float valve! raising the sealed piston creates a vacuum beneath it. assuming at least atmospheric...
TRANSCRIPT
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!