Water distribution
Energy: Part 4
water tower
Bernoulli and Water distribution systems
Physics 1010: Dr. Eleanor Hodby
reservoir
Lecture 12:
- Water distribution
Reminders:
HW5 due Monday. Easy print sheets available
Extra practice problem list on website
Lectures can be rewatched online
Reading for Tuesday: 9.2
Midterm 2: 2 weeks today
pipe
buildings
Where does the water flow?
What determines the water pressure in different homes/heights?
How fast does water flow out of a faucet?
How do you pump water out of wells?
ALL ABOUT CONSERVATION OF ENERGY!
GPE = mgh. KE = ½ mv2 PPE = PV
Pumps do work (Force x distance)
(all of the physics of water distribution system)
The super soaker (e.g. squirt guns)
pump
Pressure potential energy (PPE)
What the heck is pressure anyway?
Pressure = Force
Area
Units: 1Pascal (Pa) = 1N/m 2
Pump up the pressure inside just a little bit and squirt. If we pump it up
more, the water coming out will be:
a. going slower than before, b. going the same speed,
c. going faster.
Pressure potential energy (PPE)
• New form of potential energy for fluids
• PE is the energy of an object (or fluid) due to its CONDITION (situation,
surroundings etc)
• Water of mass m, at height h has associated GPE = mgh because of its
(vertical) position
- work (mgh) was done to get the water from ground to that height
- Physical details of how the work was done or how water is being
supported is not important
• Water of volume V at pressure P has associated PPE = PV
- Work (PV) was done to pressurize the water
- Physical details of how the work was done or how the pressure is
being maintained are not important
• Check that PV has units of energy (J)
PV = N × m 3 = Nm = J
m2
The plunger of a syringe has an area of 1cm 2.
I push the plunger with a force of 5N.
What’s the pressure exerted by the plunger on
the fluid inside?
a)
b)
c)
d)
e)
5N
5N/m
5 N/m 2
500N/m 2
50000N/m
50000N/m2
Forms of energy in Super Soaker
Pressure
1. Pumping does work
transforming chemical
energy in my arm into
PPE.
Work = force X distance.
Pressure
2. When I pull the
trigger, pressure does
work on the water.
I apply a force, compress
pump by a distance.
Converts PPE into KE = ½ mv2
3. When I fire upward,
this KE turns into GPE
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Energy in a water distribution system
• The same three forms of energy exist in a water distribution system
• If we add up energy in these forms, the sum must be constant.
• It just sloshes back and forth between forms!
Bernoulli’s Equation
PV + ½ mv 2 + mgh = Etotal
But what mass of water are we talking about, what height?
PPE + KE + GPE = Etotal (constant)
PV + ½ mv2 + mgh = Etotal
Consider one little bit of water of volume V and mass m:
Since Etotal is constant:
• If one thing changes, the other quantities must change correspondingly.
- If pressure changes (water comes out of nozzle), v changes.
- If height changes (go up in building), pressure or v changes, etc.
Replace m = rV where r is the fluid density
PV + ½ rVv 2 + rVgh = Etotal
(r = mass/volume
= 1000kg/m3 for water)
We can divide through by V to get the standard form for Bernoulli’s equation:
P + ½ rv 2 + rgh = Etotal/V (Etotal per unit volume )
• Like the cart coasting up and down hills with no
friction. Velocity and height are always connected
• If you know velocity and height at one place,
can calculate it at all others.
Just good old conservation of energy with the terms relabeled
Since Etotal per vol is constant:
Know P, v and h at one point  can calculate these quantities at another
notice we dropped the 1 atmosphere inside and out.
More on pressure
Apply Bernoulli to Squirt Gun
How is velocity of water out related to pressure inside gun?
Here’s a bucket of water with a faucet attached.
What is the pressure at a depth H?
Bernoulli’s Equation:
Pinside
voutside
P + ½ rv2 + rgh = Etpv
Compare water at surface and at depth H
P + ½
rv2
 P+½r
+ rgh = Etotal per vol
v2
Height constant so ignore GPE
= Etotal per vol (constant)
v = 0 everywhere  P + rgh = Etpv
H
• At surface: P = AP, h = 0
 Etpv = AP
P?
Inside gun: P = AP + Ppump, v= 0

AP + Ppump = Etotal per vol
Outside gun: P = AP, voutside is big

AP + ½ r voutside2 = Etotal per vol
AP + Ppump = AP + ½ r
voutside2
Ppump = ½ r voutside2
• At depth H: P = AP + Pw, height = -H
 Etpv = AP + Pw + rg(-H)
Faucet shut off, so
water is not moving.
Etpv constant  AP = AP + Pw – rgH
Pw = rgH
PH = AP + rgH
voutside = sqrt(2 Ppump / r)
More on pressure
Atmospheric pressure
Here’s a bucket of water with a faucet attached.
What is the pressure at a depth H?
