation of angular momentum
- Stationary CV

Example

- Consider a jet hitting a moving plate

- After 1 second

- Vnoz water has entered into the CV
- Plate has moved by Vplate
- In a control volume which moves with the plate, Vnoz-Vplate water has entered the CV (and exited at the bottom)

Example

- Pbm 5.24
- Thickness of slit =t, vol flow rate =Q, dia of pipe=d, density given
- Ignore gravity effects

3ft

6ft

Flux

Accumulation Rate

Mechanical

Work

done by

the system

Heat

Work done

by pressure force

Energy ConservationEnergy Conservation

- No Frictional losses
- Incompressible
- Steady
- No heat, work
- No internal energy change

Example

- Flow from a tank

Dia = d1

1

h1

0

2

Dia = d2

- Pressure = atm at the top and at the outlet

h3

3

- Velocity at 1 ~ 0

- Toricelli’s Law

- Sections 2 and 3

Example

- How long does it take to empty the tank?
- What if you had a pipe all the way upto level 3?

Dia = d1

1

h1

0

2

Dia = d2

h3

3

- Pressure @ section 2 != atm

- Pressure @ section 3 = atm

Example

- What if you had a pipe all the way upto level 3?

Dia = d1

1

h1

0

2

Dia = d2

h3

3

- More flow with the pipe

- Turbulence, friction
- Unsteady flow
- Vortex formation

Example

Height is known

- Moving reference; Aircraft

60 km/h

- Find P and r (eg from tables)

150 km/h

- Flight as Reference

1

2

3

Example

- Pbm. 6.4
- Steady flow through pipe , with friction
- Friction loss head = 10 psi
- Area, vol flow rate given
- Find temp increase

- Assume no heat transfer

Example

D2

- Pbm. 6.10
- Fluid entering from bottom,
- exiting at radial direction
- Steady, no friction

t

P2=atm

h2

D1

- Find Q, F on the top plate

P1=10 psig

Example

F

y

D1

P1=10 psig

- If the velocity distribution just below the top plate is known, then P can be found using Bernoulli’s eqn

Modifications to Eqn

- Unsteady state, for points 1 and 2 along a stream line

1

H

L

2

D

Draining of a tank- We can obtain the time it takes to drain a tank
- (i) Assume no friction in the drain pipe
- (ii) Assume you know the relationship between friction and velocity

- Le us take that the bottom location is 2 and the top fluid surface is 1
- Incompressible fluid

Draining of a tank: Quasi steady state

- Quasi steady state assumption
- Velocity at fluid surface at 1 is very small
- i.e. R >> D

- No friction : L is negligible
- P1 = P2 = Patm

Draining of a tank: Quasi steady state

- Original level of liquid is at H = H0
- Integrating above equation from t=0, H=H0 to t=tfinal, H =0, we can find the efflux time

Draining of a tank: Unsteady state

- BSL eg.7.7.1

- At any point of time, the kinetic + potential energy of the fluid in tank is converted into kinetic energy of the outgoing fluid
- We still neglect friction

- Potential Energy of a disk at height z and thickness dz

Draining of a tank: Unsteady state

- Also, using continuity equation

- Substituting, you get a 2nd order non linear ODE with two initial conditions. Please refer to BSL for solution

1

H

L

2

D

Draining of a tank (accounting for friction)- What if the flow in the tube is laminar and you want to account for friction?
- Bernoulli’s eqn is not used (friction present)
- Continuity

- Hagen-Poiseuille’s eqn

Draining of a tank (accounting for friction)

- Substituting and re arranging,

- Integrating with limits

- Note: The answer is given in terms of diameter of tube, so that it is easier to compare with the answer given in the book

Example

- A1,A2, initial height h1 known
- A1 >> A2

1

L

- Consider section 3 and 2

h1

3

2

- Pseudo Steady state ==> Toricelli’s law

Example

- Rearranging and solving, we get

- As t increases, the solution approaches the Toricelli’s equation

Appendix:Example

- Blood Flow in vessels
- Minimization of ‘work’

- Murray’s Law:

- Laminar Flow, negligible friction loss (other than that due to viscous loss in laminar flow) , steady

- Turbulent, pulsating flow

- Assume

Appendix:Example

- If the ratio of ‘smaller’ to larger capillary is constant

- And Metabolic requirement =m= power/volume

- Work for maintaining blood vessel

- Total work

- Optimum radius

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