Float valve!

Raising the sealed pistoncreates a vacuum beneathit. Assuming at least Atmospheric pressure is exerted on the reservoir of water beneath, water will be pushed up this pipe.

Similar to the manualbicycle tire pump:

Another “Positive Displacement” Pump

Much simpler to understand is the impeller pump

which can continuouslydrive water through.

OUT

IN

Momentum Transfer Pumps

Industrial impeller

d

P

Each pump stroke of distance d

and drives a volume of water forward.

does an amount of work = F d

To move this much water forward (saysmoothly at constant speed) against thepressure in the fluid requires a force

A. equal to the weight of displaced fluid.B. P A (fluid pressure cross sectional area) .C. equal to the atmospheric pressure.

d

P

Each pump stroke of distance d

and drives a volume of water forward.

does an amount of work = PA d

The volume of water moved by this single stroke of the pump is equal to

A. P A.B. A d.C. P d.

So work done by pump on water = P V

Lowpressure

Highpressure

can’t depend only on the staticpressure just where the pump is pushing.

For every cubic foot of waterpushed into the system, • a cubic foot must be displaced out of the way• and a cubic foot must exit out the other end!

Pushing a volume of water throughsome elaborate plumbing system

Just like a train of interconnected cars

where no car can move faster than thecar ahead, or slower than the car behind even if some are going uphill, others downhill

Doing work at one end of a pipe, fightsany pressure variations throughout the

entire system.

Unlike a train of interconnected carspushing through a fixed volume of fluid

through the entire system can mean:

changes in speed for different portions of water at different points along the route!

The confining pressures and the fluid speeds

can vary all along the route through the plumbing!

Kinetic Energy can be USED to do work

P1 P2

Dynamic equilibrium of fluids

• “Steady state” flow.

• Pressures are not equal in all directions.

• Fluid is moved only as it moves an equal volume out of its way.

P1 P2

doing work on the fluid in front of it

even as work is done on it by

the fluid pushing from behind.

P1 P2

Dynamic equilibrium of fluids

Work doneon a volume

of fluid

increase in its kinetic

energy

Work it doesin moving an

equivalent volumeout of its way!

mechanicallyby a pump

or naturallyby gravity

= +

Under steady state conditions, wherethe work driving the flow is constant,

this sum must remain constant.

The normal relationship we saw when unbalanced forces acted on solid objects.

P1 P2

Dynamic equilibrium of fluids

constant

Kinetic Energy ofvolume of fluid

Work it doesin moving an

equivalent volumeout of its way!

= +

PVmv += 2

21

mass of moving volume of fluid

work done by thisvolume’s own pressure

PVvV += 2)(21 ρ

Constant“energy/volume”

Pv += 2

21 ρ

Constant“energy/volume”

Pv += 2

21 ρ

A steady-state-flow carries water through an elaborate plumbing system.

It slows to an almost dead crawlat points in the system where thepressure is

A. exceedingly small.B. exceedingly large.C. balanced by atmospheric pressure.

Water courses through pipes nearthe center of the system at speed v.At the open faucet where it drains,the speed of the emerging water is

A. smaller than v.B. larger than v.C. the same as v.

It slows to an almost dead crawlat points in the system where thepressure is

A. exceedingly small.B. exceedingly large.C. balanced by atmospheric pressure.

Water flows at speed v within the pipewhere the pressure is equivalent to 3 atmospheres (45 lbs/sq.in.).

The pressure at the pointit exits from the pipe is

A. 1 atmosphere.B. 2 atmospheres.C. 3 atmospheres.D. 4 atmospheres.E. 5 atmospheres.F. 6 atmospheres.

With what speed does this water exit?

Only possible if narrow pipe withviscous forces slowing waterflow- See next chapter!

Constant“energy/volume”

Pv += 2

21 ρ

means:

insideinsidePv +2)(

21 ρ

outsideoutsidePv += 2)(

21 ρ

conditions withinthe pipeline

conditions atpipeline exit

)000,100()(2

1)000,300()(

2

1 22 PavPavoutsideinside

+=+ ρρ3 atms. 1 atm.

density of water = 1000 kg/m3

= 1 g/cm3

[ ]22 )()(2

1000,200

insideoutsidevvPa −= ρ

[ ] [ ]223 )()(

/1000

000,2002insideoutside

vvmkg

Pa −=

22 )()(400outsideinside

vv =+

outsideinsidevv =+ 2)(400

Density of water:

ρwater = 1 g/cc = 1000 kg/m3

Density of air:

depends on altitude!

ρair = 1.200 kg/m3

= 0.0012 gram/cc

depends on barometric pressure!depends on temperature!

at sea level and 20oC

Incompressible!

The pink spherewould initially

Were the wallsof the container

suddenly tovanish

A. remain in place.B. fly out horizontally.C. be pushed down, sliding diagonally.

It’s ownweightpulls itstraightdown.

The weight of the

above layer bear downalong thiscontact point.

Its alsosupportedby contactbelow andthe insidewall of the

beaker.

And this collection of forces acting on this individual sphere all balance!

All together these forces exactly balance (cancel)

the forceexertedby thewall.

Were it to be removed

the netforce wouldbe outward!

Water pressure on the bottom surface is the total weight of the water above it

A

Mg

A

gV )(ρ=A

gAh)(ρ= hgρ=

However this pressure is not just directedDOWNWARD

hgρ=P

But outward as well

3.0 meters

Swimming along the 10-foot bottom

of the poolyour back supports a ~3 meter column

of water.

With cross-sectional area

A 0.75 m2

That’s a 3 0.75 = 2.25 m3

volume of water

with a weight of ρVg

)/8.9)(25.2)(000,1( 233 smm

m

kg=

= 22,050 Newtons

5000 pounds!

Bernoulli’s Equation

P + ρv2 + ρgh = constant 12

When a stream of water either speeds up or flows uphill

the pressure it exerts drops!

When a stream of water slows or flows downhill

the pressure it exerts drops!

Speeding up or rising uses up energy!