Pressure at surface of water
= Atmospheric pressure (AP)
 100,000 Pa
Bernoulli’s Equation:
P + ½ rv2 + rgh = Etpv
Compare water at surface and at depth H
• Pressure due to air molecules hitting
surface and exerting a force
• Always present at surface of earth
• Usually only interested in CHANGES in
water pressure and AP cancels out
• If so can set zero of water pressure at AP
(like setting zero of height somewhere
convenient)
v = 0 everywhere  P + rgh = Etpv
H
• At surface: P = AP, h = 0
 Etpv = AP
AP + rgH
Faucet shut off, so
water is not moving.
• At depth H: P = AP + Pw, height = -H
 Etpv = AP + Pw + rg(-H)
Etpv constant  AP = AP + Pw – rgH
Pw = rgH
PH = AP + rgH
Faucet shut off, so
water is not moving.
This pressure is exerted equally in all directions
2
With the faucet off, the water is stopped at point C.
Rank the pressures at the three locations shown.
a)
b)
c)
d)
A
B
PA <
PA <
PA =
PA =
PB < PC
PB = PC
PB = PC
PB > PC
Now open the faucet. What is the pressure at point C,
just outside the faucet?
a) Atmospheric pressure
b) The same as at B
c) Less than atmospheric
pressure
A
H
B
C
Consider a little bit of water leaving the faucet. What
is its velocity at the faucet exit?
(Hint: Question about pressure and velocity changes in fluid
 Apply Bernoulli’s Eqn)
a)
b)
c)
d)
e)
Zero
rgH
Sqrt(rgH)
2gH
Sqrt (2gH)
C
Bernoulli’s Equation in Real Life
Total Energy per volume = P + ½ r v2 + rgh
• This is a good approximation but it cannot be perfectly correct
• What type of energy does it ignore?
 Think about a narrow pipe.
H
• Does not consider energy going into thermal energy- from friction with
walls etc.
• For example, for high speed flow in a narrow pipe, more water
molecules bounce off walls, creating significant friction and energy loss
as heat
• But for most water distribution systems, friction can be ignored BE works
very well.
Water towers are everywhere!
But what exactly are they for?
Bernoulli’s equation and water distribution systems
h
Etpv= P + ½ r v2 + rgh
htower
h=0
rwater = 1000 kg/m 3
Consider water at top of tower:
P = 0, v=0 and height = htower
Etpv= 0 + 0 + rghtower
Consider water at house:
P = Phouse, v  0, and h = 0
Etpv = Phouse + 0 +0
Cons. of energy:
 Etpv = the same everywhere in system
 Phouse = rghtower
= (1000)(9.8)htower
Strictly speaking, there is also one atmosphere of pressure at the top and at the
bottom, but only the difference in pressure matters.
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Bernoulli’s equation and water distribution systems
h
What if the pipe from the water tower takes a weird path
(say, over hills) to get to a house at the same height?
Etpv= P + ½ r v2 + rgh
htower
h=0
rwater = 1000 kg/m 3
Cons. of energy:
 Etpv = the same everywhere in system
 Phouse = rghtower
= (1000)(9.8)htower
A
Atmospheric pressure 100,000 Pa (= 14.6psi)
About how high would a water tower have to
be to provide 1 atmosphere of water pressure
to the house?
a. 2 m, b. 5 m c. 10 m d. 20 m, e. 50 m.
a) House A has higher pressure
b) House B has higher pressure
c) The pressure is the same at both houses.
Here I have a tank of water with a hose connected to the
bottom. When I take my finger off the hose, water (under
pressure) will squirt into the air. Will the water go higher
or lower than the opening in the tank (dashed line)?
a. Higher
b. Right exactly to the dashed line
c. Lower
d. Impossible to predict
e. None of the above.
B
Here I have a tank of water with a hose connected to the bottom.
When I take my finger off the hose, water (under pressure) will squirt
into the air. I can hold the hose high (at A) or low (at B). From which
location will the water squirt higher (relative to the ground)?
h2
A
h1
B
a. A, the higher location
b. B, the lower location
c. Water reaches the same height from
both locations
d. Impossible to predict
e. None of the above.
0
At A, the hose is 0.25 m below the water’s surface. At
B it is 1.0 m below. How much faster will the water
come out at B than at A?
h2
A
h1
hh
B
a. ¼ as fast at B as at A.
b. ½ as fast at B as at A
c. same speed
d. 2 times faster at B than at A
e. 4 times faster at B than at A
Where does the shower work best?
In the house below, water pressure on the1 st floor is 30 psi. What is the
velocity of water coming out of the shower on the 1st floor?
a. 2 m/s,
b. 5 m/s,
c. 20 m/s,
Ppipe = 30 psi
d. 100 m/s,
(200,000 Pa)
e. 414 m/s
0
4
Where does the shower work best?
Now in this same house, the 3rd floor shower is 10 m
higher. What’s the velocity of water at the 3rd floor
shower?
a. 2 m/s, b. 5 m/s, c. 15 m/s, d.
P = 30 psi
20 m/s, e. 50 m/s
(200 kPa)
Water distribution in skyscrapers
The skyscraper water problem:
• Less pressure on the higher floors,
• Water won’t make it to the top floor….
How can you solve this problem?
Etpv = GPE here
• Put very high pressure pump at bottom
(give water enough PPE at bottom)
• Use a series of pumps up the building
• Pump water to a tank on the roof, then
you will always have pressure on the
floors below.
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